OpenIntroStats.pdf

OpenIntro StatisticsFourth Edition

David DiezData Scientist

OpenIntro

Mine Çetinkaya-RundelAssociate Professor of the Practice, Duke University

Professional Educator, RStudio

Christopher D BarrInvestment Analyst

Varadero Capital

This book may be downloaded as a free PDF at openintro.org/os. This textbook is also availableunder a Creative Commons license, with the source files hosted on Github.

Table of Contents

1 Introduction to data 71.1 Case study: using stents to prevent strokes . . . . . . . . . . . . . . . . . . . . . . . 91.2 Data basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3 Sampling principles and strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.4 Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2 Summarizing data 392.1 Examining numerical data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 Considering categorical data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 612.3 Case study: malaria vaccine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3 Probability 793.1 Defining probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.2 Conditional probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.3 Sampling from a small population . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.4 Random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1153.5 Continuous distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

4 Distributions of random variables 1314.1 Normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.2 Geometric distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1444.3 Binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1494.4 Negative binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1584.5 Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

5 Foundations for inference 1685.1 Point estimates and sampling variability . . . . . . . . . . . . . . . . . . . . . . . . . 1705.2 Confidence intervals for a proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . 1815.3 Hypothesis testing for a proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

6 Inference for categorical data 2066.1 Inference for a single proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2086.2 Difference of two proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2176.3 Testing for goodness of fit using chi-square . . . . . . . . . . . . . . . . . . . . . . . . 2296.4 Testing for independence in two-way tables . . . . . . . . . . . . . . . . . . . . . . . 240

7 Inference for numerical data 2497.1 One-sample means with the t-distribution . . . . . . . . . . . . . . . . . . . . . . . . 2517.2 Paired data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2627.3 Difference of two means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2677.4 Power calculations for a difference of means . . . . . . . . . . . . . . . . . . . . . . . 2787.5 Comparing many means with ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . 285

4 TABLE OF CONTENTS

8 Introduction to linear regression 3038.1 Fitting a line, residuals, and correlation . . . . . . . . . . . . . . . . . . . . . . . . . 3058.2 Least squares regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3178.3 Types of outliers in linear regression . . . . . . . . . . . . . . . . . . . . . . . . . . . 3288.4 Inference for linear regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

9 Multiple and logistic regression 3419.1 Introduction to multiple regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3439.2 Model selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3539.3 Checking model conditions using graphs . . . . . . . . . . . . . . . . . . . . . . . . . 3589.4 Multiple regression case study: Mario Kart . . . . . . . . . . . . . . . . . . . . . . . 3659.5 Introduction to logistic regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

A Exercise solutions 384

B Data sets within the text 403

C Distribution tables 408

Preface

OpenIntro Statistics covers a first course in statistics, providing a rigorous introduction to appliedstatistics that is clear, concise, and accessible. This book was written with the undergraduate levelin mind, but it’s also popular in high schools and graduate courses.

We hope readers will take away three ideas from this book in addition to forming a foundationof statistical thinking and methods.

• Statistics is an applied field with a wide range of practical applications.

• You don’t have to be a math guru to learn from real, interesting data.

• Data are messy, and statistical tools are imperfect. But, when you understand the strengthsand weaknesses of these tools, you can use them to learn about the world.

Textbook overview

The chapters of this book are as follows:

1. Introduction to data. Data structures, variables, and basic data collection techniques.

2. Summarizing data. Data summaries, graphics, and a teaser of inference using randomization.

3. Probability. Basic principles of probability.

4. Distributions of random variables. The normal model and other key distributions.

5. Foundations for inference. General ideas for statistical inference in the context of estimatingthe population proportion.

6. Inference for categorical data. Inference for proportions and tables using the normal andchi-square distributions.

7. Inference for numerical data. Inference for one or two sample means using the t-distribution,statistical power for comparing two groups, and also comparisons of many means using ANOVA.

8. Introduction to linear regression. Regression for a numerical outcome with one predictorvariable. Most of this chapter could be covered after Chapter 1.

9. Multiple and logistic regression. Regression for numerical and categorical data using manypredictors.

OpenIntro Statistics supports flexibility in choosing and ordering topics. If the main goal is to reachmultiple regression (Chapter 9) as quickly as possible, then the following are the ideal prerequisites:

• Chapter 1, Sections 2.1, and Section 2.2 for a solid introduction to data structures and statis-tical summaries that are used throughout the book.

• Section 4.1 for a solid understanding of the normal distribution.

• Chapter 5 to establish the core set of inference tools.

• Section 7.1 to give a foundation for the t-distribution

• Chapter 8 for establishing ideas and principles for single predictor regression.

6 TABLE OF CONTENTS

Examples and exercises

Examples are provided to establish an understanding of how to apply methods

EXAMPLE 0.1

This is an example. When a question is asked here, where can the answer be found?

The answer can be found here, in the solution section of the example!

When we think the reader should be ready to try determining the solution to an example, we frameit as Guided Practice.

GUIDED PRACTICE 0.2

The reader may check or learn the answer to any Guided Practice problem by reviewing the fullsolution in a footnote.1

Exercises are also provided at the end of each section as well as review exercises at the end of eachchapter. Solutions are given for odd-numbered exercises in Appendix A.

Additional resources

Video overviews, slides, statistical software labs, data sets used in the textbook, and much more arereadily available at

openintro.org/os

We also have improved the ability to access data in this book through the addition of Appendix B,which provides additional information for each of the data sets used in the main text and is new in theFourth Edition. Online guides to each of these data sets are also provided at openintro.org/dataand through a companion R package.

We appreciate all feedback as well as reports of any typos through the website. A short-link toreport a new typo or review known typos is openintro.org/os/typos.

For those focused on statistics at the high school level, consider Advanced High School Statistics,which is a version of OpenIntro Statistics that has been heavily customized by Leah Dorazio for highschool courses and AP® Statistics.

Acknowledgements

This project would not be possible without the passion and dedication of many more people beyondthose on the author list. The authors would like to thank the OpenIntro Staff for their involvementand ongoing contributions. We are also very grateful to the hundreds of students and instructorswho have provided us with valuable feedback since we first started posting book content in 2009.

We also want to thank the many teachers who helped review this edition, including Laura Acion,Matthew E. Aiello-Lammens, Jonathan Akin, Stacey C. Behrensmeyer, Juan Gomez, Jo Hardin,Nicholas Horton, Danish Khan, Peter H.M. Klaren, Jesse Mostipak, Jon C. New, Mario Orsi, StevePhelps, and David Rockoff. We appreciate all of their feedback, which helped us tune the text insignificant ways and greatly improved this book.

1Guided Practice problems are intended to stretch your thinking, and you can check yourself by reviewing thefootnote solution for any Guided Practice.

Chapter 1Introduction to data

1.1 Case study: using stents to prevent strokes

1.2 Data basics

1.3 Sampling principles and strategies

1.4 Experiments

Scientists seek to answer questions using rigorous methods and careful

observations. These observations – collected from the likes of field

notes, surveys, and experiments – form the backbone of a statistical

investigation and are called data. Statistics is the study of how best

to collect, analyze, and draw conclusions from data, and in this first

chapter, we focus on both the properties of data and on the collection

of data.

For videos, slides, and other resources, please visit

www.openintro.org/os

1.1. CASE STUDY: USING STENTS TO PREVENT STROKES 9

1.1 Case study: using stents to prevent strokes

Section 1.1 introduces a classic challenge in statistics: evaluating the efficacy of a medicaltreatment. Terms in this section, and indeed much of this chapter, will all be revisited later in thetext. The plan for now is simply to get a sense of the role statistics can play in practice.

In this section we will consider an experiment that studies effectiveness of stents in treatingpatients at risk of stroke. Stents are devices put inside blood vessels that assist in patient recoveryafter cardiac events and reduce the risk of an additional heart attack or death. Many doctors havehoped that there would be similar benefits for patients at risk of stroke. We start by writing theprincipal question the researchers hope to answer:

Does the use of stents reduce the risk of stroke?

The researchers who asked this question conducted an experiment with 451 at-risk patients.Each volunteer patient was randomly assigned to one of two groups:

Treatment group. Patients in the treatment group received a stent and medical manage-ment. The medical management included medications, management of risk factors, and helpin lifestyle modification.

Control group. Patients in the control group received the same medical management as thetreatment group, but they did not receive stents.

Researchers randomly assigned 224 patients to the treatment group and 227 to the control group.In this study, the control group provides a reference point against which we can measure the medicalimpact of stents in the treatment group.

Researchers studied the effect of stents at two time points: 30 days after enrollment and 365 daysafter enrollment. The results of 5 patients are summarized in Figure 1.1. Patient outcomes arerecorded as “stroke” or “no event”, representing whether or not the patient had a stroke at the endof a time period.

Patient group 0-30 days 0-365 days1 treatment no event no event2 treatment stroke stroke3 treatment no event no event…

……

450 control no event no event451 control no event no event

Figure 1.1: Results for five patients from the stent study.

Considering data from each patient individually would be a long, cumbersome path towardsanswering the original research question. Instead, performing a statistical data analysis allows us toconsider all of the data at once. Figure 1.2 summarizes the raw data in a more helpful way. In thistable, we can quickly see what happened over the entire study. For instance, to identify the numberof patients in the treatment group who had a stroke within 30 days, we look on the left-side of thetable at the intersection of the treatment and stroke: 33.

0-30 days 0-365 daysstroke no event stroke no event

treatment 33 191 45 179control 13 214 28 199Total 46 405 73 378

Figure 1.2: Descriptive statistics for the stent study.

10 CHAPTER 1. INTRODUCTION TO DATA

GUIDED PRACTICE 1.1

Of the 224 patients in the treatment group, 45 had a stroke by the end of the first year. Using thesetwo numbers, compute the proportion of patients in the treatment group who had a stroke by theend of their first year. (Please note: answers to all Guided Practice exercises are provided usingfootnotes.)1

We can compute summary statistics from the table. A summary statistic is a single numbersummarizing a large amount of data. For instance, the primary results of the study after 1 yearcould be described by two summary statistics: the proportion of people who had a stroke in thetreatment and control groups.

Proportion who had a stroke in the treatment (stent) group: 45/224 = 0.20 = 20%.

Proportion who had a stroke in the control group: 28/227 = 0.12 = 12%.

These two summary statistics are useful in looking for differences in the groups, and we are in fora surprise: an additional 8% of patients in the treatment group had a stroke! This is importantfor two reasons. First, it is contrary to what doctors expected, which was that stents would reducethe rate of strokes. Second, it leads to a statistical question: do the data show a “real” differencebetween the groups?

This second question is subtle. Suppose you flip a coin 100 times. While the chance a coinlands heads in any given coin flip is 50%, we probably won’t observe exactly 50 heads. This type offluctuation is part of almost any type of data generating process. It is possible that the 8% differencein the stent study is due to this natural variation. However, the larger the difference we observe (fora particular sample size), the less believable it is that the difference is due to chance. So what weare really asking is the following: is the difference so large that we should reject the notion that itwas due to chance?

While we don’t yet have our statistical tools to fully address this question on our own, we cancomprehend the conclusions of the published analysis: there was compelling evidence of harm bystents in this study of stroke patients.

Be careful: Do not generalize the results of this study to all patients and all stents. Thisstudy looked at patients with very specific characteristics who volunteered to be a part of this studyand who may not be representative of all stroke patients. In addition, there are many types of stentsand this study only considered the self-expanding Wingspan stent (Boston Scientific). However, thisstudy does leave us with an important lesson: we should keep our eyes open for surprises.

1The proportion of the 224 patients who had a stroke within 365 days: 45/224 = 0.20.

1.1. CASE STUDY: USING STENTS TO PREVENT STROKES 11

Exercises

1.1 Migraine and acupuncture, Part I. A migraine is a particularly painful type of headache, which patientssometimes wish to treat with acupuncture. To determine whether acupuncture relieves migraine pain,researchers conducted a randomized controlled study where 89 females diagnosed with migraine headacheswere randomly assigned to one of two groups: treatment or control. 43 patients in the treatment groupreceived acupuncture that is specifically designed to treat migraines. 46 patients in the control groupreceived placebo acupuncture (needle insertion at non-acupoint locations). 24 hours after patients receivedacupuncture, they were asked if they were pain free. Results are summarized in the contingency table below.2

Pain freeYes No Total

Treatment 10 33 43Group

Control 2 44 46Total 12 77 89

identified on the antero-internal part of the antitragus, the

anterior part of the lobe and the upper auricular concha, onthe same side of pain. The majority of these points were

effective very rapidly (within 1 min), while the remaining

points produced a slower antalgic response, between 2 and5 min. The insertion of a semi-permanent needle in these

zones allowed stable control of the migraine pain, which

occurred within 30 min and still persisted 24 h later.Since the most active site in controlling migraine pain

was the antero-internal part of the antitragus, the aim ofthis study was to verify the therapeutic value of this elec-

tive area (appropriate point) and to compare it with an area

of the ear (representing the sciatic nerve) which is probablyinappropriate in terms of giving a therapeutic effect on

migraine attacks, since it has no somatotopic correlation

with head pain.

Materials and methods

The study enrolled 94 females, diagnosed as migraine

without aura following the International Classification ofHeadache Disorders [5], who were subsequently examined

at the Women’s Headache Centre, Department of Gynae-

cology and Obstetrics of Turin University. They were allincluded in the study during a migraine attack provided that

it started no more than 4 h previously. According to a

predetermined computer-made randomization list, the eli-gible patients were randomly and blindly assigned to the

following two groups: group A (n = 46) (average age

35.93 years, range 15–60), group B (n = 48) (average age33.2 years, range 16–58).

Before enrollment, each patient was asked to give an

informed consent to participation in the study.Migraine intensity was measured by means of a VAS

before applying NCT (T0).

In group A, a specific algometer exerting a maximumpressure of 250 g (SEDATELEC, France) was chosen to

identify the tender points with Pain–Pressure Test (PPT).

Every tender point located within the identified area by thepilot study (Fig. 1, area M) was tested with NCT for 10 s

starting from the auricle, that was ipsilateral, to the side of

prevalent cephalic pain. If the test was positive and thereduction was at least 25% in respect to basis, a semi-

permanent needle (ASP SEDATELEC, France) was

inserted after 1 min. On the contrary, if pain did not lessenafter 1 min, a further tender point was challenged in the

same area and so on. When patients became aware of an

initial decrease in the pain in all the zones of the headaffected, they were invited to use a specific diary card to

score the intensity of the pain with a VAS at the following

intervals: after 10 min (T1), after 30 min (T2), after60 min (T3), after 120 min (T4), and after 24 h (T5).

In group B, the lower branch of the anthelix was

repeatedly tested with the algometer for about 30 s toensure it was not sensitive. On both the French and Chinese

auricular maps, this area corresponds to the representation

of the sciatic nerve (Fig. 1, area S) and is specifically usedto treat sciatic pain. Four needles were inserted in this area,

two for each ear.

In all patients, the ear acupuncture was always per-formed by an experienced acupuncturist. The analysis of

the diaries collecting VAS data was conducted by an

impartial operator who did not know the group each patientwas in.

The average values of VAS in group A and B were

calculated at the different times of the study, and a statis-tical evaluation of the differences between the values

obtained in T0, T1, T2, T3 and T4 in the two groupsstudied was performed using an analysis of variance

(ANOVA) for repeated measures followed by multiple

t test of Bonferroni to identify the source of variance.Moreover, to evaluate the difference between group B

and group A, a t test for unpaired data was always per-

formed for each level of the variable ‘‘time’’. In the case ofproportions, a Chi square test was applied. All analyses

were performed using the Statistical Package for the Social

Sciences (SPSS) software program. All values given in thefollowing text are reported as arithmetic mean (±SEM).

Results

Only 89 patients out of the entire group of 94 (43 in groupA, 46 in group B) completed the experiment. Four patients

withdrew from the study, because they experienced an

unbearable exacerbation of pain in the period preceding thelast control at 24 h (two from group A and two from group

B) and were excluded from the statistical analysis since

they requested the removal of the needles. One patientfrom group A did not give her consent to the implant of the

semi-permanent needles. In group A, the mean number of

Fig. 1 The appropriate area(M) versus the inappropriatearea (S) used in the treatmentof migraine attacks

S174 Neurol Sci (2011) 32 (Suppl 1):S173–S175

123

Figure from the original pa-

per displaying the appropri-

ate area (M) versus the in-

appropriate area (S) used in

the treatment of migraine at-

tacks.

(a) What percent of patients in the treatment group were pain free 24 hours after receiving acupuncture?

(b) What percent were pain free in the control group?

(d) Your findings so far might suggest that acupuncture is an effective treatment for migraines for all peoplewho suffer from migraines. However this is not the only possible conclusion that can be drawn basedon your findings so far. What is one other possible explanation for the observed difference between thepercentages of patients that are pain free 24 hours after receiving acupuncture in the two groups?

1.2 Sinusitis and antibiotics, Part I. Researchers studying the effect of antibiotic treatment for acutesinusitis compared to symptomatic treatments randomly assigned 166 adults diagnosed with acute sinusitis toone of two groups: treatment or control. Study participants received either a 10-day course of amoxicillin (anantibiotic) or a placebo similar in appearance and taste. The placebo consisted of symptomatic treatmentssuch as acetaminophen, nasal decongestants, etc. At the end of the 10-day period, patients were asked ifthey experienced improvement in symptoms. The distribution of responses is summarized below.3

Self-reported improvementin symptoms

Yes No TotalTreatment 66 19 85

GroupControl 65 16 81Total 131 35 166

(a) What percent of patients in the treatment group experienced improvement in symptoms?

(b) What percent experienced improvement in symptoms in the control group?

(d) Your findings so far might suggest a real difference in effectiveness of antibiotic and placebo treatmentsfor improving symptoms of sinusitis. However, this is not the only possible conclusion that can be drawnbased on your findings so far. What is one other possible explanation for the observed difference betweenthe percentages of patients in the antibiotic and placebo treatment groups that experience improvementin symptoms of sinusitis?

2G. Allais et al. “Ear acupuncture in the treatment of migraine attacks: a randomized trial on the efficacy ofappropriate versus inappropriate acupoints”. In: Neurological Sci. 32.1 (2011), pp. 173–175.

3J.M. Garbutt et al. “Amoxicillin for Acute Rhinosinusitis: A Randomized Controlled Trial”. In: JAMA: TheJournal of the American Medical Association 307.7 (2012), pp. 685–692.

12 CHAPTER 1. INTRODUCTION TO DATA

1.2 Data basics

Effective organization and description of data is a first step in most analyses. This sectionintroduces the data matrix for organizing data as well as some terminology about different forms ofdata that will be used throughout this book.

1.2.1 Observations, variables, and data matrices

Figure 1.3 displays rows 1, 2, 3, and 50 of a data set for 50 randomly sampled loans offeredthrough Lending Club, which is a peer-to-peer lending company. These observations will be referredto as the loan50 data set.

Each row in the table represents a single loan. The formal name for a row is a case orobservational unit. The columns represent characteristics, called variables, for each of the loans.For example, the first row represents a loan of $7,500 with an interest rate of 7.34%, where theborrower is based in Maryland (MD) and has an income of $70,000.

GUIDED PRACTICE 1.2

What is the grade of the first loan in Figure 1.3? And what is the home ownership status of theborrower for that first loan? For these Guided Practice questions, you can check your answer in thefootnote.4

In practice, it is especially important to ask clarifying questions to ensure important aspects ofthe data are understood. For instance, it is always important to be sure we know what each variablemeans and the units of measurement. Descriptions of the loan50 variables are given in Figure 1.4.

loan amount interest rate term grade state total income homeownership

1 7500 7.34 36 A MD 70000 rent2 25000 9.43 60 B OH 254000 mortgage3 14500 6.08 36 A MO 80000 mortgage…

……

…50 3000 7.96 36 A CA 34000 rent

Figure 1.3: Four rows from the loan50 data matrix.

variable description

loan amount Amount of the loan received, in US dollars.interest rate Interest rate on the loan, in an annual percentage.term The length of the loan, which is always set as a whole number of months.grade Loan grade, which takes a values A through G and represents the quality

of the loan and its likelihood of being repaid.state US state where the borrower resides.total income Borrower’s total income, including any second income, in US dollars.homeownership Indicates whether the person owns, owns but has a mortgage, or rents.

Figure 1.4: Variables and their descriptions for the loan50 data set.

The data in Figure 1.3 represent a data matrix, which is a convenient and common way toorganize data, especially if collecting data in a spreadsheet. Each row of a data matrix correspondsto a unique case (observational unit), and each column corresponds to a variable.

4The loan’s grade is A, and the borrower rents their residence.

1.2. DATA BASICS 13

When recording data, use a data matrix unless you have a very good reason to use a differentstructure. This structure allows new cases to be added as rows or new variables as new columns.

GUIDED PRACTICE 1.3

The grades for assignments, quizzes, and exams in a course are often recorded in a gradebook thattakes the form of a data matrix. How might you organize grade data using a data matrix?5

GUIDED PRACTICE 1.4

We consider data for 3,142 counties in the United States, which includes each county’s name, thestate where it resides, its population in 2017, how its population changed from 2010 to 2017, povertyrate, and six additional characteristics. How might these data be organized in a data matrix?6

The data described in Guided Practice 1.4 represents the county data set, which is shown asa data matrix in Figure 1.5. The variables are summarized in Figure 1.6.

5There are multiple strategies that can be followed. One common strategy is to have each student represented bya row, and then add a column for each assignment, quiz, or exam. Under this setup, it is easy to review a single lineto understand a student’s grade history. There should also be columns to include student information, such as onecolumn to list student names.

6Each county may be viewed as a case, and there are eleven pieces of information recorded for each case. A tablewith 3,142 rows and 11 columns could hold these data, where each row represents a county and each column representsa particular piece of information.

14 CHAPTER 1. INTRODUCTION TO DATA

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1.2. DATA BASICS 15

1.2.2 Types of variables

Examine the unemp rate, pop, state, and median edu variables in the county data set. Eachof these variables is inherently different from the other three, yet some share certain characteristics.

First consider unemp rate, which is said to be a numerical variable since it can take a widerange of numerical values, and it is sensible to add, subtract, or take averages with those values. Onthe other hand, we would not classify a variable reporting telephone area codes as numerical sincethe average, sum, and difference of area codes doesn’t have any clear meaning.

The pop variable is also numerical, although it seems to be a little different than unemp rate.This variable of the population count can only take whole non-negative numbers (0, 1, 2, …). For thisreason, the population variable is said to be discrete since it can only take numerical values withjumps. On the other hand, the unemployment rate variable is said to be continuous.

The variable state can take up to 51 values after accounting for Washington, DC: AL, AK, …,and WY. Because the responses themselves are categories, state is called a categorical variable, andthe possible values are called the variable’s levels.

Finally, consider the median edu variable, which describes the median education level of countyresidents and takes values below hs, hs diploma, some college, or bachelors in each county. Thisvariable seems to be a hybrid: it is a categorical variable but the levels have a natural ordering.A variable with these properties is called an ordinal variable, while a regular categorical variablewithout this type of special ordering is called a nominal variable. To simplify analyses, any ordinalvariable in this book will be treated as a nominal (unordered) categorical variable.

all variables

numerical categorical

continuous discrete nominal(unordered categorical)

ordinal(ordered categorical)

Figure 1.7: Breakdown of variables into their respective types.

EXAMPLE 1.5

Data were collected about students in a statistics course. Three variables were recorded for eachstudent: number of siblings, student height, and whether the student had previously taken a statisticscourse. Classify each of the variables as continuous numerical, discrete numerical, or categorical.

The number of siblings and student height represent numerical variables. Because the number ofsiblings is a count, it is discrete. Height varies continuously, so it is a continuous numerical variable.The last variable classifies students into two categories – those who have and those who have nottaken a statistics course – which makes this variable categorical.

GUIDED PRACTICE 1.6

An experiment is evaluating the effectiveness of a new drug in treating migraines. A group variableis used to indicate the experiment group for each patient: treatment or control. The num migraines

variable represents the number of migraines the patient experienced during a 3-month period.Classify each variable as either numerical or categorical?7

7There group variable can take just one of two group names, making it categorical. The num migraines variabledescribes a count of the number of migraines, which is an outcome where basic arithmetic is sensible, which means thisis numerical outcome; more specifically, since it represents a count, num migraines is a discrete numerical variable.

16 CHAPTER 1. INTRODUCTION TO DATA

1.2.3 Relationships between variables

Many analyses are motivated by a researcher looking for a relationship between two or morevariables. A social scientist may like to answer some of the following questions:

(1) If homeownership is lower than the national average in one county, will the percent of multi-unitstructures in that county tend to be above or below the national average?

(2) Does a higher than average increase in county population tend to correspond to counties withhigher or lower median household incomes?

(3) How useful a predictor is median education level for the median household income for UScounties?

To answer these questions, data must be collected, such as the county data set shown inFigure 1.5. Examining summary statistics could provide insights for each of the three questionsabout counties. Additionally, graphs can be used to visually explore data.

Scatterplots are one type of graph used to study the relationship between two numerical vari-ables. Figure 1.8 compares the variables homeownership and multi unit, which is the percent ofunits in multi-unit structures (e.g. apartments, condos). Each point on the plot represents a singlecounty. For instance, the highlighted dot corresponds to County 413 in the county data set: Chat-tahoochee County, Georgia, which has 39.4% of units in multi-unit structures and a homeownershiprate of 31.3%. The scatterplot suggests a relationship between the two variables: counties witha higher rate of multi-units tend to have lower homeownership rates. We might brainstorm as towhy this relationship exists and investigate each idea to determine which are the most reasonableexplanations.

Hom

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hip

Rat

0% 20% 40% 60% 80% 100%

20%

40%

60%

80%

●

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Figure 1.8: A scatterplot of homeownership versus the percent of units that arein multi-unit structures for US counties. The highlighted dot represents Chatta-hoochee County, Georgia, which has a multi-unit rate of 39.4% and a homeowner-ship rate of 31.3%.

The multi-unit and homeownership rates are said to be associated because the plot shows adiscernible pattern. When two variables show some connection with one another, they are calledassociated variables. Associated variables can also be called dependent variables and vice-versa.

1.2. DATA BASICS 17

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−10%

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Figure 1.9: A scatterplot showing pop change against median hh income. OwsleyCounty of Kentucky, is highlighted, which lost 3.63% of its population from 2010to 2017 and had median household income of $22,736.

GUIDED PRACTICE 1.7

Examine the variables in the loan50 data set, which are described in Figure 1.4 on page 12. Createtwo questions about possible relationships between variables in loan50 that are of interest to you.8

EXAMPLE 1.8

This example examines the relationship between a county’s population change from 2010 to 2017and median household income, which is visualized as a scatterplot in Figure 1.9. Are these variablesassociated?

The larger the median household income for a county, the higher the population growth observedfor the county. While this trend isn’t true for every county, the trend in the plot is evident. Sincethere is some relationship between the variables, they are associated.

Because there is a downward trend in Figure 1.8 – counties with more units in multi-unitstructures are associated with lower homeownership – these variables are said to be negativelyassociated. A positive association is shown in the relationship between the median hh income

and pop change in Figure 1.9, where counties with higher median household income tend to havehigher rates of population growth.

If two variables are not associated, then they are said to be independent. That is, twovariables are independent if there is no evident relationship between the two.

ASSOCIATED OR INDEPENDENT, NOT BOTH

A pair of variables are either related in some way (associated) or not (independent). No pair ofvariables is both associated and independent.

8Two example questions: (1) What is the relationship between loan amount and total income? (2) If someone’sincome is above the average, will their interest rate tend to be above or below the average?

18 CHAPTER 1. INTRODUCTION TO DATA

1.2.4 Explanatory and response variables

When we ask questions about the relationship between two variables, we sometimes also wantto determine if the change in one variable causes a change in the other. Consider the followingrephrasing of an earlier question about the county data set:

If there is an increase in the median household income in a county, does this drive anincrease in its population?

In this question, we are asking whether one variable affects another. If this is our underlyingbelief, then median household income is the explanatory variable and the population change is theresponse variable in the hypothesized relationship.9

EXPLANATORY AND RESPONSE VARIABLES

When we suspect one variable might causally affect another, we label the first variable theexplanatory variable and the second the response variable.

might affectexplanatoryvariable

responsevariable

For many pairs of variables, there is no hypothesized relationship, and these labels would notbe applied to either variable in such cases.

Bear in mind that the act of labeling the variables in this way does nothing to guarantee thata causal relationship exists. A formal evaluation to check whether one variable causes a change inanother requires an experiment.

1.2.5 Introducing observational studies and experiments

There are two primary types of data collection: observational studies and experiments.Researchers perform an observational study when they collect data in a way that does not

directly interfere with how the data arise. For instance, researchers may collect information viasurveys, review medical or company records, or follow a cohort of many similar individuals to formhypotheses about why certain diseases might develop. In each of these situations, researchers merelyobserve the data that arise. In general, observational studies can provide evidence of a naturallyoccurring association between variables, but they cannot by themselves show a causal connection.

When researchers want to investigate the possibility of a causal connection, they conduct anexperiment. Usually there will be both an explanatory and a response variable. For instance, wemay suspect administering a drug will reduce mortality in heart attack patients over the followingyear. To check if there really is a causal connection between the explanatory variable and theresponse, researchers will collect a sample of individuals and split them into groups. The individualsin each group are assigned a treatment. When individuals are randomly assigned to a group, theexperiment is called a randomized experiment. For example, each heart attack patient in thedrug trial could be randomly assigned, perhaps by flipping a coin, into one of two groups: thefirst group receives a placebo (fake treatment) and the second group receives the drug. See thecase study in Section 1.1 for another example of an experiment, though that study did not employa placebo.

ASSOCIATION 6= CAUSATION

In general, association does not imply causation, and causation can only be inferred from arandomized experiment.

9Sometimes the explanatory variable is called the independent variable and the response variable is called thedependent variable. However, this becomes confusing since a pair of variables might be independent or dependent,so we avoid this language.

1.2. DATA BASICS 19

Exercises

1.3 Air pollution and birth outcomes, study components. Researchers collected data to examine therelationship between air pollutants and preterm births in Southern California. During the study air pollutionlevels were measured by air quality monitoring stations. Specifically, levels of carbon monoxide were recordedin parts per million, nitrogen dioxide and ozone in parts per hundred million, and coarse particulate matter(PM10) in µg/m3. Length of gestation data were collected on 143,196 births between the years 1989 and1993, and air pollution exposure during gestation was calculated for each birth. The analysis suggested thatincreased ambient PM10 and, to a lesser degree, CO concentrations may be associated with the occurrenceof preterm births.10

(a) Identify the main research question of the study.

(b) Who are the subjects in this study, and how many are included?

(c) What are the variables in the study? Identify each variable as numerical or categorical. If numerical,state whether the variable is discrete or continuous. If categorical, state whether the variable is ordinal.

1.4 Buteyko method, study components. The Buteyko method is a shallow breathing technique devel-oped by Konstantin Buteyko, a Russian doctor, in 1952. Anecdotal evidence suggests that the Buteykomethod can reduce asthma symptoms and improve quality of life. In a scientific study to determine theeffectiveness of this method, researchers recruited 600 asthma patients aged 18-69 who relied on medicationfor asthma treatment. These patients were randomnly split into two research groups: one practiced theButeyko method and the other did not. Patients were scored on quality of life, activity, asthma symptoms,and medication reduction on a scale from 0 to 10. On average, the participants in the Buteyko groupexperienced a significant reduction in asthma symptoms and an improvement in quality of life.11

(a) Identify the main research question of the study.

(b) Who are the subjects in this study, and how many are included?

1.5 Cheaters, study components. Researchers studying the relationship between honesty, age and self-control conducted an experiment on 160 children between the ages of 5 and 15. Participants reported theirage, sex, and whether they were an only child or not. The researchers asked each child to toss a fair coinin private and to record the outcome (white or black) on a paper sheet, and said they would only rewardchildren who report white. The study’s findings can be summarized as follows: “Half the students wereexplicitly told not to cheat and the others were not given any explicit instructions. In the no instructiongroup probability of cheating was found to be uniform across groups based on child’s characteristics. In thegroup that was explicitly told to not cheat, girls were less likely to cheat, and while rate of cheating didn’tvary by age for boys, it decreased with age for girls.”12

(a) Identify the main research question of the study.

(b) Who are the subjects in this study, and how many are included?

(c) How many variables were recorded for each subject in the study in order to conclude these findings?State the variables and their types.

10B. Ritz et al. “Effect of air pollution on preterm birth among children born in Southern California between 1989and 1993”. In: Epidemiology 11.5 (2000), pp. 502–511.

11J. McGowan. “Health Education: Does the Buteyko Institute Method make a difference?” In: Thorax 58 (2003).12Alessandro Bucciol and Marco Piovesan. “Luck or cheating? A field experiment on honesty with children”. In:

Journal of Economic Psychology 32.1 (2011), pp. 73–78.

20 CHAPTER 1. INTRODUCTION TO DATA

1.6 Stealers, study components. In a study of the relationship between socio-economic class and unethicalbehavior, 129 University of California undergraduates at Berkeley were asked to identify themselves ashaving low or high social-class by comparing themselves to others with the most (least) money, most (least)education, and most (least) respected jobs. They were also presented with a jar of individually wrappedcandies and informed that the candies were for children in a nearby laboratory, but that they could takesome if they wanted. After completing some unrelated tasks, participants reported the number of candiesthey had taken.13

(a) Identify the main research question of the study.

(b) Who are the subjects in this study, and how many are included?

(c) The study found that students who were identified as upper-class took more candy than others. Howmany variables were recorded for each subject in the study in order to conclude these findings? Statethe variables and their types.

1.7 Migraine and acupuncture, Part II. Exercise 1.1 introduced a study exploring whether acupuncturehad any effect on migraines. Researchers conducted a randomized controlled study where patients wererandomly assigned to one of two groups: treatment or control. The patients in the treatment group re-ceived acupuncture that was specifically designed to treat migraines. The patients in the control groupreceived placebo acupuncture (needle insertion at non-acupoint locations). 24 hours after patients receivedacupuncture, they were asked if they were pain free. What are the explanatory and response variables inthis study?

1.8 Sinusitis and antibiotics, Part II. Exercise 1.2 introduced a study exploring the effect of antibiotictreatment for acute sinusitis. Study participants either received either a 10-day course of an antibiotic(treatment) or a placebo similar in appearance and taste (control). At the end of the 10-day period, patientswere asked if they experienced improvement in symptoms. What are the explanatory and response variablesin this study?

1.9 Fisher’s irises. Sir Ronald Aylmer Fisher was an English statistician, evolutionary biologist, andgeneticist who worked on a data set that contained sepal length and width, and petal length and width fromthree species of iris flowers (setosa, versicolor and virginica). There were 50 flowers from each species in thedata set.14

(a) How many cases were included in the data?

(b) How many numerical variables are included inthe data? Indicate what they are, and if theyare continuous or discrete.

(c) How many categorical variables are included inthe data, and what are they? List the corre-sponding levels (categories).

Photo by Ryan Claussen

(http://flic.kr/p/6QTcuX)

CC BY-SA 2.0 license

1.10 Smoking habits of UK residents. A survey was conducted to study the smoking habits of UKresidents. Below is a data matrix displaying a portion of the data collected in this survey. Note that “£”stands for British Pounds Sterling, “cig” stands for cigarettes, and “N/A” refers to a missing component ofthe data.15

sex age marital grossIncome smoke amtWeekends amtWeekdays1 Female 42 Single Under £2,600 Yes 12 cig/day 12 cig/day2 Male 44 Single £10,400 to £15,600 No N/A N/A3 Male 53 Married Above £36,400 Yes 6 cig/day 6 cig/day

….

1691 Male 40 Single £2,600 to £5,200 Yes 8 cig/day 8 cig/day

(a) What does each row of the data matrix represent?

(b) How many participants were included in the survey?

(c) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin-uous or discrete. If categorical, indicate if the variable is ordinal.

13P.K. Piff et al. “Higher social class predicts increased unethical behavior”. In: Proceedings of the NationalAcademy of Sciences (2012).

14R.A Fisher. “The Use of Multiple Measurements in Taxonomic Problems”. In: Annals of Eugenics 7 (1936),pp. 179–188.

15National STEM Centre, Large Datasets from stats4schools.

1.2. DATA BASICS 21

1.11 US Airports. The visualization below shows the geographical distribution of airports in the contiguousUnited States and Washington, DC. This visualization was constructed based on a dataset where eachobservation is an airport.16

(a) List the variables used in creating this visualization.

(b) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin-uous or discrete. If categorical, indicate if the variable is ordinal.

1.12 UN Votes. The visualization below shows voting patterns in the United States, Canada, and Mexico inthe United Nations General Assembly on a variety of issues. Specifically, for a given year between 1946 and2015, it displays the percentage of roll calls in which the country voted yes for each issue. This visualizationwas constructed based on a dataset where each observation is a country/year pair.17

(a) List the variables used in creating this visualization.

(b) Indicate whether each variable in the study is numerical or categorical. If numerical, identify as contin-uous or discrete. If categorical, indicate if the variable is ordinal.

16Federal Aviation Administration, www.faa.gov/airports/airport safety/airportdata 5010.17David Robinson. unvotes: United Nations General Assembly Voting Data. R package version 0.2.0. 2017. url:

https://CRAN.R-project.org/package=unvotes.

22 CHAPTER 1. INTRODUCTION TO DATA

1.3 Sampling principles and strategies

The first step in conducting research is to identify topics or questions that are to be investigated.A clearly laid out research question is helpful in identifying what subjects or cases should be studiedand what variables are important. It is also important to consider how data are collected so thatthey are reliable and help achieve the research goals.

1.3.1 Populations and samples

Consider the following three research questions:

1. What is the average mercury content in swordfish in the Atlantic Ocean?

2. Over the last 5 years, what is the average time to complete a degree for Duke undergrads?

3. Does a new drug reduce the number of deaths in patients with severe heart disease?

Each research question refers to a target population. In the first question, the target population isall swordfish in the Atlantic ocean, and each fish represents a case. Often times, it is too expensiveto collect data for every case in a population. Instead, a sample is taken. A sample representsa subset of the cases and is often a small fraction of the population. For instance, 60 swordfish(or some other number) in the population might be selected, and this sample data may be used toprovide an estimate of the population average and answer the research question.

GUIDED PRACTICE 1.9

For the second and third questions above, identify the target population and what represents anindividual case.18

1.3.2 Anecdotal evidence

Consider the following possible responses to the three research questions:

1. A man on the news got mercury poisoning from eating swordfish, so the average mercuryconcentration in swordfish must be dangerously high.

2. I met two students who took more than 7 years to graduate from Duke, so it must take longerto graduate at Duke than at many other colleges.

3. My friend’s dad had a heart attack and died after they gave him a new heart disease drug,so the drug must not work.

Each conclusion is based on data. However, there are two problems. First, the data only representone or two cases. Second, and more importantly, it is unclear whether these cases are actuallyrepresentative of the population. Data collected in this haphazard fashion are called anecdotalevidence.

ANECDOTAL EVIDENCE

Be careful of data collected in a haphazard fashion. Such evidence may be true and verifiable,but it may only represent extraordinary cases.

18(2) The first question is only relevant to students who complete their degree; the average cannot be computedusing a student who never finished her degree. Thus, only Duke undergrads who graduated in the last five yearsrepresent cases in the population under consideration. Each such student is an individual case. (3) A person withsevere heart disease represents a case. The population includes all people with severe heart disease.

1.3. SAMPLING PRINCIPLES AND STRATEGIES 23

Figure 1.10: In February 2010, some media punditscited one large snow storm as valid evidence againstglobal warming. As comedian Jon Stewart pointedout, “It’s one storm, in one region, of one country.”

Anecdotal evidence typically is composed of unusual cases that we recall based on their strikingcharacteristics. For instance, we are more likely to remember the two people we met who took 7 yearsto graduate than the six others who graduated in four years. Instead of looking at the most unusualcases, we should examine a sample of many cases that represent the population.

1.3.3 Sampling from a population

We might try to estimate the time to graduation for Duke undergraduates in the last 5 yearsby collecting a sample of students. All graduates in the last 5 years represent the population, andgraduates who are selected for review are collectively called the sample. In general, we always seek torandomly select a sample from a population. The most basic type of random selection is equivalentto how raffles are conducted. For example, in selecting graduates, we could write each graduate’sname on a raffle ticket and draw 100 tickets. The selected names would represent a random sampleof 100 graduates. We pick samples randomly to reduce the chance we introduce biases.

all graduates

sample

Figure 1.11: In this graphic, five graduates are randomly selected from the popu-lation to be included in the sample.

EXAMPLE 1.10

Suppose we ask a student who happens to be majoring in nutrition to select several graduates forthe study. What kind of students do you think she might collect? Do you think her sample wouldbe representative of all graduates?

Perhaps she would pick a disproportionate number of graduates from health-related fields. Or per-haps her selection would be a good representation of the population. When selecting samples byhand, we run the risk of picking a biased sample, even if their bias isn’t intended.

24 CHAPTER 1. INTRODUCTION TO DATA

all graduates

sample

graduates fromhealth−related fields

Figure 1.12: Asked to pick a sample of graduates, a nutrition major might inad-vertently pick a disproportionate number of graduates from health-related majors.

If someone was permitted to pick and choose exactly which graduates were included in thesample, it is entirely possible that the sample could be skewed to that person’s interests, which maybe entirely unintentional. This introduces bias into a sample. Sampling randomly helps resolvethis problem. The most basic random sample is called a simple random sample, and which isequivalent to using a raffle to select cases. This means that each case in the population has an equalchance of being included and there is no implied connection between the cases in the sample.

The act of taking a simple random sample helps minimize bias. However, bias can crop up inother ways. Even when people are picked at random, e.g. for surveys, caution must be exercisedif the non-response rate is high. For instance, if only 30% of the people randomly sampled fora survey actually respond, then it is unclear whether the results are representative of the entirepopulation. This non-response bias can skew results.

population of interest

sample

population actuallysampled

Figure 1.13: Due to the possibility of non-response, surveys studies may only reacha certain group within the population. It is difficult, and often times impossible,to completely fix this problem.

Another common downfall is a convenience sample, where individuals who are easily ac-cessible are more likely to be included in the sample. For instance, if a political survey is doneby stopping people walking in the Bronx, this will not represent all of New York City. It is oftendifficult to discern what sub-population a convenience sample represents.

GUIDED PRACTICE 1.11

We can easily access ratings for products, sellers, and companies through websites. These ratingsare based only on those people who go out of their way to provide a rating. If 50% of online reviewsfor a product are negative, do you think this means that 50% of buyers are dissatisfied with theproduct?19

19Answers will vary. From our own anecdotal experiences, we believe people tend to rant more about productsthat fell below expectations than rave about those that perform as expected. For this reason, we suspect there is anegative bias in product ratings on sites like Amazon. However, since our experiences may not be representative, wealso keep an open mind.

1.3. SAMPLING PRINCIPLES AND STRATEGIES 25

1.3.4 Observational studies

Data where no treatment has been explicitly applied (or explicitly withheld) is called observa-tional data. For instance, the loan data and county data described in Section 1.2 are both examplesof observational data. Making causal conclusions based on experiments is often reasonable. How-ever, making the same causal conclusions based on observational data can be treacherous and is notrecommended. Thus, observational studies are generally only sufficient to show associations or formhypotheses that we later check using experiments.

GUIDED PRACTICE 1.12

Suppose an observational study tracked sunscreen use and skin cancer, and it was found that themore sunscreen someone used, the more likely the person was to have skin cancer. Does this meansunscreen causes skin cancer?20

Some previous research tells us that using sunscreen actually reduces skin cancer risk, so maybethere is another variable that can explain this hypothetical association between sunscreen usage andskin cancer. One important piece of information that is absent is sun exposure. If someone is outin the sun all day, she is more likely to use sunscreen and more likely to get skin cancer. Exposureto the sun is unaccounted for in the simple investigation.

sun exposure

use sunscreen skin cancer?

Sun exposure is what is called a confounding variable,21 which is a variable that is correlatedwith both the explanatory and response variables. While one method to justify making causalconclusions from observational studies is to exhaust the search for confounding variables, there is noguarantee that all confounding variables can be examined or measured.

GUIDED PRACTICE 1.13

Figure 1.8 shows a negative association between the homeownership rate and the percentage of multi-unit structures in a county. However, it is unreasonable to conclude that there is a causal relationshipbetween the two variables. Suggest a variable that might explain the negative relationship.22

Observational studies come in two forms: prospective and retrospective studies. A prospec-tive study identifies individuals and collects information as events unfold. For instance, medicalresearchers may identify and follow a group of patients over many years to assess the possible influ-ences of behavior on cancer risk. One example of such a study is The Nurses’ Health Study, startedin 1976 and expanded in 1989. This prospective study recruits registered nurses and then collectsdata from them using questionnaires. Retrospective studies collect data after events have takenplace, e.g. researchers may review past events in medical records. Some data sets may contain bothprospectively- and retrospectively-collected variables.

1.3.5 Four sampling methods

Almost all statistical methods are based on the notion of implied randomness. If observationaldata are not collected in a random framework from a population, these statistical methods – theestimates and errors associated with the estimates – are not reliable. Here we consider four randomsampling techniques: simple, stratified, cluster, and multistage sampling. Figures 1.14 and 1.15provide graphical representations of these techniques.

20No. See the paragraph following the exercise for an explanation.21Also called a lurking variable, confounding factor, or a confounder.22Answers will vary. Population density may be important. If a county is very dense, then this may require a

larger fraction of residents to live in multi-unit structures. Additionally, the high density may contribute to increasesin property value, making homeownership infeasible for many residents.

26 CHAPTER 1. INTRODUCTION TO DATA

Index

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Stratum 1

Stratum 2

Stratum 3

Stratum 4

Stratum 5

Stratum 6

Figure 1.14: Examples of simple random and stratified sampling. In the top panel,simple random sampling was used to randomly select the 18 cases. In the bottompanel, stratified sampling was used: cases were grouped into strata, then simplerandom sampling was employed within each stratum.

1.3. SAMPLING PRINCIPLES AND STRATEGIES 27

Simple random sampling is probably the most intuitive form of random sampling. Considerthe salaries of Major League Baseball (MLB) players, where each player is a member of one of theleague’s 30 teams. To take a simple random sample of 120 baseball players and their salaries, wecould write the names of that season’s several hundreds of players onto slips of paper, drop the slipsinto a bucket, shake the bucket around until we are sure the names are all mixed up, then draw outslips until we have the sample of 120 players. In general, a sample is referred to as “simple random”if each case in the population has an equal chance of being included in the final sample and knowingthat a case is included in a sample does not provide useful information about which other cases areincluded.

Stratified sampling is a divide-and-conquer sampling strategy. The population is dividedinto groups called strata. The strata are chosen so that similar cases are grouped together, then asecond sampling method, usually simple random sampling, is employed within each stratum. In thebaseball salary example, the teams could represent the strata, since some teams have a lot moremoney (up to 4 times as much!). Then we might randomly sample 4 players from each team for atotal of 120 players.

Stratified sampling is especially useful when the cases in each stratum are very similar withrespect to the outcome of interest. The downside is that analyzing data from a stratified sampleis a more complex task than analyzing data from a simple random sample. The analysis methodsintroduced in this book would need to be extended to analyze data collected using stratified sampling.

EXAMPLE 1.14

Why would it be good for cases within each stratum to be very similar?

We might get a more stable estimate for the subpopulation in a stratum if the cases are very similar,leading to more precise estimates within each group. When we combine these estimates into a singleestimate for the full population, that population estimate will tend to be more precise since eachindividual group estimate is itself more precise.

In a cluster sample, we break up the population into many groups, called clusters. Thenwe sample a fixed number of clusters and include all observations from each of those clusters in thesample. A multistage sample is like a cluster sample, but rather than keeping all observations ineach cluster, we collect a random sample within each selected cluster.

Sometimes cluster or multistage sampling can be more economical than the alternative samplingtechniques. Also, unlike stratified sampling, these approaches are most helpful when there is a lot ofcase-to-case variability within a cluster but the clusters themselves don’t look very different from oneanother. For example, if neighborhoods represented clusters, then cluster or multistage samplingwork best when the neighborhoods are very diverse. A downside of these methods is that moreadvanced techniques are typically required to analyze the data, though the methods in this bookcan be extended to handle such data.

EXAMPLE 1.15

Suppose we are interested in estimating the malaria rate in a densely tropical portion of ruralIndonesia. We learn that there are 30 villages in that part of the Indonesian jungle, each more orless similar to the next. Our goal is to test 150 individuals for malaria. What sampling methodshould be employed?

A simple random sample would likely draw individuals from all 30 villages, which could make datacollection extremely expensive. Stratified sampling would be a challenge since it is unclear how wewould build strata of similar individuals. However, cluster sampling or multistage sampling seemlike very good ideas. If we decided to use multistage sampling, we might randomly select half of thevillages, then randomly select 10 people from each. This would probably reduce our data collectioncosts substantially in comparison to a simple random sample, and the cluster sample would stillgive us reliable information, even if we would need to analyze the data with slightly more advancedmethods than we discuss in this book.

28 CHAPTER 1. INTRODUCTION TO DATA

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1.3. SAMPLING PRINCIPLES AND STRATEGIES 29

Exercises

1.13 Air pollution and birth outcomes, scope of inference. Exercise 1.3 introduces a study whereresearchers collected data to examine the relationship between air pollutants and preterm births in SouthernCalifornia. During the study air pollution levels were measured by air quality monitoring stations. Length ofgestation data were collected on 143,196 births between the years 1989 and 1993, and air pollution exposureduring gestation was calculated for each birth.

(a) Identify the population of interest and the sample in this study.

(b) Comment on whether or not the results of the study can be generalized to the population, and if thefindings of the study can be used to establish causal relationships.

1.14 Cheaters, scope of inference. Exercise 1.5 introduces a study where researchers studying the rela-tionship between honesty, age, and self-control conducted an experiment on 160 children between the agesof 5 and 15. The researchers asked each child to toss a fair coin in private and to record the outcome (whiteor black) on a paper sheet, and said they would only reward children who report white. Half the studentswere explicitly told not to cheat and the others were not given any explicit instructions. Differences wereobserved in the cheating rates in the instruction and no instruction groups, as well as some differences acrosschildren’s characteristics within each group.

(a) Identify the population of interest and the sample in this study.

(b) Comment on whether or not the results of the study can be generalized to the population, and if thefindings of the study can be used to establish causal relationships.

1.15 Buteyko method, scope of inference. Exercise 1.4 introduces a study on using the Buteyko shallowbreathing technique to reduce asthma symptoms and improve quality of life. As part of this study 600asthma patients aged 18-69 who relied on medication for asthma treatment were recruited and randomlyassigned to two groups: one practiced the Buteyko method and the other did not. Those in the Buteykogroup experienced, on average, a significant reduction in asthma symptoms and an improvement in qualityof life.

(a) Identify the population of interest and the sample in this study.

(b) Comment on whether or not the results of the study can be generalized to the population, and if thefindings of the study can be used to establish causal relationships.

1.16 Stealers, scope of inference. Exercise 1.6 introduces a study on the relationship between socio-economic class and unethical behavior. As part of this study 129 University of California Berkeley under-graduates were asked to identify themselves as having low or high social-class by comparing themselves toothers with the most (least) money, most (least) education, and most (least) respected jobs. They were alsopresented with a jar of individually wrapped candies and informed that the candies were for children in anearby laboratory, but that they could take some if they wanted. After completing some unrelated tasks,participants reported the number of candies they had taken. It was found that those who were identified asupper-class took more candy than others.

(a) Identify the population of interest and the sample in this study.

(b) Comment on whether or not the results of the study can be generalized to the population, and if thefindings of the study can be used to establish causal relationships.

1.17 Relaxing after work. The General Social Survey asked the question, “After an average work day,about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155Americans. The average relaxing time was found to be 1.65 hours. Determine which of the following is anobservation, a variable, a sample statistic (value calculated based on the observed sample), or a populationparameter.

(a) An American in the sample.

(b) Number of hours spent relaxing after an average work day.

(d) Average number of hours all Americans spend relaxing after an average work day.

30 CHAPTER 1. INTRODUCTION TO DATA

1.18 Cats on YouTube. Suppose you want to estimate the percentage of videos on YouTube that are catvideos. It is impossible for you to watch all videos on YouTube so you use a random video picker to select1000 videos for you. You find that 2% of these videos are cat videos. Determine which of the following is anobservation, a variable, a sample statistic (value calculated based on the observed sample), or a populationparameter.

(a) Percentage of all videos on YouTube that are cat videos.

(b) 2%.

(d) Whether or not a video is a cat video.

1.19 Course satisfaction across sections. A large college class has 160 students. All 160 studentsattend the lectures together, but the students are divided into 4 groups, each of 40 students, for lab sectionsadministered by different teaching assistants. The professor wants to conduct a survey about how satisfiedthe students are with the course, and he believes that the lab section a student is in might affect the student’soverall satisfaction with the course.

(a) What type of study is this?

(b) Suggest a sampling strategy for carrying out this study.

1.20 Housing proposal across dorms. On a large college campus first-year students and sophomores livein dorms located on the eastern part of the campus and juniors and seniors live in dorms located on thewestern part of the campus. Suppose you want to collect student opinions on a new housing structure thecollege administration is proposing and you want to make sure your survey equally represents opinions fromstudents from all years.

(a) What type of study is this?

(b) Suggest a sampling strategy for carrying out this study.

1.21 Internet use and life expectancy. The following scatterplot was created as part of a study evaluatingthe relationship between estimated life expectancy at birth (as of 2014) and percentage of internet users (asof 2009) in 208 countries for which such data were available.23

(a) Describe the relationship between life ex-pectancy and percentage of internet users.

(b) What type of study is this?

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1.22 Stressed out, Part I. A study that surveyed a random sample of otherwise healthy high school studentsfound that they are more likely to get muscle cramps when they are stressed. The study also noted thatstudents drink more coffee and sleep less when they are stressed.

(a) What type of study is this?

(b) Can this study be used to conclude a causal relationship between increased stress and muscle cramps?

(c) State possible confounding variables that might explain the observed relationship between increasedstress and muscle cramps.

1.23 Evaluate sampling methods. A university wants to determine what fraction of its undergraduatestudent body support a new $25 annual fee to improve the student union. For each proposed method below,indicate whether the method is reasonable or not.

(a) Survey a simple random sample of 500 students.

(b) Stratify students by their field of study, then sample 10% of students from each stratum.

(c) Cluster students by their ages (e.g. 18 years old in one cluster, 19 years old in one cluster, etc.), thenrandomly sample three clusters and survey all students in those clusters.

23CIA Factbook, Country Comparisons, 2014.

1.3. SAMPLING PRINCIPLES AND STRATEGIES 31

1.24 Random digit dialing. The Gallup Poll uses a procedure called random digit dialing, which createsphone numbers based on a list of all area codes in America in conjunction with the associated number ofresidential households in each area code. Give a possible reason the Gallup Poll chooses to use random digitdialing instead of picking phone numbers from the phone book.

1.25 Haters are gonna hate, study confirms. A study published in the Journal of Personality andSocial Psychology asked a group of 200 randomly sampled men and women to evaluate how they felt aboutvarious subjects, such as camping, health care, architecture, taxidermy, crossword puzzles, and Japan inorder to measure their attitude towards mostly independent stimuli. Then, they presented the participantswith information about a new product: a microwave oven. This microwave oven does not exist, but theparticipants didn’t know this, and were given three positive and three negative fake reviews. People whoreacted positively to the subjects on the dispositional attitude measurement also tended to react positivelyto the microwave oven, and those who reacted negatively tended to react negatively to it. Researchersconcluded that “some people tend to like things, whereas others tend to dislike things, and a more thoroughunderstanding of this tendency will lead to a more thorough understanding of the psychology of attitudes.”24

(a) What are the cases?

(b) What is (are) the response variable(s) in this study?

(d) Does the study employ random sampling?

(e) Is this an observational study or an experiment? Explain your reasoning.

(f) Can we establish a causal link between the explanatory and response variables?

(g) Can the results of the study be generalized to the population at large?

1.26 Family size. Suppose we want to estimate household size, where a “household” is defined as peopleliving together in the same dwelling, and sharing living accommodations. If we select students at randomat an elementary school and ask them what their family size is, will this be a good measure of householdsize? Or will our average be biased? If so, will it overestimate or underestimate the true value?

1.27 Sampling strategies. A statistics student who is curious about the relationship between the amountof time students spend on social networking sites and their performance at school decides to conduct asurvey. Various research strategies for collecting data are described below. In each, name the samplingmethod proposed and any bias you might expect.

(a) He randomly samples 40 students from the study’s population, gives them the survey, asks them to fillit out and bring it back the next day.

(b) He gives out the survey only to his friends, making sure each one of them fills out the survey.

(d) He randomly samples 5 classes and asks a random sample of students from those classes to fill out thesurvey.

1.28 Reading the paper. Below are excerpts from two articles published in the NY Times:

(a) An article titled Risks: Smokers Found More Prone to Dementia states the following:25

“Researchers analyzed data from 23,123 health plan members who participated in a voluntary exam and

health behavior survey from 1978 to 1985, when they were 50-60 years old. 23 years later, about 25% of

the group had dementia, including 1,136 with Alzheimer’s disease and 416 with vascular dementia. After

adjusting for other factors, the researchers concluded that pack-a-day smokers were 37% more likely than

nonsmokers to develop dementia, and the risks went up with increased smoking; 44% for one to two packs

a day; and twice the risk for more than two packs.”

Based on this study, can we conclude that smoking causes dementia later in life? Explain your reasoning.

(b) Another article titled The School Bully Is Sleepy states the following:26

“The University of Michigan study, collected survey data from parents on each child’s sleep habits and

asked both parents and teachers to assess behavioral concerns. About a third of the students studied were

identified by parents or teachers as having problems with disruptive behavior or bullying. The researchers

found that children who had behavioral issues and those who were identified as bullies were twice as likely

to have shown symptoms of sleep disorders.”

A friend of yours who read the article says, “The study shows that sleep disorders lead to bullying inschool children.” Is this statement justified? If not, how best can you describe the conclusion that canbe drawn from this study?

24Justin Hepler and Dolores Albarraćın. “Attitudes without objects – Evidence for a dispositional attitude, itsmeasurement, and its consequences”. In: Journal of personality and social psychology 104.6 (2013), p. 1060.

25R.C. Rabin. “Risks: Smokers Found More Prone to Dementia”. In: New York Times (2010).26T. Parker-Pope. “The School Bully Is Sleepy”. In: New York Times (2011).

32 CHAPTER 1. INTRODUCTION TO DATA

1.4 Experiments

Studies where the researchers assign treatments to cases are called experiments. When thisassignment includes randomization, e.g. using a coin flip to decide which treatment a patient receives,it is called a randomized experiment. Randomized experiments are fundamentally importantwhen trying to show a causal connection between two variables.

1.4.1 Principles of experimental design

Randomized experiments are generally built on four principles.

Controlling. Researchers assign treatments to cases, and they do their best to control any otherdifferences in the groups.27 For example, when patients take a drug in pill form, some patientstake the pill with only a sip of water while others may have it with an entire glass of water. Tocontrol for the effect of water consumption, a doctor may ask all patients to drink a 12 ounceglass of water with the pill.

Randomization. Researchers randomize patients into treatment groups to account for variablesthat cannot be controlled. For example, some patients may be more susceptible to a diseasethan others due to their dietary habits. Randomizing patients into the treatment or controlgroup helps even out such differences, and it also prevents accidental bias from entering thestudy.

Replication. The more cases researchers observe, the more accurately they can estimate the effectof the explanatory variable on the response. In a single study, we replicate by collecting asufficiently large sample. Additionally, a group of scientists may replicate an entire study toverify an earlier finding.

Blocking. Researchers sometimes know or suspect that variables, other than the treatment, influ-ence the response. Under these circumstances, they may first group individuals based on thisvariable into blocks and then randomize cases within each block to the treatment groups. Thisstrategy is often referred to as blocking. For instance, if we are looking at the effect of a drugon heart attacks, we might first split patients in the study into low-risk and high-risk blocks,then randomly assign half the patients from each block to the control group and the other halfto the treatment group, as shown in Figure 1.16. This strategy ensures each treatment grouphas an equal number of low-risk and high-risk patients.

It is important to incorporate the first three experimental design principles into any study, andthis book describes applicable methods for analyzing data from such experiments. Blocking is aslightly more advanced technique, and statistical methods in this book may be extended to analyzedata collected using blocking.

1.4.2 Reducing bias in human experiments

Randomized experiments are the gold standard for data collection, but they do not ensure anunbiased perspective into the cause and effect relationship in all cases. Human studies are perfectexamples where bias can unintentionally arise. Here we reconsider a study where a new drug wasused to treat heart attack patients. In particular, researchers wanted to know if the drug reduceddeaths in patients.

These researchers designed a randomized experiment because they wanted to draw causal con-clusions about the drug’s effect. Study volunteers28 were randomly placed into two study groups.One group, the treatment group, received the drug. The other group, called the control group,did not receive any drug treatment.

27This is a different concept than a control group, which we discuss in the second principle and in Section 1.4.2.28Human subjects are often called patients, volunteers, or study participants.

1.4. EXPERIMENTS 33

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Figure 1.16: Blocking using a variable depicting patient risk. Patients are firstdivided into low-risk and high-risk blocks, then each block is evenly separatedinto the treatment groups using randomization. This strategy ensures an equalrepresentation of patients in each treatment group from both the low-risk andhigh-risk categories.

34 CHAPTER 1. INTRODUCTION TO DATA

Put yourself in the place of a person in the study. If you are in the treatment group, youare given a fancy new drug that you anticipate will help you. On the other hand, a person in theother group doesn’t receive the drug and sits idly, hoping her participation doesn’t increase herrisk of death. These perspectives suggest there are actually two effects: the one of interest is theeffectiveness of the drug, and the second is an emotional effect that is difficult to quantify.

Researchers aren’t usually interested in the emotional effect, which might bias the study. Tocircumvent this problem, researchers do not want patients to know which group they are in. Whenresearchers keep the patients uninformed about their treatment, the study is said to be blind. Butthere is one problem: if a patient doesn’t receive a treatment, she will know she is in the controlgroup. The solution to this problem is to give fake treatments to patients in the control group.A fake treatment is called a placebo, and an effective placebo is the key to making a study trulyblind. A classic example of a placebo is a sugar pill that is made to look like the actual treatmentpill. Often times, a placebo results in a slight but real improvement in patients. This effect has beendubbed the placebo effect.

The patients are not the only ones who should be blinded: doctors and researchers can ac-cidentally bias a study. When a doctor knows a patient has been given the real treatment, shemight inadvertently give that patient more attention or care than a patient that she knows is onthe placebo. To guard against this bias, which again has been found to have a measurable effectin some instances, most modern studies employ a double-blind setup where doctors or researcherswho interact with patients are, just like the patients, unaware of who is or is not receiving thetreatment.29

GUIDED PRACTICE 1.16

Look back to the study in Section 1.1 where researchers were testing whether stents were effectiveat reducing strokes in at-risk patients. Is this an experiment? Was the study blinded? Was itdouble-blinded?30

GUIDED PRACTICE 1.17

For the study in Section 1.1, could the researchers have employed a placebo? If so, what would thatplacebo have looked like?31

You may have many questions about the ethics of sham surgeries to create a placebo afterreading Guided Practice 1.17. These questions may have even arisen in your mind when in thegeneral experiment context, where a possibly helpful treatment was withheld from individuals inthe control group; the main difference is that a sham surgery tends to create additional risk, whilewithholding a treatment only maintains a person’s risk.

There are always multiple viewpoints of experiments and placebos, and rarely is it obviouswhich is ethically “correct”. For instance, is it ethical to use a sham surgery when it creates a risk tothe patient? However, if we don’t use sham surgeries, we may promote the use of a costly treatmentthat has no real effect; if this happens, money and other resources will be diverted away from othertreatments that are known to be helpful. Ultimately, this is a difficult situation where we cannotperfectly protect both the patients who have volunteered for the study and the patients who maybenefit (or not) from the treatment in the future.

29There are always some researchers involved in the study who do know which patients are receiving which treat-ment. However, they do not interact with the study’s patients and do not tell the blinded health care professionalswho is receiving which treatment.

30The researchers assigned the patients into their treatment groups, so this study was an experiment. However,the patients could distinguish what treatment they received, so this study was not blind. The study could not bedouble-blind since it was not blind.

31Ultimately, can we make patients think they got treated from a surgery? In fact, we can, and some experimentsuse what’s called a sham surgery. In a sham surgery, the patient does undergo surgery, but the patient does notreceive the full treatment, though they will still get a placebo effect.

1.4. EXPERIMENTS 35

Exercises

1.29 Light and exam performance. A study is designed to test the effect of light level on exam performanceof students. The researcher believes that light levels might have different effects on males and females, sowants to make sure both are equally represented in each treatment. The treatments are fluorescent overheadlighting, yellow overhead lighting, no overhead lighting (only desk lamps).

(a) What is the response variable?

(b) What is the explanatory variable? What are its levels?

1.30 Vitamin supplements. To assess the effectiveness of taking large doses of vitamin C in reducingthe duration of the common cold, researchers recruited 400 healthy volunteers from staff and students at auniversity. A quarter of the patients were assigned a placebo, and the rest were evenly divided between 1gVitamin C, 3g Vitamin C, or 3g Vitamin C plus additives to be taken at onset of a cold for the following twodays. All tablets had identical appearance and packaging. The nurses who handed the prescribed pills tothe patients knew which patient received which treatment, but the researchers assessing the patients whenthey were sick did not. No significant differences were observed in any measure of cold duration or severitybetween the four groups, and the placebo group had the shortest duration of symptoms.32

(a) Was this an experiment or an observational study? Why?

(b) What are the explanatory and response variables in this study?

(d) Was this study double-blind?

(e) Participants are ultimately able to choose whether or not to use the pills prescribed to them. We mightexpect that not all of them will adhere and take their pills. Does this introduce a confounding variableto the study? Explain your reasoning.

1.31 Light, noise, and exam performance. A study is designed to test the effect of light level and noiselevel on exam performance of students. The researcher believes that light and noise levels might have differenteffects on males and females, so wants to make sure both are equally represented in each treatment. Thelight treatments considered are fluorescent overhead lighting, yellow overhead lighting, no overhead lighting(only desk lamps). The noise treatments considered are no noise, construction noise, and human chatternoise.

(a) What type of study is this?

(b) How many factors are considered in this study? Identify them, and describe their levels.

1.32 Music and learning. You would like to conduct an experiment in class to see if students learn betterif they study without any music, with music that has no lyrics (instrumental), or with music that has lyrics.Briefly outline a design for this study.

1.33 Soda preference. You would like to conduct an experiment in class to see if your classmates preferthe taste of regular Coke or Diet Coke. Briefly outline a design for this study.

1.34 Exercise and mental health. A researcher is interested in the effects of exercise on mental healthand he proposes the following study: Use stratified random sampling to ensure representative proportionsof 18-30, 31-40 and 41- 55 year olds from the population. Next, randomly assign half the subjects from eachage group to exercise twice a week, and instruct the rest not to exercise. Conduct a mental health exam atthe beginning and at the end of the study, and compare the results.

(a) What type of study is this?

(b) What are the treatment and control groups in this study?

(d) Does this study make use of blinding?

(e) Comment on whether or not the results of the study can be used to establish a causal relationshipbetween exercise and mental health, and indicate whether or not the conclusions can be generalized tothe population at large.

(f) Suppose you are given the task of determining if this proposed study should get funding. Would youhave any reservations about the study proposal?

32C. Audera et al. “Mega-dose vitamin C in treatment of the common cold: a randomised controlled trial”. In:Medical Journal of Australia 175.7 (2001), pp. 359–362.

36 CHAPTER 1. INTRODUCTION TO DATA

Chapter exercises

1.35 Pet names. The city of Seattle, WA has an open data portal that includes pets registered in the city.For each registered pet, we have information on the pet’s name and species. The following visualizationplots the proportion of dogs with a given name versus the proportion of cats with the same name. The 20most common cat and dog names are displayed. The diagonal line on the plot is the x = y line; if a nameappeared on this line, the name’s popularity would be exactly the same for dogs and cats.

(a) Are these data collected as part of anexperiment or an observational study?

(b) What is the most common dog name?What is the most common cat name?

(d) Is the relationship between the twovariables positive or negative? Whatdoes this mean in context of the data?

1.36 Stressed out, Part II. In a study evaluating the relationship between stress and muscle cramps, halfthe subjects are randomly assigned to be exposed to increased stress by being placed into an elevator thatfalls rapidly and stops abruptly and the other half are left at no or baseline stress.

(a) What type of study is this?

(b) Can this study be used to conclude a causal relationship between increased stress and muscle cramps?

1.37 Chia seeds and weight loss. Chia Pets – those terra-cotta figurines that sprout fuzzy green hair –made the chia plant a household name. But chia has gained an entirely new reputation as a diet supplement.In one 2009 study, a team of researchers recruited 38 men and divided them randomly into two groups:treatment or control. They also recruited 38 women, and they randomly placed half of these participantsinto the treatment group and the other half into the control group. One group was given 25 grams of chiaseeds twice a day, and the other was given a placebo. The subjects volunteered to be a part of the study.After 12 weeks, the scientists found no significant difference between the groups in appetite or weight loss.33

(a) What type of study is this?

(b) What are the experimental and control treatments in this study?

(d) Has blinding been used in this study?

(e) Comment on whether or not we can make a causal statement, and indicate whether or not we cangeneralize the conclusion to the population at large.

1.38 City council survey. A city council has requested a household survey be conducted in a suburbanarea of their city. The area is broken into many distinct and unique neighborhoods, some including largehomes, some with only apartments, and others a diverse mixture of housing structures. For each part below,identify the sampling methods described, and describe the statistical pros and cons of the method in thecity’s context.

(a) Randomly sample 200 households from the city.

(b) Divide the city into 20 neighborhoods, and sample 10 households from each neighborhood.

(c) Divide the city into 20 neighborhoods, randomly sample 3 neighborhoods, and then sample all householdsfrom those 3 neighborhoods.

(d) Divide the city into 20 neighborhoods, randomly sample 8 neighborhoods, and then randomly sample50 households from those neighborhoods.

(e) Sample the 200 households closest to the city council offices.

33D.C. Nieman et al. “Chia seed does not promote weight loss or alter disease risk factors in overweight adults”.In: Nutrition Research 29.6 (2009), pp. 414–418.

1.4. EXPERIMENTS 37

1.39 Flawed reasoning. Identify the flaw(s) in reasoning in the following scenarios. Explain what theindividuals in the study should have done differently if they wanted to make such strong conclusions.

(a) Students at an elementary school are given a questionnaire that they are asked to return after theirparents have completed it. One of the questions asked is, “Do you find that your work schedule makesit difficult for you to spend time with your kids after school?” Of the parents who replied, 85% said“no”. Based on these results, the school officials conclude that a great majority of the parents have nodifficulty spending time with their kids after school.

(b) A survey is conducted on a simple random sample of 1,000 women who recently gave birth, asking themabout whether or not they smoked during pregnancy. A follow-up survey asking if the children haverespiratory problems is conducted 3 years later. However, only 567 of these women are reached at thesame address. The researcher reports that these 567 women are representative of all mothers.

(c) An orthopedist administers a questionnaire to 30 of his patients who do not have any joint problemsand finds that 20 of them regularly go running. He concludes that running decreases the risk of jointproblems.

1.40 Income and education in US counties. The scatterplot below shows the relationship between percapita income (in thousands of dollars) and percent of population with a bachelor’s degree in 3,143 countiesin the US in 2010.

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ita In

com

1.41 Eat better, feel better? In a public health study on the effects of consumption of fruits and vegetableson psychological well-being in young adults, participants were randomly assigned to three groups: (1) diet-as-usual, (2) an ecological momentary intervention involving text message reminders to increase their fruitsand vegetable consumption plus a voucher to purchase them, or (3) a fruit and vegetable intervention inwhich participants were given two additional daily servings of fresh fruits and vegetables to consume on topof their normal diet. Participants were asked to take a nightly survey on their smartphones. Participantswere student volunteers at the University of Otago, New Zealand. At the end of the 14-day study, onlyparticipants in the third group showed improvements to their psychological well-being across the 14-daysrelative to the other groups.34

(a) What type of study is this?

(b) Identify the explanatory and response variables.

(d) Comment on whether the results of the study can be used to establish causal relationships.

(e) A newspaper article reporting on the study states, “The results of this study provide proof that givingyoung adults fresh fruits and vegetables to eat can have psychological benefits, even over a brief periodof time.” How would you suggest revising this statement so that it can be supported by the study?

34Tamlin S Conner et al. “Let them eat fruit! The effect of fruit and vegetable consumption on psychologicalwell-being in young adults: A randomized controlled trial”. In: PloS one 12.2 (2017), e0171206.

38 CHAPTER 1. INTRODUCTION TO DATA

1.42 Screens, teens, and psychological well-being. In a study of three nationally representative large-scale data sets from Ireland, the United States, and the United Kingdom (n = 17,247), teenagers betweenthe ages of 12 to 15 were asked to keep a diary of their screen time and answer questions about how theyfelt or acted. The answers to these questions were then used to compute a psychological well-being score.Additional data were collected and included in the analysis, such as each child’s sex and age, and on themother’s education, ethnicity, psychological distress, and employment. The study concluded that there islittle clear-cut evidence that screen time decreases adolescent well-being.35

(a) What type of study is this?

(b) Identify the explanatory variables.

(d) Comment on whether the results of the study can be generalized to the population, and why.

(e) Comment on whether the results of the study can be used to establish causal relationships.

1.43 Stanford Open Policing. The Stanford Open Policing project gathers, analyzes, and releases recordsfrom traffic stops by law enforcement agencies across the United States. Their goal is to help researchers,journalists, and policymakers investigate and improve interactions between police and the public.36 Thefollowing is an excerpt from a summary table created based off of the data collected as part of this project.

Driver’s No. of stops % of stoppedCounty State race per year cars searched drivers arrested

Apaice County Arizona Black 266 0.08 0.02Apaice County Arizona Hispanic 1008 0.05 0.02Apaice County Arizona White 6322 0.02 0.01Cochise County Arizona Black 1169 0.05 0.01Cochise County Arizona Hispanic 9453 0.04 0.01Cochise County Arizona White 10826 0.02 0.01· · · · · · · · · · · · · · · · · ·Wood County Wisconsin Black 16 0.24 0.10Wood County Wisconsin Hispanic 27 0.04 0.03Wood County Wisconsin White 1157 0.03 0.03

(a) What variables were collected on each individual traffic stop in order to create to the summary tableabove?

(b) State whether each variable is numerical or categorical. If numerical, state whether it is continuous ordiscrete. If categorical, state whether it is ordinal or not.

(c) Suppose we wanted to evaluate whether vehicle search rates are different for drivers of different races. Inthis analysis, which variable would be the response variable and which variable would be the explanatoryvariable?

1.44 Space launches. The following summary table shows the number of space launches in the US by thetype of launching agency and the outcome of the launch (success or failure).37

1957 – 1999 2000 – 2018Failure Success Failure Success

Private 13 295 10 562State 281 3751 33 711Startup – – 5 65

(a) What variables were collected on each launch in order to create to the summary table above?

(b) State whether each variable is numerical or categorical. If numerical, state whether it is continuous ordiscrete. If categorical, state whether it is ordinal or not.

(c) Suppose we wanted to study how the success rate of launches vary between launching agencies and overtime. In this analysis, which variable would be the response variable and which variable would be theexplanatory variable?

35Amy Orben and AK Baukney-Przybylski. “Screens, Teens and Psychological Well-Being: Evidence from threetime-use diary studies”. In: Psychological Science (2018).

36Emma Pierson et al. “A large-scale analysis of racial disparities in police stops across the United States”. In:arXiv preprint arXiv:1706.05678 (2017).

37JSR Launch Vehicle Database, A comprehensive list of suborbital space launches, 2019 Feb 10 Edition.

Chapter 2Summarizing data

2.1 Examining numerical data

2.2 Considering categorical data

2.3 Case study: malaria vaccine

This chapter focuses on the mechanics and construction of summary

statistics and graphs. We use statistical software for generating the

summaries and graphs presented in this chapter and book. However,

since this might be your first exposure to these concepts, we take our

time in this chapter to detail how to create them. Mastery of the

content presented in this chapter will be crucial for understanding the

methods and techniques introduced in rest of the book.

For videos, slides, and other resources, please visit

www.openintro.org/os

2.1. EXAMINING NUMERICAL DATA 41

2.1 Examining numerical data

In this section we will explore techniques for summarizing numerical variables. For example,consider the loan amount variable from the loan50 data set, which represents the loan size for all50 loans in the data set. This variable is numerical since we can sensibly discuss the numericaldifference of the size of two loans. On the other hand, area codes and zip codes are not numerical,but rather they are categorical variables.

Throughout this section and the next, we will apply these methods using the loan50 andcounty data sets, which were introduced in Section 1.2. If you’d like to review the variables fromeither data set, see Figures 1.3 and 1.5.

2.1.1 Scatterplots for paired data

A scatterplot provides a case-by-case view of data for two numerical variables. In Figure 1.8on page 16, a scatterplot was used to examine the homeownership rate against the fraction of housingunits that were part of multi-unit properties (e.g. apartments) in the county data set. Anotherscatterplot is shown in Figure 2.1, comparing the total income of a borrower (total income) andthe amount they borrowed (loan amount) for the loan50 data set. In any scatterplot, each pointrepresents a single case. Since there are 50 cases in loan50, there are 50 points in Figure 2.1.

Total Income

$0 $50k $100k $150k $200k $250k $300k$0

$10k

$20k

$30k

$40k

Loan

ount

Figure 2.1: A scatterplot of total income versus loan amount for the loan50

data set.

Looking at Figure 2.1, we see that there are many borrowers with an income below $100,000on the left side of the graph, while there are a handful of borrowers with income above $250,000.

EXAMPLE 2.1

Figure 2.2 shows a plot of median household income against the poverty rate for 3,142 counties.What can be said about the relationship between these variables?

The relationship is evidently nonlinear, as highlighted by the dashed line. This is different fromprevious scatterplots we’ve seen, which show relationships that do not show much, if any, curvaturein the trend.

42 CHAPTER 2. SUMMARIZING DATA

0% 10% 20% 30% 40% 50%

$20k

$40k

$60k

$80k

$100k

$120k

Poverty Rate (Percent)

Med

ian

Hou

seho

ld In

com

Figure 2.2: A scatterplot of the median household income against the poverty ratefor the county data set. A statistical model has also been fit to the data and isshown as a dashed line.

GUIDED PRACTICE 2.2

What do scatterplots reveal about the data, and how are they useful?1

GUIDED PRACTICE 2.3

Describe two variables that would have a horseshoe-shaped association in a scatterplot (∩ or _).2

2.1.2 Dot plots and the mean

Sometimes two variables are one too many: only one variable may be of interest. In these cases,a dot plot provides the most basic of displays. A dot plot is a one-variable scatterplot; an exampleusing the interest rate of 50 loans is shown in Figure 2.3. A stacked version of this dot plot is shownin Figure 2.4.

Interest Rate

5% 10% 15% 20% 25%

Figure 2.3: A dot plot of interest rate for the loan50 data set. The distribu-tion’s mean is shown as a red triangle.

1Answers may vary. Scatterplots are helpful in quickly spotting associations relating variables, whether thoseassociations come in the form of simple trends or whether those relationships are more complex.

2Consider the case where your vertical axis represents something “good” and your horizontal axis representssomething that is only good in moderation. Health and water consumption fit this description: we require some waterto survive, but consume too much and it becomes toxic and can kill a person.

2.1. EXAMINING NUMERICAL DATA 43

Interest Rate, Rounded to Nearest Percent

●●●

●●●●●

●●●●

●●●●●

●●●●●●●●

●●●●●

● ●●●

● ● ● ●●●

●●

● ● ● ● ●

5% 10% 15% 20% 25%

Figure 2.4: A stacked dot plot of interest rate for the loan50 data set. The rateshave been rounded to the nearest percent in this plot, and the distribution’s meanis shown as a red triangle.

The mean, often called the average, is a common way to measure the center of a distributionof data. To compute the mean interest rate, we add up all the interest rates and divide by the numberof observations:

x̄ =10.90% + 9.92% + 26.30% + · · ·+ 6.08%

50= 11.57%

The sample mean is often labeled x̄. The letter x is being used as a generic placeholder for thevariable of interest, interest rate, and the bar over the x communicates we’re looking at theaverage interest rate, which for these 50 loans was 11.57%. It is useful to think of the mean as thebalancing point of the distribution, and it’s shown as a triangle in Figures 2.3 and 2.4.

MEAN

The sample mean can be computed as the sum of the observed values divided by the numberof observations:

x̄ =x1 + x2 + · · ·+ xn

where x1, x2, . . . , xn represent the n observed values.

GUIDED PRACTICE 2.4

Examine the equation for the mean. What does x1 correspond to? And x2? Can you infer a generalmeaning to what xi might represent?3

GUIDED PRACTICE 2.5

What was n in this sample of loans?4

The loan50 data set represents a sample from a larger population of loans made throughLending Club. We could compute a mean for this population in the same way as the sample mean.However, the population mean has a special label: µ. The symbol µ is the Greek letter mu andrepresents the average of all observations in the population. Sometimes a subscript, such as x, is usedto represent which variable the population mean refers to, e.g. µx. Often times it is too expensiveto measure the population mean precisely, so we often estimate µ using the sample mean, x̄.

3×1 corresponds to the interest rate for the first loan in the sample (10.90%), x2 to the second loan’s interest rate(9.92%), and xi corresponds to the interest rate for the ith loan in the data set. For example, if i = 4, then we’reexamining x4, which refers to the fourth observation in the data set.

4The sample size was n = 50.

44 CHAPTER 2. SUMMARIZING DATA

EXAMPLE 2.6

The average interest rate across all loans in the population can be estimated using the sample data.Based on the sample of 50 loans, what would be a reasonable estimate of µx, the mean interest ratefor all loans in the full data set?

The sample mean, 11.57%, provides a rough estimate of µx. While it’s not perfect, this is our singlebest guess of the average interest rate of all the loans in the population under study.

In Chapter 5 and beyond, we will develop tools to characterize the accuracy of point estimates likethe sample mean. As you might have guessed, point estimates based on larger samples tend to bemore accurate than those based on smaller samples.

EXAMPLE 2.7

The mean is useful because it allows us to rescale or standardize a metric into something more easilyinterpretable and comparable. Provide 2 examples where the mean is useful for making comparisons.

1. We would like to understand if a new drug is more effective at treating asthma attacks than thestandard drug. A trial of 1500 adults is set up, where 500 receive the new drug, and 1000 receive astandard drug in the control group:

New drug Standard drugNumber of patients 500 1000Total asthma attacks 200 300

Comparing the raw counts of 200 to 300 asthma attacks would make it appear that the new drug isbetter, but this is an artifact of the imbalanced group sizes. Instead, we should look at the averagenumber of asthma attacks per patient in each group:

New drug: 200/500 = 0.4 Standard drug: 300/1000 = 0.3

The standard drug has a lower average number of asthma attacks per patient than the average inthe treatment group.

2. Emilio opened a food truck last year where he sells burritos, and his business has stabilizedover the last 3 months. Over that 3 month period, he has made $11,000 while working 625 hours.Emilio’s average hourly earnings provides a useful statistic for evaluating whether his venture is,at least from a financial perspective, worth it:

$11000

625 hours= $17.60 per hour

By knowing his average hourly wage, Emilio now has put his earnings into a standard unit that iseasier to compare with many other jobs that he might consider.

EXAMPLE 2.8

Suppose we want to compute the average income per person in the US. To do so, we might firstthink to take the mean of the per capita incomes across the 3,142 counties in the county data set.What would be a better approach?

The county data set is special in that each county actually represents many individual people. Ifwe were to simply average across the income variable, we would be treating counties with 5,000 and5,000,000 residents equally in the calculations. Instead, we should compute the total income for eachcounty, add up all the counties’ totals, and then divide by the number of people in all the counties.If we completed these steps with the county data, we would find that the per capita income for theUS is $30,861. Had we computed the simple mean of per capita income across counties, the resultwould have been just $26,093!

This example used what is called a weighted mean. For more information on this topic, check outthe following online supplement regarding weighted means openintro.org/d?file=stat wtd mean.

2.1. EXAMINING NUMERICAL DATA 45

2.1.3 Histograms and shape

Dot plots show the exact value for each observation. This is useful for small data sets, but theycan become hard to read with larger samples. Rather than showing the value of each observation, weprefer to think of the value as belonging to a bin. For example, in the loan50 data set, we created atable of counts for the number of loans with interest rates between 5.0% and 7.5%, then the numberof loans with rates between 7.5% and 10.0%, and so on. Observations that fall on the boundary ofa bin (e.g. 10.00%) are allocated to the lower bin. This tabulation is shown in Figure 2.5. Thesebinned counts are plotted as bars in Figure 2.6 into what is called a histogram, which resembles amore heavily binned version of the stacked dot plot shown in Figure 2.4.

Interest Rate 5.0% – 7.5% 7.5% – 10.0% 10.0% – 12.5% 12.5% – 15.0% · · · 25.0% – 27.5%

Count 11 15 8 4 · · · 1

Figure 2.5: Counts for the binned interest rate data.

Interest Rate

Fre

quen

5% 10% 15% 20% 25%

Figure 2.6: A histogram of interest rate. This distribution is strongly skewedto the right.

Histograms provide a view of the data density. Higher bars represent where the data arerelatively more common. For instance, there are many more loans with rates between 5% and 10%than loans with rates between 20% and 25% in the data set. The bars make it easy to see how thedensity of the data changes relative to the interest rate.

Histograms are especially convenient for understanding the shape of the data distribution.Figure 2.6 suggests that most loans have rates under 15%, while only a handful of loans have ratesabove 20%. When data trail off to the right in this way and has a longer right tail, the shape is saidto be right skewed.5

Data sets with the reverse characteristic – a long, thinner tail to the left – are said to be leftskewed. We also say that such a distribution has a long left tail. Data sets that show roughly equaltrailing off in both directions are called symmetric.

LONG TAILS TO IDENTIFY SKEW

When data trail off in one direction, the distribution has a long tail. If a distribution has along left tail, it is left skewed. If a distribution has a long right tail, it is right skewed.

5Other ways to describe data that are right skewed: skewed to the right, skewed to the high end, or skewedto the positive end.

46 CHAPTER 2. SUMMARIZING DATA

GUIDED PRACTICE 2.9

Take a look at the dot plots in Figures 2.3 and 2.4. Can you see the skew in the data? Is it easierto see the skew in this histogram or the dot plots?6

GUIDED PRACTICE 2.10

Besides the mean (since it was labeled), what can you see in the dot plots that you cannot see inthe histogram?7

In addition to looking at whether a distribution is skewed or symmetric, histograms can beused to identify modes. A mode is represented by a prominent peak in the distribution. There isonly one prominent peak in the histogram of loan amount.

A definition of mode sometimes taught in math classes is the value with the most occurrencesin the data set. However, for many real-world data sets, it is common to have no observations withthe same value in a data set, making this definition impractical in data analysis.

Figure 2.7 shows histograms that have one, two, or three prominent peaks. Such distributionsare called unimodal, bimodal, and multimodal, respectively. Any distribution with more than2 prominent peaks is called multimodal. Notice that there was one prominent peak in the unimodaldistribution with a second less prominent peak that was not counted since it only differs from itsneighboring bins by a few observations.

0 5 10 15

0 5 10 15 20

Figure 2.7: Counting only prominent peaks, the distributions are (left to right)unimodal, bimodal, and multimodal. Note that we’ve said the left plot is unimodalintentionally. This is because we are counting prominent peaks, not just any peak.

EXAMPLE 2.11

Figure 2.6 reveals only one prominent mode in the interest rate. Is the distribution unimodal,bimodal, or multimodal?

Unimodal. Remember that uni stands for 1 (think unicycles). Similarly, bi stands for 2 (thinkbicycles). We’re hoping a multicycle will be invented to complete this analogy.

GUIDED PRACTICE 2.12

Height measurements of young students and adult teachers at a K-3 elementary school were taken.How many modes would you expect in this height data set?8

Looking for modes isn’t about finding a clear and correct answer about the number of modes ina distribution, which is why prominent is not rigorously defined in this book. The most importantpart of this examination is to better understand your data.

6The skew is visible in all three plots, though the flat dot plot is the least useful. The stacked dot plot andhistogram are helpful visualizations for identifying skew.

7The interest rates for individual loans.8There might be two height groups visible in the data set: one of the students and one of the adults. That is, the

data are probably bimodal.

2.1. EXAMINING NUMERICAL DATA 47

2.1.4 Variance and standard deviation

The mean was introduced as a method to describe the center of a data set, and variability inthe data is also important. Here, we introduce two measures of variability: the variance and thestandard deviation. Both of these are very useful in data analysis, even though their formulas area bit tedious to calculate by hand. The standard deviation is the easier of the two to comprehend,and it roughly describes how far away the typical observation is from the mean.

We call the distance of an observation from its mean its deviation. Below are the deviationsfor the 1st, 2nd, 3rd, and 50th observations in the interest rate variable:

x1 − x̄ = 10.90− 11.57 = −0.67

x2 − x̄ = 9.92− 11.57 = −1.65

x3 − x̄ = 26.30− 11.57 = 14.73

…

x50 − x̄ = 6.08− 11.57 = −5.49

If we square these deviations and then take an average, the result is equal to the sample variance,denoted by s2:

s2 =(−0.67)2 + (−1.65)2 + (14.73)2 + · · ·+ (−5.49)2

50− 1

=0.45 + 2.72 + 216.97 + · · ·+ 30.14

49= 25.52

We divide by n − 1, rather than dividing by n, when computing a sample’s variance; there’s somemathematical nuance here, but the end result is that doing this makes this statistic slightly morereliable and useful.

Notice that squaring the deviations does two things. First, it makes large values relativelymuch larger, seen by comparing (−0.67)2, (−1.65)2, (14.73)2, and (−5.49)2. Second, it gets rid ofany negative signs.

The standard deviation is defined as the square root of the variance:

s =√

25.52 = 5.05

While often omitted, a subscript of x may be added to the variance and standard deviation, i.e.s2x and sx, if it is useful as a reminder that these are the variance and standard deviation of the

observations represented by x1, x2, …, xn.

VARIANCE AND STANDARD DEVIATION

The variance is the average squared distance from the mean. The standard deviation is thesquare root of the variance. The standard deviation is useful when considering how far the dataare distributed from the mean.

The standard deviation represents the typical deviation of observations from the mean. Usuallyabout 70% of the data will be within one standard deviation of the mean and about 95% willbe within two standard deviations. However, as seen in Figures 2.8 and 2.9, these percentagesare not strict rules.

Like the mean, the population values for variance and standard deviation have special symbols:σ2 for the variance and σ for the standard deviation. The symbol σ is the Greek letter sigma.

48 CHAPTER 2. SUMMARIZING DATA

Interest Rate, x = 11.57%, sx = 5.05%

6.5% 11.6% 16.7% 21.8% 26.9%

Figure 2.8: For the interest rate variable, 34 of the 50 loans (68%) had interestrates within 1 standard deviation of the mean, and 48 of the 50 loans (96%) hadrates within 2 standard deviations. Usually about 70% of the data are within1 standard deviation of the mean and 95% within 2 standard deviations, thoughthis is far from a hard rule.

−3 −2 −1 0 1 2 3

Figure 2.9: Three very different population distributions with the same mean µ = 0and standard deviation σ = 1.

GUIDED PRACTICE 2.13

On page 45, the concept of shape of a distribution was introduced. A good description of the shapeof a distribution should include modality and whether the distribution is symmetric or skewed toone side. Using Figure 2.9 as an example, explain why such a description is important.9

EXAMPLE 2.14

Describe the distribution of the interest rate variable using the histogram in Figure 2.6. Thedescription should incorporate the center, variability, and shape of the distribution, and it shouldalso be placed in context. Also note any especially unusual cases.

The distribution of interest rates is unimodal and skewed to the high end. Many of the rates fallnear the mean at 11.57%, and most fall within one standard deviation (5.05%) of the mean. Thereare a few exceptionally large interest rates in the sample that are above 20%.

In practice, the variance and standard deviation are sometimes used as a means to an end, wherethe “end” is being able to accurately estimate the uncertainty associated with a sample statistic.For example, in Chapter 5 the standard deviation is used in calculations that help us understandhow much a sample mean varies from one sample to the next.

9Figure 2.9 shows three distributions that look quite different, but all have the same mean, variance, and standarddeviation. Using modality, we can distinguish between the first plot (bimodal) and the last two (unimodal). Usingskewness, we can distinguish between the last plot (right skewed) and the first two. While a picture, like a histogram,tells a more complete story, we can use modality and shape (symmetry/skew) to characterize basic information abouta distribution.

2.1. EXAMINING NUMERICAL DATA 49

2.1.5 Box plots, quartiles, and the median

A box plot summarizes a data set using five statistics while also plotting unusual observations.Figure 2.10 provides a vertical dot plot alongside a box plot of the interest rate variable from theloan50 data set.

Inte

rest

Rat

10%

15%

20%

25%

lower whisker

Q1 (first quartile)

median

Q3 (third quartile)

upper whisker

max whisker reach

suspected outliers

Figure 2.10: A vertical dot plot, where points have been horizontally stacked, nextto a labeled box plot for the interest rates of the 50 loans.

The first step in building a box plot is drawing a dark line denoting the median, which splits thedata in half. Figure 2.10 shows 50% of the data falling below the median and other 50% falling abovethe median. There are 50 loans in the data set (an even number) so the data are perfectly split intotwo groups of 25. We take the median in this case to be the average of the two observations closest tothe 50th percentile, which happen to be the same value in this data set: (9.93%+9.93%)/2 = 9.93%.When there are an odd number of observations, there will be exactly one observation that splits thedata into two halves, and in such a case that observation is the median (no average needed).

MEDIAN: THE NUMBER IN THE MIDDLE

If the data are ordered from smallest to largest, the median is the observation right in themiddle. If there are an even number of observations, there will be two values in the middle, andthe median is taken as their average.

The second step in building a box plot is drawing a rectangle to represent the middle 50% ofthe data. The total length of the box, shown vertically in Figure 2.10, is called the interquartilerange (IQR, for short). It, like the standard deviation, is a measure of variability in data. The morevariable the data, the larger the standard deviation and IQR tend to be. The two boundaries of thebox are called the first quartile (the 25th percentile, i.e. 25% of the data fall below this value) andthe third quartile (the 75th percentile), and these are often labeled Q1 and Q3, respectively.

50 CHAPTER 2. SUMMARIZING DATA

INTERQUARTILE RANGE (IQR)

The IQR is the length of the box in a box plot. It is computed as

IQR = Q3 −Q1

where Q1 and Q3 are the 25th and 75th percentiles.

GUIDED PRACTICE 2.15

What percent of the data fall between Q1 and the median? What percent is between the medianand Q3?10

Extending out from the box, the whiskers attempt to capture the data outside of the box.However, their reach is never allowed to be more than 1.5× IQR. They capture everything withinthis reach. In Figure 2.10, the upper whisker does not extend to the last two points, which is beyondQ3 + 1.5× IQR, and so it extends only to the last point below this limit. The lower whisker stopsat the lowest value, 5.31%, since there is no additional data to reach; the lower whisker’s limit is notshown in the figure because the plot does not extend down to Q1−1.5× IQR. In a sense, the box islike the body of the box plot and the whiskers are like its arms trying to reach the rest of the data.

Any observation lying beyond the whiskers is labeled with a dot. The purpose of labeling thesepoints – instead of extending the whiskers to the minimum and maximum observed values – is to helpidentify any observations that appear to be unusually distant from the rest of the data. Unusuallydistant observations are called outliers. In this case, it would be reasonable to classify the interestrates of 24.85% and 26.30% as outliers since they are numerically distant from most of the data.

OUTLIERS ARE EXTREME

An outlier is an observation that appears extreme relative to the rest of the data.

Examining data for outliers serves many useful purposes, including

1. Identifying strong skew in the distribution.

2. Identifying possible data collection or data entry errors.

3. Providing insight into interesting properties of the data.

GUIDED PRACTICE 2.16

Using Figure 2.10, estimate the following values for interest rate in the loan50 data set:(a) Q1, (b) Q3, and (c) IQR.11

10Since Q1 and Q3 capture the middle 50% of the data and the median splits the data in the middle, 25% of thedata fall between Q1 and the median, and another 25% falls between the median and Q3.

11These visual estimates will vary a little from one person to the next: Q1 = 8%, Q3 = 14%, IQR = Q3−Q1 = 6%.(The true values: Q1 = 7.96%, Q3 = 13.72%, IQR = 5.76%.)

2.1. EXAMINING NUMERICAL DATA 51

2.1.6 Robust statistics

How are the sample statistics of the interest rate data set affected by the observation, 26.3%?What would have happened if this loan had instead been only 15%? What would happen to thesesummary statistics if the observation at 26.3% had been even larger, say 35%? These scenarios areplotted alongside the original data in Figure 2.11, and sample statistics are computed under eachscenario in Figure 2.12.

Interest Rate

●

5% 10% 15% 20% 25% 30% 35%

Original

●26.3% to 15%

●26.3% to 35%

Figure 2.11: Dot plots of the original interest rate data and two modified data sets.

robust not robustscenario median IQR x̄ soriginal interest rate data 9.93% 5.76% 11.57% 5.05%move 26.3% → 15% 9.93% 5.76% 11.34% 4.61%move 26.3% → 35% 9.93% 5.76% 11.74% 5.68%

Figure 2.12: A comparison of how the median, IQR, mean (x̄), and standard deviation(s) change had an extreme observations from the interest rate variable been different.

GUIDED PRACTICE 2.17

(a) Which is more affected by extreme observations, the mean or median? Figure 2.12 may behelpful. (b) Is the standard deviation or IQR more affected by extreme observations?12

The median and IQR are called robust statistics because extreme observations have littleeffect on their values: moving the most extreme value generally has little influence on these statistics.On the other hand, the mean and standard deviation are more heavily influenced by changes inextreme observations, which can be important in some situations.

EXAMPLE 2.18

The median and IQR did not change under the three scenarios in Figure 2.12. Why might this bethe case?

The median and IQR are only sensitive to numbers near Q1, the median, and Q3. Since values inthese regions are stable in the three data sets, the median and IQR estimates are also stable.

GUIDED PRACTICE 2.19

The distribution of loan amounts in the loan50 data set is right skewed, with a few large loanslingering out into the right tail. If you were wanting to understand the typical loan size, should yoube more interested in the mean or median?13

12(a) Mean is affected more. (b) Standard deviation is affected more. Complete explanations are provided in thematerial following Guided Practice 2.17.

13Answers will vary! If we’re looking to simply understand what a typical individual loan looks like, the median isprobably more useful. However, if the goal is to understand something that scales well, such as the total amount ofmoney we might need to have on hand if we were to offer 1,000 loans, then the mean would be more useful.

52 CHAPTER 2. SUMMARIZING DATA

2.1.7 Transforming data (special topic)

When data are very strongly skewed, we sometimes transform them so they are easier to model.

Population (m = millions)

0m 2m 4m 6m 8m 10m

500

1000

1500

2000

2500

3000

Fre

quen

(a)

log10(Population)

2 3 4 5 6 7

500

1000

Fre

quen

cy(b)

Figure 2.13: (a) A histogram of the populations of all US counties. (b) A histogramof log10-transformed county populations. For this plot, the x-value corresponds tothe power of 10, e.g. “4” on the x-axis corresponds to 104 = 10,000.

EXAMPLE 2.20

Consider the histogram of county populations shown in Figure 2.13(a), which shows extreme skew.What isn’t useful about this plot?

Nearly all of the data fall into the left-most bin, and the extreme skew obscures many of thepotentially interesting details in the data.

There are some standard transformations that may be useful for strongly right skewed datawhere much of the data is positive but clustered near zero. A transformation is a rescaling ofthe data using a function. For instance, a plot of the logarithm (base 10) of county populationsresults in the new histogram in Figure 2.13(b). This data is symmetric, and any potential outliersappear much less extreme than in the original data set. By reigning in the outliers and extremeskew, transformations like this often make it easier to build statistical models against the data.

Transformations can also be applied to one or both variables in a scatterplot. A scatterplot ofthe population change from 2010 to 2017 against the population in 2010 is shown in Figure 2.14(a).In this first scatterplot, it’s hard to decipher any interesting patterns because the population variableis so strongly skewed. However, if we apply a log10 transformation to the population variable, asshown in Figure 2.14(b), a positive association between the variables is revealed. In fact, we maybe interested in fitting a trend line to the data when we explore methods around fitting regressionlines in Chapter 8.

Transformations other than the logarithm can be useful, too. For instance, the square root(√

original observation) and inverse ( 1original observation ) are commonly used by data scientists. Com-

mon goals in transforming data are to see the data structure differently, reduce skew, assist inmodeling, or straighten a nonlinear relationship in a scatterplot.

2.1. EXAMINING NUMERICAL DATA 53

Pop

ulat

ion

Cha

nge

−20%

20%

40%

0m 2m 4m 6m 8m 10m

Population Before Change (m = millions)

(a)

log10(Population Before Change)2 3 4 5 6 7

−20%

20%

40%

Pop

ulat

ion

Cha

nge

(b)

Figure 2.14: (a) Scatterplot of population change against the population beforethe change. (b) A scatterplot of the same data but where the population size hasbeen log-transformed.

2.1.8 Mapping data (special topic)

The county data set offers many numerical variables that we could plot using dot plots, scatter-plots, or box plots, but these miss the true nature of the data. Rather, when we encounter geographicdata, we should create an intensity map, where colors are used to show higher and lower valuesof a variable. Figures 2.15 and 2.16 shows intensity maps for poverty rate in percent (poverty),unemployment rate (unemployment rate), homeownership rate in percent (homeownership), andmedian household income (median hh income). The color key indicates which colors correspondto which values. The intensity maps are not generally very helpful for getting precise values inany given county, but they are very helpful for seeing geographic trends and generating interestingresearch questions or hypotheses.

EXAMPLE 2.21

What interesting features are evident in the poverty and unemployment rate intensity maps?

Poverty rates are evidently higher in a few locations. Notably, the deep south shows higher povertyrates, as does much of Arizona and New Mexico. High poverty rates are evident in the Mississippiflood plains a little north of New Orleans and also in a large section of Kentucky.

The unemployment rate follows similar trends, and we can see correspondence between the twovariables. In fact, it makes sense for higher rates of unemployment to be closely related to povertyrates. One observation that stand out when comparing the two maps: the poverty rate is muchhigher than the unemployment rate, meaning while many people may be working, they are notmaking enough to break out of poverty.

GUIDED PRACTICE 2.22

What interesting features are evident in the median hh income intensity map in Figure 2.16(b)?14

14Note: answers will vary. There is some correspondence between high earning and metropolitan areas, where wecan see darker spots (higher median household income), though there are several exceptions. You might look for largecities you are familiar with and try to spot them on the map as dark spots.

54 CHAPTER 2. SUMMARIZING DATA

14%

>25%

Pov

erty

(a)

>7%U

nem

ploy

men

t Rat

(b)

Figure 2.15: (a) Intensity map of poverty rate (percent). (b) Map of the unem-ployment rate (percent).

2.1. EXAMINING NUMERICAL DATA 55

<55%

73%

91%

Hom

eow

ners

hip

Rat

(a)

$19

$47

>$75M

edia

n H

ouse

hold

Inco

(b)

Figure 2.16: (a) Intensity map of homeownership rate (percent). (b) Intensity mapof median household income ($1000s).

56 CHAPTER 2. SUMMARIZING DATA

Exercises

2.1 Mammal life spans. Data were collected on life spans (in years) and gestation lengths (in days) for 62mammals. A scatterplot of life span versus length of gestation is shown below.15

(a) What type of an association is apparent be-tween life span and length of gestation?

(b) What type of an association would you ex-pect to see if the axes of the plot were re-versed, i.e. if we plotted length of gestationversus life span?

●

●●

●

●●

●

●●

●

●●

●

●●

●

Gestation (days)Li

fe S

pan

(yea

rs)

0 200 400 6000

100

2.2 Associations. Indicate which of the plots show (a) a positive association, (b) a negative association,or (c) no association. Also determine if the positive and negative associations are linear or nonlinear. Eachpart may refer to more than one plot.

(1) (2) (3) (4)

2.3 Reproducing bacteria. Suppose that there is only sufficient space and nutrients to support one millionbacterial cells in a petri dish. You place a few bacterial cells in this petri dish, allow them to reproduce freely,and record the number of bacterial cells in the dish over time. Sketch a plot representing the relationshipbetween number of bacterial cells and time.

2.4 Office productivity. Office productivity is relatively low when the employees feel no stress about theirwork or job security. However, high levels of stress can also lead to reduced employee productivity. Sketcha plot to represent the relationship between stress and productivity.

2.5 Parameters and statistics. Identify which value represents the sample mean and which value representsthe claimed population mean.

(a) American households spent an average of about $52 in 2007 on Halloween merchandise such as costumes,decorations and candy. To see if this number had changed, researchers conducted a new survey in 2008before industry numbers were reported. The survey included 1,500 households and found that averageHalloween spending was $58 per household.

(b) The average GPA of students in 2001 at a private university was 3.37. A survey on a sample of 203students from this university yielded an average GPA of 3.59 a decade later.

2.6 Sleeping in college. A recent article in a college newspaper stated that college students get an averageof 5.5 hrs of sleep each night. A student who was skeptical about this value decided to conduct a surveyby randomly sampling 25 students. On average, the sampled students slept 6.25 hours per night. Identifywhich value represents the sample mean and which value represents the claimed population mean.

15T. Allison and D.V. Cicchetti. “Sleep in mammals: ecological and constitutional correlates”. In: Arch. Hydrobiol75 (1975), p. 442.

2.1. EXAMINING NUMERICAL DATA 57

2.7 Days off at a mining plant. Workers at a particular mining site receive an average of 35 days paidvacation, which is lower than the national average. The manager of this plant is under pressure from alocal union to increase the amount of paid time off. However, he does not want to give more days off tothe workers because that would be costly. Instead he decides he should fire 10 employees in such a way asto raise the average number of days off that are reported by his employees. In order to achieve this goal,should he fire employees who have the most number of days off, least number of days off, or those who haveabout the average number of days off?

2.8 Medians and IQRs. For each part, compare distributions (1) and (2) based on their medians and IQRs.You do not need to calculate these statistics; simply state how the medians and IQRs compare. Make sureto explain your reasoning.

(a) (1) 3, 5, 6, 7, 9(2) 3, 5, 6, 7, 20

(b) (1) 3, 5, 6, 7, 9(2) 3, 5, 7, 8, 9

(d) (1) 0, 10, 50, 60, 100(2) 0, 100, 500, 600, 1000

2.9 Means and SDs. For each part, compare distributions (1) and (2) based on their means and standarddeviations. You do not need to calculate these statistics; simply state how the means and the standarddeviations compare. Make sure to explain your reasoning. Hint: It may be useful to sketch dot plots of thedistributions.

(a) (1) 3, 5, 5, 5, 8, 11, 11, 11, 13(2) 3, 5, 5, 5, 8, 11, 11, 11, 20

(b) (1) -20, 0, 0, 0, 15, 25, 30, 30(2) -40, 0, 0, 0, 15, 25, 30, 30

(d) (1) 100, 200, 300, 400, 500(2) 0, 50, 300, 550, 600

2.10 Mix-and-match. Describe the distribution in the histograms below and match them to the box plots.

(a)

50 60 70

(b)

0 50 100

(c)

0 2 4 6

(1)

(2)

(3)

100

58 CHAPTER 2. SUMMARIZING DATA

2.11 Air quality. Daily air quality is measured by the air quality index (AQI) reported by the Environ-mental Protection Agency. This index reports the pollution level and what associated health effects mightbe a concern. The index is calculated for five major air pollutants regulated by the Clean Air Act and takesvalues from 0 to 300, where a higher value indicates lower air quality. AQI was reported for a sample of91 days in 2011 in Durham, NC. The relative frequency histogram below shows the distribution of the AQIvalues on these days.16

(a) Estimate the median AQI value of this sample.

(b) Would you expect the mean AQI value of this sampleto be higher or lower than the median? Explain yourreasoning.

(d) Would any of the days in this sample be consideredto have an unusually low or high AQI? Explain yourreasoning.

Daily AQI

10 20 30 40 50 600

0.05

0.1

0.15

0.2

2.12 Median vs. mean. Estimate the median for the 400 observations shown in the histogram, and notewhether you expect the mean to be higher or lower than the median.

40 50 60 70 80 90 100

2.13 Histograms vs. box plots. Compare the two plots below. What characteristics of the distributionare apparent in the histogram and not in the box plot? What characteristics are apparent in the box plotbut not in the histogram?

5 10 15 20 250

100

150

200

510152025

2.14 Facebook friends. Facebook data indicate that 50% of Facebook users have 100 or more friends,and that the average friend count of users is 190. What do these findings suggest about the shape of thedistribution of number of friends of Facebook users?17

2.15 Distributions and appropriate statistics, Part I. For each of the following, state whether you expectthe distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or medianwould best represent a typical observation in the data, and whether the variability of observations would bebest represented using the standard deviation or IQR. Explain your reasoning.

(a) Number of pets per household.

(b) Distance to work, i.e. number of miles between work and home.

16US Environmental Protection Agency, AirData, 2011.17Lars Backstrom. “Anatomy of Facebook”. In: Facebook Data Team’s Notes (2011).

2.1. EXAMINING NUMERICAL DATA 59

2.16 Distributions and appropriate statistics, Part II. For each of the following, state whether you expectthe distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or medianwould best represent a typical observation in the data, and whether the variability of observations would bebest represented using the standard deviation or IQR. Explain your reasoning.

(a) Housing prices in a country where 25% of the houses cost below $350,000, 50% of the houses cost below$450,000, 75% of the houses cost below $1,000,000 and there are a meaningful number of houses thatcost more than $6,000,000.

(b) Housing prices in a country where 25% of the houses cost below $300,000, 50% of the houses cost below$600,000, 75% of the houses cost below $900,000 and very few houses that cost more than $1,200,000.

(c) Number of alcoholic drinks consumed by college students in a given week. Assume that most of thesestudents don’t drink since they are under 21 years old, and only a few drink excessively.

(d) Annual salaries of the employees at a Fortune 500 company where only a few high level executives earnmuch higher salaries than all the other employees.

2.17 Income at the coffee shop. The first histogram below shows the distribution of the yearly incomes of40 patrons at a college coffee shop. Suppose two new people walk into the coffee shop: one making $225,000and the other $250,000. The second histogram shows the new income distribution. Summary statistics arealso provided.

(1)$60k $62.5k $65k $67.5k $70k

(2)$60k $110k $160k $210k $260k

(1) (2)

n 40 42Min. 60,680 60,680

1st Qu. 63,620 63,710Median 65,240 65,350

Mean 65,090 73,3003rd Qu. 66,160 66,540

Max. 69,890 250,000SD 2,122 37,321

(a) Would the mean or the median best represent what we might think of as a typical income for the 42patrons at this coffee shop? What does this say about the robustness of the two measures?

(b) Would the standard deviation or the IQR best represent the amount of variability in the incomes of the42 patrons at this coffee shop? What does this say about the robustness of the two measures?

2.18 Midrange. The midrange of a distribution is defined as the average of the maximum and the minimumof that distribution. Is this statistic robust to outliers and extreme skew? Explain your reasoning

60 CHAPTER 2. SUMMARIZING DATA

2.19 Commute times. The US census collects data on time it takes Americans to commute to work, amongmany other variables. The histogram below shows the distribution of average commute times in 3,142 UScounties in 2010. Also shown below is a spatial intensity map of the same data.

Mean work travel (in min)10 20 30 40

100

200

>33

(a) Describe the numerical distribution and comment on whether or not a log transformation may beadvisable for these data.

(b) Describe the spatial distribution of commuting times using the map below.

2.20 Hispanic population. The US census collects data on race and ethnicity of Americans, among manyother variables. The histogram below shows the distribution of the percentage of the population that isHispanic in 3,142 counties in the US in 2010. Also shown is a histogram of logs of these values.

Hispanic %0 20 40 60 80 100

500

1000

1500

2000

log(% Hispanic)−2 −1 0 1 2 3 4

100

150

200

250

>40

(a) Describe the numerical distribution and comment on why we might want to use log-transformed valuesin analyzing or modeling these data.

(b) What features of the distribution of the Hispanic population in US counties are apparent in the mapbut not in the histogram? What features are apparent in the histogram but not the map?

2.2. CONSIDERING CATEGORICAL DATA 61

2.2 Considering categorical data

In this section, we will introduce tables and other basic tools for categorical data that areused throughout this book. The loan50 data set represents a sample from a larger loan data setcalled loans. This larger data set contains information on 10,000 loans made through Lending Club.We will examine the relationship between homeownership, which for the loans data can take a valueof rent, mortgage (owns but has a mortgage), or own, and app type, which indicates whether theloan application was made with a partner or whether it was an individual application.

2.2.1 Contingency tables and bar plots

Figure 2.17 summarizes two variables: app type and homeownership. A table that summarizesdata for two categorical variables in this way is called a contingency table. Each value in the tablerepresents the number of times a particular combination of variable outcomes occurred. For example,the value 3496 corresponds to the number of loans in the data set where the borrower rents theirhome and the application type was by an individual. Row and column totals are also included. Therow totals provide the total counts across each row (e.g. 3496 + 3839 + 1170 = 8505), and columntotals are total counts down each column. We can also create a table that shows only the overallpercentages or proportions for each combination of categories, or we can create a table for a singlevariable, such as the one shown in Figure 2.18 for the homeownership variable.

homeownership

rent mortgage own Totalindividual 3496 3839 1170 8505

app typejoint 362 950 183 1495Total 3858 4789 1353 10000

Figure 2.17: A contingency table for app type and homeownership.

homeownership Countrent 3858mortgage 4789own 1353Total 10000

Figure 2.18: A table summarizing the frequencies of each value for thehomeownership variable.

A bar plot is a common way to display a single categorical variable. The left panel of Figure 2.19shows a bar plot for the homeownership variable. In the right panel, the counts are converted intoproportions, showing the proportion of observations that are in each level (e.g. 3858/10000 = 0.3858for rent).

62 CHAPTER 2. SUMMARIZING DATA

rent mortgage own

Fre

quen

1000

2000

3000

4000

Homeownership

rent mortgage own0.0

0.1

0.2

0.3

0.4

Pro

port

ion

Homeownership

Figure 2.19: Two bar plots of number. The left panel shows the counts, and theright panel shows the proportions in each group.

2.2.2 Row and column proportions

Sometimes it is useful to understand the fractional breakdown of one variable in another, and wecan modify our contingency table to provide such a view. Figure 2.20 shows the row proportionsfor Figure 2.17, which are computed as the counts divided by their row totals. The value 3496 atthe intersection of individual and rent is replaced by 3496/8505 = 0.411, i.e. 3496 divided byits row total, 8505. So what does 0.411 represent? It corresponds to the proportion of individualapplicants who rent.

rent mortgage own Totalindividual 0.411 0.451 0.138 1.000joint 0.242 0.635 0.122 1.000Total 0.386 0.479 0.135 1.000

Figure 2.20: A contingency table with row proportions for the app type andhomeownership variables. The row total is off by 0.001 for the joint row dueto a rounding error.

A contingency table of the column proportions is computed in a similar way, where each columnproportion is computed as the count divided by the corresponding column total. Figure 2.21 showssuch a table, and here the value 0.906 indicates that 90.6% of renters applied as individuals for theloan. This rate is higher compared to loans from people with mortgages (80.2%) or who own theirhome (86.5%). Because these rates vary between the three levels of homeownership (rent, mortgage,own), this provides evidence that the app type and homeownership variables are associated.

rent mortgage own Totalindividual 0.906 0.802 0.865 0.851joint 0.094 0.198 0.135 0.150Total 1.000 1.000 1.000 1.000

Figure 2.21: A contingency table with column proportions for the app type andhomeownership variables. The total for the last column is off by 0.001 due to arounding error.

We could also have checked for an association between app type and homeownership in Fig-ure 2.20 using row proportions. When comparing these row proportions, we would look downcolumns to see if the fraction of loans where the borrower rents, has a mortgage, or owns variedacross the individual to joint application types.

2.2. CONSIDERING CATEGORICAL DATA 63

GUIDED PRACTICE 2.23

(a) What does 0.451 represent in Figure 2.20?(b) What does 0.802 represent in Figure 2.21?18

GUIDED PRACTICE 2.24

(a) What does 0.122 at the intersection of joint and own represent in Figure 2.20?(b) What does 0.135 represent in the Figure 2.21?19

EXAMPLE 2.25

Data scientists use statistics to filter spam from incoming email messages. By noting specific char-acteristics of an email, a data scientist may be able to classify some emails as spam or not spam withhigh accuracy. One such characteristic is whether the email contains no numbers, small numbers, orbig numbers. Another characteristic is the email format, which indicates whether or not an emailhas any HTML content, such as bolded text. We’ll focus on email format and spam status using theemail data set, and these variables are summarized in a contingency table in Figure 2.22. Whichwould be more helpful to someone hoping to classify email as spam or regular email for this table:row or column proportions?

A data scientist would be interested in how the proportion of spam changes within each emailformat. This corresponds to column proportions: the proportion of spam in plain text emails andthe proportion of spam in HTML emails.

If we generate the column proportions, we can see that a higher fraction of plain text emails are spam(209/1195 = 17.5%) than compared to HTML emails (158/2726 = 5.8%). This information on itsown is insufficient to classify an email as spam or not spam, as over 80% of plain text emails are notspam. Yet, when we carefully combine this information with many other characteristics, we stand areasonable chance of being able to classify some emails as spam or not spam with confidence.

text HTML Totalspam 209 158 367not spam 986 2568 3554Total 1195 2726 3921

Figure 2.22: A contingency table for spam and format.

Example 2.25 points out that row and column proportions are not equivalent. Before settlingon one form for a table, it is important to consider each to ensure that the most useful table isconstructed. However, sometimes it simply isn’t clear which, if either, is more useful.

EXAMPLE 2.26

Look back to Tables 2.20 and 2.21. Are there any obvious scenarios where one might be more usefulthan the other?

None that we thought were obvious! What is distinct about app type and homeownership vs theemail example is that these two variables don’t have a clear explanatory-response variable relation-ship that we might hypothesize (see Section 1.2.4 for these terms). Usually it is most useful to“condition” on the explanatory variable. For instance, in the email example, the email format wasseen as a possible explanatory variable of whether the message was spam, so we would find it moreinteresting to compute the relative frequencies (proportions) for each email format.

18(a) 0.451 represents the proportion of individual applicants who have a mortgage. (b) 0.802 represents thefraction of applicants with mortgages who applied as individuals.

19(a) 0.122 represents the fraction of joint borrowers who own their home. (b) 0.135 represents the home-owningborrowers who had a joint application for the loan.

64 CHAPTER 2. SUMMARIZING DATA

2.2.3 Using a bar plot with two variables

Contingency tables using row or column proportions are especially useful for examining howtwo categorical variables are related. Stacked bar plots provide a way to visualize the informationin these tables.

A stacked bar plot is a graphical display of contingency table information. For example,a stacked bar plot representing Figure 2.21 is shown in Figure 2.23(a), where we have first created abar plot using the homeownership variable and then divided each group by the levels of app type.

One related visualization to the stacked bar plot is the side-by-side bar plot, where anexample is shown in Figure 2.23(b).

For the last type of bar plot we introduce, the column proportions for the app type andhomeownership contingency table have been translated into a standardized stacked bar plot inFigure 2.23(c). This type of visualization is helpful in understanding the fraction of individualor joint loan applications for borrowers in each level of homeownership. Additionally, since theproportions of joint and individual vary across the groups, we can conclude that the two variablesare associated.

rent mortgage own0

1000

2000

3000

4000

jointindividual

Fre

quen

(a)

rent mortgage own0

1000

2000

3000

4000

jointindividual

Fre

quen

(b)

rent mortgage own

0.0

0.2

0.4

0.6

0.8

1.0

jointindividual

Pro

port

ion

(c)

Figure 2.23: (a) Stacked bar plot for homeownership, where the counts have beenfurther broken down by app type. (b) Side-by-side bar plot for homeownership

and app type. (c) Standardized version of the stacked bar plot.

2.2. CONSIDERING CATEGORICAL DATA 65

EXAMPLE 2.27

Examine the three bar plots in Figure 2.23. When is the stacked, side-by-side, or standardizedstacked bar plot the most useful?

The stacked bar plot is most useful when it’s reasonable to assign one variable as the explanatoryvariable and the other variable as the response, since we are effectively grouping by one variable firstand then breaking it down by the others.

Side-by-side bar plots are more agnostic in their display about which variable, if any, represents theexplanatory and which the response variable. It is also easy to discern the number of cases in of thesix different group combinations. However, one downside is that it tends to require more horizontalspace; the narrowness of Figure 2.23(b) makes the plot feel a bit cramped. Additionally, when twogroups are of very different sizes, as we see in the own group relative to either of the other twogroups, it is difficult to discern if there is an association between the variables.

The standardized stacked bar plot is helpful if the primary variable in the stacked bar plot is relativelyimbalanced, e.g. the own category has only a third of the observations in the mortgage category,making the simple stacked bar plot less useful for checking for an association. The major downsideof the standardized version is that we lose all sense of how many cases each of the bars represents.

2.2.4 Mosaic plots

A mosaic plot is a visualization technique suitable for contingency tables that resembles astandardized stacked bar plot with the benefit that we still see the relative group sizes of the primaryvariable as well.

To get started in creating our first mosaic plot, we’ll break a square into columns for eachcategory of the homeownership variable, with the result shown in Figure 2.24(a). Each columnrepresents a level of homeownership, and the column widths correspond to the proportion of loansin each of those categories. For instance, there are fewer loans where the borrower is an owner thanwhere the borrower has a mortgage. In general, mosaic plots use box areas to represent the numberof cases in each category.

rent mortgage own

(a)

rent mortgage own

joint

indiv.

(b)

Figure 2.24: (a) The one-variable mosaic plot for homeownership. (b) Two-variablemosaic plot for both homeownership and app type.

To create a completed mosaic plot, the single-variable mosaic plot is further divided into piecesin Figure 2.24(b) using the app type variable. Each column is split proportional to the numberof loans from individual and joint borrowers. For example, the second column represents loanswhere the borrower has a mortgage, and it was divided into individual loans (upper) and joint loans(lower). As another example, the bottom segment of the third column represents loans where theborrower owns their home and applied jointly, while the upper segment of this column representsborrowers who are homeowners and filed individually. We can again use this plot to see that thehomeownership and app type variables are associated, since some columns are divided in different

66 CHAPTER 2. SUMMARIZING DATA

vertical locations than others, which was the same technique used for checking an association in thestandardized stacked bar plot.

In Figure 2.24, we chose to first split by the homeowner status of the borrower. However, wecould have instead first split by the application type, as in Figure 2.25. Like with the bar plots, it’scommon to use the explanatory variable to represent the first split in a mosaic plot, and then for theresponse to break up each level of the explanatory variable, if these labels are reasonable to attachto the variables under consideration.

indiv. joint

rent

mortgage

own

Figure 2.25: Mosaic plot where loans are grouped by the homeownership variableafter they’ve been divided into the individual and joint application types.

2.2.5 The only pie chart you will see in this book

A pie chart is shown in Figure 2.26 alongside a bar plot representing the same information.Pie charts can be useful for giving a high-level overview to show how a set of cases break down.However, it is also difficult to decipher details in a pie chart. For example, it takes a couple secondslonger to recognize that there are more loans where the borrower has a mortgage than rent whenlooking at the pie chart, while this detail is very obvious in the bar plot. While pie charts can beuseful, we prefer bar plots for their ease in comparing groups.

rent

mortgage

own

rent mortgage own

Homeownership

1000

2000

3000

4000

Fre

quen

Figure 2.26: A pie chart and bar plot of homeownership.

2.2. CONSIDERING CATEGORICAL DATA 67

2.2.6 Comparing numerical data across groups

Some of the more interesting investigations can be considered by examining numerical dataacross groups. The methods required here aren’t really new: all that’s required is to make a numericalplot for each group in the same graph. Here two convenient methods are introduced: side-by-sidebox plots and hollow histograms.

We will take a look again at the county data set and compare the median household incomefor counties that gained population from 2010 to 2017 versus counties that had no gain. While wemight like to make a causal connection here, remember that these are observational data and sosuch an interpretation would be, at best, half-baked.

There were 1,454 counties where the population increased from 2010 to 2017, and there were1,672 counties with no gain (all but one were a loss). A random sample of 100 counties from thefirst group and 50 from the second group are shown in Figure 2.27 to give a better sense of some ofthe raw median income data.

Median Income for 150 Counties, in $1000s

Population Gain No Population Gain38.2 43.6 42.2 61.5 51.1 45.7 48.3 60.3 50.744.6 51.8 40.7 48.1 56.4 41.9 39.3 40.4 40.340.6 63.3 52.1 60.3 49.8 51.7 57 47.2 45.951.1 34.1 45.5 52.8 49.1 51 42.3 41.5 46.180.8 46.3 82.2 43.6 39.7 49.4 44.9 51.7 46.475.2 40.6 46.3 62.4 44.1 51.3 29.1 51.8 50.551.9 34.7 54 42.9 52.2 45.1 27 30.9 34.961 51.4 56.5 62 46 46.4 40.7 51.8 61.1

53.8 57.6 69.2 48.4 40.5 48.6 43.4 34.7 45.753.1 54.6 55 46.4 39.9 56.7 33.1 21 3763 49.1 57.2 44.1 50 38.9 52 31.9 45.7

46.6 46.5 38.9 50.9 56 34.6 56.3 38.7 45.774.2 63 49.6 53.7 77.5 60 56.2 43 21.763.2 47.6 55.9 39.1 57.8 42.6 44.5 34.5 48.950.4 49 45.6 39 38.8 37.1 50.9 42.1 43.257.2 44.7 71.7 35.3 100.2 35.4 41.3 33.642.6 55.5 38.6 52.7 63 43.4 56.5

Figure 2.27: In this table, median household income (in $1000s) from a randomsample of 100 counties that had population gains are shown on the left. Medianincomes from a random sample of 50 counties that had no population gain areshown on the right.

68 CHAPTER 2. SUMMARIZING DATA

Change in Population

Gain No Gain

$20k

$40k

$60k

$80k

$100k

$120k

Med

ian

Hou

seho

ld In

com

Median Household Income

$20k $40k $60k $80k $100k

GainNo Gain

Figure 2.28: Side-by-side box plot (left panel) and hollow histograms (right panel)for med hh income, where the counties are split by whether there was a populationgain or loss.

The side-by-side box plot is a traditional tool for comparing across groups. An example isshown in the left panel of Figure 2.28, where there are two box plots, one for each group, placedinto one plotting window and drawn on the same scale.

Another useful plotting method uses hollow histograms to compare numerical data acrossgroups. These are just the outlines of histograms of each group put on the same plot, as shown inthe right panel of Figure 2.28.

GUIDED PRACTICE 2.28

Use the plots in Figure 2.28 to compare the incomes for counties across the two groups. What doyou notice about the approximate center of each group? What do you notice about the variabilitybetween groups? Is the shape relatively consistent between groups? How many prominent modesare there for each group?20

GUIDED PRACTICE 2.29

What components of each plot in Figure 2.28 do you find most useful?21

20Answers may vary a little. The counties with population gains tend to have higher income (median of about$45,000) versus counties without a gain (median of about $40,000). The variability is also slightly larger for thepopulation gain group. This is evident in the IQR, which is about 50% bigger in the gain group. Both distributionsshow slight to moderate right skew and are unimodal. The box plots indicate there are many observations far abovethe median in each group, though we should anticipate that many observations will fall beyond the whiskers whenexamining any data set that contain more than a couple hundred data points.

21Answers will vary. The side-by-side box plots are especially useful for comparing centers and spreads, while thehollow histograms are more useful for seeing distribution shape, skew, and potential anomalies.

2.2. CONSIDERING CATEGORICAL DATA 69

Exercises

2.21 Antibiotic use in children. The bar plot and the pie chart below show the distribution of pre-existingmedical conditions of children involved in a study on the optimal duration of antibiotic use in treatment oftracheitis, which is an upper respiratory infection.

GastrointestinalImmunocompromised

Genetic/metabolicNeuromuscular

TraumaRespiratory

CardiovascularPrematurity

Relative frequency0.00 0.10 0.20 0.30

GastrointestinalImmunocompromised

Genetic/metabolic

NeuromuscularTrauma

Respiratory

Cardiovascular

Prematurity

(a) What features are apparent in the bar plot but not in the pie chart?

(b) What features are apparent in the pie chart but not in the bar plot?

2.22 Views on immigration. 910 randomly sampled registered voters from Tampa, FL were asked if theythought workers who have illegally entered the US should be (i) allowed to keep their jobs and apply forUS citizenship, (ii) allowed to keep their jobs as temporary guest workers but not allowed to apply for UScitizenship, or (iii) lose their jobs and have to leave the country. The results of the survey by politicalideology are shown below.22

Political ideologyConservative Moderate Liberal Total

(i) Apply for citizenship 57 120 101 278(ii) Guest worker 121 113 28 262

Response(iii) Leave the country 179 126 45 350(iv) Not sure 15 4 1 20Total 372 363 175 910

(a) What percent of these Tampa, FL voters identify themselves as conservatives?

(b) What percent of these Tampa, FL voters are in favor of the citizenship option?

(c) What percent of these Tampa, FL voters identify themselves as conservatives and are in favor of thecitizenship option?

(d) What percent of these Tampa, FL voters who identify themselves as conservatives are also in favor ofthe citizenship option? What percent of moderates share this view? What percent of liberals share thisview?

(e) Do political ideology and views on immigration appear to be independent? Explain your reasoning.

22SurveyUSA, News Poll #18927, data collected Jan 27-29, 2012.

70 CHAPTER 2. SUMMARIZING DATA

2.23 Views on the DREAM Act. A random sample of registered voters from Tampa, FL were askedif they support the DREAM Act, a proposed law which would provide a path to citizenship for peoplebrought illegally to the US as children. The survey also collected information on the political ideology of therespondents. Based on the mosaic plot shown below, do views on the DREAM Act and political ideologyappear to be independent? Explain your reasoning.23

Conservative Moderate Liberal

Support

Not support

Not sure

2.24 Raise taxes. A random sample of registered voters nationally were asked whether they think it’sbetter to raise taxes on the rich or raise taxes on the poor. The survey also collected information on thepolitical party affiliation of the respondents. Based on the mosaic plot shown below, do views on raisingtaxes and political affiliation appear to be independent? Explain your reasoning.24

Democrat Republican Indep / Other

Raise taxes on the rich

Raise taxes on the poorNot sure

23SurveyUSA, News Poll #18927, data collected Jan 27-29, 2012.24Public Policy Polling, Americans on College Degrees, Classic Literature, the Seasons, and More, data collected

Feb 20-22, 2015.

2.3. CASE STUDY: MALARIA VACCINE 71

2.3 Case study: malaria vaccine

EXAMPLE 2.30

Suppose your professor splits the students in class into two groups: students on the left and studentson the right. If p̂

Land p̂

Rrepresent the proportion of students who own an Apple product on the

left and right, respectively, would you be surprised if p̂L

did not exactly equal p̂R

While the proportions would probably be close to each other, it would be unusual for them to beexactly the same. We would probably observe a small difference due to chance.

GUIDED PRACTICE 2.31

If we don’t think the side of the room a person sits on in class is related to whether the personowns an Apple product, what assumption are we making about the relationship between these twovariables?25

2.3.1 Variability within data

We consider a study on a new malaria vaccine called PfSPZ. In this study, volunteer patientswere randomized into one of two experiment groups: 14 patients received an experimental vaccineor 6 patients received a placebo vaccine. Nineteen weeks later, all 20 patients were exposed to adrug-sensitive malaria virus strain; the motivation of using a drug-sensitive strain of virus here is forethical considerations, allowing any infections to be treated effectively. The results are summarizedin Figure 2.29, where 9 of the 14 treatment patients remained free of signs of infection while all ofthe 6 patients in the control group patients showed some baseline signs of infection.

outcome

infection no infection Totalvaccine 5 9 14

treatmentplacebo 6 0 6Total 11 9 20

Figure 2.29: Summary results for the malaria vaccine experiment.

GUIDED PRACTICE 2.32

Is this an observational study or an experiment? What implications does the study type have onwhat can be inferred from the results?26

In this study, a smaller proportion of patients who received the vaccine showed signs of aninfection (35.7% versus 100%). However, the sample is very small, and it is unclear whether thedifference provides convincing evidence that the vaccine is effective.

25We would be assuming that these two variables are independent.26The study is an experiment, as patients were randomly assigned an experiment group. Since this is an experiment,

the results can be used to evaluate a causal relationship between the malaria vaccine and whether patients showedsigns of an infection.

72 CHAPTER 2. SUMMARIZING DATA

EXAMPLE 2.33

Data scientists are sometimes called upon to evaluate the strength of evidence. When looking atthe rates of infection for patients in the two groups in this study, what comes to mind as we try todetermine whether the data show convincing evidence of a real difference?

The observed infection rates (35.7% for the treatment group versus 100% for the control group)suggest the vaccine may be effective. However, we cannot be sure if the observed difference representsthe vaccine’s efficacy or is just from random chance. Generally there is a little bit of fluctuation insample data, and we wouldn’t expect the sample proportions to be exactly equal, even if the truthwas that the infection rates were independent of getting the vaccine. Additionally, with such smallsamples, perhaps it’s common to observe such large differences when we randomly split a group dueto chance alone!

Example 2.33 is a reminder that the observed outcomes in the data sample may not perfectlyreflect the true relationships between variables since there is random noise. While the observeddifference in rates of infection is large, the sample size for the study is small, making it unclear ifthis observed difference represents efficacy of the vaccine or whether it is simply due to chance. Welabel these two competing claims, H0 and HA, which are spoken as “H-nought” and “H-A”:

H0: Independence model. The variables treatment and outcome are independent. They haveno relationship, and the observed difference between the proportion of patients who developedan infection in the two groups, 64.3%, was due to chance.

HA: Alternative model. The variables are not independent. The difference in infection rates of64.3% was not due to chance, and vaccine affected the rate of infection.

What would it mean if the independence model, which says the vaccine had no influence on therate of infection, is true? It would mean 11 patients were going to develop an infection no matterwhich group they were randomized into, and 9 patients would not develop an infection no matterwhich group they were randomized into. That is, if the vaccine did not affect the rate of infection,the difference in the infection rates was due to chance alone in how the patients were randomized.

Now consider the alternative model: infection rates were influenced by whether a patient re-ceived the vaccine or not. If this was true, and especially if this influence was substantial, we wouldexpect to see some difference in the infection rates of patients in the groups.

We choose between these two competing claims by assessing if the data conflict so much withH0 that the independence model cannot be deemed reasonable. If this is the case, and the datasupport HA, then we will reject the notion of independence and conclude there was discrimination.

2.3.2 Simulating the study

We’re going to implement simulations, where we will pretend we know that the malaria vaccinebeing tested does not work. Ultimately, we want to understand if the large difference we observedis common in these simulations. If it is common, then maybe the difference we observed was purelydue to chance. If it is very uncommon, then the possibility that the vaccine was helpful seems moreplausible.

Figure 2.29 shows that 11 patients developed infections and 9 did not. For our simulation,we will suppose the infections were independent of the vaccine and we were able to rewind backto when the researchers randomized the patients in the study. If we happened to randomize thepatients differently, we may get a different result in this hypothetical world where the vaccine doesn’tinfluence the infection. Let’s complete another randomization using a simulation.

2.3. CASE STUDY: MALARIA VACCINE 73

In this simulation, we take 20 notecards to represent the 20 patients, where we write down“infection” on 11 cards and “no infection” on 9 cards. In this hypothetical world, we believe eachpatient that got an infection was going to get it regardless of which group they were in, so let’s seewhat happens if we randomly assign the patients to the treatment and control groups again. Wethoroughly shuffle the notecards and deal 14 into a vaccine pile and 6 into a placebo pile. Finally,we tabulate the results, which are shown in Figure 2.30.

outcome

infection no infection Totaltreatment vaccine 7 7 14(simulated) placebo 4 2 6

Total 11 9 20

Figure 2.30: Simulation results, where any difference in infection rates is purelydue to chance.

GUIDED PRACTICE 2.34

What is the difference in infection rates between the two simulated groups in Figure 2.30? How doesthis compare to the observed 64.3% difference in the actual data?27

2.3.3 Checking for independence

We computed one possible difference under the independence model in Guided Practice 2.34,which represents one difference due to chance. While in this first simulation, we physically dealt outnotecards to represent the patients, it is more efficient to perform this simulation using a computer.Repeating the simulation on a computer, we get another difference due to chance:

6− 9

14= −0.310

And another:

6− 8

14= −0.071

And so on until we repeat the simulation enough times that we have a good idea of what representsthe distribution of differences from chance alone. Figure 2.31 shows a stacked plot of the differencesfound from 100 simulations, where each dot represents a simulated difference between the infectionrates (control rate minus treatment rate).

Note that the distribution of these simulated differences is centered around 0. We simulatedthese differences assuming that the independence model was true, and under this condition, weexpect the difference to be near zero with some random fluctuation, where near is pretty generousin this case since the sample sizes are so small in this study.

EXAMPLE 2.35

How often would you observe a difference of at least 64.3% (0.643) according to Figure 2.31? Often,sometimes, rarely, or never?

It appears that a difference of at least 64.3% due to chance alone would only happen about 2% ofthe time according to Figure 2.31. Such a low probability indicates a rare event.

274/6 − 7/14 = 0.167 or about 16.7% in favor of the vaccine. This difference due to chance is much smaller thanthe difference observed in the actual groups.

74 CHAPTER 2. SUMMARIZING DATA

●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●

●●●

●●

Difference in Infection Rates

−0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8

Figure 2.31: A stacked dot plot of differences from 100 simulations produced underthe independence model, H0, where in these simulations infections are unaffectedby the vaccine. Two of the 100 simulations had a difference of at least 64.3%, thedifference observed in the study.

The difference of 64.3% being a rare event suggests two possible interpretations of the resultsof the study:

H0 Independence model. The vaccine has no effect on infection rate, and we just happened toobserve a difference that would only occur on a rare occasion.

HA Alternative model. The vaccine has an effect on infection rate, and the difference weobserved was actually due to the vaccine being effective at combatting malaria, which explainsthe large difference of 64.3%.

Based on the simulations, we have two options. (1) We conclude that the study results do notprovide strong evidence against the independence model. That is, we do not have sufficiently strongevidence to conclude the vaccine had an effect in this clinical setting. (2) We conclude the evidenceis sufficiently strong to reject H0 and assert that the vaccine was useful. When we conduct formalstudies, usually we reject the notion that we just happened to observe a rare event.28 So in this case,we reject the independence model in favor of the alternative. That is, we are concluding the dataprovide strong evidence that the vaccine provides some protection against malaria in this clinicalsetting.

One field of statistics, statistical inference, is built on evaluating whether such differences aredue to chance. In statistical inference, data scientists evaluate which model is most reasonable giventhe data. Errors do occur, just like rare events, and we might choose the wrong model. While wedo not always choose correctly, statistical inference gives us tools to control and evaluate how oftenthese errors occur. In Chapter 5, we give a formal introduction to the problem of model selection.We spend the next two chapters building a foundation of probability and theory necessary to makethat discussion rigorous.

28This reasoning does not generally extend to anecdotal observations. Each of us observes incredibly rare eventsevery day, events we could not possibly hope to predict. However, in the non-rigorous setting of anecdotal evidence,almost anything may appear to be a rare event, so the idea of looking for rare events in day-to-day activities istreacherous. For example, we might look at the lottery: there was only a 1 in 292 million chance that the Powerballnumbers for the largest jackpot in history (January 13th, 2016) would be (04, 08, 19, 27, 34) with a Powerball of(10), but nonetheless those numbers came up! However, no matter what numbers had turned up, they would havehad the same incredibly rare odds. That is, any set of numbers we could have observed would ultimately be incrediblyrare. This type of situation is typical of our daily lives: each possible event in itself seems incredibly rare, but if weconsider every alternative, those outcomes are also incredibly rare. We should be cautious not to misinterpret suchanecdotal evidence.

2.3. CASE STUDY: MALARIA VACCINE 75

Exercises

2.25 Side effects of Avandia. Rosiglitazone is the active ingredient in the controversial type 2 diabetesmedicine Avandia and has been linked to an increased risk of serious cardiovascular problems such as stroke,heart failure, and death. A common alternative treatment is pioglitazone, the active ingredient in a diabetesmedicine called Actos. In a nationwide retrospective observational study of 227,571 Medicare beneficiariesaged 65 years or older, it was found that 2,593 of the 67,593 patients using rosiglitazone and 5,386 ofthe 159,978 using pioglitazone had serious cardiovascular problems. These data are summarized in thecontingency table below.29

Cardiovascular problemsYes No Total

TreatmentRosiglitazone 2,593 65,000 67,593Pioglitazone 5,386 154,592 159,978Total 7,979 219,592 227,571

(a) Determine if each of the following statements is true or false. If false, explain why. Be careful: Thereasoning may be wrong even if the statement’s conclusion is correct. In such cases, the statementshould be considered false.

i. Since more patients on pioglitazone had cardiovascular problems (5,386 vs. 2,593), we can concludethat the rate of cardiovascular problems for those on a pioglitazone treatment is higher.

ii. The data suggest that diabetic patients who are taking rosiglitazone are more likely to have cardio-vascular problems since the rate of incidence was (2,593 / 67,593 = 0.038) 3.8% for patients on thistreatment, while it was only (5,386 / 159,978 = 0.034) 3.4% for patients on pioglitazone.

iii. The fact that the rate of incidence is higher for the rosiglitazone group proves that rosiglitazonecauses serious cardiovascular problems.

iv. Based on the information provided so far, we cannot tell if the difference between the rates ofincidences is due to a relationship between the two variables or due to chance.

(b) What proportion of all patients had cardiovascular problems?

(c) If the type of treatment and having cardiovascular problems were independent, about how many patientsin the rosiglitazone group would we expect to have had cardiovascular problems?

(d) We can investigate the relationship between outcome and treatment in this study using a randomizationtechnique. While in reality we would carry out the simulations required for randomization using statisti-cal software, suppose we actually simulate using index cards. In order to simulate from the independencemodel, which states that the outcomes were independent of the treatment, we write whether or not eachpatient had a cardiovascular problem on cards, shuffled all the cards together, then deal them into twogroups of size 67,593 and 159,978. We repeat this simulation 1,000 times and each time record the num-ber of people in the rosiglitazone group who had cardiovascular problems. Use the relative frequencyhistogram of these counts to answer (i)-(iii).

i. What are the claims being tested?

ii. Compared to the number calculated in part (b),which would provide more support for the alterna-tive hypothesis, more or fewer patients with car-diovascular problems in the rosiglitazone group?

iii. What do the simulation results suggest about therelationship between taking rosiglitazone and hav-ing cardiovascular problems in diabetic patients?

Simulated rosiglitazone cardiovascular events2250 2350 2450

0.1

0.2

29D.J. Graham et al. “Risk of acute myocardial infarction, stroke, heart failure, and death in elderly Medicarepatients treated with rosiglitazone or pioglitazone”. In: JAMA 304.4 (2010), p. 411. issn: 0098-7484.

76 CHAPTER 2. SUMMARIZING DATA

2.26 Heart transplants. The Stanford University Heart Transplant Study was conducted to determinewhether an experimental heart transplant program increased lifespan. Each patient entering the programwas designated an official heart transplant candidate, meaning that he was gravely ill and would most likelybenefit from a new heart. Some patients got a transplant and some did not. The variable transplant

indicates which group the patients were in; patients in the treatment group got a transplant and those in thecontrol group did not. Of the 34 patients in the control group, 30 died. Of the 69 people in the treatmentgroup, 45 died. Another variable called survived was used to indicate whether or not the patient was aliveat the end of the study.30

control treatment

alive

dead

Sur

viva

l Tim

e (d

ays)

control treatment

500

1000

1500

(a) Based on the mosaic plot, is survival independent of whether or not the patient got a transplant? Explainyour reasoning.

(b) What do the box plots below suggest about the efficacy (effectiveness) of the heart transplant treatment.

(d) One approach for investigating whether or not the treatment is effective is to use a randomizationtechnique.

i. What are the claims being tested?

ii. The paragraph below describes the set up for such approach, if we were to do it without usingstatistical software. Fill in the blanks with a number or phrase, whichever is appropriate.

We write alive on cards representing patients who were alive at the end ofthe study, and dead on cards representing patients who were not. Then,we shuffle these cards and split them into two groups: one group of sizerepresenting treatment, and another group of size representing control. Wecalculate the difference between the proportion of dead cards in the treatment and controlgroups (treatment – control) and record this value. We repeat this 100 times to build adistribution centered at . Lastly, we calculate the fraction of simulationswhere the simulated differences in proportions are . If this fraction is low,we conclude that it is unlikely to have observed such an outcome by chance and that thenull hypothesis should be rejected in favor of the alternative.

iii. What do the simulation results shown below suggest about the effectiveness of the transplant pro-gram?

simulated differences in proportions−0.25 −0.15 −0.05 0.05 0.15 0.25

●

● ●

●

30B. Turnbull et al. “Survivorship of Heart Transplant Data”. In: Journal of the American Statistical Association69 (1974), pp. 74–80.

2.3. CASE STUDY: MALARIA VACCINE 77

Chapter exercises

2.27 Make-up exam. In a class of 25 students, 24 of them took an exam in class and 1 student took amake-up exam the following day. The professor graded the first batch of 24 exams and found an averagescore of 74 points with a standard deviation of 8.9 points. The student who took the make-up the followingday scored 64 points on the exam.

(a) Does the new student’s score increase or decrease the average score?

(b) What is the new average?

2.28 Infant mortality. The infant mortality rate is defined as the number of infant deaths per 1,000 livebirths. This rate is often used as an indicator of the level of health in a country. The relative frequencyhistogram below shows the distribution of estimated infant death rates for 224 countries for which such datawere available in 2014.31

(a) Estimate Q1, the median, and Q3 from thehistogram.

(b) Would you expect the mean of this data setto be smaller or larger than the median?Explain your reasoning.

Infant Mortality (per 1000 Live Births)

Fra

ctio

n of

Cou

ntrie

0 20 40 60 80 100 1200

0.1

0.2

0.3

0.4

2.29 TV watchers. Students in an AP Statistics class were asked how many hours of television theywatch per week (including online streaming). This sample yielded an average of 4.71 hours, with a standarddeviation of 4.18 hours. Is the distribution of number of hours students watch television weekly symmetric?If not, what shape would you expect this distribution to have? Explain your reasoning.

2.30 A new statistic. The statistic x̄median

can be used as a measure of skewness. Suppose we have adistribution where all observations are greater than 0, xi > 0. What is the expected shape of the distributionunder the following conditions? Explain your reasoning.

(a) x̄median

= 1

(b) x̄median

< 1

> 1

2.31 Oscar winners. The first Oscar awards for best actor and best actress were given out in 1929. Thehistograms below show the age distribution for all of the best actor and best actress winners from 1929 to2018. Summary statistics for these distributions are also provided. Compare the distributions of ages ofbest actor and actress winners.32

Best actor

Best actress

20 40 60 80

Age (in years)

Best Actress

Mean 36.2SD 11.9n 92

Best Actor

Mean 43.8SD 8.83n 92

31CIA Factbook, Country Comparisons, 2014.32Oscar winners from 1929 – 2012, data up to 2009 from the Journal of Statistics Education data archive and more

current data from wikipedia.org.

78 CHAPTER 2. SUMMARIZING DATA

2.32 Exam scores. The average on a history exam (scored out of 100 points) was 85, with a standarddeviation of 15. Is the distribution of the scores on this exam symmetric? If not, what shape would youexpect this distribution to have? Explain your reasoning.

2.33 Stats scores. Below are the final exam scores of twenty introductory statistics students.

57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94

Create a box plot of the distribution of these scores. The five number summary provided below may beuseful.

Min Q1 Q2 (Median) Q3 Max

57 72.5 78.5 82.5 94

2.34 Marathon winners. The histogram and box plots below show the distribution of finishing times formale and female winners of the New York Marathon between 1970 and 1999.

2.0 2.4 2.8 3.20

Mar

atho

n tim

es2.0

2.4

2.8

3.2

(a) What features of the distribution are apparent in the histogram and not the box plot? What featuresare apparent in the box plot but not in the histogram?

(b) What may be the reason for the bimodal distribution? Explain.

2.0 2.4 2.8 3.2

Women

Men

(d) The time series plot shown below is another way to look at these data. Describe what is visible in thisplot but not in the others.

● ● ● ●

●● ● ●

● ●

●● ● ● ● ● ●

● ● ●

●

● ● ● ●Mar

atho

n tim

1970 1975 1980 1985 1990 1995 20002.0

2.4

2.8

3.2●

WomenMen

Chapter 3Probability

3.1 Defining probability

3.2 Conditional probability

3.3 Sampling from a small population

3.4 Random variables

3.5 Continuous distributions

Probability forms the foundation of statistics, and you’re probably

already aware of many of the ideas presented in this chapter. However,

formalization of probability concepts is likely new for most readers.

While this chapter provides a theoretical foundation for the ideas in

later chapters and provides a path to a deeper understanding, mastery

of the concepts introduced in this chapter is not required for applying

the methods introduced in the rest of this book.

For videos, slides, and other resources, please visit

www.openintro.org/os

3.1. DEFINING PROBABILITY 81

3.1 Defining probability

Statistics is based on probability, and while probability is not required for the applied techniquesin this book, it may help you gain a deeper understanding of the methods and set a better foundationfor future courses.

3.1.1 Introductory examples

Before we get into technical ideas, let’s walk through some basic examples that may feel morefamiliar.

EXAMPLE 3.1

A “die”, the singular of dice, is a cube with six faces numbered 1, 2, 3, 4, 5, and 6. What is thechance of getting 1 when rolling a die?

If the die is fair, then the chance of a 1 is as good as the chance of any other number. Since thereare six outcomes, the chance must be 1-in-6 or, equivalently, 1/6.

EXAMPLE 3.2

What is the chance of getting a 1 or 2 in the next roll?

1 and 2 constitute two of the six equally likely possible outcomes, so the chance of getting one ofthese two outcomes must be 2/6 = 1/3.

EXAMPLE 3.3

What is the chance of getting either 1, 2, 3, 4, 5, or 6 on the next roll?

100%. The outcome must be one of these numbers.

EXAMPLE 3.4

What is the chance of not rolling a 2?

Since the chance of rolling a 2 is 1/6 or 16.6̄%, the chance of not rolling a 2 must be 100%−16.6̄% =83.3̄% or 5/6.

Alternatively, we could have noticed that not rolling a 2 is the same as getting a 1, 3, 4, 5, or 6,which makes up five of the six equally likely outcomes and has probability 5/6.

EXAMPLE 3.5

Consider rolling two dice. If 1/6 of the time the first die is a 1 and 1/6 of those times the seconddie is a 1, what is the chance of getting two 1s?

If 16.6̄% of the time the first die is a 1 and 1/6 of those times the second die is also a 1, then thechance that both dice are 1 is (1/6)× (1/6) or 1/36.

82 CHAPTER 3. PROBABILITY

3.1.2 Probability

We use probability to build tools to describe and understand apparent randomness. We oftenframe probability in terms of a random process giving rise to an outcome.

Roll a die → 1, 2, 3, 4, 5, or 6Flip a coin → H or T

Rolling a die or flipping a coin is a seemingly random process and each gives rise to an outcome.

PROBABILITY

The probability of an outcome is the proportion of times the outcome would occur if weobserved the random process an infinite number of times.

Probability is defined as a proportion, and it always takes values between 0 and 1 (inclusively).It may also be displayed as a percentage between 0% and 100%.

Probability can be illustrated by rolling a die many times. Let p̂n be the proportion of outcomesthat are 1 after the first n rolls. As the number of rolls increases, p̂n will converge to the probabilityof rolling a 1, p = 1/6. Figure 3.1 shows this convergence for 100,000 die rolls. The tendency of p̂nto stabilize around p is described by the Law of Large Numbers.

n (number of rolls)

1 10 100 1,000 10,000 100,000

0.0

0.1

0.2

0.3

p̂n

Figure 3.1: The fraction of die rolls that are 1 at each stage in a simulation. Theproportion tends to get closer to the probability 1/6 ≈ 0.167 as the number of rollsincreases.

LAW OF LARGE NUMBERS

As more observations are collected, the proportion p̂n of occurrences with a particular outcomeconverges to the probability p of that outcome.

Occasionally the proportion will veer off from the probability and appear to defy the Law ofLarge Numbers, as p̂n does many times in Figure 3.1. However, these deviations become smaller asthe number of rolls increases.

Above we write p as the probability of rolling a 1. We can also write this probability as

P (rolling a 1)

As we become more comfortable with this notation, we will abbreviate it further. For instance, if itis clear that the process is “rolling a die”, we could abbreviate P (rolling a 1) as P (1).

3.1. DEFINING PROBABILITY 83

GUIDED PRACTICE 3.6

Random processes include rolling a die and flipping a coin. (a) Think of another random process.(b) Describe all the possible outcomes of that process. For instance, rolling a die is a random processwith possible outcomes 1, 2, …, 6.1

What we think of as random processes are not necessarily random, but they may just be toodifficult to understand exactly. The fourth example in the footnote solution to Guided Practice 3.6suggests a roommate’s behavior is a random process. However, even if a roommate’s behavior is nottruly random, modeling her behavior as a random process can still be useful.

3.1.3 Disjoint or mutually exclusive outcomes

Two outcomes are called disjoint or mutually exclusive if they cannot both happen. Forinstance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the otherhand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcomeof the roll is a 1. The terms disjoint and mutually exclusive are equivalent and interchangeable.

Calculating the probability of disjoint outcomes is easy. When rolling a die, the outcomes 1

and 2 are disjoint, and we compute the probability that one of these outcomes will occur by addingtheir separate probabilities:

P (1 or 2) = P (1) + P (2) = 1/6 + 1/6 = 1/3

What about the probability of rolling a 1, 2, 3, 4, 5, or 6? Here again, all of the outcomes aredisjoint so we add the probabilities:

P (1 or 2 or 3 or 4 or 5 or 6)

= P (1) + P (2) + P (3) + P (4) + P (5) + P (6)

= 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1

The Addition Rule guarantees the accuracy of this approach when the outcomes are disjoint.

ADDITION RULE OF DISJOINT OUTCOMES

If A1 and A2 represent two disjoint outcomes, then the probability that one of them occurs isgiven by

P (A1 or A2) = P (A1) + P (A2)

If there are many disjoint outcomes A1, …, Ak, then the probability that one of these outcomeswill occur is

P (A1) + P (A2) + · · ·+ P (Ak)

1Here are four examples. (i) Whether someone gets sick in the next month or not is an apparently random processwith outcomes sick and not. (ii) We can generate a random process by randomly picking a person and measuringthat person’s height. The outcome of this process will be a positive number. (iii) Whether the stock market goes upor down next week is a seemingly random process with possible outcomes up, down, and no change. Alternatively, wecould have used the percent change in the stock market as a numerical outcome. (iv) Whether your roommate cleansher dishes tonight probably seems like a random process with possible outcomes cleans dishes and leaves dishes.

84 CHAPTER 3. PROBABILITY

GUIDED PRACTICE 3.7

We are interested in the probability of rolling a 1, 4, or 5. (a) Explain why the outcomes 1, 4, and5 are disjoint. (b) Apply the Addition Rule for disjoint outcomes to determine P (1 or 4 or 5).2

GUIDED PRACTICE 3.8

In the loans data set in Chapter 2, the homeownership variable described whether the borrowerrents, has a mortgage, or owns her property. Of the 10,000 borrowers, 3858 rented, 4789 had amortgage, and 1353 owned their home.3

(a) Are the outcomes rent, mortgage, and own disjoint?

(b) Determine the proportion of loans with value mortgage and own separately.

(c) Use the Addition Rule for disjoint outcomes to compute the probability a randomly selectedloan from the data set is for someone who has a mortgage or owns her home.

Data scientists rarely work with individual outcomes and instead consider sets or collectionsof outcomes. Let A represent the event where a die roll results in 1 or 2 and B represent the eventthat the die roll is a 4 or a 6. We write A as the set of outcomes {1, 2} and B = {4, 6}. Thesesets are commonly called events. Because A and B have no elements in common, they are disjointevents. A and B are represented in Figure 3.2.

1 2 3 4 5 6

Figure 3.2: Three events, A, B, and D, consist of outcomes from rolling a die. Aand B are disjoint since they do not have any outcomes in common.

The Addition Rule applies to both disjoint outcomes and disjoint events. The probability thatone of the disjoint events A or B occurs is the sum of the separate probabilities:

P (A or B) = P (A) + P (B) = 1/3 + 1/3 = 2/3

GUIDED PRACTICE 3.9

(a) Verify the probability of event A, P (A), is 1/3 using the Addition Rule. (b) Do the same forevent B.4

GUIDED PRACTICE 3.10

(a) Using Figure 3.2 as a reference, what outcomes are represented by event D? (b) Are events Band D disjoint? (c) Are events A and D disjoint?5

GUIDED PRACTICE 3.11

In Guided Practice 3.10, you confirmed B and D from Figure 3.2 are disjoint. Compute the proba-bility that event B or event D occurs.6

2(a) The random process is a die roll, and at most one of these outcomes can come up. This means they aredisjoint outcomes. (b) P (1 or 4 or 5) = P (1) + P (4) + P (5) = 1

6+ 1

6= 3

6= 1

23(a) Yes. Each loan is categorized in only one level of homeownership. (b) Mortgage: 4789

10000= 0.479. Own:

135310000

= 0.135. (c) P (mortgage or own) = P (mortgage) + P (own) = 0.479 + 0.135 = 0.614.4(a) P (A) = P (1 or 2) = P (1) + P (2) = 1

6+ 1

6= 2

6= 1

3. (b) Similarly, P (B) = 1/3.

5(a) Outcomes 2 and 3. (b) Yes, events B and D are disjoint because they share no outcomes. (c) The events Aand D share an outcome in common, 2, and so are not disjoint.

6Since B and D are disjoint events, use the Addition Rule: P (B or D) = P (B) + P (D) = 13

+ 13

= 23

3.1. DEFINING PROBABILITY 85

3.1.4 Probabilities when events are not disjoint

Let’s consider calculations for two events that are not disjoint in the context of a regular deckof 52 cards, represented in Figure 3.3. If you are unfamiliar with the cards in a regular deck, pleasesee the footnote.7

2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣ A♣2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦ A♦2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥ A♥2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠

Figure 3.3: Representations of the 52 unique cards in a deck.

GUIDED PRACTICE 3.12

(a) What is the probability that a randomly selected card is a diamond? (b) What is the probabilitythat a randomly selected card is a face card?8

Venn diagrams are useful when outcomes can be categorized as “in” or “out” for two or threevariables, attributes, or random processes. The Venn diagram in Figure 3.4 uses a circle to representdiamonds and another to represent face cards. If a card is both a diamond and a face card, it fallsinto the intersection of the circles. If it is a diamond but not a face card, it will be in part of theleft circle that is not in the right circle (and so on). The total number of cards that are diamonds isgiven by the total number of cards in the diamonds circle: 10 + 3 = 13. The probabilities are alsoshown (e.g. 10/52 = 0.1923).

10 3 90.1923 0.0577 0.1731

There are also30 cards that areneither diamonds

nor face cards

Diamonds, 0.2500

Face cards, 0.2308

Figure 3.4: A Venn diagram for diamonds and face cards.

Let A represent the event that a randomly selected card is a diamond and B represent theevent that it is a face card. How do we compute P (A or B)? Events A and B are not disjoint – thecards J♦, Q♦, and K♦ fall into both categories – so we cannot use the Addition Rule for disjointevents. Instead we use the Venn diagram. We start by adding the probabilities of the two events:

P (A) + P (B) = P (♦) + P (face card) = 13/52 + 12/52

7The 52 cards are split into four suits: ♣ (club), ♦ (diamond), ♥ (heart), ♠ (spade). Each suit has its 13 cardslabeled: 2, 3, …, 10, J (jack), Q (queen), K (king), and A (ace). Thus, each card is a unique combination of a suit anda label, e.g. 4♥ and J♣. The 12 cards represented by the jacks, queens, and kings are called face cards. The cardsthat are ♦ or ♥ are typically colored red while the other two suits are typically colored black.

8(a) There are 52 cards and 13 diamonds. If the cards are thoroughly shuffled, each card has an equal chanceof being drawn, so the probability that a randomly selected card is a diamond is P (♦) = 13

52= 0.250. (b) Likewise,

there are 12 face cards, so P (face card) = 1252

= 313

= 0.231.

86 CHAPTER 3. PROBABILITY

However, the three cards that are in both events were counted twice, once in each probability. Wemust correct this double counting:

P (A or B) = P (♦ or face card)

= P (♦) + P (face card)− P (♦ and face card)

= 13/52 + 12/52− 3/52

= 22/52 = 11/26

This equation is an example of the General Addition Rule.

GENERAL ADDITION RULE

If A and B are any two events, disjoint or not, then the probability that at least one of themwill occur is

P (A or B) = P (A) + P (B)− P (A and B)

where P (A and B) is the probability that both events occur.

TIP: “or” is inclusive

When we write “or” in statistics, we mean “and/or” unless we explicitly state otherwise.Thus, A or B occurs means A, B, or both A and B occur.

GUIDED PRACTICE 3.13

(a) If A and B are disjoint, describe why this implies P (A and B) = 0. (b) Using part (a), verifythat the General Addition Rule simplifies to the simpler Addition Rule for disjoint events if A andB are disjoint.9

GUIDED PRACTICE 3.14

In the loans data set describing 10,000 loans, 1495 loans were from joint applications (e.g. a coupleapplied together), 4789 applicants had a mortgage, and 950 had both of these characteristics. Createa Venn diagram for this setup.10

GUIDED PRACTICE 3.15

(a) Use your Venn diagram from Guided Practice 3.14 to determine the probability a randomlydrawn loan from the loans data set is from a joint application where the couple had a mortgage.(b) What is the probability that the loan had either of these attributes?11

9(a) If A and B are disjoint, A and B can never occur simultaneously. (b) If A and B are disjoint, then the lastP (A and B) term of in the General Addition Rule formula is 0 (see part (a)) and we are left with the Addition Rulefor disjoint events.

10Both the counts and corresponding probabilities (e.g. 3839/10000 = 0.384)are shown. Notice that the number of loans represented in the left circlecorresponds to 3839 + 950 = 4789, and the number represented in the rightcircle is 950 + 545 = 1495.

applicant had a mortgage joint application

3839 950 5450.384 0.095 0.055

Other loans: 10000 − 3839 − 950 − 545 = 4666 (0.467)11(a) The solution is represented by the intersection of the two circles: 0.095. (b) This is the sum of the three

disjoint probabilities shown in the circles: 0.384 + 0.095 + 0.055 = 0.534 (off by 0.001 due to a rounding error).

3.1. DEFINING PROBABILITY 87

3.1.5 Probability distributions

A probability distribution is a table of all disjoint outcomes and their associated probabili-ties. Figure 3.5 shows the probability distribution for the sum of two dice.

Dice sum 2 3 4 5 6 7 8 9 10 11 12

Probability 136

236

336

436

536

636

536

436

336

236

136

Figure 3.5: Probability distribution for the sum of two dice.

RULES FOR PROBABILITY DISTRIBUTIONS

A probability distribution is a list of the possible outcomes with corresponding probabilitiesthat satisfies three rules:

1. The outcomes listed must be disjoint.

2. Each probability must be between 0 and 1.

3. The probabilities must total 1.

GUIDED PRACTICE 3.16

Figure 3.6 suggests three distributions for household income in the United States. Only one iscorrect. Which one must it be? What is wrong with the other two?12

Income Range $0-25k $25k-50k $50k-100k $100k+(a) 0.18 0.39 0.33 0.16(b) 0.38 -0.27 0.52 0.37(c) 0.28 0.27 0.29 0.16

Figure 3.6: Proposed distributions of US household incomes (Guided Prac-tice 3.16).

Chapter 1 emphasized the importance of plotting data to provide quick summaries. Probabilitydistributions can also be summarized in a bar plot. For instance, the distribution of US householdincomes is shown in Figure 3.7 as a bar plot. The probability distribution for the sum of two dice isshown in Figure 3.5 and plotted in Figure 3.8.

$0−25k $25k−50k $50k−100k $100k+0.00

0.05

0.10

0.15

0.20

0.25

US Household Incomes

Pro

babi

lity

Figure 3.7: The probability distribution of US household income.

12The probabilities of (a) do not sum to 1. The second probability in (b) is negative. This leaves (c), which sureenough satisfies the requirements of a distribution. One of the three was said to be the actual distribution of UShousehold incomes, so it must be (c).

88 CHAPTER 3. PROBABILITY

Dice Sum

2 3 4 5 6 7 8 9 10 11 12

0.00

0.05

0.10

0.15

Pro

babi

lity

Figure 3.8: The probability distribution of the sum of two dice.

In these bar plots, the bar heights represent the probabilities of outcomes. If the outcomesare numerical and discrete, it is usually (visually) convenient to make a bar plot that resembles ahistogram, as in the case of the sum of two dice. Another example of plotting the bars at theirrespective locations is shown in Figure 3.18 on page 115.

3.1.6 Complement of an event

Rolling a die produces a value in the set {1, 2, 3, 4, 5, 6}. This set of all possible outcomesis called the sample space (S) for rolling a die. We often use the sample space to examine thescenario where an event does not occur.

Let D = {2, 3} represent the event that the outcome of a die roll is 2 or 3. Then the com-plement of D represents all outcomes in our sample space that are not in D, which is denotedby Dc = {1, 4, 5, 6}. That is, Dc is the set of all possible outcomes not already included in D.Figure 3.9 shows the relationship between D, Dc, and the sample space S.

1 4 5 62 3

D DC

Figure 3.9: Event D = {2, 3} and its complement, Dc = {1, 4, 5, 6}. S representsthe sample space, which is the set of all possible outcomes.

GUIDED PRACTICE 3.17

(a) Compute P (Dc) = P (rolling a 1, 4, 5, or 6). (b) What is P (D) + P (Dc)?13

GUIDED PRACTICE 3.18

Events A = {1, 2} and B = {4, 6} are shown in Figure 3.2 on page 84. (a) Write out what Ac andBc represent. (b) Compute P (Ac) and P (Bc). (c) Compute P (A) + P (Ac) and P (B) + P (Bc).14

13(a) The outcomes are disjoint and each has probability 1/6, so the total probability is 4/6 = 2/3. (b) We canalso see that P (D) = 1

6+ 1

6= 1/3. Since D and Dc are disjoint, P (D) + P (Dc) = 1.

14Brief solutions: (a) Ac = {3, 4, 5, 6} and Bc = {1, 2, 3, 5}. (b) Noting that each outcome is disjoint, add theindividual outcome probabilities to get P (Ac) = 2/3 and P (Bc) = 2/3. (c) A and Ac are disjoint, and the same istrue of B and Bc. Therefore, P (A) + P (Ac) = 1 and P (B) + P (Bc) = 1.

3.1. DEFINING PROBABILITY 89

A complement of an event A is constructed to have two very important properties: (i) everypossible outcome not in A is in Ac, and (ii) A and Ac are disjoint. Property (i) implies

P (A or Ac) = 1

That is, if the outcome is not in A, it must be represented in Ac. We use the Addition Rule fordisjoint events to apply Property (ii):

P (A or Ac) = P (A) + P (Ac)

Combining the last two equations yields a very useful relationship between the probability of anevent and its complement.

COMPLEMENT

The complement of event A is denoted Ac, and Ac represents all outcomes not in A. A and Ac

are mathematically related:

P (A) + P (Ac) = 1, i.e. P (A) = 1− P (Ac)

In simple examples, computingA orAc is feasible in a few steps. However, using the complementcan save a lot of time as problems grow in complexity.

GUIDED PRACTICE 3.19

Let A represent the event where we roll two dice and their total is less than 12. (a) What does theevent Ac represent? (b) Determine P (Ac) from Figure 3.5 on page 87. (c) Determine P (A).15

GUIDED PRACTICE 3.20

Find the following probabilities for rolling two dice:16

(a) The sum of the dice is not 6.

(b) The sum is at least 4. That is, determine the probability of the event B = {4, 5, …, 12}.(c) The sum is no more than 10. That is, determine the probability of the event D = {2, 3, …, 10}.

3.1.7 Independence

Just as variables and observations can be independent, random processes can be independent,too. Two processes are independent if knowing the outcome of one provides no useful informationabout the outcome of the other. For instance, flipping a coin and rolling a die are two independentprocesses – knowing the coin was heads does not help determine the outcome of a die roll. On theother hand, stock prices usually move up or down together, so they are not independent.

Example 3.5 provides a basic example of two independent processes: rolling two dice. We wantto determine the probability that both will be 1. Suppose one of the dice is red and the other white.If the outcome of the red die is a 1, it provides no information about the outcome of the white die. Wefirst encountered this same question in Example 3.5 (page 81), where we calculated the probabilityusing the following reasoning: 1/6 of the time the red die is a 1, and 1/6 of those times the white die

15(a) The complement of A: when the total is equal to 12. (b) P (Ac) = 1/36. (c) Use the probability of thecomplement from part (b), P (Ac) = 1/36, and the equation for the complement: P (less than 12) = 1 − P (12) =1− 1/36 = 35/36.

16(a) First find P (6) = 5/36, then use the complement: P (not 6) = 1− P (6) = 31/36.(b) First find the complement, which requires much less effort: P (2 or 3) = 1/36 + 2/36 = 1/12. Then calculate

P (B) = 1− P (Bc) = 1− 1/12 = 11/12.(c) As before, finding the complement is the clever way to determine P (D). First find P (Dc) = P (11 or 12) =

2/36 + 1/36 = 1/12. Then calculate P (D) = 1− P (Dc) = 11/12.

90 CHAPTER 3. PROBABILITY

will also be 1. This is illustrated in Figure 3.10. Because the rolls are independent, the probabilitiesof the corresponding outcomes can be multiplied to get the final answer: (1/6)× (1/6) = 1/36. Thiscan be generalized to many independent processes.

All rolls

1/6th of the firstrolls are a 1.

1/6th of those times wherethe first roll is a 1 thesecond roll is also a 1.

Figure 3.10: 1/6 of the time, the first roll is a 1. Then 1/6 of those times, thesecond roll will also be a 1.

EXAMPLE 3.21

What if there was also a blue die independent of the other two? What is the probability of rollingthe three dice and getting all 1s?

The same logic applies from Example 3.5. If 1/36 of the time the white and red dice are both 1,then 1/6 of those times the blue die will also be 1, so multiply:

P (white = 1 and red = 1 and blue = 1) = P (white = 1)× P (red = 1)× P (blue = 1)

= (1/6)× (1/6)× (1/6) = 1/216

Example 3.21 illustrates what is called the Multiplication Rule for independent processes.

MULTIPLICATION RULE FOR INDEPENDENT PROCESSES

If A and B represent events from two different and independent processes, then the probabilitythat both A and B occur can be calculated as the product of their separate probabilities:

P (A and B) = P (A)× P (B)

Similarly, if there are k events A1, …, Ak from k independent processes, then the probabilitythey all occur is

P (A1)× P (A2)× · · · × P (Ak)

GUIDED PRACTICE 3.22

About 9% of people are left-handed. Suppose 2 people are selected at random from the U.S. pop-ulation. Because the sample size of 2 is very small relative to the population, it is reasonable toassume these two people are independent. (a) What is the probability that both are left-handed?(b) What is the probability that both are right-handed?17

17(a) The probability the first person is left-handed is 0.09, which is the same for the second person. We apply theMultiplication Rule for independent processes to determine the probability that both will be left-handed: 0.09×0.09 =0.0081.

(b) It is reasonable to assume the proportion of people who are ambidextrous (both right- and left-handed) is nearly0, which results in P (right-handed) = 1− 0.09 = 0.91. Using the same reasoning as in part (a), the probability thatboth will be right-handed is 0.91× 0.91 = 0.8281.

3.1. DEFINING PROBABILITY 91

GUIDED PRACTICE 3.23

Suppose 5 people are selected at random.18

(a) What is the probability that all are right-handed?

(b) What is the probability that all are left-handed?

Suppose the variables handedness and sex are independent, i.e. knowing someone’s sex pro-vides no useful information about their handedness and vice-versa. Then we can compute whethera randomly selected person is right-handed and female19 using the Multiplication Rule:

P (right-handed and female) = P (right-handed)× P (female)

= 0.91× 0.50 = 0.455

GUIDED PRACTICE 3.24

Three people are selected at random.20

(a) What is the probability that the first person is male and right-handed?

(b) What is the probability that the first two people are male and right-handed?.

(d) What is the probability that the first two people are male and right-handed and the thirdperson is female and left-handed?

Sometimes we wonder if one outcome provides useful information about another outcome. Thequestion we are asking is, are the occurrences of the two events independent? We say that twoevents A and B are independent if they satisfy P (A and B) = P (A)× P (B).

EXAMPLE 3.25

If we shuffle up a deck of cards and draw one, is the event that the card is a heart independent ofthe event that the card is an ace?

The probability the card is a heart is 1/4 and the probability that it is an ace is 1/13. The probabilitythe card is the ace of hearts is 1/52. We check whether P (A and B) = P (A)× P (B) is satisfied:

P (♥)× P (ace) =1

4× 1

13=

52= P (♥ and ace)

Because the equation holds, the event that the card is a heart and the event that the card is an aceare independent events.

18(a) The abbreviations RH and LH are used for right-handed and left-handed, respectively. Since each are indepen-dent, we apply the Multiplication Rule for independent processes:

P (all five are RH) = P (first = RH, second = RH, …, fifth = RH)

= P (first = RH)× P (second = RH)× · · · × P (fifth = RH)

= 0.91× 0.91× 0.91× 0.91× 0.91 = 0.624

(b) Using the same reasoning as in (a), 0.09× 0.09× 0.09× 0.09× 0.09 = 0.0000059(c) Use the complement, P (all five are RH), to answer this question:

P (not all RH) = 1− P (all RH) = 1− 0.624 = 0.376

19The actual proportion of the U.S. population that is female is about 50%, and so we use 0.5 for the probabilityof sampling a woman. However, this probability does differ in other countries.

20Brief answers are provided. (a) This can be written in probability notation as P (a randomly selected person ismale and right-handed) = 0.455. (b) 0.207. (c) 0.045. (d) 0.0093.

92 CHAPTER 3. PROBABILITY

Exercises

3.1 True or false. Determine if the statements below are true or false, and explain your reasoning.

(a) If a fair coin is tossed many times and the last eight tosses are all heads, then the chance that the nexttoss will be heads is somewhat less than 50%.

(b) Drawing a face card (jack, queen, or king) and drawing a red card from a full deck of playing cards aremutually exclusive events.

3.2 Roulette wheel. The game of roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chanceof capturing the ball.

(a) You watch a roulette wheel spin 3 consecutive times and the balllands on a red slot each time. What is the probability that theball will land on a red slot on the next spin?

(b) You watch a roulette wheel spin 300 consecutive times and theball lands on a red slot each time. What is the probability thatthe ball will land on a red slot on the next spin?

Photo by H̊akan Dahlström(http://flic.kr/p/93fEzp)

CC BY 2.0 license

3.3 Four games, one winner. Below are four versions of the same game. Your archnemesis gets to pickthe version of the game, and then you get to choose how many times to flip a coin: 10 times or 100 times.Identify how many coin flips you should choose for each version of the game. It costs $1 to play each game.Explain your reasoning.

(a) If the proportion of heads is larger than 0.60, you win $1.

(b) If the proportion of heads is larger than 0.40, you win $1.

(d) If the proportion of heads is smaller than 0.30, you win $1.

3.4 Backgammon. Backgammon is a board game for two players in which the playing pieces are movedaccording to the roll of two dice. Players win by removing all of their pieces from the board, so it is usuallygood to roll high numbers. You are playing backgammon with a friend and you roll two 6s in your firstroll and two 6s in your second roll. Your friend rolls two 3s in his first roll and again in his second row.Your friend claims that you are cheating, because rolling double 6s twice in a row is very unlikely. Usingprobability, show that your rolls were just as likely as his.

3.5 Coin flips. If you flip a fair coin 10 times, what is the probability of

(a) getting all tails?

(b) getting all heads?

3.6 Dice rolls. If you roll a pair of fair dice, what is the probability of

(a) getting a sum of 1?

(b) getting a sum of 5?

3.1. DEFINING PROBABILITY 93

3.7 Swing voters. A Pew Research survey asked 2,373 randomly sampled registered voters their politicalaffiliation (Republican, Democrat, or Independent) and whether or not they identify as swing voters. 35%of respondents identified as Independent, 23% identified as swing voters, and 11% identified as both.21

(a) Are being Independent and being a swing voter disjoint, i.e. mutually exclusive?

(b) Draw a Venn diagram summarizing the variables and their associated probabilities.

(d) What percent of voters are Independent or swing voters?

(e) What percent of voters are neither Independent nor swing voters?

(f) Is the event that someone is a swing voter independent of the event that someone is a political Indepen-dent?

3.8 Poverty and language. The American Community Survey is an ongoing survey that provides dataevery year to give communities the current information they need to plan investments and services. The2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7%speak a language other than English (foreign language) at home, and 4.2% fall into both categories.22

(a) Are living below the poverty line and speaking a foreign language at home disjoint?

(b) Draw a Venn diagram summarizing the variables and their associated probabilities.

(d) What percent of Americans live below the poverty line or speak a foreign language at home?

(e) What percent of Americans live above the poverty line and only speak English at home?

(f) Is the event that someone lives below the poverty line independent of the event that the person speaksa foreign language at home?

3.9 Disjoint vs. independent. In parts (a) and (b), identify whether the events are disjoint, independent,or neither (events cannot be both disjoint and independent).

(a) You and a randomly selected student from your class both earn A’s in this course.

(b) You and your class study partner both earn A’s in this course.

3.10 Guessing on an exam. In a multiple choice exam, there are 5 questions and 4 choices for eachquestion (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers.What is the probability that:

(a) the first question she gets right is the 5th question?

(b) she gets all of the questions right?

21Pew Research Center, With Voters Focused on Economy, Obama Lead Narrows, data collected between April4-15, 2012.

22U.S. Census Bureau, 2010 American Community Survey 1-Year Estimates, Characteristics of People by LanguageSpoken at Home.

94 CHAPTER 3. PROBABILITY

3.11 Educational attainment of couples. The table below shows the distribution of education levelattained by US residents by gender based on data collected in the 2010 American Community Survey.23

GenderMale Female

Less than 9th grade 0.07 0.139th to 12th grade, no diploma 0.10 0.09

Highest HS graduate (or equivalent) 0.30 0.20education Some college, no degree 0.22 0.24attained Associate’s degree 0.06 0.08

Bachelor’s degree 0.16 0.17Graduate or professional degree 0.09 0.09Total 1.00 1.00

(a) What is the probability that a randomly chosen man has at least a Bachelor’s degree?

(b) What is the probability that a randomly chosen woman has at least a Bachelor’s degree?

(c) What is the probability that a man and a woman getting married both have at least a Bachelor’s degree?Note any assumptions you must make to answer this question.

(d) If you made an assumption in part (c), do you think it was reasonable? If you didn’t make an assumption,double check your earlier answer and then return to this part.

3.12 School absences. Data collected at elementary schools in DeKalb County, GA suggest that eachyear roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more daysdue to sickness.24

(a) What is the probability that a student chosen at random doesn’t miss any days of school due to sicknessthis year?

(b) What is the probability that a student chosen at random misses no more than one day?

(d) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kidwill miss any school? Note any assumption you must make to answer this question.

(e) If a parent has two kids at a DeKalb County elementary school, what is the probability that both kidswill miss some school, i.e. at least one day? Note any assumption you make.

(f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn’t make anyassumptions, double check your earlier answers.

23U.S. Census Bureau, 2010 American Community Survey 1-Year Estimates, Educational Attainment.24S.S. Mizan et al. “Absence, Extended Absence, and Repeat Tardiness Related to Asthma Status among Elemen-

tary School Children”. In: Journal of Asthma 48.3 (2011), pp. 228–234.

3.2. CONDITIONAL PROBABILITY 95

3.2 Conditional probability

There can be rich relationships between two or more variables that are useful to understand.For example a car insurance company will consider information about a person’s driving history toassess the risk that they will be responsible for an accident. These types of relationships are therealm of conditional probabilities.

3.2.1 Exploring probabilities with a contingency table

The photo classify data set represents a classifier a sample of 1822 photos from a photosharing website. Data scientists have been working to improve a classifier for whether the photo isabout fashion or not, and these 1822 photos represent a test for their classifier. Each photo gets twoclassifications: the first is called mach learn and gives a classification from a machine learning (ML)system of either pred fashion or pred not. Each of these 1822 photos have also been classifiedcarefully by a team of people, which we take to be the source of truth; this variable is called truth

and takes values fashion and not. Figure 3.11 summarizes the results.

truth

fashion not Totalpred fashion 197 22 219

mach learnpred not 112 1491 1603Total 309 1513 1822

Figure 3.11: Contingency table summarizing the photo classify data set.

ML Predicts Fashion

Fashion Photos

0.110.01

0.06

Neither: 0.82

Figure 3.12: A Venn diagram using boxes for the photo classify data set.

EXAMPLE 3.26

If a photo is actually about fashion, what is the chance the ML classifier correctly identified thephoto as being about fashion?

We can estimate this probability using the data. Of the 309 fashion photos, the ML algorithmcorrectly classified 197 of the photos:

P (mach learn is pred fashion given truth is fashion) =197

309= 0.638

96 CHAPTER 3. PROBABILITY

EXAMPLE 3.27

We sample a photo from the data set and learn the ML algorithm predicted this photo was notabout fashion. What is the probability that it was incorrect and the photo is about fashion?

If the ML classifier suggests a photo is not about fashion, then it comes from the second row in thedata set. Of these 1603 photos, 112 were actually about fashion:

P (truth is fashion given mach learn is pred not) =112

1603= 0.070

3.2.2 Marginal and joint probabilities

Figure 3.11 includes row and column totals for each variable separately in the photo classify

data set. These totals represent marginal probabilities for the sample, which are the probabilitiesbased on a single variable without regard to any other variables. For instance, a probability basedsolely on the mach learn variable is a marginal probability:

P (mach learn is pred fashion) =219

1822= 0.12

A probability of outcomes for two or more variables or processes is called a joint probability:

P (mach learn is pred fashion and truth is fashion) =197

1822= 0.11

It is common to substitute a comma for “and” in a joint probability, although using either the word“and” or a comma is acceptable:

P (mach learn is pred fashion, truth is fashion)

means the same thing as

P (mach learn is pred fashion and truth is fashion)

MARGINAL AND JOINT PROBABILITIES

If a probability is based on a single variable, it is a marginal probability. The probability ofoutcomes for two or more variables or processes is called a joint probability.

We use table proportions to summarize joint probabilities for the photo classify sample.These proportions are computed by dividing each count in Figure 3.11 by the table’s total, 1822,to obtain the proportions in Figure 3.13. The joint probability distribution of the mach learn andtruth variables is shown in Figure 3.14.

truth: fashion truth: not Totalmach learn: pred fashion 0.1081 0.0121 0.1202mach learn: pred not 0.0615 0.8183 0.8798Total 0.1696 0.8304 1.00

Figure 3.13: Probability table summarizing the photo classify data set.

3.2. CONDITIONAL PROBABILITY 97

Joint outcome Probabilitymach learn is pred fashion and truth is fashion 0.1081mach learn is pred fashion and truth is not 0.0121mach learn is pred not and truth is fashion 0.0615mach learn is pred not and truth is not 0.8183Total 1.0000

Figure 3.14: Joint probability distribution for the photo classify data set.

GUIDED PRACTICE 3.28

Verify Figure 3.14 represents a probability distribution: events are disjoint, all probabilities arenon-negative, and the probabilities sum to 1.25

We can compute marginal probabilities using joint probabilities in simple cases. For example,the probability a randomly selected photo from the data set is about fashion is found by summingthe outcomes where truth takes value fashion:

P (truth is fashion) = P (mach learn is pred fashion and truth is fashion)

+ P (mach learn is pred not and truth is fashion)

= 0.1081 + 0.0615

= 0.1696

3.2.3 Defining conditional probability

The ML classifier predicts whether a photo is about fashion, even if it is not perfect. We wouldlike to better understand how to use information from a variable like mach learn to improve ourprobability estimation of a second variable, which in this example is truth.

The probability that a random photo from the data set is about fashion is about 0.17. If weknew the machine learning classifier predicted the photo was about fashion, could we get a betterestimate of the probability the photo is actually about fashion? Absolutely. To do so, we limit ourview to only those 219 cases where the ML classifier predicted that the photo was about fashion andlook at the fraction where the photo was actually about fashion:

P (truth is fashion given mach learn is pred fashion) =197

219= 0.900

We call this a conditional probability because we computed the probability under a condition:the ML classifier prediction said the photo was about fashion.

There are two parts to a conditional probability, the outcome of interest and the condition.It is useful to think of the condition as information we know to be true, and this information usuallycan be described as a known outcome or event. We generally separate the text inside our probabilitynotation into the outcome of interest and the condition with a vertical bar:

P (truth is fashion given mach learn is pred fashion)

= P (truth is fashion | mach learn is pred fashion) =197

219= 0.900

The vertical bar “|” is read as given.

25Each of the four outcome combination are disjoint, all probabilities are indeed non-negative, and the sum of theprobabilities is 0.1081 + 0.0121 + 0.0615 + 0.8183 = 1.00.

98 CHAPTER 3. PROBABILITY

In the last equation, we computed the probability a photo was about fashion based on thecondition that the ML algorithm predicted it was about fashion as a fraction:

P (truth is fashion | mach learn is pred fashion)

=# cases where truth is fashion and mach learn is pred fashion

# cases where mach learn is pred fashion

=197

219= 0.900

We considered only those cases that met the condition, mach learn is pred fashion, and then wecomputed the ratio of those cases that satisfied our outcome of interest, photo was actually aboutfashion.

Frequently, marginal and joint probabilities are provided instead of count data. For example,disease rates are commonly listed in percentages rather than in a count format. We would like tobe able to compute conditional probabilities even when no counts are available, and we use the lastequation as a template to understand this technique.

We considered only those cases that satisfied the condition, where the ML algorithm predictedfashion. Of these cases, the conditional probability was the fraction representing the outcome ofinterest, that the photo was about fashion. Suppose we were provided only the information in Fig-ure 3.13, i.e. only probability data. Then if we took a sample of 1000 photos, we would anticipateabout 12.0% or 0.120 × 1000 = 120 would be predicted to be about fashion (mach learn is pred

fashion). Similarly, we would expect about 10.8% or 0.108 × 1000 = 108 to meet both the in-formation criteria and represent our outcome of interest. Then the conditional probability can becomputed as

P (truth is fashion | mach learn is pred fashion)

=# (truth is fashion and mach learn is pred fashion)

# (mach learn is pred fashion)

=108

120=

0.108

0.120= 0.90

Here we are examining exactly the fraction of two probabilities, 0.108 and 0.120, which we can writeas

P (truth is fashion and mach learn is pred fashion) and P (mach learn is pred fashion).

The fraction of these probabilities is an example of the general formula for conditional probability.

CONDITIONAL PROBABILITY

The conditional probability of outcome A given condition B is computed as the following:

P (A|B) =P (A and B)

P (B)

GUIDED PRACTICE 3.29

(a) Write out the following statement in conditional probability notation: “The probability that theML prediction was correct, if the photo was about fashion”. Here the condition is now based on thephoto’s truth status, not the ML algorithm.

(b) Determine the probability from part (a). Table 3.13 on page 96 may be helpful.26

26(a) If the photo is about fashion and the ML algorithm prediction was correct, then the ML algorithm my havea value of pred fashion:

P (mach learn is pred fashion | truth is fashion)

(b) The equation for conditional probability indicates we should first findP (mach learn is pred fashion and truth is fashion) = 0.1081 and P (truth is fashion) = 0.1696.Then the ratio represents the conditional probability: 0.1081/0.1696 = 0.6374.

3.2. CONDITIONAL PROBABILITY 99

GUIDED PRACTICE 3.30

(a) Determine the probability that the algorithm is incorrect if it is known the photo is about fashion.

(b) Using the answers from part (a) and Guided Practice 3.29(b), compute

P (mach learn is pred fashion | truth is fashion)

+ P (mach learn is pred not | truth is fashion)

3.2.4 Smallpox in Boston, 1721

The smallpox data set provides a sample of 6,224 individuals from the year 1721 who wereexposed to smallpox in Boston. Doctors at the time believed that inoculation, which involvesexposing a person to the disease in a controlled form, could reduce the likelihood of death.

Each case represents one person with two variables: inoculated and result. The variableinoculated takes two levels: yes or no, indicating whether the person was inoculated or not. Thevariable result has outcomes lived or died. These data are summarized in Tables 3.15 and 3.16.

inoculatedyes no Total

lived 238 5136 5374result

died 6 844 850Total 244 5980 6224

Figure 3.15: Contingency table for the smallpox data set.

inoculatedyes no Total

lived 0.0382 0.8252 0.8634result

died 0.0010 0.1356 0.1366Total 0.0392 0.9608 1.0000

Figure 3.16: Table proportions for the smallpox data, computed by dividing eachcount by the table total, 6224.

GUIDED PRACTICE 3.31

Write out, in formal notation, the probability a randomly selected person who was not inoculateddied from smallpox, and find this probability.28

GUIDED PRACTICE 3.32

Determine the probability that an inoculated person died from smallpox. How does this resultcompare with the result of Guided Practice 3.31?29

27(a) This probability isP (mach learn is pred not, truth is fashion)

P (truth is fashion)= 0.0615

0.1696= 0.3626. (b) The total equals 1. (c) Under

the condition the photo is about fashion, the ML algorithm must have either predicted it was about fashion orpredicted it was not about fashion. The complement still works for conditional probabilities, provided the probabilitiesare conditioned on the same information.

28P (result = died | inoculated = no) =P (result = died and inoculated = no)

P (inoculated = no)= 0.1356

0.9608= 0.1411.

29P (result = died | inoculated = yes) =P (result = died and inoculated = yes)

P (inoculated = yes)= 0.0010

0.0392= 0.0255 (if we avoided

rounding errors, we’d get 6/244 = 0.0246). The death rate for individuals who were inoculated is only about 1 in 40while the death rate is about 1 in 7 for those who were not inoculated.

100 CHAPTER 3. PROBABILITY

GUIDED PRACTICE 3.33

The people of Boston self-selected whether or not to be inoculated. (a) Is this study observationalor was this an experiment? (b) Can we infer any causal connection using these data? (c) What aresome potential confounding variables that might influence whether someone lived or died and alsoaffect whether that person was inoculated?30

3.2.5 General multiplication rule

Section 3.1.7 introduced the Multiplication Rule for independent processes. Here we providethe General Multiplication Rule for events that might not be independent.

GENERAL MULTIPLICATION RULE

If A and B represent two outcomes or events, then

P (A and B) = P (A|B)× P (B)

It is useful to think of A as the outcome of interest and B as the condition.

This General Multiplication Rule is simply a rearrangement of the conditional probability equation.

EXAMPLE 3.34

Consider the smallpox data set. Suppose we are given only two pieces of information: 96.08%of residents were not inoculated, and 85.88% of the residents who were not inoculated ended upsurviving. How could we compute the probability that a resident was not inoculated and lived?

We will compute our answer using the General Multiplication Rule and then verify it using Fig-ure 3.16. We want to determine

P (result = lived and inoculated = no)

and we are given that

P (result = lived | inoculated = no) = 0.8588 P (inoculated = no) = 0.9608

Among the 96.08% of people who were not inoculated, 85.88% survived:

P (result = lived and inoculated = no) = 0.8588× 0.9608 = 0.8251

This is equivalent to the General Multiplication Rule. We can confirm this probability in Figure 3.16at the intersection of no and lived (with a small rounding error).

GUIDED PRACTICE 3.35

Use P (inoculated = yes) = 0.0392 and P (result = lived | inoculated = yes) = 0.9754 todetermine the probability that a person was both inoculated and lived.31

GUIDED PRACTICE 3.36

If 97.54% of the inoculated people lived, what proportion of inoculated people must have died?32

30Brief answers: (a) Observational. (b) No, we cannot infer causation from this observational study. (c) Accessi-bility to the latest and best medical care. There are other valid answers for part (c).

31The answer is 0.0382, which can be verified using Figure 3.16.32There were only two possible outcomes: lived or died. This means that 100% – 97.54% = 2.46% of the people

who were inoculated died.

3.2. CONDITIONAL PROBABILITY 101

SUM OF CONDITIONAL PROBABILITIES

Let A1, …, Ak represent all the disjoint outcomes for a variable or process. Then if B is anevent, possibly for another variable or process, we have:

P (A1|B) + · · ·+ P (Ak|B) = 1

The rule for complements also holds when an event and its complement are conditioned on thesame information:

P (A|B) = 1− P (Ac|B)

GUIDED PRACTICE 3.37

Based on the probabilities computed above, does it appear that inoculation is effective at reducingthe risk of death from smallpox?33

3.2.6 Independence considerations in conditional probability

If two events are independent, then knowing the outcome of one should provide no informationabout the other. We can show this is mathematically true using conditional probabilities.

GUIDED PRACTICE 3.38

Let X and Y represent the outcomes of rolling two dice.34

(a) What is the probability that the first die, X, is 1?

(b) What is the probability that both X and Y are 1?

(d) What is P (Y = 1)? Is this different from the answer from part (c)? Explain.

We can show in Guided Practice 3.38(c) that the conditioning information has no influence byusing the Multiplication Rule for independence processes:

P (Y = 1 | X = 1) =P (Y = 1 and X = 1)

P (X = 1)

=P (Y = 1)× P (X = 1)

P (X = 1)

= P (Y = 1)

GUIDED PRACTICE 3.39

Ron is watching a roulette table in a casino and notices that the last five outcomes were black. Hefigures that the chances of getting black six times in a row is very small (about 1/64) and puts hispaycheck on red. What is wrong with his reasoning?35

33The samples are large relative to the difference in death rates for the “inoculated” and “not inoculated” groups,so it seems there is an association between inoculated and outcome. However, as noted in the solution to GuidedPractice 3.33, this is an observational study and we cannot be sure if there is a causal connection. (Further researchhas shown that inoculation is effective at reducing death rates.)

34Brief solutions: (a) 1/6. (b) 1/36. (c)P (Y= 1 and X= 1)

P (X= 1)=

1/361/6

= 1/6. (d) The probability is the same as in

part (c): P (Y = 1) = 1/6. The probability that Y = 1 was unchanged by knowledge about X, which makes sense asX and Y are independent.

35He has forgotten that the next roulette spin is independent of the previous spins. Casinos do employ this practice,posting the last several outcomes of many betting games to trick unsuspecting gamblers into believing the odds arein their favor. This is called the gambler’s fallacy.

102 CHAPTER 3. PROBABILITY

3.2.7 Tree diagrams

Tree diagrams are a tool to organize outcomes and probabilities around the structure of thedata. They are most useful when two or more processes occur in a sequence and each process isconditioned on its predecessors.

The smallpox data fit this description. We see the population as split by inoculation: yes

and no. Following this split, survival rates were observed for each group. This structure is reflectedin the tree diagram shown in Figure 3.17. The first branch for inoculation is said to be theprimary branch while the other branches are secondary.

Inoculated Result

yes, 0.0392

lived, 0.97540.0392*0.9754 = 0.03824

died, 0.02460.0392*0.0246 = 0.00096

no, 0.9608

lived, 0.85890.9608*0.8589 = 0.82523

died, 0.14110.9608*0.1411 = 0.13557

Figure 3.17: A tree diagram of the smallpox data set.

Tree diagrams are annotated with marginal and conditional probabilities, as shown in Fig-ure 3.17. This tree diagram splits the smallpox data by inoculation into the yes and no groupswith respective marginal probabilities 0.0392 and 0.9608. The secondary branches are conditionedon the first, so we assign conditional probabilities to these branches. For example, the top branch inFigure 3.17 is the probability that result = lived conditioned on the information that inoculated= yes. We may (and usually do) construct joint probabilities at the end of each branch in our treeby multiplying the numbers we come across as we move from left to right. These joint probabilitiesare computed using the General Multiplication Rule:

P (inoculated = yes and result = lived)

= P (inoculated = yes)× P (result = lived|inoculated = yes)

= 0.0392× 0.9754 = 0.0382

3.2. CONDITIONAL PROBABILITY 103

EXAMPLE 3.40

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on themidterm. Of those students who earned an A on the midterm, 47% received an A on the final, and11% of the students who earned lower than an A on the midterm received an A on the final. Yourandomly pick up a final exam and notice the student received an A. What is the probability thatthis student earned an A on the midterm?

The end-goal is to find P (midterm = A|final = A). To calculate this conditional probability, weneed the following probabilities:

P (midterm = A and final = A) and P (final = A)

However, this information is not provided, and it is not obvious how to calculate these probabilities.Since we aren’t sure how to proceed, it is useful to organize the information into a tree diagram:

Midterm Final

A, 0.13

A, 0.470.13*0.47 = 0.0611

other, 0.530.13*0.53 = 0.0689

other, 0.87

A, 0.110.87*0.11 = 0.0957

other, 0.890.87*0.89 = 0.7743

When constructing a tree diagram, variables provided with marginal probabilities are often used tocreate the tree’s primary branches; in this case, the marginal probabilities are provided for midtermgrades. The final grades, which correspond to the conditional probabilities provided, will be shownon the secondary branches.

With the tree diagram constructed, we may compute the required probabilities:

P (midterm = A and final = A) = 0.0611

P (final = A)

= P (midterm = other and final = A) + P (midterm = A and final = A)

= 0.0957 + 0.0611 = 0.1568

The marginal probability, P (final = A), was calculated by adding up all the joint probabilities onthe right side of the tree that correspond to final = A. We may now finally take the ratio of thetwo probabilities:

P (midterm = A|final = A) =P (midterm = A and final = A)

P (final = A)

=0.0611

0.1568= 0.3897

The probability the student also earned an A on the midterm is about 0.39.

104 CHAPTER 3. PROBABILITY

GUIDED PRACTICE 3.41

After an introductory statistics course, 78% of students can successfully construct tree diagrams.Of those who can construct tree diagrams, 97% passed, while only 57% of those students who couldnot construct tree diagrams passed. (a) Organize this information into a tree diagram. (b) What isthe probability that a randomly selected student passed? (c) Compute the probability a student isable to construct a tree diagram if it is known that she passed.36

3.2.8 Bayes’ Theorem

In many instances, we are given a conditional probability of the form

P (statement about variable 1 | statement about variable 2)

but we would really like to know the inverted conditional probability:

P (statement about variable 2 | statement about variable 1)

Tree diagrams can be used to find the second conditional probability when given the first. However,sometimes it is not possible to draw the scenario in a tree diagram. In these cases, we can apply avery useful and general formula: Bayes’ Theorem.

We first take a critical look at an example of inverting conditional probabilities where we stillapply a tree diagram.

36(a) The tree diagram is shown to the right.(b) Identify which two joint probabilities represent students who passed, and add them: P (passed) = 0.7566+0.1254 =0.8820.(c) P (construct tree diagram | passed) = 0.7566

0.8820= 0.8578.

Able to constructtree diagrams

Pass class

yes, 0.78

pass, 0.970.78*0.97 = 0.7566

fail, 0.030.78*0.03 = 0.0234

no, 0.22

pass, 0.570.22*0.57 = 0.1254

fail, 0.430.22*0.43 = 0.0946

3.2. CONDITIONAL PROBABILITY 105

EXAMPLE 3.42

In Canada, about 0.35% of women over 40 will develop breast cancer in any given year. A commonscreening test for cancer is the mammogram, but this test is not perfect. In about 11% of patientswith breast cancer, the test gives a false negative: it indicates a woman does not have breastcancer when she does have breast cancer. Similarly, the test gives a false positive in 7% of patientswho do not have breast cancer: it indicates these patients have breast cancer when they actually donot. If we tested a random woman over 40 for breast cancer using a mammogram and the test cameback positive – that is, the test suggested the patient has cancer – what is the probability that thepatient actually has breast cancer?

Notice that we are given sufficient information to quickly compute the probability of testing positiveif a woman has breast cancer (1.00 − 0.11 = 0.89). However, we seek the inverted probability ofcancer given a positive test result. (Watch out for the non-intuitive medical language: a positivetest result suggests the possible presence of cancer in a mammogram screening.) This invertedprobability may be broken into two pieces:

P (has BC | mammogram+) =P (has BC and mammogram+)

P (mammogram+)

where “has BC” is an abbreviation for the patient having breast cancer and “mammogram+” meansthe mammogram screening was positive. We can construct a tree diagram for these probabilities:

Truth Mammogram

cancer, 0.0035

positive, 0.890.0035*0.89 = 0.00312

negative, 0.110.0035*0.11 = 0.00038

no cancer, 0.9965

positive, 0.070.9965*0.07 = 0.06976

negative, 0.930.9965*0.93 = 0.92675

The probability the patient has breast cancer and the mammogram is positive is

P (has BC and mammogram+) = P (mammogram+ | has BC)P (has BC)

= 0.89× 0.0035 = 0.00312

The probability of a positive test result is the sum of the two corresponding scenarios:

P (mammogram+) = P (mammogram+ and has BC)

+ P (mammogram+ and no BC)

= P (has BC)P (mammogram+ | has BC)

+ P (no BC)P (mammogram+ | no BC)

= 0.0035× 0.89 + 0.9965× 0.07 = 0.07288

Then if the mammogram screening is positive for a patient, the probability the patient has breastcancer is

P (has BC | mammogram+) =P (has BC and mammogram+)

P (mammogram+)

=0.00312

0.07288≈ 0.0428

That is, even if a patient has a positive mammogram screening, there is still only a 4% chance thatshe has breast cancer.

106 CHAPTER 3. PROBABILITY

Example 3.42 highlights why doctors often run more tests regardless of a first positive testresult. When a medical condition is rare, a single positive test isn’t generally definitive.

Consider again the last equation of Example 3.42. Using the tree diagram, we can see that thenumerator (the top of the fraction) is equal to the following product:

P (has BC and mammogram+) = P (mammogram+ | has BC)P (has BC)

The denominator – the probability the screening was positive – is equal to the sum of probabilitiesfor each positive screening scenario:

P (mammogram+) = P (mammogram+ and no BC) + P (mammogram+ and has BC)

In the example, each of the probabilities on the right side was broken down into a product of aconditional probability and marginal probability using the tree diagram.

P (mammogram+) = P (mammogram+ and no BC) + P (mammogram+ and has BC)

= P (mammogram+ | no BC)P (no BC)

+ P (mammogram+ | has BC)P (has BC)

We can see an application of Bayes’ Theorem by substituting the resulting probability expressionsinto the numerator and denominator of the original conditional probability.

P (has BC | mammogram+)

=P (mammogram+ | has BC)P (has BC)

P (mammogram+ | no BC)P (no BC) + P (mammogram+ | has BC)P (has BC)

BAYES’ THEOREM: INVERTING PROBABILITIES

Consider the following conditional probability for variable 1 and variable 2:

P (outcome A1 of variable 1 | outcome B of variable 2)

Bayes’ Theorem states that this conditional probability can be identified as the following frac-tion:

P (B|A1)P (A1)

P (B|A1)P (A1) + P (B|A2)P (A2) + · · ·+ P (B|Ak)P (Ak)

where A2, A3, …, and Ak represent all other possible outcomes of the first variable.

Bayes’ Theorem is a generalization of what we have done using tree diagrams. The numeratoridentifies the probability of getting both A1 and B. The denominator is the marginal probability ofgetting B. This bottom component of the fraction appears long and complicated since we have toadd up probabilities from all of the different ways to get B. We always completed this step whenusing tree diagrams. However, we usually did it in a separate step so it didn’t seem as complex.To apply Bayes’ Theorem correctly, there are two preparatory steps:

(1) First identify the marginal probabilities of each possible outcome of the first variable: P (A1),P (A2), …, P (Ak).

(2) Then identify the probability of the outcome B, conditioned on each possible scenario for thefirst variable: P (B|A1), P (B|A2), …, P (B|Ak).

Once each of these probabilities are identified, they can be applied directly within the formula.Bayes’ Theorem tends to be a good option when there are so many scenarios that drawing a treediagram would be complex.

3.2. CONDITIONAL PROBABILITY 107

GUIDED PRACTICE 3.43

Jose visits campus every Thursday evening. However, some days the parking garage is full, oftendue to college events. There are academic events on 35% of evenings, sporting events on 20% ofevenings, and no events on 45% of evenings. When there is an academic event, the garage fills upabout 25% of the time, and it fills up 70% of evenings with sporting events. On evenings whenthere are no events, it only fills up about 5% of the time. If Jose comes to campus and finds thegarage full, what is the probability that there is a sporting event? Use a tree diagram to solve thisproblem.37

EXAMPLE 3.44

Here we solve the same problem presented in Guided Practice 3.43, except this time we use Bayes’Theorem.

The outcome of interest is whether there is a sporting event (call this A1), and the condition is thatthe lot is full (B). Let A2 represent an academic event and A3 represent there being no event oncampus. Then the given probabilities can be written as

P (A1) = 0.2 P (A2) = 0.35 P (A3) = 0.45

P (B|A1) = 0.7 P (B|A2) = 0.25 P (B|A3) = 0.05

Bayes’ Theorem can be used to compute the probability of a sporting event (A1) under the conditionthat the parking lot is full (B):

P (A1|B) =P (B|A1)P (A1)

P (B|A1)P (A1) + P (B|A2)P (A2) + P (B|A3)P (A3)

=(0.7)(0.2)

(0.7)(0.2) + (0.25)(0.35) + (0.05)(0.45)

= 0.56

Based on the information that the garage is full, there is a 56% probability that a sporting event isbeing held on campus that evening.

37The tree diagram, with threeprimary branches, is shown tothe right. Next, we identify twoprobabilities from the tree dia-gram. (1) The probability thatthere is a sporting event andthe garage is full: 0.14. (2) Theprobability the garage is full:0.0875 + 0.14 + 0.0225 = 0.25.Then the solution is the ratio ofthese probabilities: 0.14

0.25= 0.56.

If the garage is full, there is a56% probability that there is asporting event.

Event Garage full

Academic, 0.35Full, 0.25

0.35*0.25 = 0.0875

Spaces Available, 0.750.35*0.75 = 0.2625

Sporting, 0.20Full, 0.7

0.2*0.7 = 0.14

Spaces Available, 0.30.2*0.3 = 0.06

None, 0.45Full, 0.05

0.45*0.05 = 0.0225

Spaces Available, 0.950.45*0.95 = 0.4275

108 CHAPTER 3. PROBABILITY

GUIDED PRACTICE 3.45

Use the information in the previous exercise and example to verify the probability that there is anacademic event conditioned on the parking lot being full is 0.35.38

GUIDED PRACTICE 3.46

In Guided Practice 3.43 and 3.45, you found that if the parking lot is full, the probability there is asporting event is 0.56 and the probability there is an academic event is 0.35. Using this information,compute P (no event | the lot is full).39

The last several exercises offered a way to update our belief about whether there is a sportingevent, academic event, or no event going on at the school based on the information that the parkinglot was full. This strategy of updating beliefs using Bayes’ Theorem is actually the foundation of anentire section of statistics called Bayesian statistics. While Bayesian statistics is very importantand useful, we will not have time to cover much more of it in this book.

38Short answer:

P (A2|B) =P (B|A2)P (A2)

P (B|A1)P (A1) + P (B|A2)P (A2) + P (B|A3)P (A3)

=(0.25)(0.35)

(0.7)(0.2) + (0.25)(0.35) + (0.05)(0.45)

= 0.35

39Each probability is conditioned on the same information that the garage is full, so the complement may be used:1.00− 0.56− 0.35 = 0.09.

3.2. CONDITIONAL PROBABILITY 109

Exercises

3.13 Joint and conditional probabilities. P(A) = 0.3, P(B) = 0.7

(a) Can you compute P(A and B) if you only know P(A) and P(B)?

(b) Assuming that events A and B arise from independent random processes,

i. what is P(A and B)?

ii. what is P(A or B)?

iii. what is P(A|B)?

(d) If we are given that P(A and B) = 0.1, what is P(A|B)?

3.14 PB & J. Suppose 80% of people like peanut butter, 89% like jelly, and 78% like both. Given that arandomly sampled person likes peanut butter, what’s the probability that he also likes jelly?

3.15 Global warming. A Pew Research poll asked 1,306 Americans “From what you’ve read and heard,is there solid evidence that the average temperature on earth has been getting warmer over the past fewdecades, or not?”. The table below shows the distribution of responses by party and ideology, where thecounts have been replaced with relative frequencies.40

ResponseEarth is Not Don’t Knowwarming warming Refuse Total

Conservative Republican 0.11 0.20 0.02 0.33Party and Mod/Lib Republican 0.06 0.06 0.01 0.13Ideology Mod/Cons Democrat 0.25 0.07 0.02 0.34

Liberal Democrat 0.18 0.01 0.01 0.20Total 0.60 0.34 0.06 1.00

(a) Are believing that the earth is warming and being a liberal Democrat mutually exclusive?

(b) What is the probability that a randomly chosen respondent believes the earth is warming or is a liberalDemocrat?

(c) What is the probability that a randomly chosen respondent believes the earth is warming given that heis a liberal Democrat?

(d) What is the probability that a randomly chosen respondent believes the earth is warming given that heis a conservative Republican?

(e) Does it appear that whether or not a respondent believes the earth is warming is independent of theirparty and ideology? Explain your reasoning.

(f) What is the probability that a randomly chosen respondent is a moderate/liberal Republican given thathe does not believe that the earth is warming?

40Pew Research Center, Majority of Republicans No Longer See Evidence of Global Warming, data collected onOctober 27, 2010.

110 CHAPTER 3. PROBABILITY

3.16 Health coverage, relative frequencies. The Behavioral Risk Factor Surveillance System (BRFSS)is an annual telephone survey designed to identify risk factors in the adult population and report emerginghealth trends. The following table displays the distribution of health status of respondents to this survey(excellent, very good, good, fair, poor) and whether or not they have health insurance.

Health StatusExcellent Very good Good Fair Poor Total

Health No 0.0230 0.0364 0.0427 0.0192 0.0050 0.1262Coverage Yes 0.2099 0.3123 0.2410 0.0817 0.0289 0.8738

Total 0.2329 0.3486 0.2838 0.1009 0.0338 1.0000

(a) Are being in excellent health and having health coverage mutually exclusive?

(b) What is the probability that a randomly chosen individual has excellent health?

(d) What is the probability that a randomly chosen individual has excellent health given that he doesn’thave health coverage?

(e) Do having excellent health and having health coverage appear to be independent?

3.17 Burger preferences. A 2010 SurveyUSA poll asked 500 Los Angeles residents, “What is the besthamburger place in Southern California? Five Guys Burgers? In-N-Out Burger? Fat Burger? Tommy’sHamburgers? Umami Burger? Or somewhere else?” The distribution of responses by gender is shownbelow.41

GenderMale Female Total

Five Guys Burgers 5 6 11In-N-Out Burger 162 181 343

Best Fat Burger 10 12 22hamburger Tommy’s Hamburgers 27 27 54place Umami Burger 5 1 6

Other 26 20 46Not Sure 13 5 18Total 248 252 500

(a) Are being female and liking Five Guys Burgers mutually exclusive?

(b) What is the probability that a randomly chosen male likes In-N-Out the best?

(d) What is the probability that a man and a woman who are dating both like In-N-Out the best? Noteany assumption you make and evaluate whether you think that assumption is reasonable.

(e) What is the probability that a randomly chosen person likes Umami best or that person is female?

41SurveyUSA, Results of SurveyUSA News Poll #17718, data collected on December 2, 2010.

3.2. CONDITIONAL PROBABILITY 111

3.18 Assortative mating. Assortative mating is a nonrandom mating pattern where individuals with simi-lar genotypes and/or phenotypes mate with one another more frequently than what would be expected undera random mating pattern. Researchers studying this topic collected data on eye colors of 204 Scandinavianmen and their female partners. The table below summarizes the results. For simplicity, we only includeheterosexual relationships in this exercise.42

Partner (female)Blue Brown Green Total

Blue 78 23 13 114

Self (male)Brown 19 23 12 54Green 11 9 16 36Total 108 55 41 204

(a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?

(b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blueeyes?

(c) What is the probability that a randomly chosen male respondent with brown eyes has a partner withblue eyes? What about the probability of a randomly chosen male respondent with green eyes having apartner with blue eyes?

(d) Does it appear that the eye colors of male respondents and their partners are independent? Explainyour reasoning.

3.19 Drawing box plots. After an introductory statistics course, 80% of students can successfully constructbox plots. Of those who can construct box plots, 86% passed, while only 65% of those students who couldnot construct box plots passed.

(a) Construct a tree diagram of this scenario.

(b) Calculate the probability that a student is able to construct a box plot if it is known that he passed.

3.20 Predisposition for thrombosis. A genetic test is used to determine if people have a predispositionfor thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of bloodthrough the circulatory system. It is believed that 3% of people actually have this predisposition. Thegenetic test is 99% accurate if a person actually has the predisposition, meaning that the probability of apositive test result when a person actually has the predisposition is 0.99. The test is 98% accurate if aperson does not have the predisposition. What is the probability that a randomly selected person who testspositive for the predisposition by the test actually has the predisposition?

3.21 It’s never lupus. Lupus is a medical phenomenon where antibodies that are supposed to attackforeign cells to prevent infections instead see plasma proteins as foreign bodies, leading to a high risk ofblood clotting. It is believed that 2% of the population suffer from this disease. The test is 98% accurateif a person actually has the disease. The test is 74% accurate if a person does not have the disease. Thereis a line from the Fox television show House that is often used after a patient tests positive for lupus: “It’snever lupus.” Do you think there is truth to this statement? Use appropriate probabilities to support youranswer.

3.22 Exit poll. Edison Research gathered exit poll results from several sources for the Wisconsin recallelection of Scott Walker. They found that 53% of the respondents voted in favor of Scott Walker. Addition-ally, they estimated that of those who did vote in favor for Scott Walker, 37% had a college degree, while44% of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a personwho participated in the exit poll and found that he had a college degree. What is the probability that hevoted in favor of Scott Walker?43

42B. Laeng et al. “Why do blue-eyed men prefer women with the same eye color?” In: Behavioral Ecology andSociobiology 61.3 (2007), pp. 371–384.

43New York Times, Wisconsin recall exit polls.

112 CHAPTER 3. PROBABILITY

3.3 Sampling from a small population

When we sample observations from a population, usually we’re only sampling a small fraction of thepossible individuals or cases. However, sometimes our sample size is large enough or the populationis small enough that we sample more than 10% of a population44 without replacement (meaningwe do not have a chance of sampling the same cases twice). Sampling such a notable fraction of apopulation can be important for how we analyze the sample.

EXAMPLE 3.47

Professors sometimes select a student at random to answer a question. If each student has an equalchance of being selected and there are 15 people in your class, what is the chance that she will pickyou for the next question?

If there are 15 people to ask and none are skipping class, then the probability is 1/15, or about0.067.

EXAMPLE 3.48

If the professor asks 3 questions, what is the probability that you will not be selected? Assume thatshe will not pick the same person twice in a given lecture.

For the first question, she will pick someone else with probability 14/15. When she asks the secondquestion, she only has 14 people who have not yet been asked. Thus, if you were not picked on thefirst question, the probability you are again not picked is 13/14. Similarly, the probability you areagain not picked on the third question is 12/13, and the probability of not being picked for any ofthe three questions is

P (not picked in 3 questions)

= P (Q1 = not picked, Q2 = not picked, Q3 = not picked.)

=14

15× 13

14× 12

13=

15= 0.80

GUIDED PRACTICE 3.49

What rule permitted us to multiply the probabilities in Example 3.48?45

44The 10% guideline is a rule of thumb cutoff for when these considerations become more important.45The three probabilities we computed were actually one marginal probability, P (Q1=not picked), and two condi-

tional probabilities:

P (Q2 = not picked | Q1 = not picked)

P (Q3 = not picked | Q1 = not picked, Q2 = not picked)

Using the General Multiplication Rule, the product of these three probabilities is the probability of not being pickedin 3 questions.

3.3. SAMPLING FROM A SMALL POPULATION 113

EXAMPLE 3.50

Suppose the professor randomly picks without regard to who she already selected, i.e. students canbe picked more than once. What is the probability that you will not be picked for any of the threequestions?

Each pick is independent, and the probability of not being picked for any individual question is14/15. Thus, we can use the Multiplication Rule for independent processes.

P (not picked in 3 questions)

= P (Q1 = not picked, Q2 = not picked, Q3 = not picked.)

=14

15× 14

15= 0.813

You have a slightly higher chance of not being picked compared to when she picked a new personfor each question. However, you now may be picked more than once.

GUIDED PRACTICE 3.51

Under the setup of Example 3.50, what is the probability of being picked to answer all three ques-tions?46

If we sample from a small population without replacement, we no longer have independencebetween our observations. In Example 3.48, the probability of not being picked for the second ques-tion was conditioned on the event that you were not picked for the first question. In Example 3.50,the professor sampled her students with replacement: she repeatedly sampled the entire classwithout regard to who she already picked.

GUIDED PRACTICE 3.52

Your department is holding a raffle. They sell 30 tickets and offer seven prizes. (a) They place thetickets in a hat and draw one for each prize. The tickets are sampled without replacement, i.e. theselected tickets are not placed back in the hat. What is the probability of winning a prize if you buyone ticket? (b) What if the tickets are sampled with replacement?47

GUIDED PRACTICE 3.53

Compare your answers in Guided Practice 3.52. How much influence does the sampling methodhave on your chances of winning a prize?48

Had we repeated Guided Practice 3.52 with 300 tickets instead of 30, we would have foundsomething interesting: the results would be nearly identical. The probability would be 0.0233without replacement and 0.0231 with replacement. When the sample size is only a small fractionof the population (under 10%), observations are nearly independent even when sampling withoutreplacement.

46P (being picked to answer all three questions) =(

115

)3= 0.00030.

47(a) First determine the probability of not winning. The tickets are sampled without replacement, which meansthe probability you do not win on the first draw is 29/30, 28/29 for the second, …, and 23/24 for the seventh. Theprobability you win no prize is the product of these separate probabilities: 23/30. That is, the probability of winninga prize is 1− 23/30 = 7/30 = 0.233. (b) When the tickets are sampled with replacement, there are seven independentdraws. Again we first find the probability of not winning a prize: (29/30)7 = 0.789. Thus, the probability of winning(at least) one prize when drawing with replacement is 0.211.

48There is about a 10% larger chance of winning a prize when using sampling without replacement. However, atmost one prize may be won under this sampling procedure.

114 CHAPTER 3. PROBABILITY

Exercises

3.23 Marbles in an urn. Imagine you have an urn containing 5 red, 3 blue, and 2 orange marbles in it.

(a) What is the probability that the first marble you draw is blue?

(b) Suppose you drew a blue marble in the first draw. If drawing with replacement, what is the probabilityof drawing a blue marble in the second draw?

(c) Suppose you instead drew an orange marble in the first draw. If drawing with replacement, what is theprobability of drawing a blue marble in the second draw?

(d) If drawing with replacement, what is the probability of drawing two blue marbles in a row?

(e) When drawing with replacement, are the draws independent? Explain.

3.24 Socks in a drawer. In your sock drawer you have 4 blue, 5 gray, and 3 black socks. Half asleep onemorning you grab 2 socks at random and put them on. Find the probability you end up wearing

(a) 2 blue socks

(b) no gray socks

(d) a green sock

(e) matching socks

3.25 Chips in a bag. Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips.

(a) Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability thenext is also blue?

(b) Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. Whatis the probability this second chip is blue?

(d) When drawing without replacement, are the draws independent? Explain.

3.26 Books on a bookshelf. The table below shows the distribution of books on a bookcase based onwhether they are nonfiction or fiction and hardcover or paperback.

FormatHardcover Paperback Total

TypeFiction 13 59 72Nonfiction 15 8 23Total 28 67 95

(a) Find the probability of drawing a hardcover book first then a paperback fiction book second whendrawing without replacement.

(b) Determine the probability of drawing a fiction book first and then a hardcover book second, whendrawing without replacement.

(c) Calculate the probability of the scenario in part (b), except this time complete the calculations underthe scenario where the first book is placed back on the bookcase before randomly drawing the secondbook.

(d) The final answers to parts (b) and (c) are very similar. Explain why this is the case.

3.27 Student outfits. In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing shorts, 8are wearing skirts, and the rest are wearing leggings. If we randomly select 3 students without replacement,what is the probability that one of the selected students is wearing leggings and the other two are wearingjeans? Note that these are mutually exclusive clothing options.

3.28 The birthday problem. Suppose we pick three people at random. For each of the following questions,ignore the special case where someone might be born on February 29th, and assume that births are evenlydistributed throughout the year.

(a) What is the probability that the first two people share a birthday?

(b) What is the probability that at least two people share a birthday?

3.4. RANDOM VARIABLES 115

3.4 Random variables

It’s often useful to model a process using what’s called a random variable. Such a model allows usto apply a mathematical framework and statistical principles for better understanding and predictingoutcomes in the real world.

EXAMPLE 3.54

Two books are assigned for a statistics class: a textbook and its corresponding study guide. Theuniversity bookstore determined 20% of enrolled students do not buy either book, 55% buy thetextbook only, and 25% buy both books, and these percentages are relatively constant from oneterm to another. If there are 100 students enrolled, how many books should the bookstore expectto sell to this class?

Around 20 students will not buy either book (0 books total), about 55 will buy one book (55 bookstotal), and approximately 25 will buy two books (totaling 50 books for these 25 students). Thebookstore should expect to sell about 105 books for this class.

GUIDED PRACTICE 3.55

Would you be surprised if the bookstore sold slightly more or less than 105 books?49

EXAMPLE 3.56

The textbook costs $137 and the study guide $33. How much revenue should the bookstore expectfrom this class of 100 students?

About 55 students will just buy a textbook, providing revenue of

$137× 55 = $7, 535

The roughly 25 students who buy both the textbook and the study guide would pay a total of

($137 + $33)× 25 = $170× 25 = $4, 250

Thus, the bookstore should expect to generate about $7, 535 + $4, 250 = $11, 785 from these 100students for this one class. However, there might be some sampling variability so the actual amountmay differ by a little bit.

Cost$0 $137 $170

0.0

0.2

0.4

Pro

babi

lity

Figure 3.18: Probability distribution for the bookstore’s revenue from one student.The triangle represents the average revenue per student.

49If they sell a little more or a little less, this should not be a surprise. Hopefully Chapter 1 helped make clearthat there is natural variability in observed data. For example, if we would flip a coin 100 times, it will not usuallycome up heads exactly half the time, but it will probably be close.

116 CHAPTER 3. PROBABILITY

EXAMPLE 3.57

What is the average revenue per student for this course?

The expected total revenue is $11,785, and there are 100 students. Therefore the expected revenueper student is $11, 785/100 = $117.85.

3.4.1 Expectation

We call a variable or process with a numerical outcome a random variable, and we usuallyrepresent this random variable with a capital letter such as X, Y , or Z. The amount of money asingle student will spend on her statistics books is a random variable, and we represent it by X.

RANDOM VARIABLE

A random process or variable with a numerical outcome.

The possible outcomes of X are labeled with a corresponding lower case letter x and subscripts.For example, we write x1 = $0, x2 = $137, and x3 = $170, which occur with probabilities 0.20, 0.55,and 0.25. The distribution of X is summarized in Figure 3.18 and Figure 3.19.

i 1 2 3 Totalxi $0 $137 $170 –P (X = xi) 0.20 0.55 0.25 1.00

Figure 3.19: The probability distribution for the random variable X, representingthe bookstore’s revenue from a single student.

We computed the average outcome of X as $117.85 in Example 3.57. We call this average theexpected value of X, denoted by E(X). The expected value of a random variable is computed byadding each outcome weighted by its probability:

E(X) = 0× P (X = 0) + 137× P (X = 137) + 170× P (X = 170)

= 0× 0.20 + 137× 0.55 + 170× 0.25 = 117.85

EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE

If X takes outcomes x1, …, xk with probabilities P (X = x1), …, P (X = xk), the expectedvalue of X is the sum of each outcome multiplied by its corresponding probability:

E(X) = x1 × P (X = x1) + · · ·+ xk × P (X = xk)

k∑i=1

xiP (X = xi)

The Greek letter µ may be used in place of the notation E(X).

3.4. RANDOM VARIABLES 117

0 137 170

117.85

Figure 3.20: A weight system representing the probability distribution for X. Thestring holds the distribution at the mean to keep the system balanced.

Figure 3.21: A continuous distribution can also be balanced at its mean.

The expected value for a random variable represents the average outcome. For example, E(X) =117.85 represents the average amount the bookstore expects to make from a single student, whichwe could also write as µ = 117.85.

It is also possible to compute the expected value of a continuous random variable (see Sec-tion 3.5). However, it requires a little calculus and we save it for a later class.50

In physics, the expectation holds the same meaning as the center of gravity. The distributioncan be represented by a series of weights at each outcome, and the mean represents the balancingpoint. This is represented in Figures 3.18 and 3.20. The idea of a center of gravity also expandsto continuous probability distributions. Figure 3.21 shows a continuous probability distributionbalanced atop a wedge placed at the mean.

50µ =∫xf(x)dx where f(x) represents a function for the density curve.

118 CHAPTER 3. PROBABILITY

3.4.2 Variability in random variables

Suppose you ran the university bookstore. Besides how much revenue you expect to generate,you might also want to know the volatility (variability) in your revenue.

The variance and standard deviation can be used to describe the variability of a random variable.Section 2.1.4 introduced a method for finding the variance and standard deviation for a data set. Wefirst computed deviations from the mean (xi−µ), squared those deviations, and took an average toget the variance. In the case of a random variable, we again compute squared deviations. However,we take their sum weighted by their corresponding probabilities, just like we did for the expectation.This weighted sum of squared deviations equals the variance, and we calculate the standard deviationby taking the square root of the variance, just as we did in Section 2.1.4.

GENERAL VARIANCE FORMULA

If X takes outcomes x1, …, xk with probabilities P (X = x1), …, P (X = xk) and expected valueµ = E(X), then the variance of X, denoted by V ar(X) or the symbol σ2, is

σ2 = (x1 − µ)2 × P (X = x1) + · · ·· · ·+ (xk − µ)2 × P (X = xk)

k∑j=1

(xj − µ)2P (X = xj)

The standard deviation of X, labeled σ, is the square root of the variance.

EXAMPLE 3.58

Compute the expected value, variance, and standard deviation of X, the revenue of a single statisticsstudent for the bookstore.

It is useful to construct a table that holds computations for each outcome separately, then add upthe results.

i 1 2 3 Totalxi $0 $137 $170P (X = xi) 0.20 0.55 0.25xi × P (X = xi) 0 75.35 42.50 117.85

Thus, the expected value is µ = 117.85, which we computed earlier. The variance can be constructedby extending this table:

i 1 2 3 Totalxi $0 $137 $170P (X = xi) 0.20 0.55 0.25xi × P (X = xi) 0 75.35 42.50 117.85xi − µ -117.85 19.15 52.15(xi − µ)2 13888.62 366.72 2719.62(xi − µ)2 × P (X = xi) 2777.7 201.7 679.9 3659.3

The variance of X is σ2 = 3659.3, which means the standard deviation is σ =√

3659.3 = $60.49.

3.4. RANDOM VARIABLES 119

GUIDED PRACTICE 3.59

The bookstore also offers a chemistry textbook for $159 and a book supplement for $41. From pastexperience, they know about 25% of chemistry students just buy the textbook while 60% buy boththe textbook and supplement.51

(a) What proportion of students don’t buy either book? Assume no students buy the supplementwithout the textbook.

(b) Let Y represent the revenue from a single student. Write out the probability distribution ofY , i.e. a table for each outcome and its associated probability.

(d) Find the standard deviation to describe the variability associated with the revenue from asingle student.

3.4.3 Linear combinations of random variables

So far, we have thought of each variable as being a complete story in and of itself. Sometimesit is more appropriate to use a combination of variables. For instance, the amount of time a personspends commuting to work each week can be broken down into several daily commutes. Similarly,the total gain or loss in a stock portfolio is the sum of the gains and losses in its components.

EXAMPLE 3.60

John travels to work five days a week. We will use X1 to represent his travel time on Monday, X2 torepresent his travel time on Tuesday, and so on. Write an equation using X1, …, X5 that representshis travel time for the week, denoted by W .

His total weekly travel time is the sum of the five daily values:

W = X1 +X2 +X3 +X4 +X5

Breaking the weekly travel time W into pieces provides a framework for understanding each sourceof randomness and is useful for modeling W .

51(a) 100% – 25% – 60% = 15% of students do not buy any books for the class. Part (b) is represented by the firsttwo lines in the table below. The expectation for part (c) is given as the total on the line yi × P (Y = yi). The result

of part (d) is the square-root of the variance listed on in the total on the last line: σ =√V ar(Y ) = $69.28.

i (scenario) 1 (noBook) 2 (textbook) 3 (both) Totalyi 0.00 159.00 200.00

P (Y = yi) 0.15 0.25 0.60yi × P (Y = yi) 0.00 39.75 120.00 E(Y ) = 159.75

yi − E(Y ) -159.75 -0.75 40.25(yi − E(Y ))2 25520.06 0.56 1620.06

(yi − E(Y ))2 × P (Y ) 3828.0 0.1 972.0 V ar(Y ) ≈ 4800

120 CHAPTER 3. PROBABILITY

EXAMPLE 3.61

It takes John an average of 18 minutes each day to commute to work. What would you expect hisaverage commute time to be for the week?

We were told that the average (i.e. expected value) of the commute time is 18 minutes per day:E(Xi) = 18. To get the expected time for the sum of the five days, we can add up the expectedtime for each individual day:

E(W ) = E(X1 +X2 +X3 +X4 +X5)

= E(X1) + E(X2) + E(X3) + E(X4) + E(X5)

= 18 + 18 + 18 + 18 + 18 = 90 minutes

The expectation of the total time is equal to the sum of the expected individual times. Moregenerally, the expectation of a sum of random variables is always the sum of the expectation foreach random variable.

GUIDED PRACTICE 3.62

Elena is selling a TV at a cash auction and also intends to buy a toaster oven in the auction. IfX represents the profit for selling the TV and Y represents the cost of the toaster oven, write anequation that represents the net change in Elena’s cash.52

GUIDED PRACTICE 3.63

Based on past auctions, Elena figures she should expect to make about $175 on the TV and payabout $23 for the toaster oven. In total, how much should she expect to make or spend?53

GUIDED PRACTICE 3.64

Would you be surprised if John’s weekly commute wasn’t exactly 90 minutes or if Elena didn’t makeexactly $152? Explain.54

Two important concepts concerning combinations of random variables have so far been in-troduced. First, a final value can sometimes be described as the sum of its parts in an equation.Second, intuition suggests that putting the individual average values into this equation gives theaverage value we would expect in total. This second point needs clarification – it is guaranteed tobe true in what are called linear combinations of random variables.

A linear combination of two random variables X and Y is a fancy phrase to describe acombination

aX + bY

where a and b are some fixed and known numbers. For John’s commute time, there were five randomvariables – one for each work day – and each random variable could be written as having a fixedcoefficient of 1:

1X1 + 1X2 + 1X3 + 1X4 + 1X5

For Elena’s net gain or loss, the X random variable had a coefficient of +1 and the Y randomvariable had a coefficient of -1.

52She will make X dollars on the TV but spend Y dollars on the toaster oven: X − Y .53E(X − Y ) = E(X)− E(Y ) = 175− 23 = $152. She should expect to make about $152.54No, since there is probably some variability. For example, the traffic will vary from one day to next, and auction

prices will vary depending on the quality of the merchandise and the interest of the attendees.

3.4. RANDOM VARIABLES 121

When considering the average of a linear combination of random variables, it is safe to plugin the mean of each random variable and then compute the final result. For a few examples ofnonlinear combinations of random variables – cases where we cannot simply plug in the means – seethe footnote.55

LINEAR COMBINATIONS OF RANDOM VARIABLES AND THE AVERAGE RESULT

If X and Y are random variables, then a linear combination of the random variables is given by

aX + bY

where a and b are some fixed numbers. To compute the average value of a linear combinationof random variables, plug in the average of each individual random variable and compute theresult:

a× E(X) + b× E(Y )

Recall that the expected value is the same as the mean, e.g. E(X) = µX .

EXAMPLE 3.65

Leonard has invested $6000 in Caterpillar Inc (stock ticker: CAT) and $2000 in Exxon Mobil Corp(XOM). If X represents the change in Caterpillar’s stock next month and Y represents the changein Exxon Mobil’s stock next month, write an equation that describes how much money will be madeor lost in Leonard’s stocks for the month.

For simplicity, we will suppose X and Y are not in percents but are in decimal form (e.g. ifCaterpillar’s stock increases 1%, then X = 0.01; or if it loses 1%, then X = −0.01). Then we canwrite an equation for Leonard’s gain as

$6000×X + $2000× Y

If we plug in the change in the stock value for X and Y , this equation gives the change in value ofLeonard’s stock portfolio for the month. A positive value represents a gain, and a negative valuerepresents a loss.

GUIDED PRACTICE 3.66

Caterpillar stock has recently been rising at 2.0% and Exxon Mobil’s at 0.2% per month, respectively.Compute the expected change in Leonard’s stock portfolio for next month.56

GUIDED PRACTICE 3.67

You should have found that Leonard expects a positive gain in Guided Practice 3.66. However,would you be surprised if he actually had a loss this month?57

55If X and Y are random variables, consider the following combinations: X1+Y , X × Y , X/Y . In such cases,plugging in the average value for each random variable and computing the result will not generally lead to an accurateaverage value for the end result.

56E($6000×X + $2000× Y ) = $6000× 0.020 + $2000× 0.002 = $124.57No. While stocks tend to rise over time, they are often volatile in the short term.

122 CHAPTER 3. PROBABILITY

3.4.4 Variability in linear combinations of random variables

Quantifying the average outcome from a linear combination of random variables is helpful, butit is also important to have some sense of the uncertainty associated with the total outcome of thatcombination of random variables. The expected net gain or loss of Leonard’s stock portfolio wasconsidered in Guided Practice 3.66. However, there was no quantitative discussion of the volatilityof this portfolio. For instance, while the average monthly gain might be about $124 according tothe data, that gain is not guaranteed. Figure 3.22 shows the monthly changes in a portfolio likeLeonard’s during a three year period. The gains and losses vary widely, and quantifying thesefluctuations is important when investing in stocks.

Monthly Returns Over 3 Years

−1000 −500 0 500 1000

Figure 3.22: The change in a portfolio like Leonard’s for 36 months, where $6000is in Caterpillar’s stock and $2000 is in Exxon Mobil’s.

Just as we have done in many previous cases, we use the variance and standard deviation todescribe the uncertainty associated with Leonard’s monthly returns. To do so, the variances of eachstock’s monthly return will be useful, and these are shown in Figure 3.23. The stocks’ returns arenearly independent.

Here we use an equation from probability theory to describe the uncertainty of Leonard’smonthly returns; we leave the proof of this method to a dedicated probability course. The varianceof a linear combination of random variables can be computed by plugging in the variances of theindividual random variables and squaring the coefficients of the random variables:

V ar(aX + bY ) = a2 × V ar(X) + b2 × V ar(Y )

It is important to note that this equality assumes the random variables are independent; if inde-pendence doesn’t hold, then a modification to this equation would be required that we leave as atopic for a future course to cover. This equation can be used to compute the variance of Leonard’smonthly return:

V ar(6000×X + 2000× Y ) = 60002 × V ar(X) + 20002 × V ar(Y )

= 36, 000, 000× 0.0057 + 4, 000, 000× 0.0021

≈ 213, 600

The standard deviation is computed as the square root of the variance:√

213, 600 = $463. While anaverage monthly return of $124 on an $8000 investment is nothing to scoff at, the monthly returnsare so volatile that Leonard should not expect this income to be very stable.

Mean (x̄) Standard deviation (s) Variance (s2)CAT 0.0204 0.0757 0.0057XOM 0.0025 0.0455 0.0021

Figure 3.23: The mean, standard deviation, and variance of the CAT and XOMstocks. These statistics were estimated from historical stock data, so notation usedfor sample statistics has been used.

3.4. RANDOM VARIABLES 123

VARIABILITY OF LINEAR COMBINATIONS OF RANDOM VARIABLES

The variance of a linear combination of random variables may be computed by squaring theconstants, substituting in the variances for the random variables, and computing the result:

V ar(aX + bY ) = a2 × V ar(X) + b2 × V ar(Y )

This equation is valid as long as the random variables are independent of each other. Thestandard deviation of the linear combination may be found by taking the square root of thevariance.

EXAMPLE 3.68

Suppose John’s daily commute has a standard deviation of 4 minutes. What is the uncertainty inhis total commute time for the week?

The expression for John’s commute time was

X1 +X2 +X3 +X4 +X5

Each coefficient is 1, and the variance of each day’s time is 42 = 16. Thus, the variance of the totalweekly commute time is

variance = 12 × 16 + 12 × 16 + 12 × 16 + 12 × 16 + 12 × 16 = 5× 16 = 80

standard deviation =√

variance =√

80 = 8.94

The standard deviation for John’s weekly work commute time is about 9 minutes.

GUIDED PRACTICE 3.69

The computation in Example 3.68 relied on an important assumption: the commute time for eachday is independent of the time on other days of that week. Do you think this is valid? Explain.58

GUIDED PRACTICE 3.70

Consider Elena’s two auctions from Guided Practice 3.62 on page 120. Suppose these auctions areapproximately independent and the variability in auction prices associated with the TV and toasteroven can be described using standard deviations of $25 and $8. Compute the standard deviation ofElena’s net gain.59

Consider again Guided Practice 3.70. The negative coefficient for Y in the linear combinationwas eliminated when we squared the coefficients. This generally holds true: negatives in a linearcombination will have no impact on the variability computed for a linear combination, but they doimpact the expected value computations.

58One concern is whether traffic patterns tend to have a weekly cycle (e.g. Fridays may be worse than other days).If that is the case, and John drives, then the assumption is probably not reasonable. However, if John walks to work,then his commute is probably not affected by any weekly traffic cycle.

59The equation for Elena can be written as

(1)×X + (−1)× Y

The variances of X and Y are 625 and 64. We square the coefficients and plug in the variances:

(1)2 × V ar(X) + (−1)2 × V ar(Y ) = 1× 625 + 1× 64 = 689

The variance of the linear combination is 689, and the standard deviation is the square root of 689: about $26.25.

124 CHAPTER 3. PROBABILITY

Exercises

3.29 College smokers. At a university, 13% of students smoke.

(a) Calculate the expected number of smokers in a random sample of 100 students from this university.

(b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are27 students outside the gym waiting for it to open. Should you use the same approach from part (a) tocalculate the expected number of smokers among these 27 students?

3.30 Ace of clubs wins. Consider the following card game with a well-shuffled deck of cards. If you drawa red card, you win nothing. If you get a spade, you win $5. For any club, you win $10 plus an extra $20for the ace of clubs.

(a) Create a probability model for the amount you win at this game. Also, find the expected winnings fora single game and the standard deviation of the winnings.

(b) What is the maximum amount you would be willing to pay to play this game? Explain your reasoning.

3.31 Hearts win. In a new card game, you start with a well-shuffled full deck and draw 3 cards withoutreplacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any otherdraws, you win nothing.

(a) Create a probability model for the amount you win at this game, and find the expected winnings. Alsocompute the standard deviation of this distribution.

(b) If the game costs $5 to play, what would be the expected value and standard deviation of the net profit(or loss)? (Hint: profit = winnings − cost; X − 5)

3.32 Is it worth it? Andy is always looking for ways to make money fast. Lately, he has been trying tomake money by gambling. Here is the game he is considering playing: The game costs $2 to play. He drawsa card from a deck. If he gets a number card (2-10), he wins nothing. For any face card ( jack, queen orking), he wins $3. For any ace, he wins $5, and he wins an extra $20 if he draws the ace of clubs.

(a) Create a probability model and find Andy’s expected profit per game.

(b) Would you recommend this game to Andy as a good way to make money? Explain.

3.33 Portfolio return. A portfolio’s value increases by 18% during a financial boom and by 9% duringnormal times. It decreases by 12% during a recession. What is the expected return on this portfolio if eachscenario is equally likely?

3.34 Baggage fees. An airline charges the following baggage fees: $25 for the first bag and $35 for thesecond. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and12% have two pieces. We suppose a negligible portion of people check more than two bags.

(a) Build a probability model, compute the average revenue per passenger, and compute the correspondingstandard deviation.

(b) About how much revenue should the airline expect for a flight of 120 passengers? With what standarddeviation? Note any assumptions you make and if you think they are justified.

3.35 American roulette. The game of American roulette involves spinning a wheel with 38 slots: 18 red,18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot hasan equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on theircolor, they double their money. If it lands on another color, they lose their money. Suppose you bet $1 onred. What’s the expected value and standard deviation of your winnings?

3.36 European roulette. The game of European roulette involves spinning a wheel with 37 slots: 18 red,18 black, and 1 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot hasan equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on theircolor, they double their money. If it lands on another color, they lose their money.

(a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standarddeviation of your total winnings?

(b) Suppose you bet $1 in three different rounds. What is the expected value and standard deviation ofyour total winnings?

3.5. CONTINUOUS DISTRIBUTIONS 125

3.5 Continuous distributions

So far in this chapter we’ve discussed cases where the outcome of a variable is discrete. In thissection, we consider a context where the outcome is a continuous numerical variable.

EXAMPLE 3.71

Figure 3.24 shows a few different hollow histograms for the heights of US adults. How does changingthe number of bins allow you to make different interpretations of the data?

Adding more bins provides greater detail. This sample is extremely large, which is why much smallerbins still work well. Usually we do not use so many bins with smaller sample sizes since small countsper bin mean the bin heights are very volatile.

height (cm)140 160 180 200

height (cm)

freq

uenc

140 160 180 200

height (cm)140 160 180 200

height (cm)

freq

uenc

140 160 180 200

Figure 3.24: Four hollow histograms of US adults heights with varying bin widths.

EXAMPLE 3.72

What proportion of the sample is between 180 cm and 185 cm tall (about 5’11” to 6’1”)?

We can add up the heights of the bins in the range 180 cm and 185 and divide by the sample size.For instance, this can be done with the two shaded bins shown in Figure 3.25. The two bins inthis region have counts of 195,307 and 156,239 people, resulting in the following estimate of theprobability:

195307 + 156239

3,000,000= 0.1172

This fraction is the same as the proportion of the histogram’s area that falls in the range 180 to 185

cm.

126 CHAPTER 3. PROBABILITY

height (cm)140 160 180 200

Figure 3.25: A histogram with bin sizes of 2.5 cm. The shaded region representsindividuals with heights between 180 and 185 cm.

3.5.1 From histograms to continuous distributions

Examine the transition from a boxy hollow histogram in the top-left of Figure 3.24 to the muchsmoother plot in the lower-right. In this last plot, the bins are so slim that the hollow histogram isstarting to resemble a smooth curve. This suggests the population height as a continuous numericalvariable might best be explained by a curve that represents the outline of extremely slim bins.

This smooth curve represents a probability density function (also called a density ordistribution), and such a curve is shown in Figure 3.26 overlaid on a histogram of the sample. Adensity has a special property: the total area under the density’s curve is 1.

height (cm)140 160 180 200

Figure 3.26: The continuous probability distribution of heights for US adults.

3.5. CONTINUOUS DISTRIBUTIONS 127

3.5.2 Probabilities from continuous distributions

We computed the proportion of individuals with heights 180 to 185 cm in Example 3.72 as afraction:

number of people between 180 and 185

total sample size

We found the number of people with heights between 180 and 185 cm by determining the fractionof the histogram’s area in this region. Similarly, we can use the area in the shaded region under thecurve to find a probability (with the help of a computer):

P (height between 180 and 185) = area between 180 and 185 = 0.1157

The probability that a randomly selected person is between 180 and 185 cm is 0.1157. This is veryclose to the estimate from Example 3.72: 0.1172.

height (cm)140 160 180 200

Figure 3.27: Density for heights in the US adult population with the area between180 and 185 cm shaded. Compare this plot with Figure 3.25.

GUIDED PRACTICE 3.73

Three US adults are randomly selected. The probability a single adult is between 180 and 185 cmis 0.1157.60

(a) What is the probability that all three are between 180 and 185 cm tall?

(b) What is the probability that none are between 180 and 185 cm?

EXAMPLE 3.74

What is the probability that a randomly selected person is exactly 180 cm? Assume you canmeasure perfectly.

This probability is zero. A person might be close to 180 cm, but not exactly 180 cm tall. This alsomakes sense with the definition of probability as area; there is no area captured between 180 cmand 180 cm.

GUIDED PRACTICE 3.75

Suppose a person’s height is rounded to the nearest centimeter. Is there a chance that a randomperson’s measured height will be 180 cm?61

60Brief answers: (a) 0.1157× 0.1157× 0.1157 = 0.0015. (b) (1− 0.1157)3 = 0.69261This has positive probability. Anyone between 179.5 cm and 180.5 cm will have a measured height of 180 cm.

This is probably a more realistic scenario to encounter in practice versus Example 3.74.

128 CHAPTER 3. PROBABILITY

Exercises

3.37 Cat weights. The histogram shown below represents the weights (in kg) of 47 female and 97 malecats.62

(a) What fraction of these cats weigh less than 2.5kg?

(b) What fraction of these cats weigh between 2.5and 2.75 kg?

Body weightF

requ

ency

2.0 2.5 3.0 3.5 4.0

3.38 Income and gender. The relative frequency table below displays the distribution of annual totalpersonal income (in 2009 inflation-adjusted dollars) for a representative sample of 96,420,486 Americans.These data come from the American Community Survey for 2005-2009. This sample is comprised of 59%males and 41% females.63

(a) Describe the distribution of total personal income.

(b) What is the probability that a randomly chosen US residentmakes less than $50,000 per year?

(c) What is the probability that a randomly chosen US residentmakes less than $50,000 per year and is female? Note anyassumptions you make.

(d) The same data source indicates that 71.8% of females makeless than $50,000 per year. Use this value to determinewhether or not the assumption you made in part (c) is valid.

Income Total

$1 to $9,999 or loss 2.2%$10,000 to $14,999 4.7%$15,000 to $24,999 15.8%$25,000 to $34,999 18.3%$35,000 to $49,999 21.2%$50,000 to $64,999 13.9%$65,000 to $74,999 5.8%$75,000 to $99,999 8.4%$100,000 or more 9.7%

62W. N. Venables and B. D. Ripley. Modern Applied Statistics with S. Fourth Edition.www.stats.ox.ac.uk/pub/MASS4. New York: Springer, 2002.

63U.S. Census Bureau, 2005-2009 American Community Survey.

3.5. CONTINUOUS DISTRIBUTIONS 129

Chapter exercises

3.39 Grade distributions. Each row in the table below is a proposed grade distribution for a class. Identifyeach as a valid or invalid probability distribution, and explain your reasoning.

GradesA B C D F

(a) 0.3 0.3 0.3 0.2 0.1(b) 0 0 1 0 0(c) 0.3 0.3 0.3 0 0(d) 0.3 0.5 0.2 0.1 -0.1(e) 0.2 0.4 0.2 0.1 0.1(f) 0 -0.1 1.1 0 0

3.40 Health coverage, frequencies. The Behavioral Risk Factor Surveillance System (BRFSS) is anannual telephone survey designed to identify risk factors in the adult population and report emerging healthtrends. The following table summarizes two variables for the respondents: health status and health coverage,which describes whether each respondent had health insurance.64

Health StatusExcellent Very good Good Fair Poor Total

Health No 459 727 854 385 99 2,524Coverage Yes 4,198 6,245 4,821 1,634 578 17,476

Total 4,657 6,972 5,675 2,019 677 20,000

(a) If we draw one individual at random, what is the probability that the respondent has excellent healthand doesn’t have health coverage?

(b) If we draw one individual at random, what is the probability that the respondent has excellent healthor doesn’t have health coverage?

3.41 HIV in Swaziland. Swaziland has the highest HIV prevalence in the world: 25.9% of this country’spopulation is infected with HIV.65 The ELISA test is one of the first and most accurate tests for HIV. Forthose who carry HIV, the ELISA test is 99.7% accurate. For those who do not carry HIV, the test is 92.6%accurate. If an individual from Swaziland has tested positive, what is the probability that he carries HIV?

3.42 Twins. About 30% of human twins are identical, and the rest are fraternal. Identical twins arenecessarily the same sex – half are males and the other half are females. One-quarter of fraternal twins areboth male, one-quarter both female, and one-half are mixes: one male, one female. You have just become aparent of twins and are told they are both girls. Given this information, what is the probability that theyare identical?

3.43 Cost of breakfast. Sally gets a cup of coffee and a muffin every day for breakfast from one of themany coffee shops in her neighborhood. She picks a coffee shop each morning at random and independentlyof previous days. The average price of a cup of coffee is $1.40 with a standard deviation of 30¢ ($0.30), theaverage price of a muffin is $2.50 with a standard deviation of 15¢, and the two prices are independent ofeach other.

(a) What is the mean and standard deviation of the amount she spends on breakfast daily?

(b) What is the mean and standard deviation of the amount she spends on breakfast weekly (7 days)?

64Office of Surveillance, Epidemiology, and Laboratory Services Behavioral Risk Factor Surveillance System, BRFSS2010 Survey Data.

65Source: CIA Factbook, Country Comparison: HIV/AIDS – Adult Prevalence Rate.

130 CHAPTER 3. PROBABILITY

3.44 Scooping ice cream. Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice creamscoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as wellas the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and theamount scooped out as Y . Suppose these random variables have the following means, standard deviations,and variances:

mean SD variance

X 48 1 1Y 2 0.25 0.0625

(a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice creamdo you expect to have been served at this party? What is the standard deviation of the amount of icecream served?

(b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream?That is, find the expected value of X − Y . What is the standard deviation of the amount left in thebox?

3.45 Variance of a mean, Part I. Suppose we have independent observations X1 and X2 from a distributionwith mean µ and standard deviation σ. What is the variance of the mean of the two values: X1+X2

3.46 Variance of a mean, Part II. Suppose we have 3 independent observations X1, X2, X3 from a dis-tribution with mean µ and standard deviation σ. What is the variance of the mean of these 3 values:X1+X2+X3

3.47 Variance of a mean, Part III. Suppose we have n independent observations X1, X2, …, Xn from adistribution with mean µ and standard deviation σ. What is the variance of the mean of these n values:X1+X2+···+Xn

131

Chapter 4Distributions of randomvariables

4.1 Normal distribution

4.2 Geometric distribution

4.3 Binomial distribution

4.4 Negative binomial distribution

4.5 Poisson distribution

132

In this chapter, we discuss statistical distributions that frequently arise

in the context of data analysis or statistical inference. We start with

the normal distribution in the first section, which is used frequently in

later chapters of this book. The remaining sections will occasionally be

referenced but may be considered optional for the content in this book.

For videos, slides, and other resources, please visit

www.openintro.org/os

4.1. NORMAL DISTRIBUTION 133

4.1 Normal distribution

Among all the distributions we see in practice, one is overwhelmingly the most common. Thesymmetric, unimodal, bell curve is ubiquitous throughout statistics. Indeed it is so common, thatpeople often know it as the normal curve or normal distribution,1 shown in Figure 4.1. Variablessuch as SAT scores and heights of US adult males closely follow the normal distribution.

Figure 4.1: A normal curve.

NORMAL DISTRIBUTION FACTS

Many variables are nearly normal, but none are exactly normal. Thus the normal distribution,while not perfect for any single problem, is very useful for a variety of problems. We will use itin data exploration and to solve important problems in statistics.

4.1.1 Normal distribution model

The normal distribution always describes a symmetric, unimodal, bell-shaped curve. How-ever, these curves can look different depending on the details of the model. Specifically, the normaldistribution model can be adjusted using two parameters: mean and standard deviation. As youcan probably guess, changing the mean shifts the bell curve to the left or right, while changing thestandard deviation stretches or constricts the curve. Figure 4.2 shows the normal distribution withmean 0 and standard deviation 1 in the left panel and the normal distributions with mean 19 andstandard deviation 4 in the right panel. Figure 4.3 shows these distributions on the same axis.

−3 −2 −1 0 1 2 3

7 11 15 19 23 27 31

Figure 4.2: Both curves represent the normal distribution. However, they differ intheir center and spread.

If a normal distribution has mean µ and standard deviation σ, we may write the distributionas N(µ, σ). The two distributions in Figure 4.3 may be written as

N(µ = 0, σ = 1) and N(µ = 19, σ = 4)

Because the mean and standard deviation describe a normal distribution exactly, they are calledthe distribution’s parameters. The normal distribution with mean µ = 0 and standard deviationσ = 1 is called the standard normal distribution.

1It is also introduced as the Gaussian distribution after Frederic Gauss, the first person to formalize its mathe-matical expression.

134 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

0 10 20 30

Figure 4.3: The normal distributions shown in Figure 4.2 but plotted together andon the same scale.

GUIDED PRACTICE 4.1

Write down the short-hand for a normal distribution with2

(a) mean 5 and standard deviation 3,(b) mean -100 and standard deviation 10, and(c) mean 2 and standard deviation 9.

4.1.2 Standardizing with Z-scores

We often want to put data onto a standardized scale, which can make comparisons more reasonable.

EXAMPLE 4.2

Table 4.4 shows the mean and standard deviation for total scores on the SAT and ACT. Thedistribution of SAT and ACT scores are both nearly normal. Suppose Ann scored 1300 on her SATand Tom scored 24 on his ACT. Who performed better?

We use the standard deviation as a guide. Ann is 1 standard deviation above average on the SAT:1100 + 200 = 1300. Tom is 0.5 standard deviations above the mean on the ACT: 21 + 0.5× 6 = 24.In Figure 4.5, we can see that Ann tends to do better with respect to everyone else than Tom did,so her score was better.

SAT ACTMean 1100 21SD 200 6

Figure 4.4: Mean and standard deviation for the SAT and ACT.

Example 4.2 used a standardization technique called a Z-score, a method most commonlyemployed for nearly normal observations but that may be used with any distribution. The Z-scoreof an observation is defined as the number of standard deviations it falls above or below the mean.If the observation is one standard deviation above the mean, its Z-score is 1. If it is 1.5 standarddeviations below the mean, then its Z-score is -1.5. If x is an observation from a distribution N(µ, σ),we define the Z-score mathematically as

Z =x− µσ

Using µSAT = 1100, σSAT = 200, and xAnn

= 1300, we find Ann’s Z-score:

ZAnn

Ann− µ

SAT

σSAT

=1300− 1100

200= 1

2(a) N(µ = 5, σ = 3). (b) N(µ = −100, σ = 10). (c) N(µ = 2, σ = 9).

4.1. NORMAL DISTRIBUTION 135

700 900 1100 1300 1500

Ann

9 15 21 27 33

Tom

Figure 4.5: Ann’s and Tom’s scores shown against the SAT and ACT distributions.

THE Z-SCORE

The Z-score of an observation is the number of standard deviations it falls above or below themean. We compute the Z-score for an observation x that follows a distribution with mean µand standard deviation σ using

Z =x− µσ

GUIDED PRACTICE 4.3

Use Tom’s ACT score, 24, along with the ACT mean and standard deviation to find his Z-score.3

Observations above the mean always have positive Z-scores, while those below the mean alwayshave negative Z-scores. If an observation is equal to the mean, such as an SAT score of 1100, thenthe Z-score is 0.

GUIDED PRACTICE 4.4

Let X represent a random variable from N(µ = 3, σ = 2), and suppose we observe x = 5.19.(a) Find the Z-score of x.(b) Use the Z-score to determine how many standard deviations above or below the mean x falls.4

GUIDED PRACTICE 4.5

Head lengths of brushtail possums follow a normal distribution with mean 92.6 mm and standarddeviation 3.6 mm. Compute the Z-scores for possums with head lengths of 95.4 mm and 85.8 mm.5

We can use Z-scores to roughly identify which observations are more unusual than others. Anobservation x1 is said to be more unusual than another observation x2 if the absolute value of its Z-score is larger than the absolute value of the other observation’s Z-score: |Z1| > |Z2|. This techniqueis especially insightful when a distribution is symmetric.

GUIDED PRACTICE 4.6

Which of the observations in Guided Practice 4.5 is more unusual?6

3ZTom = xTom−µACTσACT

= 24−216

= 0.54(a) Its Z-score is given by Z = x−µ

σ= 5.19−3

2= 2.19/2 = 1.095. (b) The observation x is 1.095 standard

deviations above the mean. We know it must be above the mean since Z is positive.5For x1 = 95.4 mm: Z1 = x1−µ

σ= 95.4−92.6

3.6= 0.78. For x2 = 85.8 mm: Z2 = 85.8−92.6

3.6= −1.89.

6Because the absolute value of Z-score for the second observation is larger than that of the first, the secondobservation has a more unusual head length.

136 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

4.1.3 Finding tail areas

It’s very useful in statistics to be able to identify tail areas of distributions. For instance, howmany people have an SAT score below Ann’s score of 1300? This is the same as Ann’s percentile,which is the fraction of cases that have lower scores than Ann. We can visualize such a tail area likethe curve and shading shown in Figure 4.6.

500 700 900 1100 1300 1500 1700

Figure 4.6: The area to the left of Z represents the percentile of the observation.

There are many techniques for doing this, and we’ll discuss three of the options.

1. The most common approach in practice is to use statistical software. For example, in theprogram R, we could find the area shown in Figure 4.6 using the following command, whichtakes in the Z-score and returns the lower tail area:…..> pnorm(1)

…..[1] 0.8413447

According to this calculation, the region shaded that is below 1300 represents the proportion0.841 (84.1%) of SAT test takers who had Z-scores below Z = 1. More generally, we can alsospecify the cutoff explicitly if we also note the mean and standard deviation:…..> pnorm(1300, mean = 1100, sd = 200)

…..[1] 0.8413447

There are many other software options, such as Python or SAS; even spreadsheet programssuch as Excel and Google Sheets support these calculations.

2. A common strategy in classrooms is to use a graphing calculator, such as a TI or Casio calcula-tor. These calculators require a series of button presses that are less concisely described. Youcan find instructions on using these calculators for finding tail areas of a normal distributionin the OpenIntro video library:

www.openintro.org/videos

3. The last option for finding tail areas is to use what’s called a probability table; these areoccasionally used in classrooms but rarely in practice. Appendix C.1 contains such a table anda guide for how to use it.

We will solve normal distribution problems in this section by always first finding the Z-score. Thereason is that we will encounter close parallels called test statistics beginning in Chapter 5; theseare, in many instances, an equivalent of a Z-score.

4.1. NORMAL DISTRIBUTION 137

4.1.4 Normal probability examples

Cumulative SAT scores are approximated well by a normal model, N(µ = 1100, σ = 200).

EXAMPLE 4.7

Shannon is a randomly selected SAT taker, and nothing is known about Shannon’s SAT aptitude.What is the probability Shannon scores at least 1190 on her SATs?

First, always draw and label a picture of the normal distribution. (Drawings need not be exact tobe useful.) We are interested in the chance she scores above 1190, so we shade this upper tail:

700 1100 1500

The picture shows the mean and the values at 2 standard deviations above and below the mean.The simplest way to find the shaded area under the curve makes use of the Z-score of the cutoffvalue. With µ = 1100, σ = 200, and the cutoff value x = 1190, the Z-score is computed as

Z =x− µσ

=1190− 1100

200=

200= 0.45

Using statistical software (or another preferred method), we can area left of Z = 0.45 as 0.6736. Tofind the area above Z = 0.45, we compute one minus the area of the lower tail:

1.0000 0.6736 0.3264 =

The probability Shannon scores at least 1190 on the SAT is 0.3264.

ALWAYS DRAW A PICTURE FIRST, AND FIND THE Z-SCORE SECOND

For any normal probability situation, always always always draw and label the normal curveand shade the area of interest first. The picture will provide an estimate of the probability.After drawing a figure to represent the situation, identify the Z-score for the value of interest.

GUIDED PRACTICE 4.8

If the probability of Shannon scoring at least 1190 is 0.3264, then what is the probability she scoresless than 1190? Draw the normal curve representing this exercise, shading the lower region insteadof the upper one.7

7We found this probability in Example 4.7: 0.6736.

700 1100 1500

138 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

EXAMPLE 4.9

Edward earned a 1030 on his SAT. What is his percentile?

First, a picture is needed. Edward’s percentile is the proportion of people who do not get as highas a 1030. These are the scores to the left of 1030.

700 1100 1500

Identifying the mean µ = 1100, the standard deviation σ = 200, and the cutoff for the tail areax = 1030 makes it easy to compute the Z-score:

Z =x− µσ

=1030− 1100

200= −0.35

Using statistical software, we get a tail area of 0.3632. Edward is at the 36th percentile.

GUIDED PRACTICE 4.10

Use the results of Example 4.9 to compute the proportion of SAT takers who did better than Edward.Also draw a new picture.8

FINDING AREAS TO THE RIGHT

Many software programs return the area to the left when given a Z-score. If you would like thearea to the right, first find the area to the left and then subtract this amount from one.

GUIDED PRACTICE 4.11

Stuart earned an SAT score of 1500. Draw a picture for each part.(a) What is his percentile?(b) What percent of SAT takers did better than Stuart?9

Based on a sample of 100 men, the heights of male adults in the US is nearly normal with mean70.0” and standard deviation 3.3”.

GUIDED PRACTICE 4.12

Mike is 5’7” and Jose is 6’4”, and they both live in the US.(a) What is Mike’s height percentile?(b) What is Jose’s height percentile?Also draw one picture for each part.10

8If Edward did better than 36% of SAT takers, then about 64% must have done better than him.

700 1100 1500

9We leave the drawings to you. (a) Z = 1500−1100200

= 2→ 0.9772. (b) 1− 0.9772 = 0.0228.10First put the heights into inches: 67 and 76 inches. Figures are shown below.

(a) ZMike = 67−703.3

= −0.91 → 0.1814. (b) ZJose = 76−703.3

= 1.82 → 0.9656.

67 70

Mike

70 76

Jose

4.1. NORMAL DISTRIBUTION 139

The last several problems have focused on finding the percentile (or upper tail) for a particularobservation. What if you would like to know the observation corresponding to a particular percentile?

EXAMPLE 4.13

Erik’s height is at the 40th percentile. How tall is he?

As always, first draw the picture.

63.4 70 76.6

40%(0.40)

In this case, the lower tail probability is known (0.40), which can be shaded on the diagram. We wantto find the observation that corresponds to this value. As a first step in this direction, we determinethe Z-score associated with the 40th percentile. Using software, we can obtain the correspondingZ-score of about -0.25.

Knowing ZErik = −0.25 and the population parameters µ = 70 and σ = 3.3 inches, the Z-scoreformula can be set up to determine Erik’s unknown height, labeled x

Erik:

−0.25 = ZErik

Erik− µσ

Erik− 70

3.3

Solving for xErik

yields a height of 69.18 inches. That is, Erik is about 5’9”.

EXAMPLE 4.14

What is the adult male height at the 82nd percentile?

Again, we draw the figure first.

63.4 70 76.6

82%(0.82)

18%(0.18)

Next, we want to find the Z-score at the 82nd percentile, which will be a positive value and can befound using software as Z = 0.92. Finally, the height x is found using the Z-score formula with theknown mean µ, standard deviation σ, and Z-score Z = 0.92:

0.92 = Z =x− µσ

=x− 70

3.3

This yields 73.04 inches or about 6’1” as the height at the 82nd percentile.

GUIDED PRACTICE 4.15

The SAT scores follow N(1100, 200).11

(a) What is the 95th percentile for SAT scores?(b) What is the 97.5th percentile for SAT scores?

11Short answers: (a) Z95 = 1.65→ 1430 SAT score. (b) Z97.5 = 1.96→ 1492 SAT score.

140 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

GUIDED PRACTICE 4.16

Adult male heights follow N(70.0”, 3.3”).12

(a) What is the probability that a randomly selected male adult is at least 6’2” (74 inches)?(b) What is the probability that a male adult is shorter than 5’9” (69 inches)?

EXAMPLE 4.17

What is the probability that a random adult male is between 5’9” and 6’2”?

These heights correspond to 69 inches and 74 inches. First, draw the figure. The area of interest isno longer an upper or lower tail.

63.4 70.0 76.6

The total area under the curve is 1. If we find the area of the two tails that are not shaded (fromGuided Practice 4.16, these areas are 0.3821 and 0.1131), then we can find the middle area:

1.0000 0.3821 0.1131 0.5048 =

That is, the probability of being between 5’9” and 6’2” is 0.5048.

GUIDED PRACTICE 4.18

SAT scores follow N(1100, 200). What percent of SAT takers get between 1100 and 1400?13

GUIDED PRACTICE 4.19

Adult male heights follow N(70.0”, 3.3”). What percent of adult males are between 5’5” and 5’7”?14

12Short answers: (a) Z = 1.21→ 0.8869, then subtract this value from 1 to get 0.1131. (b) Z = −0.30→ 0.3821.13This is an abbreviated solution. (Be sure to draw a figure!) First find the percent who get below 1100 and the

percent that get above 1400: Z1100 = 0.00→ 0.5000 (area below), Z1400 = 1.5→ 0.0668 (area above). Final answer:1.0000− 0.5000− 0.0668 = 0.4332.

145’5” is 65 inches (Z = −1.52). 5’7” is 67 inches (Z = −0.91). Numerical solution: 1.000−0.0643−0.8186 = 0.1171,i.e. 11.71%.

4.1. NORMAL DISTRIBUTION 141

4.1.5 68-95-99.7 rule

Here, we present a useful rule of thumb for the probability of falling within 1, 2, and 3 standarddeviations of the mean in the normal distribution. This will be useful in a wide range of practicalsettings, especially when trying to make a quick estimate without a calculator or Z-table.

µ − 3σ µ − 2σ µ − σ µ µ + σ µ + 2σ µ + 3σ

99.7%

95%

68%

Figure 4.7: Probabilities for falling within 1, 2, and 3 standard deviations of themean in a normal distribution.

GUIDED PRACTICE 4.20

Use software, a calculator, or a probability table to confirm that about 68%, 95%, and 99.7% ofobservations fall within 1, 2, and 3, standard deviations of the mean in the normal distribution,respectively. For instance, first find the area that falls between Z = −1 and Z = 1, which shouldhave an area of about 0.68. Similarly there should be an area of about 0.95 between Z = −2 andZ = 2.15

It is possible for a normal random variable to fall 4, 5, or even more standard deviations fromthe mean. However, these occurrences are very rare if the data are nearly normal. The probability ofbeing further than 4 standard deviations from the mean is about 1-in-15,000. For 5 and 6 standarddeviations, it is about 1-in-2 million and 1-in-500 million, respectively.

GUIDED PRACTICE 4.21

SAT scores closely follow the normal model with mean µ = 1100 and standard deviation σ = 200.16

(a) About what percent of test takers score 700 to 1500?(b) What percent score between 1100 and 1500?

15First draw the pictures. Using software, we get 0.6827 within 1 standard deviation, 0.9545 within 2 standarddeviations, and 0.9973 within 3 standard deviations.

16(a) 700 and 1500 represent two standard deviations below and above the mean, which means about 95% of testtakers will score between 700 and 1500. (b) We found that 700 to 1500 represents about 95% of test takers. These test

takers would be evenly split by the center of the distribution, 1100, so 95%2

= 47.5% of all test takers score between1100 and 1500.

142 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Exercises

4.1 Area under the curve, Part I. What percent of a standard normal distribution N(µ = 0, σ = 1) isfound in each region? Be sure to draw a graph.

(a) Z < −1.35 (b) Z > 1.48 (c) −0.4 < Z < 1.5 (d) |Z| > 2

4.2 Area under the curve, Part II. What percent of a standard normal distribution N(µ = 0, σ = 1) isfound in each region? Be sure to draw a graph.

(a) Z > −1.13 (b) Z < 0.18 (c) Z > 8 (d) |Z| < 0.5

4.3 GRE scores, Part I. Sophia who took the Graduate Record Examination (GRE) scored 160 on the Ver-bal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal Reasoningsection for all test takers was 151 with a standard deviation of 7, and the mean score for the QuantitativeReasoning was 153 with a standard deviation of 7.67. Suppose that both distributions are nearly normal.

(a) Write down the short-hand for these two normal distributions.

(b) What is Sophia’s Z-score on the Verbal Reasoning section? On the Quantitative Reasoning section?Draw a standard normal distribution curve and mark these two Z-scores.

(d) Relative to others, which section did she do better on?

(e) Find her percentile scores for the two exams.

(f) What percent of the test takers did better than her on the Verbal Reasoning section? On the QuantitativeReasoning section?

(g) Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusionas to which section a student did better on.

(h) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b)- (f) change? Explain your reasoning.

4.4 Triathlon times, Part I. In triathlons, it is common for racers to be placed into age and gender groups.Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages30 – 34 group while Mary competed in the Women, Ages 25 – 29 group. Leo completed the race in 1:22:28(4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster,but they are curious about how they did within their respective groups. Can you help them? Here is someinformation on the performance of their groups:

• The finishing times of the Men, Ages 30 – 34 group has a mean of 4313 seconds with a standarddeviation of 583 seconds.

• The finishing times of the Women, Ages 25 – 29 group has a mean of 5261 seconds with a standarddeviation of 807 seconds.

• The distributions of finishing times for both groups are approximately Normal.

Remember: a better performance corresponds to a faster finish.

(a) Write down the short-hand for these two normal distributions.

(b) What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?

(d) What percent of the triathletes did Leo finish faster than in his group?

(e) What percent of the triathletes did Mary finish faster than in her group?

(f) If the distributions of finishing times are not nearly normal, would your answers to parts (b) – (e) change?Explain your reasoning.

4.5 GRE scores, Part II. In Exercise 4.3 we saw two distributions for GRE scores: N(µ = 151, σ = 7) forthe verbal part of the exam and N(µ = 153, σ = 7.67) for the quantitative part. Use this information tocompute each of the following:

(a) The score of a student who scored in the 80th percentile on the Quantitative Reasoning section.

(b) The score of a student who scored worse than 70% of the test takers in the Verbal Reasoning section.

4.1. NORMAL DISTRIBUTION 143

4.6 Triathlon times, Part II. In Exercise 4.4 we saw two distributions for triathlon times: N(µ = 4313, σ =583) for Men, Ages 30 – 34 and N(µ = 5261, σ = 807) for the Women, Ages 25 – 29 group. Times are listedin seconds. Use this information to compute each of the following:

(a) The cutoff time for the fastest 5% of athletes in the men’s group, i.e. those who took the shortest 5%of time to finish.

(b) The cutoff time for the slowest 10% of athletes in the women’s group.

4.7 LA weather, Part I. The average daily high temperature in June in LA is 77◦F with a standard deviationof 5◦F. Suppose that the temperatures in June closely follow a normal distribution.

(a) What is the probability of observing an 83◦F temperature or higher in LA during a randomly chosenday in June?

(b) How cool are the coldest 10% of the days (days with lowest average high temperature) during June inLA?

4.8 CAPM. The Capital Asset Pricing Model (CAPM) is a financial model that assumes returns on aportfolio are normally distributed. Suppose a portfolio has an average annual return of 14.7% (i.e. anaverage gain of 14.7%) with a standard deviation of 33%. A return of 0% means the value of the portfoliodoesn’t change, a negative return means that the portfolio loses money, and a positive return means thatthe portfolio gains money.

(a) What percent of years does this portfolio lose money, i.e. have a return less than 0%?

(b) What is the cutoff for the highest 15% of annual returns with this portfolio?

4.9 LA weather, Part II. Exercise 4.7 states that average daily high temperature in June in LA is 77◦F witha standard deviation of 5◦F, and it can be assumed that they to follow a normal distribution. We use thefollowing equation to convert ◦F (Fahrenheit) to ◦C (Celsius):

C = (F − 32)× 5

(a) Write the probability model for the distribution of temperature in ◦C in June in LA.

(b) What is the probability of observing a 28◦C (which roughly corresponds to 83◦F) temperature or higherin June in LA? Calculate using the ◦C model from part (a).

(c) Did you get the same answer or different answers in part (b) of this question and part (a) of Exercise 4.7?Are you surprised? Explain.

(d) Estimate the IQR of the temperatures (in ◦C) in June in LA.

4.10 Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distributionwith mean 185 milligrams per deciliter (mg/dl). Women with cholesterol levels above 220 mg/dl are con-sidered to have high cholesterol and about 18.5% of women fall into this category. What is the standarddeviation of the distribution of cholesterol levels for women aged 20 to 34?

144 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

4.2 Geometric distribution

How long should we expect to flip a coin until it turns up heads? Or how many times shouldwe expect to roll a die until we get a 1? These questions can be answered using the geometricdistribution. We first formalize each trial – such as a single coin flip or die toss – using the Bernoullidistribution, and then we combine these with our tools from probability (Chapter 3) to constructthe geometric distribution.

4.2.1 Bernoulli distribution

Many health insurance plans in the United States have a deductible, where the insured individ-ual is responsible for costs up to the deductible, and then the costs above the deductible are sharedbetween the individual and insurance company for the remainder of the year.

Suppose a health insurance company found that 70% of the people they insure stay below theirdeductible in any given year. Each of these people can be thought of as a trial. We label a person asuccess if her healthcare costs do not exceed the deductible. We label a person a failure if she doesexceed her deductible in the year. Because 70% of the individuals will not hit their deductible, wedenote the probability of a success as p = 0.7. The probability of a failure is sometimes denotedwith q = 1− p, which would be 0.3 for the insurance example.

When an individual trial only has two possible outcomes, often labeled as success or failure,it is called a Bernoulli random variable. We chose to label a person who does not hit herdeductible as a “success” and all others as “failures”. However, we could just as easily have reversedthese labels. The mathematical framework we will build does not depend on which outcome islabeled a success and which a failure, as long as we are consistent.

Bernoulli random variables are often denoted as 1 for a success and 0 for a failure. In additionto being convenient in entering data, it is also mathematically handy. Suppose we observe ten trials:

1 1 1 0 1 0 0 1 1 0

Then the sample proportion, p̂, is the sample mean of these observations:

p̂ =# of successes

# of trials=

1 + 1 + 1 + 0 + 1 + 0 + 0 + 1 + 1 + 0

10= 0.6

This mathematical inquiry of Bernoulli random variables can be extended even further. Because 0

and 1 are numerical outcomes, we can define the mean and standard deviation of a Bernoulli randomvariable. (See Exercises 4.15 and 4.16.)

BERNOULLI RANDOM VARIABLE

If X is a random variable that takes value 1 with probability of success p and 0 with probability1− p, then X is a Bernoulli random variable with mean and standard deviation

µ = p σ =√p(1− p)

In general, it is useful to think about a Bernoulli random variable as a random process withonly two outcomes: a success or failure. Then we build our mathematical framework using thenumerical labels 1 and 0 for successes and failures, respectively.

4.2. GEOMETRIC DISTRIBUTION 145

4.2.2 Geometric distribution

The geometric distribution is used to describe how many trials it takes to observe a success.Let’s first look at an example.

EXAMPLE 4.22

Suppose we are working at the insurance company and need to find a case where the person didnot exceed her (or his) deductible as a case study. If the probability a person will not exceed herdeductible is 0.7 and we are drawing people at random, what are the chances that the first personwill not have exceeded her deductible, i.e. be a success? The second person? The third? Whatabout we pull n− 1 cases before we find the first success, i.e. the first success is the nth person? (Ifthe first success is the fifth person, then we say n = 5.)

The probability of stopping after the first person is just the chance the first person will not hit her(or his) deductible: 0.7. The probability the second person is the first to hit her deductible is

P (second person is the first to hit deductible)

= P (the first won’t, the second will) = (0.3)(0.7) = 0.21

Likewise, the probability it will be the third case is (0.3)(0.3)(0.7) = 0.063.

If the first success is on the nth person, then there are n − 1 failures and finally 1 success, whichcorresponds to the probability (0.3)n−1(0.7). This is the same as (1− 0.7)n−1(0.7).

Example 4.22 illustrates what the geometric distribution, which describes the waiting timeuntil a success for independent and identically distributed (iid) Bernoulli random variables.In this case, the independence aspect just means the individuals in the example don’t affect eachother, and identical means they each have the same probability of success.

The geometric distribution from Example 4.22 is shown in Figure 4.8. In general, the proba-bilities for a geometric distribution decrease exponentially fast.

Pro

babi

lity

Number of Trials Until a Success for p = 0.71 2 3 4 5 6 7 8

0.0

0.2

0.4

0.6

Figure 4.8: The geometric distribution when the probability of success is p = 0.7.

While this text will not derive the formulas for the mean (expected) number of trials neededto find the first success or the standard deviation or variance of this distribution, we present generalformulas for each.

146 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

GEOMETRIC DISTRIBUTION

If the probability of a success in one trial is p and the probability of a failure is 1− p, then theprobability of finding the first success in the nth trial is given by

(1− p)n−1p

The mean (i.e. expected value), variance, and standard deviation of this wait time are given by

µ =1

pσ2 =

1− pp2

σ =

√1− pp2

It is no accident that we use the symbol µ for both the mean and expected value. The meanand the expected value are one and the same.

It takes, on average, 1/p trials to get a success under the geometric distribution. This mathe-matical result is consistent with what we would expect intuitively. If the probability of a success ishigh (e.g. 0.8), then we don’t usually wait very long for a success: 1/0.8 = 1.25 trials on average.If the probability of a success is low (e.g. 0.1), then we would expect to view many trials before wesee a success: 1/0.1 = 10 trials.

GUIDED PRACTICE 4.23

The probability that a particular case would not exceed their deductible is said to be 0.7. If we wereto examine cases until we found one that where the person did not hit her deductible, how manycases should we expect to check?17

EXAMPLE 4.24

What is the chance that we would find the first success within the first 3 cases?

This is the chance it is the first (n = 1), second (n = 2), or third (n = 3) case is the first success,which are three disjoint outcomes. Because the individuals in the sample are randomly sampledfrom a large population, they are independent. We compute the probability of each case and addthe separate results:

P (n = 1, 2, or 3)

= P (n = 1) + P (n = 2) + P (n = 3)

= (0.3)1−1(0.7) + (0.3)2−1(0.7) + (0.3)3−1(0.7)

= 0.973

There is a probability of 0.973 that we would find a successful case within 3 cases.

GUIDED PRACTICE 4.25

Determine a more clever way to solve Example 4.24. Show that you get the same result.18

17We would expect to see about 1/0.7 ≈ 1.43 individuals to find the first success.18First find the probability of the complement: P (no success in first 3 trials) = 0.33 = 0.027. Next, compute one

minus this probability: 1− P (no success in 3 trials) = 1− 0.027 = 0.973.

4.2. GEOMETRIC DISTRIBUTION 147

EXAMPLE 4.26

Suppose a car insurer has determined that 88% of its drivers will not exceed their deductible in agiven year. If someone at the company were to randomly draw driver files until they found one thathad not exceeded their deductible, what is the expected number of drivers the insurance employeemust check? What is the standard deviation of the number of driver files that must be drawn?

In this example, a success is again when someone will not exceed the insurance deductible, whichhas probability p = 0.88. The expected number of people to be checked is 1/p = 1/0.88 = 1.14 andthe standard deviation is

√(1− p)/p2 = 0.39.

GUIDED PRACTICE 4.27

Using the results from Example 4.26, µ = 1.14 and σ = 0.39, would it be appropriate to use thenormal model to find what proportion of experiments would end in 3 or fewer trials?19

The independence assumption is crucial to the geometric distribution’s accurate descriptionof a scenario. Mathematically, we can see that to construct the probability of the success on thenth trial, we had to use the Multiplication Rule for Independent Processes. It is no simple task togeneralize the geometric model for dependent trials.

19No. The geometric distribution is always right skewed and can never be well-approximated by the normal model.

148 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Exercises

4.11 Is it Bernoulli? Determine if each trial can be considered an independent Bernoulli trial for thefollowing situations.

(a) Cards dealt in a hand of poker.

(b) Outcome of each roll of a die.

4.12 With and without replacement. In the following situations assume that half of the specified popula-tion is male and the other half is female.

(a) Suppose you’re sampling from a room with 10 people. What is the probability of sampling two females ina row when sampling with replacement? What is the probability when sampling without replacement?

(b) Now suppose you’re sampling from a stadium with 10,000 people. What is the probability of samplingtwo females in a row when sampling with replacement? What is the probability when sampling withoutreplacement?

(c) We often treat individuals who are sampled from a large population as independent. Using your findingsfrom parts (a) and (b), explain whether or not this assumption is reasonable.

4.13 Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible fortheir children to have brown eyes (probability 0.75), blue eyes (0.125), or green eyes (0.125).

(a) What is the probability the first blue-eyed child they have is their third child? Assume that the eyecolors of the children are independent of each other.

(b) On average, how many children would such a pair of parents have before having a blue-eyed child? Whatis the standard deviation of the number of children they would expect to have until the first blue-eyedchild?

4.14 Defective rate. A machine that produces a special type of transistor (a component of computers) hasa 2% defective rate. The production is considered a random process where each transistor is independent ofthe others.

(a) What is the probability that the 10th transistor produced is the first with a defect?

(b) What is the probability that the machine produces no defective transistors in a batch of 100?

(c) On average, how many transistors would you expect to be produced before the first with a defect? Whatis the standard deviation?

(d) Another machine that also produces transistors has a 5% defective rate where each transistor is producedindependent of the others. On average how many transistors would you expect to be produced with thismachine before the first with a defect? What is the standard deviation?

(e) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect themean and standard deviation of the wait time until success?

4.15 Bernoulli, the mean. Use the probability rules from Section 3.4 to derive the mean of a Bernoullirandom variable, i.e. a random variable X that takes value 1 with probability p and value 0 with probability1− p. That is, compute the expected value of a generic Bernoulli random variable.

4.16 Bernoulli, the standard deviation. Use the probability rules from Section 3.4 to derive the standarddeviation of a Bernoulli random variable, i.e. a random variable X that takes value 1 with probability pand value 0 with probability 1− p. That is, compute the square root of the variance of a generic Bernoullirandom variable.

4.3. BINOMIAL DISTRIBUTION 149

4.3 Binomial distribution

The binomial distribution is used to describe the number of successes in a fixed number oftrials. This is different from the geometric distribution, which described the number of trials wemust wait before we observe a success.

4.3.1 The binomial distribution

Let’s again imagine ourselves back at the insurance agency where 70% of individuals do not exceedtheir deductible.

EXAMPLE 4.28

Suppose the insurance agency is considering a random sample of four individuals they insure. Whatis the chance exactly one of them will exceed the deductible and the other three will not? Let’s callthe four people Ariana (A), Brittany (B), Carlton (C), and Damian (D) for convenience.

Let’s consider a scenario where one person exceeds the deductible:

P (A = exceed, B = not, C = not, D = not)

= P (A = exceed) P (B = not) P (C = not) P (D = not)

= (0.3)(0.7)(0.7)(0.7)

= (0.7)3(0.3)1

= 0.103

But there are three other scenarios: Brittany, Carlton, or Damian could have been the one to exceedthe deductible. In each of these cases, the probability is again (0.7)3(0.3)1. These four scenariosexhaust all the possible ways that exactly one of these four people could have exceeded the deductible,so the total probability is 4× (0.7)3(0.3)1 = 0.412.

GUIDED PRACTICE 4.29

Verify that the scenario where Brittany is the only one exceed the deductible has probability(0.7)3(0.3)1. 20

The scenario outlined in Example 4.28 is an example of a binomial distribution scenario. Thebinomial distribution describes the probability of having exactly k successes in n independentBernoulli trials with probability of a success p (in Example 4.28, n = 4, k = 3, p = 0.7). We wouldlike to determine the probabilities associated with the binomial distribution more generally, i.e. wewant a formula where we can use n, k, and p to obtain the probability. To do this, we reexamineeach part of Example 4.28.

There were four individuals who could have been the one to exceed the deductible, and each ofthese four scenarios had the same probability. Thus, we could identify the final probability as

[# of scenarios]× P (single scenario)

The first component of this equation is the number of ways to arrange the k = 3 successes amongthe n = 4 trials. The second component is the probability of any of the four (equally probable)scenarios.

20 P (A = not, B = exceed, C = not, D = not) = (0.7)(0.3)(0.7)(0.7) = (0.7)3(0.3)1.

150 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Consider P (single scenario) under the general case of k successes and n − k failures in the ntrials. In any such scenario, we apply the Multiplication Rule for independent events:

pk(1− p)n−k

This is our general formula for P (single scenario).Secondly, we introduce a general formula for the number of ways to choose k successes in n

trials, i.e. arrange k successes and n− k failures:(n

k!(n− k)!

The quantity(nk

)is read n choose k.21 The exclamation point notation (e.g. k!) denotes a factorial

expression.

0! = 1

1! = 1

2! = 2× 1 = 2

3! = 3× 2× 1 = 6

4! = 4× 3× 2× 1 = 24

…

n! = n× (n− 1)× …× 3× 2× 1

Using the formula, we can compute the number of ways to choose k = 3 successes in n = 4 trials:(4

3!(4− 3)!=

3!1!=

4× 3× 2× 1

(3× 2× 1)(1)= 4

This result is exactly what we found by carefully thinking of each possible scenario in Example 4.28.Substituting n choose k for the number of scenarios and pk(1 − p)n−k for the single scenario

probability yields the general binomial formula.

BINOMIAL DISTRIBUTION

Suppose the probability of a single trial being a success is p. Then the probability of observingexactly k successes in n independent trials is given by(

)pk(1− p)n−k =

k!(n− k)!pk(1− p)n−k

The mean, variance, and standard deviation of the number of observed successes are

µ = np σ2 = np(1− p) σ =√np(1− p)

IS IT BINOMIAL? FOUR CONDITIONS TO CHECK.

(1) The trials are independent.(2) The number of trials, n, is fixed.(3) Each trial outcome can be classified as a success or failure.(4) The probability of a success, p, is the same for each trial.

21Other notation for n choose k includes nCk, Ckn, and C(n, k).

4.3. BINOMIAL DISTRIBUTION 151

EXAMPLE 4.30

What is the probability that 3 of 8 randomly selected individuals will have exceeded the insurancedeductible, i.e. that 5 of 8 will not exceed the deductible? Recall that 70% of individuals will notexceed the deductible.

We would like to apply the binomial model, so we check the conditions. The number of trials isfixed (n = 8) (condition 2) and each trial outcome can be classified as a success or failure (condition3). Because the sample is random, the trials are independent (condition 1) and the probability of asuccess is the same for each trial (condition 4).

In the outcome of interest, there are k = 5 successes in n = 8 trials (recall that a success is anindividual who does not exceed the deductible, and the probability of a success is p = 0.7. So theprobability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by(

)(0.7)5(1− 0.7)8−5 =

5!(8− 5)!(0.7)5(1− 0.7)8−5

=8!

5!3!(0.7)5(0.3)3

Dealing with the factorial part:

5!3!=

8× 7× 6× 5× 4× 3× 2× 1

(5× 4× 3× 2× 1)(3× 2× 1)=

8× 7× 6

3× 2× 1= 56

Using (0.7)5(0.3)3 ≈ 0.00454, the final probability is about 56× 0.00454 ≈ 0.254.

COMPUTING BINOMIAL PROBABILITIES

The first step in using the binomial model is to check that the model is appropriate. The secondstep is to identify n, p, and k. As the last stage use software or the formulas to determine theprobability, then interpret the results.

If you must do calculations by hand, it’s often useful to cancel out as many terms as possiblein the top and bottom of the binomial coefficient.

GUIDED PRACTICE 4.31

If we randomly sampled 40 case files from the insurance agency discussed earlier, how many of thecases would you expect to not have exceeded the deductible in a given year? What is the standarddeviation of the number that would not have exceeded the deductible?22

GUIDED PRACTICE 4.32

The probability that a random smoker will develop a severe lung condition in his or her lifetime isabout 0.3. If you have 4 friends who smoke, are the conditions for the binomial model satisfied?23

22We are asked to determine the expected number (the mean) and the standard deviation, both of which can be

directly computed from the formulas: µ = np = 40× 0.7 = 28 and σ =√np(1− p) =

√40× 0.7× 0.3 = 2.9. Because

very roughly 95% of observations fall within 2 standard deviations of the mean (see Section 2.1.4), we would probablyobserve at least 22 but fewer than 34 individuals in our sample who would not exceed the deductible.

23One possible answer: if the friends know each other, then the independence assumption is probably not satisfied.For example, acquaintances may have similar smoking habits, or those friends might make a pact to quit together.

152 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

GUIDED PRACTICE 4.33

Suppose these four friends do not know each other and we can treat them as if they were a randomsample from the population. Is the binomial model appropriate? What is the probability that24

(a) None of them will develop a severe lung condition?

(b) One will develop a severe lung condition?

GUIDED PRACTICE 4.34

What is the probability that at least 2 of your 4 smoking friends will develop a severe lung conditionin their lifetimes?25

GUIDED PRACTICE 4.35

Suppose you have 7 friends who are smokers and they can be treated as a random sample of smok-ers.26

(a) How many would you expect to develop a severe lung condition, i.e. what is the mean?

(b) What is the probability that at most 2 of your 7 friends will develop a severe lung condition.

Next we consider the first term in the binomial probability, n choose k under some specialscenarios.

GUIDED PRACTICE 4.36

Why is it true that(n0

)= 1 and

(nn

)= 1 for any number n?27

GUIDED PRACTICE 4.37

How many ways can you arrange one success and n−1 failures in n trials? How many ways can youarrange n− 1 successes and one failure in n trials?28

24To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we cantreat the friends as a random sample, they are independent. (ii) We have a fixed number of trials (n = 4). (iii) Eachoutcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like arandom sample (p = 0.3 if we say a “success” is someone getting a lung condition, a morbid choice). Compute parts (a)

and (b) using the binomial formula: P (0) =(40

)(0.3)0(0.7)4 = 1× 1× 0.74 = 0.2401, P (1) =

(41

)(0.3)1(0.7)3 = 0.4116.

Note: 0! = 1. Part (c) can be computed as the sum of parts (a) and (b): P (0) + P (1) = 0.2401 + 0.4116 = 0.6517.That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lungcondition.

25The complement (no more than one will develop a severe lung condition) as computed in Guided Practice 4.33as 0.6517, so we compute one minus this value: 0.3483.

26(a) µ = 0.3×7 = 2.1. (b) P (0, 1, or 2 develop severe lung condition) = P (k = 0)+P (k = 1)+P (k = 2) = 0.6471.27Frame these expressions into words. How many different ways are there to arrange 0 successes and n failures in

n trials? (1 way.) How many different ways are there to arrange n successes and 0 failures in n trials? (1 way.)28One success and n− 1 failures: there are exactly n unique places we can put the success, so there are n ways to

arrange one success and n− 1 failures. A similar argument is used for the second question. Mathematically, we showthese results by verifying the following two equations:(n

)= n,

( n

n− 1

)= n

4.3. BINOMIAL DISTRIBUTION 153

4.3.2 Normal approximation to the binomial distribution

The binomial formula is cumbersome when the sample size (n) is large, particularly when weconsider a range of observations. In some cases we may use the normal distribution as an easier andfaster way to estimate binomial probabilities.

EXAMPLE 4.38

Approximately 15% of the US population smokes cigarettes. A local government believed theircommunity had a lower smoker rate and commissioned a survey of 400 randomly selected individuals.The survey found that only 42 of the 400 participants smoke cigarettes. If the true proportion ofsmokers in the community was really 15%, what is the probability of observing 42 or fewer smokersin a sample of 400 people?

We leave the usual verification that the four conditions for the binomial model are valid as anexercise.

The question posed is equivalent to asking, what is the probability of observing k = 0, 1, 2, …, or42 smokers in a sample of n = 400 when p = 0.15? We can compute these 43 different probabilitiesand add them together to find the answer:

P (k = 0 or k = 1 or · · · or k = 42)

= P (k = 0) + P (k = 1) + · · ·+ P (k = 42)

= 0.0054

If the true proportion of smokers in the community is p = 0.15, then the probability of observing 42or fewer smokers in a sample of n = 400 is 0.0054.

The computations in Example 4.38 are tedious and long. In general, we should avoid suchwork if an alternative method exists that is faster, easier, and still accurate. Recall that calculatingprobabilities of a range of values is much easier in the normal model. We might wonder, is itreasonable to use the normal model in place of the binomial distribution? Surprisingly, yes, ifcertain conditions are met.

GUIDED PRACTICE 4.39

Here we consider the binomial model when the probability of a success is p = 0.10. Figure 4.9 showsfour hollow histograms for simulated samples from the binomial distribution using four differentsample sizes: n = 10, 30, 100, 300. What happens to the shape of the distributions as the samplesize increases? What distribution does the last hollow histogram resemble?29

NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION

The binomial distribution with probability of success p is nearly normal when the sample sizen is sufficiently large that np and n(1 − p) are both at least 10. The approximate normaldistribution has parameters corresponding to the mean and standard deviation of the binomialdistribution:

µ = np σ =√np(1− p)

The normal approximation may be used when computing the range of many possible successes.For instance, we may apply the normal distribution to the setting of Example 4.38.

29The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normaldistribution in last hollow histogram.

154 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

n = 10

0 2 4 6

n = 30

0 2 4 6 8 10

n = 100

0 5 10 15 20

n = 300

10 20 30 40 50

Figure 4.9: Hollow histograms of samples from the binomial model when p = 0.10.The sample sizes for the four plots are n = 10, 30, 100, and 300, respectively.

EXAMPLE 4.40

How can we use the normal approximation to estimate the probability of observing 42 or fewersmokers in a sample of 400, if the true proportion of smokers is p = 0.15?

Showing that the binomial model is reasonable was a suggested exercise in Example 4.38. We alsoverify that both np and n(1− p) are at least 10:

np = 400× 0.15 = 60 n(1− p) = 400× 0.85 = 340

With these conditions checked, we may use the normal approximation in place of the binomialdistribution using the mean and standard deviation from the binomial model:

µ = np = 60 σ =√np(1− p) = 7.14

We want to find the probability of observing 42 or fewer smokers using this model.

GUIDED PRACTICE 4.41

Use the normal model N(µ = 60, σ = 7.14) to estimate the probability of observing 42 or fewersmokers. Your answer should be approximately equal to the solution of Example 4.38: 0.0054. 30

30Compute the Z-score first: Z = 42−607.14

= −2.52. The corresponding left tail area is 0.0059.

4.3. BINOMIAL DISTRIBUTION 155

4.3.3 The normal approximation breaks down on small intervals

The normal approximation to the binomial distribution tends to perform poorly when estimat-ing the probability of a small range of counts, even when the conditions are met.

Suppose we wanted to compute the probability of observing 49, 50, or 51 smokers in 400 whenp = 0.15. With such a large sample, we might be tempted to apply the normal approximationand use the range 49 to 51. However, we would find that the binomial solution and the normalapproximation notably differ:

Binomial: 0.0649 Normal: 0.0421

We can identify the cause of this discrepancy using Figure 4.10, which shows the areas representingthe binomial probability (outlined) and normal approximation (shaded). Notice that the width ofthe area under the normal distribution is 0.5 units too slim on both sides of the interval.

40 50 60 70 80

Figure 4.10: A normal curve with the area between 49 and 51 shaded. The outlinedarea represents the exact binomial probability.

IMPROVING THE NORMAL APPROXIMATION FOR THE BINOMIAL DISTRIBUTION

The normal approximation to the binomial distribution for intervals of values is usually improvedif cutoff values are modified slightly. The cutoff values for the lower end of a shaded region shouldbe reduced by 0.5, and the cutoff value for the upper end should be increased by 0.5.

The tip to add extra area when applying the normal approximation is most often useful whenexamining a range of observations. In the example above, the revised normal distribution estimateis 0.0633, much closer to the exact value of 0.0649. While it is possible to also apply this correctionwhen computing a tail area, the benefit of the modification usually disappears since the total intervalis typically quite wide.

156 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Exercises

4.17 Underage drinking, Part I. Data collected by the Substance Abuse and Mental Health ServicesAdministration (SAMSHA) suggests that 69.7% of 18-20 year olds consumed alcoholic beverages in anygiven year.31

(a) Suppose a random sample of ten 18-20 year olds is taken. Is the use of the binomial distributionappropriate for calculating the probability that exactly six consumed alcoholic beverages? Explain.

(b) Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20 year olds consumed analcoholic drink.

(d) What is the probability that at most 2 out of 5 randomly sampled 18-20 year olds have consumedalcoholic beverages?

(e) What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumedalcoholic beverages?

4.18 Chicken pox, Part I. The National Vaccine Information Center estimates that 90% of Americans havehad chickenpox by the time they reach adulthood.32

(a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distributionappropriate for calculating the probability that exactly 97 out of 100 randomly sampled Americanadults had chickenpox during childhood? Explain.

(b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpoxduring childhood.

(c) What is the probability that exactly 3 out of a new sample of 100 American adults have not hadchickenpox in their childhood?

(d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?

(e) What is the probability that at most 3 out of 10 randomly sampled American adults have not hadchickenpox?

4.19 Underage drinking, Part II. We learned in Exercise 4.17 that about 70% of 18-20 year olds consumedalcoholic beverages in any given year. We now consider a random sample of fifty 18-20 year olds.

(a) How many people would you expect to have consumed alcoholic beverages? And with what standarddeviation?

(b) Would you be surprised if there were 45 or more people who have consumed alcoholic beverages?

(c) What is the probability that 45 or more people in this sample have consumed alcoholic beverages? Howdoes this probability relate to your answer to part (b)?

4.20 Chickenpox, Part II. We learned in Exercise 4.18 that about 90% of American adults had chickenpoxbefore adulthood. We now consider a random sample of 120 American adults.

(a) How many people in this sample would you expect to have had chickenpox in their childhood? Andwith what standard deviation?

(b) Would you be surprised if there were 105 people who have had chickenpox in their childhood?

(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?How does this probability relate to your answer to part (b)?

4.21 Game of dreidel. A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, andshin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose youspin a dreidel three times. Calculate the probability of getting

(a) at least one nun?

(b) exactly 2 nuns?

(d) at most 2 gimels?

Photo by Staccabees, cropped

(http://flic.kr/p/7gLZTf)

CC BY 2.0 license

31SAMHSA, Office of Applied Studies, National Survey on Drug Use and Health, 2007 and 2008.32National Vaccine Information Center, Chickenpox, The Disease & The Vaccine Fact Sheet.

4.3. BINOMIAL DISTRIBUTION 157

4.22 Arachnophobia. A Gallup Poll found that 7% of teenagers (ages 13 to 17) suffer from arachnophobiaand are extremely afraid of spiders. At a summer camp there are 10 teenagers sleeping in each tent. Assumethat these 10 teenagers are independent of each other.33

(a) Calculate the probability that at least one of them suffers from arachnophobia.

(b) Calculate the probability that exactly 2 of them suffer from arachnophobia.

(d) If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, doesit seem reasonable for him to randomly assign teenagers to tents?

4.23 Eye color, Part II. Exercise 4.13 introduces a husband and wife with brown eyes who have 0.75probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and0.125 probability of having children with green eyes.

(a) What is the probability that their first child will have green eyes and the second will not?

(b) What is the probability that exactly one of their two children will have green eyes?

(d) If they have six children, what is the probability that at least one will have green eyes?

(e) What is the probability that the first green eyed child will be the 4th child?

(f) Would it be considered unusual if only 2 out of their 6 children had brown eyes?

4.24 Sickle cell anemia. Sickle cell anemia is a genetic blood disorder where red blood cells lose theirflexibility and assume an abnormal, rigid, “sickle” shape, which results in a risk of various complications. Ifboth parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance ofbeing a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who arecarriers of the disease have 3 children, what is the probability that

(a) two will have the disease?

(b) none will have the disease?

(d) the first child with the disease will the be 3rd child?

4.25 Exploring permutations. The formula for the number of ways to arrange n objects is n! = n× (n−1)× · · · × 2× 1. This exercise walks you through the derivation of this formula for a couple of special cases.

A small company has five employees: Anna, Ben, Carl, Damian, and Eddy. There are five parkingspots in a row at the company, none of which are assigned, and each day the employees pull into a randomparking spot. That is, all possible orderings of the cars in the row of spots are equally likely.

(a) On a given day, what is the probability that the employees park in alphabetical order?

(b) If the alphabetical order has an equal chance of occurring relative to all other possible orderings, howmany ways must there be to arrange the five cars?

4.26 Male children. While it is often assumed that the probabilities of having a boy or a girl are the same,the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 kids.

(a) Use the binomial model to calculate the probability that two of them will be boys.

(b) Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate thesame probability from part (a) but using the addition rule for disjoint outcomes. Confirm that youranswers from parts (a) and (b) match.

(c) If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, brieflydescribe why the approach from part (b) would be more tedious than the approach from part (a).

33Gallup Poll, What Frightens America’s Youth?, March 29, 2005.

158 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

4.4 Negative binomial distribution

The geometric distribution describes the probability of observing the first success on the nth

trial. The negative binomial distribution is more general: it describes the probability of observ-ing the kth success on the nth trial.

EXAMPLE 4.42

Each day a high school football coach tells his star kicker, Brian, that he can go home after hesuccessfully kicks four 35 yard field goals. Suppose we say each kick has a probability p of beingsuccessful. If p is small – e.g. close to 0.1 – would we expect Brian to need many attempts beforehe successfully kicks his fourth field goal?

We are waiting for the fourth success (k = 4). If the probability of a success (p) is small, then thenumber of attempts (n) will probably be large. This means that Brian is more likely to need manyattempts before he gets k = 4 successes. To put this another way, the probability of n being smallis low.

To identify a negative binomial case, we check 4 conditions. The first three are common to thebinomial distribution.

IS IT NEGATIVE BINOMIAL? FOUR CONDITIONS TO CHECK

(1) The trials are independent.(2) Each trial outcome can be classified as a success or failure.(3) The probability of a success (p) is the same for each trial.(4) The last trial must be a success.

GUIDED PRACTICE 4.43

Suppose Brian is very diligent in his attempts and he makes each 35 yard field goal with probabilityp = 0.8. Take a guess at how many attempts he would need before making his fourth kick.34

EXAMPLE 4.44

In yesterday’s practice, it took Brian only 6 tries to get his fourth field goal. Write out each of thepossible sequence of kicks.

Because it took Brian six tries to get the fourth success, we know the last kick must have been asuccess. That leaves three successful kicks and two unsuccessful kicks (we label these as failures)that make up the first five attempts. There are ten possible sequences of these first five kicks, whichare shown in Figure 4.11. If Brian achieved his fourth success (k = 4) on his sixth attempt (n = 6),then his order of successes and failures must be one of these ten possible sequences.

GUIDED PRACTICE 4.45

Each sequence in Figure 4.11 has exactly two failures and four successes with the last attempt alwaysbeing a success. If the probability of a success is p = 0.8, find the probability of the first sequence.35

34One possible answer: since he is likely to make each field goal attempt, it will take him at least 4 attempts butprobably not more than 6 or 7.

35The first sequence: 0.2× 0.2× 0.8× 0.8× 0.8× 0.8 = 0.0164.

4.4. NEGATIVE BINOMIAL DISTRIBUTION 159

Kick Attempt1 2 3 4 5 6

1 F F1

2 F1

S F2

3 F1

S F3

4 F1

S F4

S F F2

S F2

S F3

S F2

S F4

S F F3

S F3

S F4

101

S F F4

Figure 4.11: The ten possible sequences when the fourth successful kick is on thesixth attempt.

If the probability Brian kicks a 35 yard field goal is p = 0.8, what is the probability it takesBrian exactly six tries to get his fourth successful kick? We can write this as

P (it takes Brian six tries to make four field goals)

= P (Brian makes three of his first five field goals, and he makes the sixth one)

= P (1st sequence OR 2nd sequence OR … OR 10th sequence)

where the sequences are from Figure 4.11. We can break down this last probability into the sum often disjoint possibilities:

P (1st sequence OR 2nd sequence OR … OR 10th sequence)

= P (1st sequence) + P (2nd sequence) + · · ·+ P (10th sequence)

The probability of the first sequence was identified in Guided Practice 4.45 as 0.0164, and each of theother sequences have the same probability. Since each of the ten sequence has the same probability,the total probability is ten times that of any individual sequence.

The way to compute this negative binomial probability is similar to how the binomial problemswere solved in Section 4.3. The probability is broken into two pieces:

P (it takes Brian six tries to make four field goals)

= [Number of possible sequences]× P (Single sequence)

Each part is examined separately, then we multiply to get the final result.We first identify the probability of a single sequence. One particular case is to first observe all

the failures (n− k of them) followed by the k successes:

P (Single sequence)

= P (n− k failures and then k successes)

= (1− p)n−kpk

160 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

We must also identify the number of sequences for the general case. Above, ten sequences wereidentified where the fourth success came on the sixth attempt. These sequences were identified byfixing the last observation as a success and looking for all the ways to arrange the other observations.In other words, how many ways could we arrange k− 1 successes in n− 1 trials? This can be foundusing the n choose k coefficient but for n− 1 and k − 1 instead:(

n− 1

k − 1

(n− 1)!

(k − 1)! ((n− 1)− (k − 1))!=

(n− 1)!

(k − 1)! (n− k)!

This is the number of different ways we can order k − 1 successes and n− k failures in n− 1 trials.If the factorial notation (the exclamation point) is unfamiliar, see page 150.

NEGATIVE BINOMIAL DISTRIBUTION

The negative binomial distribution describes the probability of observing the kth success on thenth trial, where all trials are independent:

P (the kth success on the nth trial) =

(n− 1

k − 1

)pk(1− p)n−k

The value p represents the probability that an individual trial is a success.

EXAMPLE 4.46

Show using the formula for the negative binomial distribution that the probability Brian kicks hisfourth successful field goal on the sixth attempt is 0.164.

The probability of a single success is p = 0.8, the number of successes is k = 4, and the number ofnecessary attempts under this scenario is n = 6.(

n− 1

k − 1

)pk(1− p)n−k =

3!2!(0.8)4(0.2)2 = 10× 0.0164 = 0.164

GUIDED PRACTICE 4.47

The negative binomial distribution requires that each kick attempt by Brian is independent. Do youthink it is reasonable to suggest that each of Brian’s kick attempts are independent?36

GUIDED PRACTICE 4.48

Assume Brian’s kick attempts are independent. What is the probability that Brian will kick hisfourth field goal within 5 attempts?37

36Answers may vary. We cannot conclusively say they are or are not independent. However, many statisticalreviews of athletic performance suggests such attempts are very nearly independent.

37If his fourth field goal (k = 4) is within five attempts, it either took him four or five tries (n = 4 or n = 5). Wehave p = 0.8 from earlier. Use the negative binomial distribution to compute the probability of n = 4 tries and n = 5tries, then add those probabilities together:

P (n = 4 OR n = 5) = P (n = 4) + P (n = 5)

=(4− 1

4− 1

)0.84 +

(5− 1

4− 1

)(0.8)4(1− 0.8) = 1× 0.41 + 4× 0.082 = 0.41 + 0.33 = 0.74

4.4. NEGATIVE BINOMIAL DISTRIBUTION 161

BINOMIAL VERSUS NEGATIVE BINOMIAL

In the binomial case, we typically have a fixed number of trials and instead consider the numberof successes. In the negative binomial case, we examine how many trials it takes to observe afixed number of successes and require that the last observation be a success.

GUIDED PRACTICE 4.49

On 70% of days, a hospital admits at least one heart attack patient. On 30% of the days, no heartattack patients are admitted. Identify each case below as a binomial or negative binomial case, andcompute the probability.38

(a) What is the probability the hospital will admit a heart attack patient on exactly three days thisweek?

(b) What is the probability the second day with a heart attack patient will be the fourth day of theweek?

38In each part, p = 0.7. (a) The number of days is fixed, so this is binomial. The parameters are k = 3 andn = 7: 0.097. (b) The last “success” (admitting a heart attack patient) is fixed to the last day, so we should applythe negative binomial distribution. The parameters are k = 2, n = 4: 0.132. (c) This problem is negative binomialwith k = 1 and n = 5: 0.006. Note that the negative binomial case when k = 1 is the same as using the geometricdistribution.

162 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Exercises

4.27 Rolling a die. Calculate the following probabilities and indicate which probability distribution modelis appropriate in each case. You roll a fair die 5 times. What is the probability of rolling

(a) the first 6 on the fifth roll?

(b) exactly three 6s?

4.28 Playing darts. Calculate the following probabilities and indicate which probability distribution modelis appropriate in each case. A very good darts player can hit the bull’s eye (red circle in the center of thedart board) 65% of the time. What is the probability that he

(a) hits the bullseye for the 10th time on the 15th try?

(b) hits the bullseye 10 times in 15 tries?

4.29 Sampling at school. For a sociology class project you are asked to conduct a survey on 20 studentsat your school. You decide to stand outside of your dorm’s cafeteria and conduct the survey on a randomsample of 20 students leaving the cafeteria after dinner one evening. Your dorm is comprised of 45% malesand 55% females.

(a) Which probability model is most appropriate for calculating the probability that the 4th person yousurvey is the 2nd female? Explain.

(b) Compute the probability from part (a).

{M,M,F, F}, {M,F,M,F}, {F,M,M,F}

One common feature among these scenarios is that the last trial is always female. In the first threetrials there are 2 males and 1 female. Use the binomial coefficient to confirm that there are 3 ways ofordering 2 males and 1 female.

(d) Use the findings presented in part (c) to explain why the formula for the coefficient for the negativebinomial is

(n−1k−1

)while the formula for the binomial coefficient is

(nk

4.30 Serving in volleyball. A not-so-skilled volleyball player has a 15% chance of making the serve, whichinvolves hitting the ball so it passes over the net on a trajectory such that it will land in the opposing team’scourt. Suppose that her serves are independent of each other.

(a) What is the probability that on the 10th try she will make her 3rd successful serve?

(b) Suppose she has made two successful serves in nine attempts. What is the probability that her 10th

serve will be successful?

(c) Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should bedifferent. Can you explain the reason for this discrepancy?

4.5. POISSON DISTRIBUTION 163

4.5 Poisson distribution

EXAMPLE 4.50

There are about 8 million individuals in New York City. How many individuals might we expect tobe hospitalized for acute myocardial infarction (AMI), i.e. a heart attack, each day? According tohistorical records, the average number is about 4.4 individuals. However, we would also like to knowthe approximate distribution of counts. What would a histogram of the number of AMI occurrenceseach day look like if we recorded the daily counts over an entire year?

A histogram of the number of occurrences of AMI on 365 days for NYC is shown in Figure 4.12.39

The sample mean (4.38) is similar to the historical average of 4.4. The sample standard deviationis about 2, and the histogram indicates that about 70% of the data fall between 2.4 and 6.4. Thedistribution’s shape is unimodal and skewed to the right.

0 5 10

AMI Events (by Day)

Fre

quen

Figure 4.12: A histogram of the number of occurrences of AMI on 365 separatedays in NYC.

The Poisson distribution is often useful for estimating the number of events in a largepopulation over a unit of time. For instance, consider each of the following events:

• having a heart attack,

• getting married, and

• getting struck by lightning.

The Poisson distribution helps us describe the number of such events that will occur in a day for afixed population if the individuals within the population are independent. The Poisson distributioncould also be used over another unit of time, such as an hour or a week.

The histogram in Figure 4.12 approximates a Poisson distribution with rate equal to 4.4. Therate for a Poisson distribution is the average number of occurrences in a mostly-fixed population perunit of time. In Example 4.50, the time unit is a day, the population is all New York City residents,and the historical rate is 4.4. The parameter in the Poisson distribution is the rate – or how manyevents we expect to observe – and it is typically denoted by λ (the Greek letter lambda) or µ. Usingthe rate, we can describe the probability of observing exactly k events in a single unit of time.

39These data are simulated. In practice, we should check for an association between successive days.

164 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

POISSON DISTRIBUTION

Suppose we are watching for events and the number of observed events follows a Poisson distri-bution with rate λ. Then

P (observe k events) =λke−λ

where k may take a value 0, 1, 2, and so on, and k! represents k-factorial, as described onpage 150. The letter e ≈ 2.718 is the base of the natural logarithm. The mean and standarddeviation of this distribution are λ and

√λ, respectively.

We will leave a rigorous set of conditions for the Poisson distribution to a later course. However,we offer a few simple guidelines that can be used for an initial evaluation of whether the Poissonmodel would be appropriate.

A random variable may follow a Poisson distribution if we are looking for the number of events,the population that generates such events is large, and the events occur independently of each other.

Even when events are not really independent – for instance, Saturdays and Sundays are es-pecially popular for weddings – a Poisson model may sometimes still be reasonable if we allow itto have a different rate for different times. In the wedding example, the rate would be modeled ashigher on weekends than on weekdays. The idea of modeling rates for a Poisson distribution againsta second variable such as the day of week forms the foundation of some more advanced methods thatfall in the realm of generalized linear models. In Chapters 8 and 9, we will discuss a foundationof linear models.

4.5. POISSON DISTRIBUTION 165

Exercises

4.31 Customers at a coffee shop. A coffee shop serves an average of 75 customers per hour during themorning rush.

(a) Which distribution have we studied that is most appropriate for calculating the probability of a givennumber of customers arriving within one hour during this time of day?

(b) What are the mean and the standard deviation of the number of customers this coffee shop serves inone hour during this time of day?

(c) Would it be considered unusually low if only 60 customers showed up to this coffee shop in one hourduring this time of day?

(d) Calculate the probability that this coffee shop serves 70 customers in one hour during this time of day.

4.32 Stenographer’s typos. A very skilled court stenographer makes one typographical error (typo) perhour on average.

(a) What probability distribution is most appropriate for calculating the probability of a given number oftypos this stenographer makes in an hour?

(b) What are the mean and the standard deviation of the number of typos this stenographer makes?

(d) Calculate the probability that this stenographer makes at most 2 typos in a given hour.

4.33 How many cars show up? For Monday through Thursday when there isn’t a holiday, the averagenumber of vehicles that visit a particular retailer between 2pm and 3pm each afternoon is 6.5, and thenumber of cars that show up on any given day follows a Poisson distribution.

(a) What is the probability that exactly 5 cars will show up next Monday?

(b) What is the probability that 0, 1, or 2 cars will show up next Monday between 2pm and 3pm?

(c) There is an average of 11.7 people who visit during those same hours from vehicles. Is it likely that thenumber of people visiting by car during this hour is also Poisson? Explain.

4.34 Lost baggage. Occasionally an airline will lose a bag. Suppose a small airline has found it canreasonably model the number of bags lost each weekday using a Poisson model with a mean of 2.2 bags.

(a) What is the probability that the airline will lose no bags next Monday?

(b) What is the probability that the airline will lose 0, 1, or 2 bags on next Monday?

(c) Suppose the airline expands over the course of the next 3 years, doubling the number of flights it makes,and the CEO asks you if it’s reasonable for them to continue using the Poisson model with a meanof 2.2. What is an appropriate recommendation? Explain.

166 CHAPTER 4. DISTRIBUTIONS OF RANDOM VARIABLES

Chapter exercises

4.35 Roulette winnings. In the game of roulette, a wheel is spun and you place bets on where it will stop.One popular bet is that it will stop on a red slot; such a bet has an 18/38 chance of winning. If it stops onred, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, eachtime with a $1 bet. Let Y represent the total amount won or lost. Write a probability model for Y.

4.36 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of4.78 miles/hour.40

(a) What percent of passenger vehicles travel slower than 80 miles/hour?

(b) What percent of passenger vehicles travel between 60 and 80 miles/hour?

(d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passengervehicles travel above the speed limit on this stretch of the I-5.

4.37 University admissions. Suppose a university announced that it admitted 2,500 students for thefollowing year’s freshman class. However, the university has dorm room spots for only 1,786 freshmanstudents. If there is a 70% chance that an admitted student will decide to accept the offer and attend thisuniversity, what is the approximate probability that the university will not have enough dormitory roomspots for the freshman class?

4.38 Speeding on the I-5, Part II. Exercise 4.36 states that the distribution of speeds of cars traveling onthe Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standarddeviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour.

(a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars passand none are speeding? Assume that the speeds of the cars are independent of each other.

(b) On average, how many cars would the highway patrol officer expect to watch until the first car that isspeeding? What is the standard deviation of the number of cars he would expect to watch?

4.39 Auto insurance premiums. Suppose a newspaper article states that the distribution of auto insurancepremiums for residents of California is approximately normal with a mean of $1,650. The article also statesthat 25% of California residents pay more than $1,800.

(a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normaldistribution?

(b) What is the mean insurance cost? What is the cutoff for the 75th percentile?

4.40 SAT scores. SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standarddeviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probabilitythis student’s score will be at least 1500? (The material covered in Section 3.2 on conditional probabilitywould be useful for this question.)

4.41 Married women. The American Community Survey estimates that 47.1% of women ages 15 yearsand over are married.41

(a) We randomly select three women between these ages. What is the probability that the third womanselected is the only one who is married?

(b) What is the probability that all three randomly selected women are married?

(c) On average, how many women would you expect to sample before selecting a married woman? What isthe standard deviation?

(d) If the proportion of married women was actually 30%, how many women would you expect to samplebefore selecting a married woman? What is the standard deviation?

(e) Based on your answers to parts (c) and (d), how does decreasing the probability of an event affect themean and standard deviation of the wait time until success?

40S. Johnson and D. Murray. “Empirical Analysis of Truck and Automobile Speeds on Rural Interstates: Impactof Posted Speed Limits”. In: Transportation Research Board 89th Annual Meeting. 2010.

41U.S. Census Bureau, 2010 American Community Survey, Marital Status.

4.5. POISSON DISTRIBUTION 167

4.42 Survey response rate. Pew Research reported that the typical response rate to their surveys is only9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500will agree to respond?42

4.43 Overweight baggage. Suppose weights of the checked baggage of airline passengers follow a nearlynormal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee forbaggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

4.44 Heights of 10 year olds, Part I. Heights of 10 year olds, regardless of gender, closely follow a normaldistribution with mean 55 inches and standard deviation 6 inches.

(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?

(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?

4.45 Buying books on Ebay. Suppose you’re considering buying your expensive chemistry textbook onEbay. Looking at past auctions suggests that the prices of this textbook follow an approximately normaldistribution with mean $89 and standard deviation $15.

(a) What is the probability that a randomly selected auction for this book closes at more than $100?

(b) Ebay allows you to set your maximum bid price so that if someone outbids you on an auction youcan automatically outbid them, up to the maximum bid price you set. If you are only bidding on oneauction, what are the advantages and disadvantages of setting a bid price too high or too low? What ifyou are bidding on multiple auctions?

(c) If you watched 10 auctions, roughly what percentile might you use for a maximum bid cutoff to besomewhat sure that you will win one of these ten auctions? Is it possible to find a cutoff point that willensure that you win an auction?

(d) If you are willing to track up to ten auctions closely, about what price might you use as your maximumbid price if you want to be somewhat sure that you will buy one of these ten books?

4.46 Heights of 10 year olds, Part II. Heights of 10 year olds, regardless of gender, closely follow a normaldistribution with mean 55 inches and standard deviation 6 inches.

(a) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percentof 10 year olds cannot go on this ride?

(b) Suppose there are four 10 year olds. What is the chance that at least two of them will be able to rideBatman the Ride?

(c) Suppose you work at the park to help them better understand their customers’ demographics, and youare counting people as they enter the park. What is the chance that the first 10 year old you see whocan ride Batman the Ride is the 3rd 10 year old who enters the park?

(d) What is the chance that the fifth 10 year old you see who can ride Batman the Ride is the 12th 10 yearold who enters the park?

4.47 Heights of 10 year olds, Part III. Heights of 10 year olds, regardless of gender, closely follow a normaldistribution with mean 55 inches and standard deviation 6 inches.

(a) What fraction of 10 year olds are taller than 76 inches?

(b) If there are 2,000 10 year olds entering Six Flags Magic Mountain in a single day, then compute theexpected number of 10 year olds who are at least 76 inches tall. (You may assume the heights of the10-year olds are independent.)

(c) Using the binomial distribution, compute the probability that 0 of the 2,000 10 year olds will be at least76 inches tall.

(d) The number of 10 year olds who enter Six Flags Magic Mountain and are at least 76 inches tall in agiven day follows a Poisson distribution with mean equal to the value found in part (b). Use the Poissondistribution to identify the probability no 10 year old will enter the park who is 76 inches or taller.

4.48 Multiple choice quiz. In a multiple choice quiz there are 5 questions and 4 choices for each question(a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What isthe probability that

(a) the first question she gets right is the 3rd question?

(b) she gets exactly 3 or exactly 4 questions right?

42Pew Research Center, Assessing the Representativeness of Public Opinion Surveys, May 15, 2012.

168

Chapter 5Foundations for inference

5.1 Point estimates and sampling variability

5.2 Confidence intervals for a proportion

5.3 Hypothesis testing for a proportion

169

Statistical inference is primarily concerned with understanding and

quantifying the uncertainty of parameter estimates. While the equa-

tions and details change depending on the setting, the foundations for

inference are the same throughout all of statistics.

We start with a familiar topic: the idea of using a sample proportion

to estimate a population proportion. Next, we create what’s called a

confidence interval, which is a range of plausible values where we may

find the true population value. Finally, we introduce the hypothesis

testing framework, which allows us to formally evaluate claims about

the population, such as whether a survey provides strong evidence that

a candidate has the support of a majority of the voting population.

For videos, slides, and other resources, please visit

www.openintro.org/os

170 CHAPTER 5. FOUNDATIONS FOR INFERENCE

5.1 Point estimates and sampling variability

Companies such as Pew Research frequently conduct polls as a way to understand the stateof public opinion or knowledge on many topics, including politics, scientific understanding, brandrecognition, and more. The ultimate goal in taking a poll is generally to use the responses to estimatethe opinion or knowledge of the broader population.

5.1.1 Point estimates and error

Suppose a poll suggested the US President’s approval rating is 45%. We would consider 45% tobe a point estimate of the approval rating we might see if we collected responses from the entirepopulation. This entire-population response proportion is generally referred to as the parameterof interest. When the parameter is a proportion, it is often denoted by p, and we often refer to thesample proportion as p̂ (pronounced p-hat1). Unless we collect responses from every individual inthe population, p remains unknown, and we use p̂ as our estimate of p. The difference we observefrom the poll versus the parameter is called the error in the estimate. Generally, the error consistsof two aspects: sampling error and bias.

Sampling error, sometimes called sampling uncertainty, describes how much an estimate willtend to vary from one sample to the next. For instance, the estimate from one sample might be 1%too low while in another it may be 3% too high. Much of statistics, including much of this book,is focused on understanding and quantifying sampling error, and we will find it useful to consider asample’s size to help us quantify this error; the sample size is often represented by the letter n.

Bias describes a systematic tendency to over- or under-estimate the true population value.For example, if we were taking a student poll asking about support for a new college stadium, we’dprobably get a biased estimate of the stadium’s level of student support by wording the question as,Do you support your school by supporting funding for the new stadium? We try to minimize biasthrough thoughtful data collection procedures, which were discussed in Chapter 1 and are the topicof many other books.

5.1.2 Understanding the variability of a point estimate

Suppose the proportion of American adults who support the expansion of solar energy is p =0.88, which is our parameter of interest.2 If we were to take a poll of 1000 American adults on thistopic, the estimate would not be perfect, but how close might we expect the sample proportion inthe poll would be to 88%? We want to understand, how does the sample proportion p̂ behave whenthe true population proportion is 0.88.3 Let’s find out! We can simulate responses we would getfrom a simple random sample of 1000 American adults, which is only possible because we know theactual support for expanding solar energy is 0.88. Here’s how we might go about constructing sucha simulation:

1. There were about 250 million American adults in 2018. On 250 million pieces of paper, write“support” on 88% of them and “not” on the other 12%.

2. Mix up the pieces of paper and pull out 1000 pieces to represent our sample of 1000 Americanadults.

3. Compute the fraction of the sample that say “support”.

Any volunteers to conduct this simulation? Probably not. Running this simulation with 250 millionpieces of paper would be time-consuming and very costly, but we can simulate it using computer

1Not to be confused with phat, the slang term used for something cool, like this book.2We haven’t actually conducted a census to measure this value perfectly. However, a very large sample has

suggested the actual level of support is about 88%.388% written as a proportion would be 0.88. It is common to switch between proportion and percent. However,

formulas presented in this book always refer to the proportion, not the percent.

5.1. POINT ESTIMATES AND SAMPLING VARIABILITY 171

code; we’ve written a short program in Figure 5.1 in case you are curious what the computer codelooks like. In this simulation, the sample gave a point estimate of p̂1 = 0.894. We know thepopulation proportion for the simulation was p = 0.88, so we know the estimate had an error of0.894− 0.88 = +0.014.

# 1. Create a set of 250 million entries, where 88% of them are "support"

# and 12% are "not".

pop size <- 250000000

possible entries <- c(rep("support", 0.88 * pop size), rep("not", 0.12 * pop size))

# 2. Sample 1000 entries without replacement.

sampled entries <- sample(possible entries, size = 1000)

# 3. Compute p-hat: count the number that are "support", then divide by

# the sample size.

sum(sampled entries == "support") / 1000

Figure 5.1: For those curious, this is code for a single p̂ simulation using the statis-tical software called R. Each line that starts with # is a code comment, which isused to describe in regular language what the code is doing. We’ve provided soft-ware labs in R at openintro.org/stat/labs for anyone interested in learning more.

One simulation isn’t enough to get a great sense of the distribution of estimates we might expectin the simulation, so we should run more simulations. In a second simulation, we get p̂2 = 0.885,which has an error of +0.005. In another, p̂3 = 0.878 for an error of -0.002. And in another, anestimate of p̂4 = 0.859 with an error of -0.021. With the help of a computer, we’ve run the simulation10,000 times and created a histogram of the results from all 10,000 simulations in Figure 5.2. Thisdistribution of sample proportions is called a sampling distribution. We can characterize thissampling distribution as follows:

Center. The center of the distribution is x̄p̂ = 0.880, which is the same as the parameter. Noticethat the simulation mimicked a simple random sample of the population, which is a straight-forward sampling strategy that helps avoid sampling bias.

Spread. The standard deviation of the distribution is sp̂ = 0.010. When we’re talking about asampling distribution or the variability of a point estimate, we typically use the term standarderror rather than standard deviation, and the notation SEp̂ is used for the standard errorassociated with the sample proportion.

Shape. The distribution is symmetric and bell-shaped, and it resembles a normal distribution.

These findings are encouraging! When the population proportion is p = 0.88 and the sample size isn = 1000, the sample proportion p̂ tends to give a pretty good estimate of the population proportion.We also have the interesting observation that the histogram resembles a normal distribution.

SAMPLING DISTRIBUTIONS ARE NEVER OBSERVED, BUT WE KEEP THEM IN MIND

In real-world applications, we never actually observe the sampling distribution, yet it is useful toalways think of a point estimate as coming from such a hypothetical distribution. Understandingthe sampling distribution will help us characterize and make sense of the point estimates thatwe do observe.

EXAMPLE 5.1

If we used a much smaller sample size of n = 50, would you guess that the standard error for p̂would be larger or smaller than when we used n = 1000?

Intuitively, it seems like more data is better than less data, and generally that is correct! The typicalerror when p = 0.88 and n = 50 would be larger than the error we would expect when n = 1000.

Example 5.1 highlights an important property we will see again and again: a bigger sampletends to provide a more precise point estimate than a smaller sample.

172 CHAPTER 5. FOUNDATIONS FOR INFERENCE

Sample Proportions

0.84 0.86 0.88 0.90 0.920

250

500

750

Fre

quen

Figure 5.2: A histogram of 10,000 sample proportions, where each sample is takenfrom a population where the population proportion is 0.88 and the sample size isn = 1000.

5.1.3 Central Limit Theorem

The distribution in Figure 5.2 looks an awful lot like a normal distribution. That is no anomaly;it is the result of a general principle called the Central Limit Theorem.

CENTRAL LIMIT THEOREM AND THE SUCCESS-FAILURE CONDITION

When observations are independent and the sample size is sufficiently large, the sample propor-tion p̂ will tend to follow a normal distribution with the following mean and standard error:

µp̂ = p SEp̂ =

√p(1− p)

In order for the Central Limit Theorem to hold, the sample size is typically considered sufficientlylarge when np ≥ 10 and n(1− p) ≥ 10, which is called the success-failure condition.

The Central Limit Theorem is incredibly important, and it provides a foundation for muchof statistics. As we begin applying the Central Limit Theorem, be mindful of the two technicalconditions: the observations must be independent, and the sample size must be sufficiently largesuch that np ≥ 10 and n(1− p) ≥ 10.

EXAMPLE 5.2

Earlier we estimated the mean and standard error of p̂ using simulated data when p = 0.88 andn = 1000. Confirm that the Central Limit Theorem applies and the sampling distribution is ap-proximately normal.

Independence. There are n = 1000 observations for each sample proportion p̂, and each of thoseobservations are independent draws. The most common way for observations to be consideredindependent is if they are from a simple random sample.

Success-failure condition. We can confirm the sample size is sufficiently large by checking thesuccess-failure condition and confirming the two calculated values are greater than 10:

np = 1000× 0.88 = 880 ≥ 10 n(1− p) = 1000× (1− 0.88) = 120 ≥ 10

The independence and success-failure conditions are both satisfied, so the Central Limit Theoremapplies, and it’s reasonable to model p̂ using a normal distribution.

5.1. POINT ESTIMATES AND SAMPLING VARIABILITY 173

HOW TO VERIFY SAMPLE OBSERVATIONS ARE INDEPENDENT

Subjects in an experiment are considered independent if they undergo random assignment tothe treatment groups.

If the observations are from a simple random sample, then they are independent.

If a sample is from a seemingly random process, e.g. an occasional error on an assembly line,checking independence is more difficult. In this case, use your best judgement.

An additional condition that is sometimes added for samples from a population is that theyare no larger than 10% of the population. When the sample exceeds 10% of the population size, themethods we discuss tend to overestimate the sampling error slightly versus what we would get usingmore advanced methods.4 This is very rarely an issue, and when it is an issue, our methods tend tobe conservative, so we consider this additional check as optional.

EXAMPLE 5.3

Compute the theoretical mean and standard error of p̂ when p = 0.88 and n = 1000, according tothe Central Limit Theorem.

The mean of the p̂’s is simply the population proportion: µp̂ = 0.88.

The calculation of the standard error of p̂ uses the following formula:

SEp̂ =

√p(1− p)

√0.88(1− 0.88)

1000= 0.010

EXAMPLE 5.4

Estimate how frequently the sample proportion p̂ should be within 0.02 (2%) of the populationvalue, p = 0.88. Based on Examples 5.2 and 5.3, we know that the distribution is approximatelyN(µp̂ = 0.88, SEp̂ = 0.010).

After so much practice in Section 4.1, this normal distribution example will hopefully feel familiar!We would like to understand the fraction of p̂’s between 0.86 and 0.90:

0.86 0.88 0.90

With µp̂ = 0.88 and SEp̂ = 0.010, we can compute the Z-score for both the left and right cutoffs:

Z0.86 =0.86− 0.88

0.010= −2 Z0.90 =

0.90− 0.88

0.010= 2

We can use either statistical software, a graphing calculator, or a table to find the areas to the tails,and in any case we will find that they are each 0.0228. The total tail areas are 2× 0.0228 = 0.0456,which leaves the shaded area of 0.9544. That is, about 95.44% of the sampling distribution inFigure 5.2 is within ±0.02 of the population proportion, p = 0.88.

4For example, we could use what’s called the finite population correction factor: if the sample is of size n

and the population size is N , then we can multiple the typical standard error formula by√N−nN−1

to obtain a smaller,

more precise estimate of the actual standard error. When n < 0.1×N , this correction factor is relatively small.

174 CHAPTER 5. FOUNDATIONS FOR INFERENCE

GUIDED PRACTICE 5.5

In Example 5.1 we discussed how a smaller sample would tend to produce a less reliable estimate.

Explain how this intuition is reflected in the formula for SEp̂ =√

p(1−p)n .5

5.1.4 Applying the Central Limit Theorem to a real-world setting

We do not actually know the population proportion unless we conduct an expensive poll of allindividuals in the population. Our earlier value of p = 0.88 was based on a Pew Research conducteda poll of 1000 American adults that found p̂ = 0.887 of them favored expanding solar energy. Theresearchers might have wondered: does the sample proportion from the poll approximately follow anormal distribution? We can check the conditions from the Central Limit Theorem:

Independence. The poll is a simple random sample of American adults, which means that theobservations are independent.

Success-failure condition. To check this condition, we need the population proportion, p, tocheck if both np and n(1− p) are greater than 10. However, we do not actually know p, whichis exactly why the pollsters would take a sample! In cases like these, we often use p̂ as ournext best way to check the success-failure condition:

np̂ = 1000× 0.887 = 887 n(1− p̂) = 1000× (1− 0.887) = 113

The sample proportion p̂ acts as a reasonable substitute for p during this check, and each valuein this case is well above the minimum of 10.

This substitution approximation of using p̂ in place of p is also useful when computing thestandard error of the sample proportion:

SEp̂ =

√p(1− p)

n≈√p̂(1− p̂)

√0.887(1− 0.887)

1000= 0.010

This substitution technique is sometimes referred to as the “plug-in principle”. In this case, SEp̂didn’t change enough to be detected using only 3 decimal places versus when we completed thecalculation with 0.88 earlier. The computed standard error tends to be reasonably stable even whenobserving slightly different proportions in one sample or another.

5Since the sample size n is in the denominator (on the bottom) of the fraction, a bigger sample size means theentire expression when calculated will tend to be smaller. That is, a larger sample size would correspond to a smallerstandard error.

5.1. POINT ESTIMATES AND SAMPLING VARIABILITY 175

5.1.5 More details regarding the Central Limit Theorem

We’ve applied the Central Limit Theorem in numerous examples so far this chapter:

When observations are independent and the sample size is sufficiently large, the distri-bution of p̂ resembles a normal distribution with

µp̂ = p SEp̂ =

√p(1− p)

The sample size is considered sufficiently large when np ≥ 10 and n(1− p) ≥ 10.

In this section, we’ll explore the success-failure condition and seek to better understand the CentralLimit Theorem.

An interesting question to answer is, what happens when np < 10 or n(1− p) < 10? As we didin Section 5.1.2, we can simulate drawing samples of different sizes where, say, the true proportionis p = 0.25. Here’s a sample of size 10:

no, no, yes, yes, no, no, no, no, no, no

In this sample, we observe a sample proportion of yeses of p̂ = 210 = 0.2. We can simulate many such

proportions to understand the sampling distribution of p̂ when n = 10 and p = 0.25, which we’veplotted in Figure 5.3 alongside a normal distribution with the same mean and variability. Thesedistributions have a number of important differences.

Sample Proportions

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

1000

2000

Fre

quen

−0.2 0.0 0.2 0.4 0.6

Figure 5.3: Left: simulations of p̂ when the sample size is n = 10 and the populationproportion is p = 0.25. Right: a normal distribution with the same mean (0.25)and standard deviation (0.137).

Unimodal? Smooth? Symmetric?Normal: N(0.25, 0.14) Yes Yes Yesn = 10, p = 0.25 Yes No No

Notice that the success-failure condition was not satisfied when n = 10 and p = 0.25:

np = 10× 0.25 = 2.5 n(1− p) = 10× 0.75 = 7.5

This single sampling distribution does not show that the success-failure condition is the perfectguideline, but we have found that the guideline did correctly identify that a normal distributionmight not be appropriate.

We can complete several additional simulations, shown in Figures 5.4 and 5.5, and we can seesome trends:

1. When either np or n(1− p) is small, the distribution is more discrete, i.e. not continuous.

2. When np or n(1− p) is smaller than 10, the skew in the distribution is more noteworthy.

3. The larger both np and n(1−p), the more normal the distribution. This may be a little harderto see for the larger sample size in these plots as the variability also becomes much smaller.

4. When np and n(1 − p) are both very large, the distribution’s discreteness is hardly evident,and the distribution looks much more like a normal distribution.

176 CHAPTER 5. FOUNDATIONS FOR INFERENCE

n = 10 n = 25

p = 0.1

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

p = 0.2

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

p = 0.5

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

p = 0.8

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

p = 0.9

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

Figure 5.4: Sampling distributions for several scenarios of p and n.Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90.Columns: n = 10 and n = 25.

5.1. POINT ESTIMATES AND SAMPLING VARIABILITY 177

n = 50 n = 100 n = 250

0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0

Figure 5.5: Sampling distributions for several scenarios of p and n.Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90.Columns: n = 50, n = 100, and n = 250.

178 CHAPTER 5. FOUNDATIONS FOR INFERENCE

So far we’ve only focused on the skew and discreteness of the distributions. We haven’t con-sidered how the mean and standard error of the distributions change. Take a moment to look backat the graphs, and pay attention to three things:

1. The centers of the distribution are always at the population proportion, p, that was used togenerate the simulation. Because the sampling distribution of p̂ is always centered at thepopulation parameter p, it means the sample proportion p̂ is unbiased when the data areindependent and drawn from such a population.

2. For a particular population proportion p, the variability in the sampling distribution decreasesas the sample size n becomes larger. This will likely align with your intuition: an estimatebased on a larger sample size will tend to be more accurate.

3. For a particular sample size, the variability will be largest when p = 0.5. The differences maybe a little subtle, so take a close look. This reflects the role of the proportion p in the standard

error formula: SE =√

p(1−p)n . The standard error is largest when p = 0.5.

At no point will the distribution of p̂ look perfectly normal, since p̂ will always be take discretevalues (x/n). It is always a matter of degree, and we will use the standard success-failure conditionwith minimums of 10 for np and n(1− p) as our guideline within this book.

5.1.6 Extending the framework for other statistics

The strategy of using a sample statistic to estimate a parameter is quite common, and it’sa strategy that we can apply to other statistics besides a proportion. For instance, if we want toestimate the average salary for graduates from a particular college, we could survey a random sampleof recent graduates; in that example, we’d be using a sample mean x̄ to estimate the populationmean µ for all graduates. As another example, if we want to estimate the difference in productprices for two websites, we might take a random sample of products available on both sites, checkthe prices on each, and use then compute the average difference; this strategy certainly would giveus some idea of the actual difference through a point estimate.

While this chapter emphases a single proportion context, we’ll encounter many different con-texts throughout this book where these methods will be applied. The principles and general ideasare the same, even if the details change a little. We’ve also sprinkled some other contexts into theexercises to help you start thinking about how the ideas generalize.

5.1. POINT ESTIMATES AND SAMPLING VARIABILITY 179

Exercises

5.1 Identify the parameter, Part I. For each of the following situations, state whether the parameter ofinterest is a mean or a proportion. It may be helpful to examine whether individual responses are numericalor categorical.

(a) In a survey, one hundred college students are asked how many hours per week they spend on the Internet.

(b) In a survey, one hundred college students are asked: “What percentage of the time you spend on theInternet is part of your course work?”

(c) In a survey, one hundred college students are asked whether or not they cited information from Wikipediain their papers.

(d) In a survey, one hundred college students are asked what percentage of their total weekly spending ison alcoholic beverages.

(e) In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a jobwithin one year of their graduation date.

5.2 Identify the parameter, Part II. For each of the following situations, state whether the parameter ofinterest is a mean or a proportion.

(a) A poll shows that 64% of Americans personally worry a great deal about federal spending and the budgetdeficit.

(b) A survey reports that local TV news has shown a 17% increase in revenue within a two year periodwhile newspaper revenues decreased by 6.4% during this time period.

(c) In a survey, high school and college students are asked whether or not they use geolocation services ontheir smart phones.

(d) In a survey, smart phone users are asked whether or not they use a web-based taxi service.

(e) In a survey, smart phone users are asked how many times they used a web-based taxi service over thelast year.

5.3 Quality control. As part of a quality control process for computer chips, an engineer at a factoryrandomly samples 212 chips during a week of production to test the current rate of chips with severe defects.She finds that 27 of the chips are defective.

(a) What population is under consideration in the data set?

(b) What parameter is being estimated?

(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?

(e) Compute the value from part (d) for this context.

(f) The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defectsduring the current week?

(g) Suppose the true population value was found to be 10%. If we use this proportion to recompute thevalue in part (e) using p = 0.1 instead of p̂, does the resulting value change much?

5.4 Unexpected expense. In a random sample 765 adults in the United States, 322 say they could notcover a $400 unexpected expense without borrowing money or going into debt.

(a) What population is under consideration in the data set?

(b) What parameter is being estimated?

(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?

(e) Compute the value from part (d) for this context.

(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?

(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute thevalue in part (e) using p = 0.4 instead of p̂, does the resulting value change much?

180 CHAPTER 5. FOUNDATIONS FOR INFERENCE

5.5 Repeated water samples. A nonprofit wants to understand the fraction of households that haveelevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels oflead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrievewater samples, and they compute the fraction of these homes with elevated lead levels. They repeat this1,000 times and build a distribution of sample proportions.

(a) What is this distribution called?

(b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explainyour reasoning.

(d) What is the formal name of the value you computed in (c)?

(e) Suppose the researchers’ budget is reduced, and they are only able to collect 250 observations per sample,but they can still collect 1,000 samples. They build a new distribution of sample proportions. How willthe variability of this new distribution compare to the variability of the distribution when each samplecontained 800 observations?

5.6 Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the currentyear. As part of a class project, students randomly sample 40 students and check if those students madethe list. They repeat this 1,000 times and build a distribution of sample proportions.

(a) What is this distribution called?

(b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explainyour reasoning.

(d) What is the formal name of the value you computed in (c)?

(e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they againcollect 1,000 samples. They build a new distribution of sample proportions. How will the variabilityof this new distribution compare to the variability of the distribution when each sample contained 40observations?

5.2. CONFIDENCE INTERVALS FOR A PROPORTION 181

5.2 Confidence intervals for a proportion

The sample proportion p̂ provides a single plausible value for the population proportion p.However, the sample proportion isn’t perfect and will have some standard error associated with it.When stating an estimate for the population proportion, it is better practice to provide a plausiblerange of values instead of supplying just the point estimate.

5.2.1 Capturing the population parameter

Using only a point estimate is like fishing in a murky lake with a spear. We can throw a spearwhere we saw a fish, but we will probably miss. On the other hand, if we toss a net in that area,we have a good chance of catching the fish. A confidence interval is like fishing with a net, andit represents a range of plausible values where we are likely to find the population parameter.

If we report a point estimate p̂, we probably will not hit the exact population proportion. Onthe other hand, if we report a range of plausible values, representing a confidence interval, we havea good shot at capturing the parameter.

GUIDED PRACTICE 5.6

If we want to be very certain we capture the population proportion in an interval, should we use awider interval or a smaller interval?6

5.2.2 Constructing a 95% confidence interval

Our sample proportion p̂ is the most plausible value of the population proportion, so it makessense to build a confidence interval around this point estimate. The standard error provides a guidefor how large we should make the confidence interval.

The standard error represents the standard deviation of the point estimate, and when the Cen-tral Limit Theorem conditions are satisfied, the point estimate closely follows a normal distribution.In a normal distribution, 95% of the data is within 1.96 standard deviations of the mean. Using thisprinciple, we can construct a confidence interval that extends 1.96 standard errors from the sampleproportion to be 95% confident that the interval captures the population proportion:

point estimate ± 1.96× SE

p̂ ± 1.96×√p(1− p)

But what does “95% confident” mean? Suppose we took many samples and built a 95% confidenceinterval from each. Then about 95% of those intervals would contain the parameter, p. Figure 5.6shows the process of creating 25 intervals from 25 samples from the simulation in Section 5.1.2, where24 of the resulting confidence intervals contain the simulation’s population proportion of p = 0.88,and one interval does not.

6If we want to be more certain we will capture the fish, we might use a wider net. Likewise, we use a widerconfidence interval if we want to be more certain that we capture the parameter.

182 CHAPTER 5. FOUNDATIONS FOR INFERENCE

p = 0.88

●

●●●

●

Figure 5.6: Twenty-five point estimates and confidence intervals from the sim-ulations in Section 5.1.2. These intervals are shown relative to the populationproportion p = 0.88. Only 1 of these 25 intervals did not capture the populationproportion, and this interval has been bolded.

EXAMPLE 5.7

In Figure 5.6, one interval does not contain p = 0.88. Does this imply that the population proportionused in the simulation could not have been p = 0.88?

Just as some observations naturally occur more than 1.96 standard deviations from the mean, somepoint estimates will be more than 1.96 standard errors from the parameter of interest. A confidenceinterval only provides a plausible range of values. While we might say other values are implausiblebased on the data, this does not mean they are impossible.

95% CONFIDENCE INTERVAL FOR A PARAMETER

When the distribution of a point estimate qualifies for the Central Limit Theorem and thereforeclosely follows a normal distribution, we can construct a 95% confidence interval as

point estimate± 1.96× SE

EXAMPLE 5.8

In Section 5.1 we learned about a Pew Research poll where 88.7% of a random sample of 1000 Amer-ican adults supported expanding the role of solar power. Compute and interpret a 95% confidenceinterval for the population proportion.

We earlier confirmed that p̂ follows a normal distribution and has a standard error of SEp̂ = 0.010.To compute the 95% confidence interval, plug the point estimate p̂ = 0.887 and standard error intothe 95% confidence interval formula:

p̂± 1.96× SEp̂ → 0.887± 1.96× 0.010 → (0.8674, 0.9066)

We are 95% confident that the actual proportion of American adults who support expanding solarpower is between 86.7% and 90.7%. (It’s common to round to the nearest percentage point or nearesttenth of a percentage point when reporting a confidence interval.)

5.2. CONFIDENCE INTERVALS FOR A PROPORTION 183

5.2.3 Changing the confidence level

Suppose we want to consider confidence intervals where the confidence level is higher than 95%,such as a confidence level of 99%. Think back to the analogy about trying to catch a fish: if we wantto be more sure that we will catch the fish, we should use a wider net. To create a 99% confidencelevel, we must also widen our 95% interval. On the other hand, if we want an interval with lowerconfidence, such as 90%, we could use a slightly narrower interval than our original 95% interval.

The 95% confidence interval structure provides guidance in how to make intervals with differentconfidence levels. The general 95% confidence interval for a point estimate that follows a normaldistribution is

point estimate ± 1.96× SE

There are three components to this interval: the point estimate, “1.96”, and the standard error.The choice of 1.96 × SE was based on capturing 95% of the data since the estimate is within 1.96standard errors of the parameter about 95% of the time. The choice of 1.96 corresponds to a 95%confidence level.

GUIDED PRACTICE 5.9

If X is a normally distributed random variable, what is the probability of the value X being within2.58 standard deviations of the mean?7

Guided Practice 5.9 highlights that 99% of the time a normal random variable will be within2.58 standard deviations of the mean. To create a 99% confidence interval, change 1.96 in the 95%confidence interval formula to be 2.58. That is, the formula for a 99% confidence interval is

point estimate ± 2.58× SE

Standard Deviations from the Mean

−3 −2 −1 0 1 2 3

95%, extends −1.96 to 1.96

99%, extends −2.58 to 2.58

Figure 5.7: The area between -z? and z? increases as z? becomes larger. If theconfidence level is 99%, we choose z? such that 99% of a normal normal distributionis between -z? and z?, which corresponds to 0.5% in the lower tail and 0.5% in theupper tail: z? = 2.58.

7This is equivalent to asking how often the Z-score will be larger than -2.58 but less than 2.58. For a picture, seeFigure 5.7. To determine this probability, we can use statistical software, a calculator, or a table to look up -2.58and 2.58 for a normal distribution: 0.0049 and 0.9951. Thus, there is a 0.9951 − 0.0049 ≈ 0.99 probability that anunobserved normal random variable X will be within 2.58 standard deviations of µ.

184 CHAPTER 5. FOUNDATIONS FOR INFERENCE

This approach – using the Z-scores in the normal model to compute confidence levels – isappropriate when a point estimate such as p̂ is associated with a normal distribution. For some otherpoint estimates, a normal model is not a good fit; in these cases, we’ll use alternative distributionsthat better represent the sampling distribution.

CONFIDENCE INTERVAL USING ANY CONFIDENCE LEVEL

If a point estimate closely follows a normal model with standard error SE, then a confidenceinterval for the population parameter is

point estimate ± z? × SE

where z? corresponds to the confidence level selected.

Figure 5.7 provides a picture of how to identify z? based on a confidence level. We select z? sothat the area between -z? and z? in the standard normal distribution, N(0, 1), corresponds to theconfidence level.

MARGIN OF ERROR

In a confidence interval, z? × SE is called the margin of error.

EXAMPLE 5.10

Use the data in Example 5.8 to create a 90% confidence interval for the proportion of American adultsthat support expanding the use of solar power. We have already verified conditions for normality.

We first find z? such that 90% of the distribution falls between -z? and z? in the standard normaldistribution, N(µ = 0, σ = 1). We can do this using a graphing calculator, statistical software, or aprobability table by looking for an upper tail of 5% (the other 5% is in the lower tail): z? = 1.65.The 90% confidence interval can then be computed as

p̂ ± 1.65× SEp̂ → 0.887 ± 1.65× 0.0100 → (0.8705, 0.9035)

That is, we are 90% confident that 87.1% to 90.4% of American adults supported the expansion ofsolar power in 2018.

CONFIDENCE INTERVAL FOR A SINGLE PROPORTION

Once you’ve determined a one-proportion confidence interval would be helpful for an application,there are four steps to constructing the interval:

Prepare. Identify p̂ and n, and determine what confidence level you wish to use.

Check. Verify the conditions to ensure p̂ is nearly normal. For one-proportion confidenceintervals, use p̂ in place of p to check the success-failure condition.

Calculate. If the conditions hold, compute SE using p̂, find z?, and construct the interval.

Conclude. Interpret the confidence interval in the context of the problem.

5.2. CONFIDENCE INTERVALS FOR A PROPORTION 185

5.2.4 More case studies

In New York City on October 23rd, 2014, a doctor who had recently been treating Ebolapatients in Guinea went to the hospital with a slight fever and was subsequently diagnosed withEbola. Soon thereafter, an NBC 4 New York/The Wall Street Journal/Marist Poll found that 82%of New Yorkers favored a “mandatory 21-day quarantine for anyone who has come in contact with anEbola patient”. This poll included responses of 1,042 New York adults between Oct 26th and 28th,2014.

EXAMPLE 5.11

What is the point estimate in this case, and is it reasonable to use a normal distribution to modelthat point estimate?

The point estimate, based on a sample of size n = 1042, is p̂ = 0.82. To check whether p̂ canbe reasonably modeled using a normal distribution, we check independence (the poll is based on asimple random sample) and the success-failure condition (1042× p̂ ≈ 854 and 1042× (1− p̂) ≈ 188,both easily greater than 10). With the conditions met, we are assured that the sampling distributionof p̂ can be reasonably modeled using a normal distribution.

EXAMPLE 5.12

Estimate the standard error of p̂ = 0.82 from the Ebola survey.

We’ll use the substitution approximation of p ≈ p̂ = 0.82 to compute the standard error:

SEp̂ =

√p(1− p)

n≈√

0.82(1− 0.82)

1042= 0.012

EXAMPLE 5.13

Construct a 95% confidence interval for p, the proportion of New York adults who supported aquarantine for anyone who has come into contact with an Ebola patient.

Using the standard error SE = 0.012 from Example 5.12, the point estimate 0.82, and z? = 1.96 fora 95% confidence level, the confidence interval is

point estimate ± z? × SE → 0.82 ± 1.96× 0.012 → (0.796, 0.844)

We are 95% confident that the proportion of New York adults in October 2014 who supported aquarantine for anyone who had come into contact with an Ebola patient was between 0.796 and0.844.

GUIDED PRACTICE 5.14

Answer the following two questions about the confidence interval from Example 5.13:8

(a) What does 95% confident mean in this context?

(b) Do you think the confidence interval is still valid for the opinions of New Yorkers today?

8(a) If we took many such samples and computed a 95% confidence interval for each, then about 95% of thoseintervals would contain the actual proportion of New York adults who supported a quarantine for anyone who hascome into contact with an Ebola patient.(b) Not necessarily. The poll was taken at a time where there was a huge public safety concern. Now that people havehad some time to step back, they may have changed their opinions. We would need to run a new poll if we wantedto get an estimate of the current proportion of New York adults who would support such a quarantine period.

186 CHAPTER 5. FOUNDATIONS FOR INFERENCE

GUIDED PRACTICE 5.15

In the Pew Research poll about solar energy, they also inquired about other forms of energy, and84.8% of the 1000 respondents supported expanding the use of wind turbines.9

(a) Is it reasonable to model the proportion of US adults who support expanding wind turbinesusing a normal distribution?

(b) Create a 99% confidence interval for the level of American support for expanding the use ofwind turbines for power generation.

We can also construct confidence intervals for other parameters, such as a population mean. Inthese cases, a confidence interval would be computed in a similar way to that of a single proportion:a point estimate plus/minus some margin of error. We’ll dive into these details in later chapters.

5.2.5 Interpreting confidence intervals

In each of the examples, we described the confidence intervals by putting them into the contextof the data and also using somewhat formal language:

Solar. We are 90% confident that 87.1% to 90.4% of American adults support the expansion ofsolar power in 2018.

Ebola. We are 95% confident that the proportion of New York adults in October 2014 who sup-ported a quarantine for anyone who had come into contact with an Ebola patient was between0.796 and 0.844.

Wind Turbine. We are 99% confident the proportion of Americans adults that support expandingthe use of wind turbines is between 81.9% and 87.7% in 2018.

First, notice that the statements are always about the population parameter, which considers allAmerican adults for the energy polls or all New York adults for the quarantine poll.

We also avoided another common mistake: incorrect language might try to describe the confi-dence interval as capturing the population parameter with a certain probability. Making a proba-bility interpretation is a common error: while it might be useful to think of it as a probability, theconfidence level only quantifies how plausible it is that the parameter is in the given interval.

Another important consideration of confidence intervals is that they are only about the popula-tion parameter. A confidence interval says nothing about individual observations or point estimates.Confidence intervals only provide a plausible range for population parameters.

Lastly, keep in mind the methods we discussed only apply to sampling error, not to bias. Ifa data set is collected in a way that will tend to systematically under-estimate (or over-estimate)the population parameter, the techniques we have discussed will not address that problem. Instead,we rely on careful data collection procedures to help protect against bias in the examples we haveconsidered, which is a common practice employed by data scientists to combat bias.

GUIDED PRACTICE 5.16

Consider the 90% confidence interval for the solar energy survey: 87.1% to 90.4%. If we ran thesurvey again, can we say that we’re 90% confident that the new survey’s proportion will be between87.1% and 90.4%?10

9(a) The survey was a random sample and counts are both ≥ 10 (1000× 0.848 = 848 and 1000× 0.152 = 152), soindependence and the success-failure condition are satisfied, and p̂ = 0.848 can be modeled using a normal distribution.(b) Guided Practice 5.15 confirmed that p̂ closely follows a normal distribution, so we can use the C.I. formula:

point estimate± z? × SE

In this case, the point estimate is p̂ = 0.848. For a 99% confidence interval, z? = 2.58. Computing the standard error:

SEp̂ =√

0.848(1−0.848)1000

= 0.0114. Finally, we compute the interval as 0.848± 2.58× 0.0114→ (0.8186, 0.8774). It is

also important to always provide an interpretation for the interval: we are 99% confident the proportion of Americanadults that support expanding the use of wind turbines in 2018 is between 81.9% and 87.7%.

10 No, a confidence interval only provides a range of plausible values for a parameter, not future point estimates.

5.2. CONFIDENCE INTERVALS FOR A PROPORTION 187

Exercises

5.7 Chronic illness, Part I. In 2013, the Pew Research Foundation reported that “45% of U.S. adults reportthat they live with one or more chronic conditions”.11 However, this value was based on a sample, so it maynot be a perfect estimate for the population parameter of interest on its own. The study reported a standarderror of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidenceinterval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret theconfidence interval in the context of the study.

5.8 Twitter users and news, Part I. A poll conducted in 2013 found that 52% of U.S. adult Twitter users getat least some news on Twitter.12. The standard error for this estimate was 2.4%, and a normal distributionmay be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S.adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

5.9 Chronic illness, Part II. In 2013, the Pew Research Foundation reported that “45% of U.S. adultsreport that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%.Identify each of the following statements as true or false. Provide an explanation to justify each of youranswers.

(a) We can say with certainty that the confidence interval from Exercise 5.7 contains the true percentageof U.S. adults who suffer from a chronic illness.

(b) If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, thenapproximately 950 of those confidence intervals would contain the true fraction of U.S. adults who sufferfrom chronic illnesses.

(c) The poll provides statistically significant evidence (at the α = 0.05 level) that the percentage of U.S.adults who suffer from chronic illnesses is below 50%.

(d) Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty abouttheir answer.

5.10 Twitter users and news, Part II. A poll conducted in 2013 found that 52% of U.S. adult Twitter usersget at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of thefollowing statements as true or false. Provide an explanation to justify each of your answers.

(a) The data provide statistically significant evidence that more than half of U.S. adult Twitter users getsome news through Twitter. Use a significance level of α = 0.01.

(b) Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users wereincluded in the study.

(d) If we construct a 90% confidence interval for the percentage of U.S. adults Twitter users who getsome news through Twitter, this confidence interval will be wider than a corresponding 99% confidenceinterval.

11Pew Research Center, Washington, D.C. The Diagnosis Difference, November 26, 2013.12Pew Research Center, Washington, D.C. Twitter News Consumers: Young, Mobile and Educated, November 4,

2013.

188 CHAPTER 5. FOUNDATIONS FOR INFERENCE

5.11 Waiting at an ER, Part I. A hospital administrator hoping to improve wait times decides to estimatethe average emergency room waiting time at her hospital. She collects a simple random sample of 64 patientsand determines the time (in minutes) between when they checked in to the ER until they were first seen bya doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based onthe normal model for the mean. Determine whether the following statements are true or false, and explainyour reasoning.

(a) We are 95% confident that the average waiting time of these 64 emergency room patients is between128 and 147 minutes.

(b) We are 95% confident that the average waiting time of all patients at this hospital’s emergency room isbetween 128 and 147 minutes.

(d) A 99% confidence interval would be narrower than the 95% confidence interval since we need to be moresure of our estimate.

(e) The margin of error is 9.5 and the sample mean is 137.5.

(f) In order to decrease the margin of error of a 95% confidence interval to half of what it is now, we wouldneed to double the sample size.

5.12 Mental health. The General Social Survey asked the question: “For how many days during the past30 days was your mental health, which includes stress, depression, and problems with emotions, not good?”Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24days in 2010.

(a) Interpret this interval in context of the data.

(b) What does “95% confident” mean? Explain in the context of the application.

(c) Suppose the researchers think a 99% confidence level would be more appropriate for this interval. Willthis new interval be smaller or wider than the 95% confidence interval?

(d) If a new survey were to be done with 500 Americans, do you think the standard error of the estimatebe larger, smaller, or about the same.

5.13 Website registration. A website is trying to increase registration for first-time visitors, exposing 1%of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the newdesign, 64 registered.

(a) Check any conditions required for constructing a confidence interval.

(b) Compute the standard error.

(c) Construct and interpret a 90% confidence interval for the fraction of first-time visitors of the site whowould register under the new design (assuming stable behaviors by new visitors over time).

5.14 Coupons driving visits. A store randomly samples 603 shoppers over the course of a year andfinds that 142 of them made their visit because of a coupon they’d received in the mail. Construct a 95%confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they’dreceived in the mail.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 189

5.3 Hypothesis testing for a proportion

The following question comes from a book written by Hans Rosling, Anna Rosling Rönnlund,and Ola Rosling called Factfulness:

How many of the world’s 1 year old children today have been vaccinated against somedisease:

a. 20%

b. 50%

c. 80%

Write down what your answer (or guess), and when you’re ready, find the answer in the footnote.13

In this section, we’ll be exploring how people with a 4-year college degree perform on this andother world health questions as we learn about hypothesis tests, which are a framework used torigorously evaluate competing ideas and claims.

5.3.1 Hypothesis testing framework

We’re interested in understanding how much people know about world health and development.If we take a multiple choice world health question, then we might like to understand if

H0: People never learn these particular topics and their responses are simply equivalent to randomguesses.

HA: People have knowledge that helps them do better than random guessing, or perhaps, they havefalse knowledge that leads them to actually do worse than random guessing.

These competing ideas are called hypotheses. We callH0 the null hypothesis andHA the alternativehypothesis. When there is a subscript 0 like in H0, data scientists pronounce it as “nought” (e.g. H0

is pronounced “H-nought”).

NULL AND ALTERNATIVE HYPOTHESES

The null hypothesis (H0) often represents a skeptical perspective or a claim to be tested.The alternative hypothesis (HA) represents an alternative claim under consideration and isoften represented by a range of possible parameter values.

Our job as data scientists is to play the role of a skeptic: before we buy into the alternativehypothesis, we need to see strong supporting evidence.

The null hypothesis often represents a skeptical position or a perspective of “no difference”. Inour first example, we’ll consider whether the typical person does any different than random guessingon Roslings’ question about infant vaccinations.

The alternative hypothesis generally represents a new or stronger perspective. In the case ofthe question about infant vaccinations, it would certainly be interesting to learn whether people dobetter than random guessing, since that would mean that the typical person knows something aboutworld health statistics. It would also be very interesting if we learned that people do worse thanrandom guessing, which would suggest people believe incorrect information about world health.

The hypothesis testing framework is a very general tool, and we often use it without a secondthought. If a person makes a somewhat unbelievable claim, we are initially skeptical. However,if there is sufficient evidence that supports the claim, we set aside our skepticism and reject the nullhypothesis in favor of the alternative. The hallmarks of hypothesis testing are also found in theUS court system.

13The correct answer is (c): 80% of the world’s 1 year olds have been vaccinated against some disease.

190 CHAPTER 5. FOUNDATIONS FOR INFERENCE

GUIDED PRACTICE 5.17

A US court considers two possible claims about a defendant: she is either innocent or guilty. If weset these claims up in a hypothesis framework, which would be the null hypothesis and which thealternative?14

Jurors examine the evidence to see whether it convincingly shows a defendant is guilty. Evenif the jurors leave unconvinced of guilt beyond a reasonable doubt, this does not mean they believethe defendant is innocent. This is also the case with hypothesis testing: even if we fail to reject thenull hypothesis, we typically do not accept the null hypothesis as true. Failing to find strong evidencefor the alternative hypothesis is not equivalent to accepting the null hypothesis.

When considering Roslings’ question about infant vaccination, the null hypothesis representsthe notion that the people we will be considering – college-educated adults – are as accurate asrandom guessing. That is, the proportion p of respondents who pick the correct answer, that 80%of 1 year olds have been vaccinated against some disease, is about 33.3% (or 1-in-3 if wanting tobe perfectly precise). The alternative hypothesis is that this proportion is something other than33.3%. While it’s helpful to write these hypotheses in words, it can be useful to write them usingmathematical notation:

H0: p = 0.333

HA: p 6= 0.333

In this hypothesis setup, we want to make a conclusion about the population parameter p. Thevalue we are comparing the parameter to is called the null value, which in this case is 0.333. It’scommon to label the null value with the same symbol as the parameter but with a subscript ‘0’.That is, in this case, the null value is p0 = 0.333 (pronounced “p-nought equals 0.333”).

EXAMPLE 5.18

It may seem impossible that the proportion of people who get the correct answer is exactly 33.3%.If we don’t believe the null hypothesis, should we simply reject it?

No. While we may not buy into the notion that the proportion is exactly 33.3%, the hypothesistesting framework requires that there be strong evidence before we reject the null hypothesis andconclude something more interesting.

After all, even if we don’t believe the proportion is exactly 33.3%, that doesn’t really tell us anythinguseful! We would still be stuck with the original question: do people do better or worse than randomguessing on Roslings’ question? Without data that strongly points in one direction or the other, itis both uninteresting and pointless to reject H0.

GUIDED PRACTICE 5.19

Another example of a real-world hypothesis testing situation is evaluating whether a new drug isbetter or worse than an existing drug at treating a particular disease. What should we use for thenull and alternative hypotheses in this case?15

14The jury considers whether the evidence is so convincing (strong) that there is no reasonable doubt regarding theperson’s guilt; in such a case, the jury rejects innocence (the null hypothesis) and concludes the defendant is guilty(alternative hypothesis).

15The null hypothesis (H0) in this case is the declaration of no difference: the drugs are equally effective. Thealternative hypothesis (HA) is that the new drug performs differently than the original, i.e. it could perform betteror worse.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 191

5.3.2 Testing hypotheses using confidence intervals

We will use the rosling responses data set to evaluate the hypothesis test evaluating whethercollege-educated adults who get the question about infant vaccination correct is different from 33.3%.This data set summarizes the answers of 50 college-educated adults. Of these 50 adults, 24% ofrespondents got the question correct that 80% of 1 year olds have been vaccinated against somedisease.

Up until now, our discussion has been philosophical. However, now that we have data, we mightask ourselves: does the data provide strong evidence that the proportion of all college-educated adultswho would answer this question correctly is different than 33.3%?

We learned in Section 5.1 that there is fluctuation from one sample to another, and it isunlikely that our sample proportion, p̂, will exactly equal p, but we want to make a conclusionabout p. We have a nagging concern: is this deviation of 24% from 33.3% simply due to chance,or does the data provide strong evidence that the population proportion is different from 33.3%?

In Section 5.2, we learned how to quantify the uncertainty in our estimate using confidenceintervals. The same method for measuring variability can be useful for the hypothesis test.

EXAMPLE 5.20

Check whether it is reasonable to construct a confidence interval for p using the sample data, and ifso, construct a 95% confidence interval.

The conditions are met for p̂ to be approximately normal: the data come from a simple randomsample (satisfies independence), and np̂ = 12 and n(1− p̂) = 38 are both at least 10 (success-failurecondition).

To construct the confidence interval, we will need to identify the point estimate (p̂ = 0.24), the criticalvalue for the 95% confidence level (z? = 1.96), and the standard error of p̂ (SEp̂ =

√p̂(1− p̂)/n =

0.060). With those pieces, the confidence interval for p can be constructed:

p̂± z? × SEp̂0.24± 1.96× 0.060

(0.122, 0.358)

We are 95% confident that the proportion of all college-educated adults to correctly answer thisparticular question about infant vaccination is between 12.2% and 35.8%.

Because the null value in the hypothesis test is p0 = 0.333, which falls within the range ofplausible values from the confidence interval, we cannot say the null value is implausible.16 Thatis, the data do not provide sufficient evidence to reject the notion that the performance of college-educated adults was different than random guessing, and we do not reject the null hypothesis, H0.

EXAMPLE 5.21

Explain why we cannot conclude that college-educated adults simply guessed on the infant vaccina-tion question.

While we failed to reject H0, that does not necessarily mean the null hypothesis is true. Perhapsthere was an actual difference, but we were not able to detect it with the relatively small sampleof 50.

DOUBLE NEGATIVES CAN SOMETIMES BE USED IN STATISTICS

In many statistical explanations, we use double negatives. For instance, we might say that thenull hypothesis is not implausible or we failed to reject the null hypothesis. Double negativesare used to communicate that while we are not rejecting a position, we are also not saying it iscorrect.

16Arguably this method is slightly imprecise. As we’ll see in a few pages, the standard error is often computedslightly differently in the context of a hypothesis test for a proportion.

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GUIDED PRACTICE 5.22

Let’s move onto a second question posed by the Roslings:

There are 2 billion children in the world today aged 0-15 years old, how many childrenwill there be in year 2100 according to the United Nations?

a. 4 billion.

b. 3 billion.

c. 2 billion.

Set up appropriate hypotheses to evaluate whether college-educated adults are better than randomguessing on this question. Also, see if you can guess the correct answer before checking the answerin the footnote!17

GUIDED PRACTICE 5.23

This time we took a larger sample of 228 college-educated adults, 34 (14.9%) selected the correctanswer to the question in Guided Practice 5.22: 2 billion. Can we model the sample proportionusing a normal distribution and construct a confidence interval?18

EXAMPLE 5.24

Compute a 95% confidence interval for the fraction of college-educated adults who answered thechildren-in-2100 question correctly, and evaluate the hypotheses in Guided Practice 5.22.

To compute the standard error, we’ll again use p̂ in place of p for the calculation:

SEp̂ =

√p̂(1− p̂)

√0.149(1− 0.149)

228= 0.024

In Guided Practice 5.23, we found that p̂ can be modeled using a normal distribution, which ensuresa 95% confidence interval may be accurately constructed as

p̂ ± z? × SE → 0.149 ± 1.96× 0.024 → (0.103, 0.195)

Because the null value, p0 = 0.333, is not in the confidence interval, a population proportion of 0.333is implausible and we reject the null hypothesis. That is, the data provide statistically significantevidence that the actual proportion of college adults who get the children-in-2100 question correctis different from random guessing. Because the entire 95% confidence interval is below 0.333, we canconclude college-educated adults do worse than random guessing on this question.

One subtle consideration is that we used a 95% confidence interval. What if we had used a 99%confidence level? Or even a 99.9% confidence level? It’s possible to come to a different conclusionif using a different confidence level. Therefore, when we make a conclusion based on confidenceinterval, we should also be sure it is clear what confidence level we used.

The worse-than-random performance on this last question is not a fluke: there are many suchworld health questions where people do worse than random guessing. In general, the answers suggestthat people tend to be more pessimistic about progress than reality suggests. This topic is discussedin much greater detail in the Roslings’ book, Factfulness.

17The appropriate hypotheses are:H0: the proportion who get the answer correct is the same as random guessing: 1-in-3, or p = 0.333.HA: the proportion who get the answer correct is different than random guessing, p 6= 0.333.The correct answer to the question is 2 billion. While the world population is projected to increase, the average

age is also expected to rise. That is, the majority of the population growth will happen in older age groups, meaningpeople are projected to live longer in the future across much of the world.

18We check both conditions, which are satisfied, so it is reasonable to use a normal distribution for p̂:Independence. Since the data are from a simple random sample, the observations are independent.Success-failure. We’ll use p̂ in place of p to check: np̂ = 34 and n(1 − p̂) = 194. Both are greater than 10, so thesuccess-failure condition is satisfied.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 193

5.3.3 Decision errors

Hypothesis tests are not flawless: we can make an incorrect decision in a statistical hypothesistest based on the data. For example, in the court system innocent people are sometimes wronglyconvicted and the guilty sometimes walk free. One key distinction with statistical hypothesis testsis that we have the tools necessary to probabilistically quantify how often we make errors in ourconclusions.

Recall that there are two competing hypotheses: the null and the alternative. In a hypothesistest, we make a statement about which one might be true, but we might choose incorrectly. Thereare four possible scenarios, which are summarized in Figure 5.8.

Test conclusion

do not reject H0 reject H0 in favor of HA

H0 true okay Type 1 ErrorTruth

HA true Type 2 Error okay

Figure 5.8: Four different scenarios for hypothesis tests.

A Type 1 Error is rejecting the null hypothesis when H0 is actually true. A Type 2 Erroris failing to reject the null hypothesis when the alternative is actually true.

GUIDED PRACTICE 5.25

In a US court, the defendant is either innocent (H0) or guilty (HA). What does a Type 1 Errorrepresent in this context? What does a Type 2 Error represent? Figure 5.8 may be useful.19

EXAMPLE 5.26

How could we reduce the Type 1 Error rate in US courts? What influence would this have on theType 2 Error rate?

To lower the Type 1 Error rate, we might raise our standard for conviction from “beyond a reasonabledoubt” to “beyond a conceivable doubt” so fewer people would be wrongly convicted. However, thiswould also make it more difficult to convict the people who are actually guilty, so we would makemore Type 2 Errors.

GUIDED PRACTICE 5.27

How could we reduce the Type 2 Error rate in US courts? What influence would this have on theType 1 Error rate?20

Exercises 5.25-5.27 provide an important lesson: if we reduce how often we make one type oferror, we generally make more of the other type.

Hypothesis testing is built around rejecting or failing to reject the null hypothesis. That is, wedo not reject H0 unless we have strong evidence. But what precisely does strong evidence mean? Asa general rule of thumb, for those cases where the null hypothesis is actually true, we do not wantto incorrectly reject H0 more than 5% of the time. This corresponds to a significance level of0.05. That is, if the null hypothesis is true, the significance level indicates how often the data leadus to incorrectly reject H0. We often write the significance level using α (the Greek letter alpha):α = 0.05. We discuss the appropriateness of different significance levels in Section 5.3.5.

19If the court makes a Type 1 Error, this means the defendant is innocent (H0 true) but wrongly convicted. Notethat a Type 1 Error is only possible if we’ve rejected the null hypothesis.

A Type 2 Error means the court failed to reject H0 (i.e. failed to convict the person) when she was in fact guilty(HA true). Note that a Type 2 Error is only possible if we have failed to reject the null hypothesis.

20To lower the Type 2 Error rate, we want to convict more guilty people. We could lower the standards forconviction from “beyond a reasonable doubt” to “beyond a little doubt”. Lowering the bar for guilt will also resultin more wrongful convictions, raising the Type 1 Error rate.

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If we use a 95% confidence interval to evaluate a hypothesis test and the null hypothesis happensto be true, we will make an error whenever the point estimate is at least 1.96 standard errors awayfrom the population parameter. This happens about 5% of the time (2.5% in each tail). Similarly,using a 99% confidence interval to evaluate a hypothesis is equivalent to a significance level ofα = 0.01.

A confidence interval is very helpful in determining whether or not to reject the null hypothesis.However, the confidence interval approach isn’t always sustainable. In several sections, we willencounter situations where a confidence interval cannot be constructed. For example, if we wantedto evaluate the hypothesis that several proportions are equal, it isn’t clear how to construct andcompare many confidence intervals altogether.

Next we will introduce a statistic called the p-value to help us expand our statistical toolkit,which will enable us to both better understand the strength of evidence and work in more complexdata scenarios in later sections.

5.3.4 Formal testing using p-values

The p-value is a way of quantifying the strength of the evidence against the null hypothesisand in favor of the alternative hypothesis. Statistical hypothesis testing typically uses the p-valuemethod rather than making a decision based on confidence intervals.

P-VALUE

The p-value is the probability of observing data at least as favorable to the alternative hy-pothesis as our current data set, if the null hypothesis were true. We typically use a summarystatistic of the data, in this section the sample proportion, to help compute the p-value andevaluate the hypotheses.

EXAMPLE 5.28

Pew Research asked a random sample of 1000 American adults whether they supported the increasedusage of coal to produce energy. Set up hypotheses to evaluate whether a majority of Americanadults support or oppose the increased usage of coal.

The uninteresting result is that there is no majority either way: half of Americans support and theother half oppose expanding the use of coal to produce energy. The alternative hypothesis would bethat there is a majority support or oppose (though we do not known which one!) expanding the useof coal. If p represents the proportion supporting, then we can write the hypotheses as

H0: p = 0.5

HA: p 6= 0.5

In this case, the null value is p0 = 0.5.

When evaluating hypotheses for proportions using the p-value method, we will slightly modifyhow we check the success-failure condition and compute the standard error for the single proportioncase. These changes aren’t dramatic, but pay close attention to how we use the null value, p0.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 195

EXAMPLE 5.29

Pew Research’s sample show that 37% of American adults support increased usage of coal. We nowwonder, does 37% represent a real difference from the null hypothesis of 50%? What would thesampling distribution of p̂ look like if the null hypothesis were true?

If the null hypothesis were true, the population proportion would be the null value, 0.5. We previ-ously learned that the sampling distribution of p̂ will be normal when two conditions are met:

Independence. The poll was based on a simple random sample, so independence is satisfied.

Success-failure. Based on the poll’s sample size of n = 1000, the success-failure condition is met,since

npH0= 1000× 0.5 = 500 n(1− p) H0= 1000× (1− 0.5) = 500

are both at least 10. Note that the success-failure condition was checked using the null value,p0 = 0.5; this is the first procedural difference from confidence intervals.

If the null hypothesis were true, the sampling distribution indicates that a sample proportion basedon n = 1000 observations would be normally distributed. Next, we can compute the standard error,where we will again use the null value p0 = 0.5 in the calculation:

SEp̂ =

√p(1− p)

H0=

√0.5× (1− 0.5)

1000= 0.016

This marks the other procedural difference from confidence intervals: since the sampling distribu-tion is determined under the null proportion, the null value p0 was used for the proportion in thecalculation rather than p̂.

Ultimately, if the null hypothesis were true, then the sample proportion should follow a normaldistribution with mean 0.5 and a standard error of 0.016. This distribution is shown in Figure 5.9.

0.37 0.50

Observed p̂ = 0.37

Figure 5.9: If the null hypothesis were true, this normal distribution describes thedistribution of p̂.

CHECKING SUCCESS-FAILURE AND COMPUTING SEP̂ FOR A HYPOTHESIS TEST

When using the p-value method to evaluate a hypothesis test, we check the conditions for p̂ andconstruct the standard error using the null value, p0, instead of using the sample proportion.

In a hypothesis test with a p-value, we are supposing the null hypothesis is true, which is adifferent mindset than when we compute a confidence interval. This is why we use p0 insteadof p̂ when we check conditions and compute the standard error in this context.

When we identify the sampling distribution under the null hypothesis, it has a special name:the null distribution. The p-value represents the probability of the observed p̂, or a p̂ that is moreextreme, if the null hypothesis were true. To find the p-value, we generally find the null distribution,and then we find a tail area in that distribution corresponding to our point estimate.

196 CHAPTER 5. FOUNDATIONS FOR INFERENCE

EXAMPLE 5.30

If the null hypothesis were true, determine the chance of finding p̂ at least as far into the tails as0.37 under the null distribution, which is a normal distribution with mean µ = 0.5 and SE = 0.016.

This is a normal probability problem where x = 0.37. First, we draw a simple graph to representthe situation, similar to what is shown in Figure 5.9. Since p̂ is so far out in the tail, we know thetail area is going to be very small. To find it, we start by computing the Z-score using the mean of0.5 and the standard error of 0.016:

Z =0.37− 0.5

0.016= −8.125

We can use software to find the tail area: 2.2 × 10−16 (0.00000000000000022). If using the normalprobability table in Appendix C.1, we’d find that Z = −8.125 is off the table, so we would use thesmallest area listed: 0.0002.

The potential p̂’s in the upper tail beyond 0.63, which are shown in Figure 5.10, also representobservations at least as extreme as the observed value of 0.37. To account for these values that arealso more extreme under the hypothesis setup, we double the lower tail to get an estimate of thep-value: 4.4× 10−16 (or if using the table method, 0.0004).

The p-value represents the probability of observing such an extreme sample proportion by chance,if the null hypothesis were true.

0.37 0.50 0.63

Tail Area for p̂ Equally unlikely if H0 is true

Figure 5.10: If H0 were true, then the values above 0.63 are just as unlikely asvalues below 0.37.

EXAMPLE 5.31

How should we evaluate the hypotheses using the p-value of 4.4×10−16? Use the standard significancelevel of α = 0.05.

If the null hypothesis were true, there’s only an incredibly small chance of observing such an extremedeviation of p̂ from 0.5. This means one of the following must be true:

1. The null hypothesis is true, and we just happened to get observe something so extreme thatonly happens about once in every 23 quadrillion times (1 quadrillion = 1 million × 1 billion).

2. The alternative hypothesis is true, which would be consistent with observing a sample propor-tion far from 0.5.

The first scenario is laughably improbable, while the second scenario seems much more plausible.

Formally, when we evaluate a hypothesis test, we compare the p-value to the significance level, whichin this case is α = 0.05. Since the p-value is less than α, we reject the null hypothesis. That is,the data provide strong evidence against H0. The data indicate the direction of the difference: amajority of Americans do not support expanding the use of coal-powered energy.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 197

COMPARE THE P-VALUE TO ααα TO EVALUATEH0H0H0

When the p-value is less than the significance level, α, reject H0. We would report a conclusionthat the data provide strong evidence supporting the alternative hypothesis.

When the p-value is greater than α, do not reject H0, and report that we do not have sufficientevidence to reject the null hypothesis.

In either case, it is important to describe the conclusion in the context of the data.

GUIDED PRACTICE 5.32

Do a majority of Americans support or oppose nuclear arms reduction? Set up hypotheses toevaluate this question.21

EXAMPLE 5.33

A simple random sample of 1028 US adults in March 2013 show that 56% support nuclear armsreduction. Does this provide convincing evidence that a majority of Americans supported nucleararms reduction at the 5% significance level?

First, check conditions:

Independence. The poll was of a simple random sample of US adults, meaning the observationsare independent.

Success-failure. In a one-proportion hypothesis test, this condition is checked using the null pro-portion, which is p0 = 0.5 in this context: np0 = n(1− p0) = 1028× 0.5 = 514 ≥ 10.

With these conditions verified, we can model p̂ using a normal model.

Next the standard error can be computed. The null value p0 is used again here, because this is ahypothesis test for a single proportion.

SEp̂ =

√p0(1− p0)

√0.5(1− 0.5)

1028= 0.0156

Based on the normal model, the test statistic can be computed as the Z-score of the point estimate:

Z =point estimate− null value

SE=

0.56− 0.50

0.0156= 3.85

It’s generally helpful to draw null distribution and the tail areas of interest for computing the p-value:

0.5 0.56

upper taillower tail

The upper tail area is about 0.0001, and we double this tail area to get the p-value: 0.0002. Becausethe p-value is smaller than 0.05, we reject H0. The poll provides convincing evidence that a majorityof Americans supported nuclear arms reduction efforts in March 2013.

21We would like to understand if a majority supports or opposes, or ultimately, if there is no difference. If p is theproportion of Americans who support nuclear arms reduction, then H0: p = 0.50 and HA: p 6= 0.50.

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HYPOTHESIS TESTING FOR A SINGLE PROPORTION

Once you’ve determined a one-proportion hypothesis test is the correct procedure, there arefour steps to completing the test:

Prepare. Identify the parameter of interest, list hypotheses, identify the significance level, andidentify p̂ and n.

Check. Verify conditions to ensure p̂ is nearly normal under H0. For one-proportion hypothesistests, use the null value to check the success-failure condition.

Calculate. If the conditions hold, compute the standard error, again using p0, compute theZ-score, and identify the p-value.

Conclude. Evaluate the hypothesis test by comparing the p-value to α, and provide a conclu-sion in the context of the problem.

5.3.5 Choosing a significance level

Choosing a significance level for a test is important in many contexts, and the traditional levelis α = 0.05. However, it can be helpful to adjust the significance level based on the application.We may select a level that is smaller or larger than 0.05 depending on the consequences of anyconclusions reached from the test.

If making a Type 1 Error is dangerous or especially costly, we should choose a small significancelevel (e.g. 0.01). Under this scenario we want to be very cautious about rejecting the null hypothesis,so we demand very strong evidence favoring HA before we would reject H0.

If a Type 2 Error is relatively more dangerous or much more costly than a Type 1 Error, thenwe might choose a higher significance level (e.g. 0.10). Here we want to be cautious about failing toreject H0 when the alternative hypothesis is actually true.

Additionally, if the cost of collecting data is small relative to the cost of a Type 2 Error, thenit may also be a good strategy to collect more data. Under this strategy, the Type 2 Error can bereduced while not affecting the Type 1 Error rate. Of course, collecting extra data is often costly,so there is typically a cost-benefit analysis to be considered.

EXAMPLE 5.34

A car manufacturer is considering switching to a new, higher quality piece of equipment that con-structs vehicle door hinges. They figure that they will save money in the long run if this new machineproduces hinges that have flaws less than 0.2% of the time. However, if the hinges are flawed morethan 0.2% of the time, they wouldn’t get a good enough return-on-investment from the new pieceof equipment, and they would lose money. Is there good reason to modify the significance level insuch a hypothesis test?

The null hypothesis would be that the rate of flawed hinges is 0.2%, while the alternative is that itthe rate is different than 0.2%. This decision is just one of many that have a marginal impact onthe car and company. A significance level of 0.05 seems reasonable since neither a Type 1 or Type 2Error should be dangerous or (relatively) much more expensive.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 199

EXAMPLE 5.35

The same car manufacturer is considering a slightly more expensive supplier for parts related tosafety, not door hinges. If the durability of these safety components is shown to be better than thecurrent supplier, they will switch manufacturers. Is there good reason to modify the significancelevel in such an evaluation?

The null hypothesis would be that the suppliers’ parts are equally reliable. Because safety is involved,the car company should be eager to switch to the slightly more expensive manufacturer (reject H0),even if the evidence of increased safety is only moderately strong. A slightly larger significance level,such as α = 0.10, might be appropriate.

GUIDED PRACTICE 5.36

A part inside of a machine is very expensive to replace. However, the machine usually functionsproperly even if this part is broken, so the part is replaced only if we are extremely certain it isbroken based on a series of measurements. Identify appropriate hypotheses for this test (in plainlanguage) and suggest an appropriate significance level.22

WHY IS 0.05 THE DEFAULT?

The α = 0.05 threshold is most common. But why? Maybe the standard level should besmaller, or perhaps larger. If you’re a little puzzled, you’re reading with an extra critical eye –good job! We’ve made a 5-minute task to help clarify why 0.05 :

www.openintro.org/why05

5.3.6 Statistical significance versus practical significance

When the sample size becomes larger, point estimates become more precise and any real differ-ences in the mean and null value become easier to detect and recognize. Even a very small differencewould likely be detected if we took a large enough sample. Sometimes researchers will take such largesamples that even the slightest difference is detected, even differences where there is no practicalvalue. In such cases, we still say the difference is statistically significant, but it is not practi-cally significant. For example, an online experiment might identify that placing additional ads ona movie review website statistically significantly increases viewership of a TV show by 0.001%, butthis increase might not have any practical value.

One role of a data scientist in conducting a study often includes planning the size of the study.The data scientist might first consult experts or scientific literature to learn what would be thesmallest meaningful difference from the null value. She also would obtain other information, suchas a very rough estimate of the true proportion p, so that she could roughly estimate the standarderror. From here, she can suggest a sample size that is sufficiently large that, if there is a realdifference that is meaningful, we could detect it. While larger sample sizes may still be used, thesecalculations are especially helpful when considering costs or potential risks, such as possible healthimpacts to volunteers in a medical study.

22Here the null hypothesis is that the part is not broken, and the alternative is that it is broken. If we don’t havesufficient evidence to reject H0, we would not replace the part. It sounds like failing to fix the part if it is broken(H0 false, HA true) is not very problematic, and replacing the part is expensive. Thus, we should require very strongevidence against H0 before we replace the part. Choose a small significance level, such as α = 0.01.

200 CHAPTER 5. FOUNDATIONS FOR INFERENCE

5.3.7 One-sided hypothesis tests (special topic)

So far we’ve only considered what are called two-sided hypothesis tests, where we care aboutdetecting whether p is either above or below some null value p0. There is a second type of hypothesistest called a one-sided hypothesis test. For a one-sided hypothesis test, the hypotheses take oneof the following forms:

1. There’s only value in detecting if the population parameter is less than some value p0. In thiscase, the alternative hypothesis is written as p < p0 for some null value p0.

2. There’s only value in detecting if the population parameter is more than some value p0: In thiscase, the alternative hypothesis is written as p > p0.

While we adjust the form of the alternative hypothesis, we continue to write the null hypothesisusing an equals-sign in the one-sided hypothesis test case.

In the entire hypothesis testing procedure, there is only one difference in evaluating a one-sided hypothesis test vs a two-sided hypothesis test: how to compute the p-value. In a one-sidedhypothesis test, we compute the p-value as the tail area in the direction of the alternative hypothesisonly, meaning it is represented by a single tail area. Herein lies the reason why one-sided tests aresometimes interesting: if we don’t have to double the tail area to get the p-value, then the p-valueis smaller and the level of evidence required to identify an interesting finding in the direction of thealternative hypothesis goes down. However, one-sided tests aren’t all sunshine and rainbows: theheavy price paid is that any interesting findings in the opposite direction must be disregarded.

EXAMPLE 5.37

In Section 1.1, we encountered an example where doctors were interested in determining whetherstents would help people who had a high risk of stroke. The researchers believed the stents wouldhelp. Unfortunately, the data showed the opposite: patients who received stents actually did worse.Why was using a two-sided test so important in this context?

Before the study, researchers had reason to believe that stents would help patients since existingresearch suggested stents helped in patients with heart attacks. It would surely have been temptingto use a one-sided test in this situation, and had they done this, they would have limited their abilityto identify potential harm to patients.

Example 5.37 highlights that using a one-sided hypothesis creates a risk of overlooking datasupporting the opposite conclusion. We could have made a similar error when reviewing the Roslings’question data this section; if we had a pre-conceived notion that college-educated people wouldn’t doworse than random guessing and so used a one-sided test, we would have missed the really interestingfinding that many people have incorrect knowledge about global public health.

When might a one-sided test be appropriate to use? Very rarely. Should you ever find yourselfconsidering using a one-sided test, carefully answer the following question:

What would I, or others, conclude if the data happens to go clearly in the opposite direc-tion than my alternative hypothesis?

If you or others would find any value in making a conclusion about the data that goes in theopposite direction of a one-sided test, then a two-sided hypothesis test should actually be used.These considerations can be subtle, so exercise caution. We will only apply two-sided tests in therest of this book.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 201

EXAMPLE 5.38

Why can’t we simply run a one-sided test that goes in the direction of the data?

We’ve been building a careful framework that controls for the Type 1 Error, which is the significancelevel α in a hypothesis test. We’ll use the α = 0.05 below to keep things simple.

Imagine we could pick the one-sided test after we saw the data. What will go wrong?

• If p̂ is smaller than the null value, then a one-sided test where p < p0 would mean that anyobservation in the lower 5% tail of the null distribution would lead to us rejecting H0.

• If p̂ is larger than the null value, then a one-sided test where p > p0 would mean that anyobservation in the upper 5% tail of the null distribution would lead to us rejecting H0.

Then if H0 were true, there’s a 10% chance of being in one of the two tails, so our testing error isactually α = 0.10, not 0.05. That is, not being careful about when to use one-sided tests effectivelyundermines the methods we’re working so hard to develop and utilize.

202 CHAPTER 5. FOUNDATIONS FOR INFERENCE

Exercises

5.15 Identify hypotheses, Part I. Write the null and alternative hypotheses in words and then symbols foreach of the following situations.

(a) A tutoring company would like to understand if most students tend to improve their grades (or not)after they use their services. They sample 200 of the students who used their service in the past yearand ask them if their grades have improved or declined from the previous year.

(b) Employers at a firm are worried about the effect of March Madness, a basketball championship held eachspring in the US, on employee productivity. They estimate that on a regular business day employeesspend on average 15 minutes of company time checking personal email, making personal phone calls,etc. They also collect data on how much company time employees spend on such non-business activitiesduring March Madness. They want to determine if these data provide convincing evidence that employeeproductivity changed during March Madness.

5.16 Identify hypotheses, Part II. Write the null and alternative hypotheses in words and using symbolsfor each of the following situations.

(a) Since 2008, chain restaurants in California have been required to display calorie counts of each menuitem. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on thenumber of calories consumed at this restaurant from a random sample of diners. Do these data provideconvincing evidence of a difference in the average calorie intake of a diners at this restaurant?

(b) The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcoholin the last year, specifically if the rate is different from the national rate of 70%. To help them answer thisquestion, they conduct a random sample of 852 residents and ask them about their alcohol consumption.

5.17 Online communication. A study suggests that 60% of college student spend 10 or more hours perweek communicating with others online. You believe that this is incorrect and decide to collect your ownsample for a hypothesis test. You randomly sample 160 students from your dorm and find that 70% spent10 or more hours a week communicating with others online. A friend of yours, who offers to help you withthe hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see.

H0 : p̂ < 0.6

HA : p̂ > 0.7

5.18 Married at 25. A study suggests that the 25% of 25 year olds have gotten married. You believe thatthis is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25year olds in census data with size 776, you find that 24% of them are married. A friend of yours offersto help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate anyerrors you see.

H0 : p̂ = 0.24

HA : p̂ 6= 0.24

5.19 Cyberbullying rates. Teens were surveyed about cyberbullying, and 54% to 64% reported experienc-ing cyberbullying (95% confidence interval).23 Answer the following questions based on this interval.

(a) A newspaper claims that a majority of teens have experienced cyberbullying. Is this claim supportedby the confidence interval? Explain your reasoning.

(b) A researcher conjectured that 70% of teens have experienced cyberbullying. Is this claim supported bythe confidence interval? Explain your reasoning.

(c) Without actually calculating the interval, determine if the claim of the researcher from part (b) wouldbe supported based on a 90% confidence interval?

23Pew Research Center, A Majority of Teens Have Experienced Some Form of Cyberbullying. September 27, 2018.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 203

5.20 Waiting at an ER, Part II. Exercise 5.11 provides a 95% confidence interval for the mean waiting timeat an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on thisinterval.

(a) A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claimsupported by the confidence interval? Explain your reasoning.

(b) The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supportedby the confidence interval? Explain your reasoning.

(c) Without actually calculating the interval, determine if the claim of the Dean from part (b) would besupported based on a 99% confidence interval?

5.21 Minimum wage, Part I. Do a majority of US adults believe raising the minimum wage will help theeconomy, or is there a majority who do not believe this? A Rasmussen Reports survey of 1,000 US adultsfound that 42% believe it will help the economy.24 Conduct an appropriate hypothesis test to help answerthe research question.

5.22 Getting enough sleep. 400 students were randomly sampled from a large university, and 289 saidthey did not get enough sleep. Conduct a hypothesis test to check whether this represents a statisticallysignificant difference from 50%, and use a significance level of 0.01.

5.23 Working backwards, Part I. You are given the following hypotheses:

H0 : p = 0.3

HA : p 6= 0.3

We know the sample size is 90. For what sample proportion would the p-value be equal to 0.05? Assumethat all conditions necessary for inference are satisfied.

5.24 Working backwards, Part II. You are given the following hypotheses:

H0 : p = 0.9

HA : p 6= 0.9

We know that the sample size is 1,429. For what sample proportion would the p-value be equal to 0.01?Assume that all conditions necessary for inference are satisfied.

5.25 Testing for Fibromyalgia. A patient named Diana was diagnosed with Fibromyalgia, a long-termsyndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’tinitially believe that anti-depressants would help her symptoms. However after a couple months of being onthe medication she decides that the anti-depressants are working, because she feels like her symptoms arein fact getting better.

(a) Write the hypotheses in words for Diana’s skeptical position when she started taking the anti-depressants.

(b) What is a Type 1 Error in this context?

5.26 Which is higher? In each part below, there is a value of interest and two scenarios (I and II). Foreach part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equalunder the scenarios.

(a) The standard error of p̂ when (I) n = 125 or (II) n = 500.

(b) The margin of error of a confidence interval when the confidence level is (I) 90% or (II) 80%.

(c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with n = 500 or based on a (II) samplewith n = 1000.

(d) The probability of making a Type 2 Error when the alternative hypothesis is true and the significancelevel is (I) 0.05 or (II) 0.10.

24Rasmussen Reports survey, Most Favor Minimum Wage of $10.50 Or Higher, April 16, 2019.

204 CHAPTER 5. FOUNDATIONS FOR INFERENCE

Chapter exercises

5.27 Relaxing after work. The General Social Survey asked the question: “After an average work day,about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of1,155 Americans.25 A 95% confidence interval for the mean number of hours spent relaxing or pursuingactivities they enjoy was (1.38, 1.92).

(a) Interpret this interval in context of the data.

(b) Suppose another set of researchers reported a confidence interval with a larger margin of error based onthe same sample of 1,155 Americans. How does their confidence level compare to the confidence levelof the interval stated above?

(c) Suppose next year a new survey asking the same question is conducted, and this time the sample sizeis 2,500. Assuming that the population characteristics, with respect to how much time people spendrelaxing after work, have not changed much within a year. How will the margin of error of the 95%confidence interval constructed based on data from the new survey compare to the margin of error ofthe interval stated above?

5.28 Minimum wage, Part II. In Exercise 5.21, we learned that a Rasmussen Reports survey of 1,000 USadults found that 42% believe raising the minimum wage will help the economy. Construct a 99% confidenceinterval for the true proportion of US adults who believe this.

5.29 Testing for food safety. A food safety inspector is called upon to investigate a restaurant with a fewcustomer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing frameworkto evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its licenseto serve food will be revoked.

(a) Write the hypotheses in words.

(b) What is a Type 1 Error in this context?

(d) Which error is more problematic for the restaurant owner? Why?

(e) Which error is more problematic for the diners? Why?

(f) As a diner, would you prefer that the food safety inspector requires strong evidence or very strongevidence of health concerns before revoking a restaurant’s license? Explain your reasoning.

5.30 True or false. Determine if the following statements are true or false, and explain your reasoning. Iffalse, state how it could be corrected.

(a) If a given value (for example, the null hypothesized value of a parameter) is within a 95% confidenceinterval, it will also be within a 99% confidence interval.

(b) Decreasing the significance level (α) will increase the probability of making a Type 1 Error.

(c) Suppose the null hypothesis is p = 0.5 and we fail to reject H0. Under this scenario, the true populationproportion is 0.5.

(d) With large sample sizes, even small differences between the null value and the observed point estimate,a difference often called the effect size, will be identified as statistically significant.

5.31 Unemployment and relationship problems. A USA Today/Gallup poll asked a group of unemployedand underemployed Americans if they have had major problems in their relationships with their spouse oranother close family member as a result of not having a job (if unemployed) or not having a full-time job (ifunderemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondentssaid they had major problems in relationships as a result of their employment status.

(a) What are the hypotheses for evaluating if the proportions of unemployed and underemployed peoplewho had relationship problems were different?

(b) The p-value for this hypothesis test is approximately 0.35. Explain what this means in context of thehypothesis test and the data.

25National Opinion Research Center, General Social Survey, 2018.

5.3. HYPOTHESIS TESTING FOR A PROPORTION 205

5.32 Nearsighted. It is believed that nearsightedness affects about 8% of all children. In a random sampleof 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these dataprovide evidence that the 8% value is inaccurate?

5.33 Nutrition labels. The nutrition label on a bag of potato chips says that a one ounce (28 gram) servingof potato chips has 130 calories and contains ten grams of fat, with three grams of saturated fat. A randomsample of 35 bags yielded a confidence interval for the number of calories per bag of 128.2 to 139.8 calories.Is there evidence that the nutrition label does not provide an accurate measure of calories in the bags ofpotato chips?

5.34 CLT for proportions. Define the term “sampling distribution” of the sample proportion, and describehow the shape, center, and spread of the sampling distribution change as the sample size increases whenp = 0.1.

5.35 Practical vs. statistical significance. Determine whether the following statement is true or false,and explain your reasoning: “With large sample sizes, even small differences between the null value and theobserved point estimate can be statistically significant.”

5.36 Same observation, different sample size. Suppose you conduct a hypothesis test based on a samplewhere the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes anddiscover that you made a careless mistake, the sample size should have been n = 500. Will your p-valueincrease, decrease, or stay the same? Explain.

5.37 Gender pay gap in medicine. A study examined the average pay for men and women entering theworkforce as doctors for 21 different positions.26

(a) If each gender was equally paid, then we would expect about half of those positions to have men paidmore than women and women would be paid more than men in the other half of positions. Writeappropriate hypotheses to test this scenario.

(b) Men were, on average, paid more in 19 of those 21 positions. Complete a hypothesis test using yourhypotheses from part (a).

26Lo Sasso AT et al. “The $16,819 Pay Gap For Newly Trained Physicians: The Unexplained Trend Of MenEarning More Than Women”. In: Health Affairs 30.2 (2011).

206

Chapter 6Inference for categorical data

6.1 Inference for a single proportion

6.2 Difference of two proportions

6.3 Testing for goodness of fit using chi-square

6.4 Testing for independence in two-way tables

207

In this chapter, we apply the methods and ideas from Chapter 5 in

several contexts for categorical data. We’ll start by revisiting what we

learned for a single proportion, where the normal distribution can be

used to model the uncertainty in the sample proportion. Next, we apply

these same ideas to analyze the difference of two proportions using the

normal model. Later in the chapter, we apply inference techniques to

contingency tables; while we will use a different distribution in this

context, the core ideas of hypothesis testing remain the same.

For videos, slides, and other resources, please visit

www.openintro.org/os

208 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.1 Inference for a single proportion

We encountered inference methods for a single proportion in Chapter 5, exploring point esti-mates, confidence intervals, and hypothesis tests. In this section, we’ll do a review of these topics andalso how to choose an appropriate sample size when collecting data for single proportion contexts.

6.1.1 Identifying when the sample proportion is nearly normal

A sample proportion p̂ can be modeled using a normal distribution when the sample observationsare independent and the sample size is sufficiently large.

SAMPLING DISTRIBUTION OF p̂̂p̂p

The sampling distribution for p̂ based on a sample of size n from a population with a trueproportion p is nearly normal when:

1. The sample’s observations are independent, e.g. are from a simple random sample.

2. We expected to see at least 10 successes and 10 failures in the sample, i.e. np ≥ 10 andn(1− p) ≥ 10. This is called the success-failure condition.

When these conditions are met, then the sampling distribution of p̂ is nearly normal with mean

p and standard error SE =√

p(1−p)n .

Typically we don’t know the true proportion p, so we substitute some value to check conditionsand estimate the standard error. For confidence intervals, the sample proportion p̂ is used to checkthe success-failure condition and compute the standard error. For hypothesis tests, typically thenull value – that is, the proportion claimed in the null hypothesis – is used in place of p.

6.1.2 Confidence intervals for a proportion

A confidence interval provides a range of plausible values for the parameter p, and when p̂ canbe modeled using a normal distribution, the confidence interval for p takes the form

p̂± z? × SE

EXAMPLE 6.1

A simple random sample of 826 payday loan borrowers was surveyed to better understand theirinterests around regulation and costs. 70% of the responses supported new regulations on paydaylenders. Is it reasonable to model p̂ = 0.70 using a normal distribution?

The data are a random sample, so the observations are independent and representative of thepopulation of interest.

We also must check the success-failure condition, which we do using p̂ in place of p when computinga confidence interval:

Support: np ≈ 826× 0.70 = 578 Not: n(1− p) ≈ 826× (1− 0.70) = 248

Since both values are at least 10, we can use the normal distribution to model p̂.

6.1. INFERENCE FOR A SINGLE PROPORTION 209

GUIDED PRACTICE 6.2

Estimate the standard error of p̂ = 0.70. Because p is unknown and the standard error is for aconfidence interval, use p̂ in place of p in the formula.1

EXAMPLE 6.3

Construct a 95% confidence interval for p, the proportion of payday borrowers who support increasedregulation for payday lenders.

Using the point estimate 0.70, z? = 1.96 for a 95% confidence interval, and the standard errorSE = 0.016 from Guided Practice 6.2, the confidence interval is

point estimate ± z? × SE → 0.70 ± 1.96× 0.016 → (0.669, 0.731)

We are 95% confident that the true proportion of payday borrowers who supported regulation atthe time of the poll was between 0.669 and 0.731.

CONFIDENCE INTERVAL FOR A SINGLE PROPORTION

Once you’ve determined a one-proportion confidence interval would be helpful for an application,there are four steps to constructing the interval:

Prepare. Identify p̂ and n, and determine what confidence level you wish to use.

Check. Verify the conditions to ensure p̂ is nearly normal. For one-proportion confidenceintervals, use p̂ in place of p to check the success-failure condition.

Calculate. If the conditions hold, compute SE using p̂, find z?, and construct the interval.

Conclude. Interpret the confidence interval in the context of the problem.

For additional one-proportion confidence interval examples, see Section 5.2.

6.1.3 Hypothesis testing for a proportion

One possible regulation for payday lenders is that they would be required to do a credit checkand evaluate debt payments against the borrower’s finances. We would like to know: would borrowerssupport this form of regulation?

GUIDED PRACTICE 6.4

Set up hypotheses to evaluate whether borrowers have a majority support or majority oppositionfor this type of regulation.2

To apply the normal distribution framework in the context of a hypothesis test for a proportion,the independence and success-failure conditions must be satisfied. In a hypothesis test, the success-failure condition is checked using the null proportion: we verify np0 and n(1 − p0) are at least 10,where p0 is the null value.

1SE =√p(1−p)n

≈√

0.70(1−0.70)826

= 0.016.2H0: p = 0.50. HA: p 6= 0.50.

210 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

GUIDED PRACTICE 6.5

Do payday loan borrowers support a regulation that would require lenders to pull their credit reportand evaluate their debt payments? From a random sample of 826 borrowers, 51% said they wouldsupport such a regulation. Is it reasonable to model p̂ = 0.51 using a normal distribution for ahypothesis test here?3

EXAMPLE 6.6

Using the hypotheses and data from Guided Practice 6.4 and 6.5, evaluate whether the poll providesconvincing evidence that a majority of payday loan borrowers support a new regulation that wouldrequire lenders to pull credit reports and evaluate debt payments.

With hypotheses already set up and conditions checked, we can move onto calculations. The standarderror in the context of a one-proportion hypothesis test is computed using the null value, p0:

SE =

√p0(1− p0)

√0.5(1− 0.5)

826= 0.017

A picture of the normal model is shown below with the p-value represented by the shaded region.

0.45 0.47 0.48 0.5 0.52 0.53 0.55

Based on the normal model, the test statistic can be computed as the Z-score of the point estimate:

Z =point estimate− null value

SE=

0.51− 0.50

0.017= 0.59

The single tail area is 0.2776, and the p-value, represented by both tail areas together, is 0.5552.Because the p-value is larger than 0.05, we do not reject H0. The poll does not provide convincingevidence that a majority of payday loan borrowers support or oppose regulations around creditchecks and evaluation of debt payments.

HYPOTHESIS TESTING FOR A SINGLE PROPORTION

Once you’ve determined a one-proportion hypothesis test is the correct procedure, there arefour steps to completing the test:

Prepare. Identify the parameter of interest, list hypotheses, identify the significance level, andidentify p̂ and n.

Check. Verify conditions to ensure p̂ is nearly normal under H0. For one-proportion hypothesistests, use the null value to check the success-failure condition.

Calculate. If the conditions hold, compute the standard error, again using p0, compute theZ-score, and identify the p-value.

Conclude. Evaluate the hypothesis test by comparing the p-value to α, and provide a conclu-sion in the context of the problem.

For additional one-proportion hypothesis test examples, see Section 5.3.

3Independence holds since the poll is based on a random sample. The success-failure condition also holds, whichis checked using the null value (p0 = 0.5) from H0: np0 = 826× 0.5 = 413, n(1− p0) = 826× 0.5 = 413.

6.1. INFERENCE FOR A SINGLE PROPORTION 211

6.1.4 When one or more conditions aren’t met

We’ve spent a lot of time discussing conditions for when p̂ can be reasonably modeled by anormal distribution. What happens when the success-failure condition fails? What about when theindependence condition fails? In either case, the general ideas of confidence intervals and hypothesistests remain the same, but the strategy or technique used to generate the interval or p-value change.

When the success-failure condition isn’t met for a hypothesis test, we can simulate the nulldistribution of p̂ using the null value, p0. The simulation concept is similar to the ideas used in themalaria case study presented in Section 2.3, and an online section outlines this strategy:

www.openintro.org/r?go=stat sim prop ht

For a confidence interval when the success-failure condition isn’t met, we can use what’s called theClopper-Pearson interval. The details are beyond the scope of this book. However, there aremany internet resources covering this topic.

The independence condition is a more nuanced requirement. When it isn’t met, it is importantto understand how and why it isn’t met. For example, if we took a cluster sample (see Section 1.3),suitable statistical methods are available but would be beyond the scope of even most second orthird courses in statistics. On the other hand, we’d be stretched to find any method that we couldconfidently apply to correct the inherent biases of data from a convenience sample.

While this book is scoped to well-constrained statistical problems, do remember that this isjust the first book in what is a large library of statistical methods that are suitable for a very widerange of data and contexts.

212 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.1.5 Choosing a sample size when estimating a proportion

When collecting data, we choose a sample size suitable for the purpose of the study. Oftentimes this means choosing a sample size large enough that the margin of error – which is thepart we add and subtract from the point estimate in a confidence interval – is sufficiently small thatthe sample is useful. For example, our task might be to find a sample size n so that the sampleproportion is within ±0.04 of the actual proportion in a 95% confidence interval.

EXAMPLE 6.7

A university newspaper is conducting a survey to determine what fraction of students support a$200 per year increase in fees to pay for a new football stadium. How big of a sample is required toensure the margin of error is smaller than 0.04 using a 95% confidence level?

The margin of error for a sample proportion is

z?√p(1− p)

Our goal is to find the smallest sample size n so that this margin of error is smaller than 0.04. Fora 95% confidence level, the value z? corresponds to 1.96:

1.96×√p(1− p)

n< 0.04

There are two unknowns in the equation: p and n. If we have an estimate of p, perhaps from a priorsurvey, we could enter in that value and solve for n. If we have no such estimate, we must use someother value for p. It turns out that the margin of error is largest when p is 0.5, so we typically usethis worst case value if no estimate of the proportion is available:

1.96×√

0.5(1− 0.5)

n< 0.04

1.962 × 0.5(1− 0.5)

n< 0.042

1.962 × 0.5(1− 0.5)

0.042< n

600.25 < n

We would need over 600.25 participants, which means we need 601 participants or more, to ensurethe sample proportion is within 0.04 of the true proportion with 95% confidence.

When an estimate of the proportion is available, we use it in place of the worst case proportionvalue, 0.5.

6.1. INFERENCE FOR A SINGLE PROPORTION 213

GUIDED PRACTICE 6.8

A manager is about to oversee the mass production of a new tire model in her factory, and she wouldlike to estimate what proportion of these tires will be rejected through quality control. The qualitycontrol team has monitored the last three tire models produced by the factory, failing 1.7% of tiresin the first model, 6.2% of the second model, and 1.3% of the third model. The manager would liketo examine enough tires to estimate the failure rate of the new tire model to within about 1% witha 90% confidence level. There are three different failure rates to choose from. Perform the samplesize computation for each separately, and identify three sample sizes to consider.4

EXAMPLE 6.9

The sample sizes vary widely in Guided Practice 6.8. Which of the three would you suggest using?What would influence your choice?

We could examine which of the old models is most like the new model, then choose the correspondingsample size. Or if two of the previous estimates are based on small samples while the other is basedon a larger sample, we might consider the value corresponding to the larger sample. There are alsoother reasonable approaches.

Also observe that the success-failure condition would need to be checked in the final sample. Forinstance, if we sampled n = 1584 tires and found a failure rate of 0.5%, the normal approximationwould not be reasonable, and we would require more advanced statistical methods for creating theconfidence interval.

GUIDED PRACTICE 6.10

Suppose we want to continually track the support of payday borrowers for regulation on lenders,where we would conduct a new poll every month. Running such frequent polls is expensive, so wedecide a wider margin of error of 5% for each individual survey would be acceptable. Based onthe original sample of borrowers where 70% supported some form of regulation, how big should ourmonthly sample be for a margin of error of 0.05 with 95% confidence?5

4For a 90% confidence interval, z? = 1.65, and since an estimate of the proportion 0.017 is available, we’ll use itin the margin of error formula:

1.65×√

0.017(1− 0.017)

n< 0.01 →

0.017(1− 0.017)

(0.01

1.65

→ 454.96 < n

For sample size calculations, we always round up, so the first tire model suggests 455 tires would be sufficient.A similar computation can be accomplished using 0.062 and 0.013 for p, and you should verify that using these

proportions results in minimum sample sizes of 1584 and 350 tires, respectively.5We complete the same computations as before, except now we use 0.70 instead of 0.5 for p:

1.96×√p(1− p)

n≈ 1.96×

√0.70(1− 0.70)

n≤ 0.05 → n ≥ 322.7

A sample size of 323 or more would be reasonable. (Reminder: always round up for sample size calculations!) Giventhat we plan to track this poll over time, we also may want to periodically repeat these calculations to ensure thatwe’re being thoughtful in our sample size recommendations in case the baseline rate fluctuates.

214 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

Exercises

6.1 Vegetarian college students. Suppose that 8% of college students are vegetarians. Determine if thefollowing statements are true or false, and explain your reasoning.

(a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximatelynormal since n ≥ 30.

(b) The distribution of the sample proportions of vegetarian college students in random samples of size 50is right skewed.

(d) A random sample of 250 college students where 12% are vegetarians would be considered unusual.

(e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250.

6.2 Young Americans, Part I. About 77% of young adults think they can achieve the American dream.Determine if the following statements are true or false, and explain your reasoning.6

(a) The distribution of sample proportions of young Americans who think they can achieve the Americandream in samples of size 20 is left skewed.

(b) The distribution of sample proportions of young Americans who think they can achieve the Americandream in random samples of size 40 is approximately normal since n ≥ 30.

(c) A random sample of 60 young Americans where 85% think they can achieve the American dream wouldbe considered unusual.

(d) A random sample of 120 young Americans where 85% think they can achieve the American dream wouldbe considered unusual.

6.3 Orange tabbies. Suppose that 90% of orange tabby cats are male. Determine if the following state-ments are true or false, and explain your reasoning.

(a) The distribution of sample proportions of random samples of size 30 is left skewed.

(b) Using a sample size that is 4 times as large will reduce the standard error of the sample proportion byone-half.

(d) The distribution of sample proportions of random samples of size 280 is approximately normal.

6.4 Young Americans, Part II. About 25% of young Americans have delayed starting a family due tothe continued economic slump. Determine if the following statements are true or false, and explain yourreasoning.7

(a) The distribution of sample proportions of young Americans who have delayed starting a family due tothe continued economic slump in random samples of size 12 is right skewed.

(b) In order for the distribution of sample proportions of young Americans who have delayed starting afamily due to the continued economic slump to be approximately normal, we need random sampleswhere the sample size is at least 40.

(c) A random sample of 50 young Americans where 20% have delayed starting a family due to the continuedeconomic slump would be considered unusual.

(d) A random sample of 150 young Americans where 20% have delayed starting a family due to the continuedeconomic slump would be considered unusual.

(e) Tripling the sample size will reduce the standard error of the sample proportion by one-third.

6A. Vaughn. “Poll finds young adults optimistic, but not about money”. In: Los Angeles Times (2011).7Demos.org. “The State of Young America: The Poll”. In: (2011).

6.1. INFERENCE FOR A SINGLE PROPORTION 215

6.5 Gender equality. The General Social Survey asked a random sample of 1,390 Americans the followingquestion: “On the whole, do you think it should or should not be the government’s responsibility to promoteequality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level,this sample has 2% margin of error. Based on this information, determine if the following statements aretrue or false, and explain your reasoning.8

(a) We are 95% confident that between 80% and 84% of Americans in this sample think it’s the government’sresponsibility to promote equality between men and women.

(b) We are 95% confident that between 80% and 84% of all Americans think it’s the government’s respon-sibility to promote equality between men and women.

(c) If we considered many random samples of 1,390 Americans, and we calculated 95% confidence intervalsfor each, 95% of these intervals would include the true population proportion of Americans who thinkit’s the government’s responsibility to promote equality between men and women.

(d) In order to decrease the margin of error to 1%, we would need to quadruple (multiply by 4) the samplesize.

(e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of Americansthink it’s the government’s responsibility to promote equality between men and women.

6.6 Elderly drivers. The Marist Poll published a report stating that 66% of adults nationally think licenseddrivers should be required to retake their road test once they reach 65 years of age. It was also reportedthat interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95%confidence level.9

(a) Verify the margin of error reported by The Marist Poll.

(b) Based on a 95% confidence interval, does the poll provide convincing evidence that more than 70% ofthe population think that licensed drivers should be required to retake their road test once they turn65?

6.7 Fireworks on July 4th. A local news outlet reported that 56% of 600 randomly sampled Kansasresidents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimateusing a 95% confidence level.10

6.8 Life rating in Greece. Greece has faced a severe economic crisis since the end of 2009. A Gallup pollsurveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate theirlives poorly enough to be considered “suffering”.11

(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?

(b) Check if the conditions required for constructing a confidence interval based on these data are met.

(d) Without doing any calculations, describe what would happen to the confidence interval if we decided touse a higher confidence level.

(e) Without doing any calculations, describe what would happen to the confidence interval if we used alarger sample.

6.9 Study abroad. A survey on 1,509 high school seniors who took the SAT and who completed an optionalweb survey shows that 55% of high school seniors are fairly certain that they will participate in a studyabroad program in college.12

(a) Is this sample a representative sample from the population of all high school seniors in the US? Explainyour reasoning.

(b) Let’s suppose the conditions for inference are met. Even if your answer to part (a) indicated that thisapproach would not be reliable, this analysis may still be interesting to carry out (though not report).Construct a 90% confidence interval for the proportion of high school seniors (of those who took theSAT) who are fairly certain they will participate in a study abroad program in college, and interpretthis interval in context.

(d) Based on this interval, would it be appropriate to claim that the majority of high school seniors arefairly certain that they will participate in a study abroad program in college?

8National Opinion Research Center, General Social Survey, 2018.9Marist Poll, Road Rules: Re-Testing Drivers at Age 65?, March 4, 2011.

10Survey USA, News Poll #19333, data collected on June 27, 2012.11Gallup World, More Than One in 10 “Suffering” Worldwide, data collected throughout 2011.12studentPOLL, College-Bound Students’ Interests in Study Abroad and Other International Learning Activities,

January 2008.

216 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.10 Legalization of marijuana, Part I. The General Social Survey asked 1,578 US residents: “Do youthink the use of marijuana should be made legal, or not?” 61% of the respondents said it should be madelegal.13

(a) Is 61% a sample statistic or a population parameter? Explain.

(b) Construct a 95% confidence interval for the proportion of US residents who think marijuana should bemade legal, and interpret it in the context of the data.

(c) A critic points out that this 95% confidence interval is only accurate if the statistic follows a normaldistribution, or if the normal model is a good approximation. Is this true for these data? Explain.

(d) A news piece on this survey’s findings states, “Majority of Americans think marijuana should be legal-ized.” Based on your confidence interval, is this news piece’s statement justified?

6.11 National Health Plan, Part I. A Kaiser Family Foundation poll for US adults in 2019 found that 79%of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”.There were 347 Democrats, 298 Republicans, and 617 Independents surveyed.14

(a) A political pundit on TV claims that a majority of Independents support a National Health Plan. Dothese data provide strong evidence to support this type of statement?

(b) Would you expect a confidence interval for the proportion of Independents who oppose the public optionplan to include 0.5? Explain.

6.12 Is college worth it? Part I. Among a simple random sample of 331 American adults who do not havea four-year college degree and are not currently enrolled in school, 48% said they decided not to go to collegebecause they could not afford school.15

(a) A newspaper article states that only a minority of the Americans who decide not to go to college doso because they cannot afford it and uses the point estimate from this survey as evidence. Conduct ahypothesis test to determine if these data provide strong evidence supporting this statement.

(b) Would you expect a confidence interval for the proportion of American adults who decide not to go tocollege because they cannot afford it to include 0.5? Explain.

6.13 Taste test. Some people claim that they can tell the difference between a diet soda and a regular sodain the first sip. A researcher wanting to test this claim randomly sampled 80 such people. He then filled 80plain white cups with soda, half diet and half regular through random assignment, and asked each personto take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identifiedthe soda.

(a) Do these data provide strong evidence that these people are any better or worse than random guessingat telling the difference between diet and regular soda?

(b) Interpret the p-value in this context.

6.14 Is college worth it? Part II. Exercise 6.12 presents the results of a poll where 48% of 331 Americanswho decide to not go to college do so because they cannot afford it.

(a) Calculate a 90% confidence interval for the proportion of Americans who decide to not go to collegebecause they cannot afford it, and interpret the interval in context.

(b) Suppose we wanted the margin of error for the 90% confidence level to be about 1.5%. How large of asurvey would you recommend?

6.15 National Health Plan, Part II. Exercise 6.11 presents the results of a poll evaluating support for ageneric “National Health Plan” in the US in 2019, reporting that 55% of Independents are supportive. If wewanted to estimate this number to within 1% with 90% confidence, what would be an appropriate samplesize?

6.16 Legalize Marijuana, Part II. As discussed in Exercise 6.10, the General Social Survey reported asample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit themargin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey?

13National Opinion Research Center, General Social Survey, 2018.14Kaiser Family Foundation, The Public On Next Steps For The ACA And Proposals To Expand Coverage, data

collected between Jan 9-14, 2019.15Pew Research Center Publications, Is College Worth It?, data collected between March 15-29, 2011.

6.2. DIFFERENCE OF TWO PROPORTIONS 217

6.2 Difference of two proportions

We would like to extend the methods from Section 6.1 to apply confidence intervals and hy-pothesis tests to differences in population proportions: p1 − p2. In our investigations, we’ll identifya reasonable point estimate of p1 − p2 based on the sample, and you may have already guessed itsform: p̂1 − p̂2. Next, we’ll apply the same processes we used in the single-proportion context: weverify that the point estimate can be modeled using a normal distribution, we compute the estimate’sstandard error, and we apply our inferential framework.

6.2.1 Sampling distribution of the difference of two proportions

Like with p̂, the difference of two sample proportions p̂1 − p̂2 can be modeled using a normaldistribution when certain conditions are met. First, we require a broader independence condition,and secondly, the success-failure condition must be met by both groups.

CONDITIONS FOR THE SAMPLING DISTRIBUTION OF p̂1 − p̂2p̂1 − p̂2p̂1 − p̂2 TO BE NORMAL

The difference p̂1 − p̂2 can be modeled using a normal distribution when

• Independence, extended. The data are independent within and between the two groups.Generally this is satisfied if the data come from two independent random samples or ifthe data come from a randomized experiment.

• Success-failure condition. The success-failure condition holds for both groups, where wecheck successes and failures in each group separately.

When these conditions are satisfied, the standard error of p̂1 − p̂2 is

SE =

√p1(1− p1)

n1+p2(1− p2)

where p1 and p2 represent the population proportions, and n1 and n2 represent the sample sizes.

6.2.2 Confidence intervals for p1 − p2p1 − p2p1 − p2

We can apply the generic confidence interval formula for a difference of two proportions, where weuse p̂1 − p̂2 as the point estimate and substitute the SE formula:

point estimate ± z? × SE → p̂1 − p̂2 ± z? ×

√p1(1− p1)

n1+p2(1− p2)

We can also follow the same Prepare, Check, Calculate, Conclude steps for computing a confidenceinterval or completing a hypothesis test. The details change a little, but the general approach remainthe same. Think about these steps when you apply statistical methods.

218 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

EXAMPLE 6.11

We consider an experiment for patients who underwent cardiopulmonary resuscitation (CPR) for aheart attack and were subsequently admitted to a hospital. These patients were randomly dividedinto a treatment group where they received a blood thinner or the control group where they did notreceive a blood thinner. The outcome variable of interest was whether the patients survived for atleast 24 hours. The results are shown in Figure 6.1. Check whether we can model the difference insample proportions using the normal distribution.

We first check for independence: since this is a randomized experiment, this condition is satisfied.

Next, we check the success-failure condition for each group. We have at least 10 successes and 10failures in each experiment arm (11, 14, 39, 26), so this condition is also satisfied.

With both conditions satisfied, the difference in sample proportions can be reasonably modeled usinga normal distribution for these data.

Survived Died TotalControl 11 39 50Treatment 14 26 40Total 25 65 90

Figure 6.1: Results for the CPR study. Patients in the treatment group were givena blood thinner, and patients in the control group were not.

EXAMPLE 6.12

Create and interpret a 90% confidence interval of the difference for the survival rates in the CPRstudy.

We’ll use pt for the survival rate in the treatment group and pc for the control group:

p̂t − p̂c =14

40− 11

50= 0.35− 0.22 = 0.13

We use the standard error formula provided on page 217. As with the one-sample proportion case,we use the sample estimates of each proportion in the formula in the confidence interval context:

SE ≈√

0.35(1− 0.35)

40+

0.22(1− 0.22)

50= 0.095

For a 90% confidence interval, we use z? = 1.65:

point estimate ± z? × SE → 0.13 ± 1.65× 0.095 → (−0.027, 0.287)

We are 90% confident that blood thinners have a difference of -2.7% to +28.7% percentage pointimpact on survival rate for patients who are like those in the study. Because 0% is contained inthe interval, we do not have enough information to say whether blood thinners help or harm heartattack patients who have been admitted after they have undergone CPR.

6.2. DIFFERENCE OF TWO PROPORTIONS 219

GUIDED PRACTICE 6.13

A 5-year experiment was conducted to evaluate the effectiveness of fish oils on reducing cardiovascularevents, where each subject was randomized into one of two treatment groups. We’ll consider heartattack outcomes in these patients:

heart attack no event Totalfish oil 145 12788 12933placebo 200 12738 12938

Create a 95% confidence interval for the effect of fish oils on heart attacks for patients who arewell-represented by those in the study. Also interpret the interval in the context of the study.16

6.2.3 Hypothesis tests for the difference of two proportions

A mammogram is an X-ray procedure used to check for breast cancer. Whether mammogramsshould be used is part of a controversial discussion, and it’s the topic of our next example where welearn about 2-proportion hypothesis tests when H0 is p1 − p2 = 0 (or equivalently, p1 = p2).

A 30-year study was conducted with nearly 90,000 female participants. During a 5-year screen-ing period, each woman was randomized to one of two groups: in the first group, women receivedregular mammograms to screen for breast cancer, and in the second group, women received regularnon-mammogram breast cancer exams. No intervention was made during the following 25 years ofthe study, and we’ll consider death resulting from breast cancer over the full 30-year period. Resultsfrom the study are summarized in Figure 6.2.

If mammograms are much more effective than non-mammogram breast cancer exams, then wewould expect to see additional deaths from breast cancer in the control group. On the other hand, ifmammograms are not as effective as regular breast cancer exams, we would expect to see an increasein breast cancer deaths in the mammogram group.

Death from breast cancer?Yes No

Mammogram 500 44,425Control 505 44,405

Figure 6.2: Summary results for breast cancer study.

GUIDED PRACTICE 6.14

Is this study an experiment or an observational study?17

16 Because the patients were randomized, the subjects are independent, both within and between the two groups.The success-failure condition is also met for both groups as all counts are at least 10. This satisfies the conditionsnecessary to model the difference in proportions using a normal distribution.

Compute the sample proportions (p̂fish oil = 0.0112, p̂placebo = 0.0155), point estimate of the difference (0.0112 −0.0155 = −0.0043), and standard error (SE =

√0.0112×0.9888

12933+ 0.0155×0.9845

12938= 0.00145). Next, plug the values into

the general formula for a confidence interval, where we’ll use a 95% confidence level with z? = 1.96:

−0.0043± 1.96× 0.00145 → (−0.0071,−0.0015)

We are 95% confident that fish oils decreases heart attacks by 0.15 to 0.71 percentage points (off of a baseline of about1.55%) over a 5-year period for subjects who are similar to those in the study. Because the interval is entirely below 0,the data provide strong evidence that fish oil supplements reduce heart attacks in patients like those in the study.

17This is an experiment. Patients were randomized to receive mammograms or a standard breast cancer exam. Wewill be able to make causal conclusions based on this study.

220 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

GUIDED PRACTICE 6.15

Set up hypotheses to test whether there was a difference in breast cancer deaths in the mammogramand control groups.18

In Example 6.16, we will check the conditions for using a normal distribution to analyze theresults of the study. The details are very similar to that of confidence intervals. However, when thenull hypothesis is that p1 − p2 = 0, we use a special proportion called the pooled proportion tocheck the success-failure condition:

p̂pooled =# of patients who died from breast cancer in the entire study

# of patients in the entire study

=500 + 505

500 + 44,425 + 505 + 44,405

= 0.0112

This proportion is an estimate of the breast cancer death rate across the entire study, and it’s ourbest estimate of the proportions pmgm and pctrl if the null hypothesis is true that pmgm = pctrl.We will also use this pooled proportion when computing the standard error.

EXAMPLE 6.16

Is it reasonable to model the difference in proportions using a normal distribution in this study?

Because the patients are randomized, they can be treated as independent, both within and betweengroups. We also must check the success-failure condition for each group. Under the null hypothesis,the proportions pmgm and pctrl are equal, so we check the success-failure condition with our bestestimate of these values under H0, the pooled proportion from the two samples, p̂pooled = 0.0112:

p̂pooled × nmgm = 0.0112× 44,925 = 503 (1− p̂pooled)× nmgm = 0.9888× 44,925 = 44,422

p̂pooled × nctrl = 0.0112× 44,910 = 503 (1− p̂pooled)× nctrl = 0.9888× 44,910 = 44,407

The success-failure condition is satisfied since all values are at least 10. With both conditionssatisfied, we can safely model the difference in proportions using a normal distribution.

USE THE POOLED PROPORTION WHENH0H0H0 IS p1 − p2 = 0p1 − p2 = 0p1 − p2 = 0

When the null hypothesis is that the proportions are equal, use the pooled proportion (p̂pooled)to verify the success-failure condition and estimate the standard error:

p̂pooled =number of “successes”

number of cases=p̂1n1 + p̂2n2

n1 + n2

Here p̂1n1 represents the number of successes in sample 1 since

p̂1 =number of successes in sample 1

Similarly, p̂2n2 represents the number of successes in sample 2.

In Example 6.16, the pooled proportion was used to check the success-failure condition.19 In thenext example, we see the second place where the pooled proportion comes into play: the standarderror calculation.

18H0: the breast cancer death rate for patients screened using mammograms is the same as the breast cancer deathrate for patients in the control, pmgm − pctrl = 0.HA: the breast cancer death rate for patients screened using mammograms is different than the breast cancer deathrate for patients in the control, pmgm − pctrl 6= 0.

19For an example of a two-proportion hypothesis test that does not require the success-failure condition to be met,see Section 2.3.

6.2. DIFFERENCE OF TWO PROPORTIONS 221

EXAMPLE 6.17

Compute the point estimate of the difference in breast cancer death rates in the two groups, anduse the pooled proportion p̂pooled = 0.0112 to calculate the standard error.

The point estimate of the difference in breast cancer death rates is

p̂mgm − p̂ctrl =500

500 + 44, 425− 505

505 + 44, 405

= 0.01113− 0.01125

= −0.00012

The breast cancer death rate in the mammogram group was 0.012% less than in the control group.Next, the standard error is calculated using the pooled proportion, p̂pooled:

SE =

√p̂pooled(1− p̂pooled)

nmgm+p̂pooled(1− p̂pooled)

nctrl= 0.00070

EXAMPLE 6.18

Using the point estimate p̂mgm − p̂ctrl = −0.00012 and standard error SE = 0.00070, calculate ap-value for the hypothesis test and write a conclusion.

Just like in past tests, we first compute a test statistic and draw a picture:

Z =point estimate− null value

SE=−0.00012− 0

0.00070= −0.17

−0.0014 0 0.0014

The lower tail area is 0.4325, which we double to get the p-value: 0.8650. Because this p-value islarger than 0.05, we do not reject the null hypothesis. That is, the difference in breast cancer deathrates is reasonably explained by chance, and we do not observe benefits or harm from mammogramsrelative to a regular breast exam.

Can we conclude that mammograms have no benefits or harm? Here are a few considerationsto keep in mind when reviewing the mammogram study as well as any other medical study:

• We do not accept the null hypothesis, which means we don’t have sufficient evidence to concludethat mammograms reduce or increase breast cancer deaths.

• If mammograms are helpful or harmful, the data suggest the effect isn’t very large.

• Are mammograms more or less expensive than a non-mammogram breast exam? If one optionis much more expensive than the other and doesn’t offer clear benefits, then we should leantowards the less expensive option.

• The study’s authors also found that mammograms led to overdiagnosis of breast cancer, whichmeans some breast cancers were found (or thought to be found) but that these cancers wouldnot cause symptoms during patients’ lifetimes. That is, something else would kill the patientbefore breast cancer symptoms appeared. This means some patients may have been treated forbreast cancer unnecessarily, and this treatment is another cost to consider. It is also importantto recognize that overdiagnosis can cause unnecessary physical or emotional harm to patients.

These considerations highlight the complexity around medical care and treatment recommendations.Experts and medical boards who study medical treatments use considerations like those above toprovide their best recommendation based on the current evidence.

222 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.2.4 More on 2-proportion hypothesis tests (special topic)

When we conduct a 2-proportion hypothesis test, usually H0 is p1 − p2 = 0. However, thereare rare situations where we want to check for some difference in p1 and p2 that is some value otherthan 0. For example, maybe we care about checking a null hypothesis where p1 − p2 = 0.1. Incontexts like these, we generally use p̂1 and p̂2 to check the success-failure condition and constructthe standard error.

GUIDED PRACTICE 6.19

A quadcopter company is considering a new manufacturer for rotor blades. The new manufacturerwould be more expensive, but they claim their higher-quality blades are more reliable, with 3% moreblades passing inspection than their competitor. Set up appropriate hypotheses for the test.20

Figure 6.3: A Phantom quadcopter.—————————–Photo by David J (http://flic.kr/p/oiWLNu). CC-BY 2.0 license.

This photo has been cropped and a border has been added.

20H0: The higher-quality blades will pass inspection 3% more frequently than the standard-quality blades. phighQ−pstandard = 0.03. HA: The higher-quality blades will pass inspection some amount different than 3% more often thanthe standard-quality blades. phighQ − pstandard 6= 0.03.

6.2. DIFFERENCE OF TWO PROPORTIONS 223

EXAMPLE 6.20

The quality control engineer from Guided Practice 6.19 collects a sample of blades, examining 1000blades from each company, and she finds that 899 blades pass inspection from the current supplierand 958 pass inspection from the prospective supplier. Using these data, evaluate the hypothesesfrom Guided Practice 6.19 with a significance level of 5%.

First, we check the conditions. The sample is not necessarily random, so to proceed we must assumethe blades are all independent; for this sample we will suppose this assumption is reasonable, butthe engineer would be more knowledgeable as to whether this assumption is appropriate. Thesuccess-failure condition also holds for each sample. Thus, the difference in sample proportions,0.958− 0.899 = 0.059, can be said to come from a nearly normal distribution.

The standard error is computed using the two sample proportions since we do not use a pooledproportion for this context:

SE =

√0.958(1− 0.958)

1000+

0.899(1− 0.899)

1000= 0.0114

In this hypothesis test, because the null is that p1 − p2 = 0.03, the sample proportions were usedfor the standard error calculation rather than a pooled proportion.

Next, we compute the test statistic and use it to find the p-value, which is depicted in Figure 6.4.

Z =point estimate− null value

SE=

0.059− 0.03

0.0114= 2.54

Using a standard normal distribution for this test statistic, we identify the right tail area as 0.006,and we double it to get the p-value: 0.012. We reject the null hypothesis because 0.012 is lessthan 0.05. Since we observed a larger-than-3% increase in blades that pass inspection, we havestatistically significant evidence that the higher-quality blades pass inspection more than 3% asoften as the currently used blades, exceeding the company’s claims.

0.03 0.059(null value)

0.006

Figure 6.4: Distribution of the test statistic if the null hypothesis was true. Thep-value is represented by the shaded areas.

224 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.2.5 Examining the standard error formula (special topic)

This subsection covers more theoretical topics that offer deeper insights into the origins of thestandard error formula for the difference of two proportions. Ultimately, all of the standard errorformulas we encounter in this chapter and in Chapter 7 can be derived from the probability principlesof Section 3.4.

The formula for the standard error of the difference in two proportions can be deconstructedinto the formulas for the standard errors of the individual sample proportions. Recall that thestandard error of the individual sample proportions p̂1 and p̂2 are

SEp̂1 =

√p1(1− p1)

n1SEp̂2 =

√p2(1− p2)

The standard error of the difference of two sample proportions can be deconstructed from thestandard errors of the separate sample proportions:

SEp̂1−p̂2 =√SE2

p̂1+ SE2

p̂2=

√p1(1− p1)

n1+p2(1− p2)

This special relationship follows from probability theory.

GUIDED PRACTICE 6.21

Prerequisite: Section 3.4. We can rewrite the equation above in a different way:

SE2p̂1−p̂2 = SE2

p̂1 + SE2p̂2

Explain where this formula comes from using the formula for the variability of the sum of two randomvariables.21

21The standard error squared represents the variance of the estimate. If X and Y are two random variables withvariances σ2

x and σ2y , then the variance of X − Y is σ2

x + σ2y . Likewise, the variance corresponding to p̂1 − p̂2 is

σ2p̂1

+σ2p̂2

. Because σ2p̂1

and σ2p̂2

are just another way of writing SE2p̂1

and SE2p̂2

, the variance associated with p̂1− p̂2

may be written as SE2p̂1

+ SE2p̂2

6.2. DIFFERENCE OF TWO PROPORTIONS 225

Exercises

6.17 Social experiment, Part I. A “social experiment” conducted by a TV program questioned whatpeople do when they see a very obviously bruised woman getting picked on by her boyfriend. On twodifferent occasions at the same restaurant, the same couple was depicted. In one scenario the woman wasdressed “provocatively” and in the other scenario the woman was dressed “conservatively”. The table belowshows how many restaurant diners were present under each scenario, and whether or not they intervened.

ScenarioProvocative Conservative Total

InterveneYes 5 15 20No 15 10 25Total 20 25 45

Explain why the sampling distribution of the difference between the proportions of interventions underprovocative and conservative scenarios does not follow an approximately normal distribution.

6.18 Heart transplant success. The Stanford University Heart Transplant Study was conducted to de-termine whether an experimental heart transplant program increased lifespan. Each patient entering theprogram was officially designated a heart transplant candidate, meaning that he was gravely ill and mightbenefit from a new heart. Patients were randomly assigned into treatment and control groups. Patients inthe treatment group received a transplant, and those in the control group did not. The table below displayshow many patients survived and died in each group.22

control treatment

alive 4 24dead 30 45

Suppose we are interested in estimating the difference in survival rate between the control and treatmentgroups using a confidence interval. Explain why we cannot construct such an interval using the normalapproximation. What might go wrong if we constructed the confidence interval despite this problem?

6.19 Gender and color preference. A study asked 1,924 male and 3,666 female undergraduate collegestudents their favorite color. A 95% confidence interval for the difference between the proportions of malesand females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on thisinformation, determine if the following statements are true or false, and explain your reasoning for eachstatement you identify as false.23

(a) We are 95% confident that the true proportion of males whose favorite color is black is 2% lower to 6%higher than the true proportion of females whose favorite color is black.

(b) We are 95% confident that the true proportion of males whose favorite color is black is 2% to 6% higherthan the true proportion of females whose favorite color is black.

(c) 95% of random samples will produce 95% confidence intervals that include the true difference betweenthe population proportions of males and females whose favorite color is black.

(d) We can conclude that there is a significant difference between the proportions of males and femaleswhose favorite color is black and that the difference between the two sample proportions is too large toplausibly be due to chance.

(e) The 95% confidence interval for (pfemale − pmale) cannot be calculated with only the information givenin this exercise.

22B. Turnbull et al. “Survivorship of Heart Transplant Data”. In: Journal of the American Statistical Association69 (1974), pp. 74–80.

23L Ellis and C Ficek. “Color preferences according to gender and sexual orientation”. In: Personality andIndividual Differences 31.8 (2001), pp. 1375–1379.

226 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.20 Government shutdown. The United States federal government shutdown of 2018–2019 occurredfrom December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomlysampled Americans during this time period reported that 48% of those who make less than $40,000 peryear and 55% of those who make $40,000 or more per year said the government shutdown has not at allaffected them personally. A 95% confidence interval for (p<40K − p≥40K), where p is the proportion of thosewho said the government shutdown has not at all affected them personally, is (-0.16, 0.02). Based on thisinformation, determine if the following statements are true or false, and explain your reasoning if you identifythe statement as false.24

(a) At the 5% significance level, the data provide convincing evidence of a real difference in the proportionwho are not affected personally between Americans who make less than $40,000 annually and Americanswho make $40,000 annually.

(b) We are 95% confident that 16% more to 2% fewer Americans who make less than $40,000 per year arenot at all personally affected by the government shutdown compared to those who make $40,000 or moreper year.

(d) A 95% confidence interval for (p≥40K − p<40K) is (-0.02, 0.16).

6.21 National Health Plan, Part III. Exercise 6.11 presents the results of a poll evaluating support for agenerically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617Independents support a National Health Plan.

(a) Calculate a 95% confidence interval for the difference between the proportion of Democrats and Inde-pendents who support a National Health Plan (pD − pI), and interpret it in this context. We havealready checked conditions for you.

(b) True or false: If we had picked a random Democrat and a random Independent at the time of this poll,it is more likely that the Democrat would support the National Health Plan than the Independent.

6.22 Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers forDisease Control and Prevention, the proportion of California residents who reported insufficient rest or sleepduring each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. Thesedata are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95%confidence interval for the difference between the proportions of Californians and Oregonians who are sleepdeprived and interpret it in context of the data.25

6.23 Offshore drilling, Part I. A survey asked 827 randomly sampled registered voters in California “Doyou support? Or do you oppose? Drilling for oil and natural gas off the Coast of California? Or do younot know enough to say?” Below is the distribution of responses, separated based on whether or not therespondent graduated from college.26

(a) What percent of college graduates and what percent of thenon-college graduates in this sample do not know enough tohave an opinion on drilling for oil and natural gas off theCoast of California?

(b) Conduct a hypothesis test to determine if the data providestrong evidence that the proportion of college graduates whodo not have an opinion on this issue is different than that ofnon-college graduates.

College GradYes No

Support 154 132Oppose 180 126Do not know 104 131Total 438 389

6.24 Sleep deprivation, CA vs. OR, Part II. Exercise 6.22 provides data on sleep deprivation rates ofCalifornians and Oregonians. The proportion of California residents who reported insufficient rest or sleepduring each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These dataare based on simple random samples of 11,545 California and 4,691 Oregon residents.

(a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivationis different for the two states. (Reminder: Check conditions)

(b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error wasmade?

24Survey USA, News Poll #24568, data collected on April 21, 2019.25CDC, Perceived Insufficient Rest or Sleep Among Adults — United States, 2008.26Survey USA, Election Poll #16804, data collected July 8-11, 2010.

6.2. DIFFERENCE OF TWO PROPORTIONS 227

6.25 Offshore drilling, Part II. Results of a poll evaluating support for drilling for oil and natural gas offthe coast of California were introduced in Exercise 6.23.

College GradYes No

Support 154 132Oppose 180 126Do not know 104 131Total 438 389

(a) What percent of college graduates and what percent of the non-college graduates in this sample supportdrilling for oil and natural gas off the Coast of California?

(b) Conduct a hypothesis test to determine if the data provide strong evidence that the proportion of collegegraduates who support off-shore drilling in California is different than that of non-college graduates.

6.26 Full body scan, Part I. A news article reports that “Americans have differing views on two potentiallyinconvenient and invasive practices that airports could implement to uncover potential terrorist attacks.”This news piece was based on a survey conducted among a random sample of 1,137 adults nationwide, whereone of the questions on the survey was “Some airports are now using ‘full-body’ digital x-ray machines toelectronically screen passengers in airport security lines. Do you think these new x-ray machines should orshould not be used at airports?” Below is a summary of responses based on party affiliation.27

Party AffiliationRepublican Democrat Independent

AnswerShould 264 299 351Should not 38 55 77Don’t know/No answer 16 15 22Total 318 369 450

(a) Conduct an appropriate hypothesis test evaluating whether there is a difference in the proportion ofRepublicans and Democrats who think the full- body scans should be applied in airports. Assume thatall relevant conditions are met.

(b) The conclusion of the test in part (a) may be incorrect, meaning a testing error was made. If an errorwas made, was it a Type 1 or a Type 2 Error? Explain.

6.27 Sleep deprived transportation workers. The National Sleep Foundation conducted a survey onthe sleep habits of randomly sampled transportation workers and a control sample of non-transportationworkers. The results of the survey are shown below.28

Transportation ProfessionalsTruck Train Bus/Taxi/Limo

Control Pilots Drivers Operators DriversLess than 6 hours of sleep 35 19 35 29 216 to 8 hours of sleep 193 132 117 119 131More than 8 hours 64 51 51 32 58Total 292 202 203 180 210

Conduct a hypothesis test to evaluate if these data provide evidence of a difference between the proportionsof truck drivers and non-transportation workers (the control group) who get less than 6 hours of sleep perday, i.e. are considered sleep deprived.

27S. Condon. “Poll: 4 in 5 Support Full-Body Airport Scanners”. In: CBS News (2010).28National Sleep Foundation, 2012 Sleep in America Poll: Transportation Workers’ Sleep, 2012.

228 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.28 Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use andautism surveyed the mothers of a random sample of children aged 24 – 60 months with autism and conductedanother separate random sample for children with typical development. The table below shows the number ofmothers in each group who did and did not use prenatal vitamins during the three months before pregnancy(periconceptional period).29

AutismAutism Typical development Total

Periconceptional No vitamin 111 70 181prenatal vitamin Vitamin 143 159 302

Total 254 229 483

(a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the threemonths before pregnancy and autism.

(b) Complete the hypothesis test and state an appropriate conclusion. (Reminder: Verify any necessaryconditions for the test.)

(c) A New York Times article reporting on this study was titled “Prenatal Vitamins May Ward Off Autism”.Do you find the title of this article to be appropriate? Explain your answer. Additionally, propose analternative title.30

6.29 HIV in sub-Saharan Africa. In July 2008 the US National Institutes of Health announced that it wasstopping a clinical study early because of unexpected results. The study population consisted of HIV-infectedwomen in sub-Saharan Africa who had been given single dose Nevaripine (a treatment for HIV) while givingbirth, to prevent transmission of HIV to the infant. The study was a randomized comparison of continuedtreatment of a woman (after successful childbirth) with Nevaripine vs Lopinavir, a second drug used to treatHIV. 240 women participated in the study; 120 were randomized to each of the two treatments. Twenty-four weeks after starting the study treatment, each woman was tested to determine if the HIV infection wasbecoming worse (an outcome called virologic failure). Twenty-six of the 120 women treated with Nevaripineexperienced virologic failure, while 10 of the 120 women treated with the other drug experienced virologicfailure.31

(a) Create a two-way table presenting the results of this study.

(b) State appropriate hypotheses to test for difference in virologic failure rates between treatment groups.

(c) Complete the hypothesis test and state an appropriate conclusion. (Reminder: Verify any necessaryconditions for the test.)

6.30 An apple a day keeps the doctor away. A physical education teacher at a high school wanting toincrease awareness on issues of nutrition and health asked her students at the beginning of the semesterwhether they believed the expression “an apple a day keeps the doctor away”, and 40% of the studentsresponded yes. Throughout the semester she started each class with a brief discussion of a study highlightingpositive effects of eating more fruits and vegetables. She conducted the same apple-a-day survey at the endof the semester, and this time 60% of the students responded yes. Can she used a two-proportion methodfrom this section for this analysis? Explain your reasoning.

29R.J. Schmidt et al. “Prenatal vitamins, one-carbon metabolism gene variants, and risk for autism”. In: Epidemi-ology 22.4 (2011), p. 476.

30R.C. Rabin. “Patterns: Prenatal Vitamins May Ward Off Autism”. In: New York Times (2011).31S. Lockman et al. “Response to antiretroviral therapy after a single, peripartum dose of nevirapine”. In: Obstet-

rical & gynecological survey 62.6 (2007), p. 361.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 229

6.3 Testing for goodness of fit using chi-square

In this section, we develop a method for assessing a null model when the data are binned. Thistechnique is commonly used in two circumstances:

• Given a sample of cases that can be classified into several groups, determine if the sample isrepresentative of the general population.

• Evaluate whether data resemble a particular distribution, such as a normal distribution or ageometric distribution.

Each of these scenarios can be addressed using the same statistical test: a chi-square test.In the first case, we consider data from a random sample of 275 jurors in a small county. Jurors

identified their racial group, as shown in Figure 6.5, and we would like to determine if these jurorsare racially representative of the population. If the jury is representative of the population, thenthe proportions in the sample should roughly reflect the population of eligible jurors, i.e. registeredvoters.

Race White Black Hispanic Other TotalRepresentation in juries 205 26 25 19 275Registered voters 0.72 0.07 0.12 0.09 1.00

Figure 6.5: Representation by race in a city’s juries and population.

While the proportions in the juries do not precisely represent the population proportions, it isunclear whether these data provide convincing evidence that the sample is not representative. If thejurors really were randomly sampled from the registered voters, we might expect small differencesdue to chance. However, unusually large differences may provide convincing evidence that the jurieswere not representative.

A second application, assessing the fit of a distribution, is presented at the end of this section.Daily stock returns from the S&P500 for 25 years are used to assess whether stock activity each dayis independent of the stock’s behavior on previous days.

In these problems, we would like to examine all bins simultaneously, not simply compare oneor two bins at a time, which will require us to develop a new test statistic.

6.3.1 Creating a test statistic for one-way tables

EXAMPLE 6.22

Of the people in the city, 275 served on a jury. If the individuals are randomly selected to serve on ajury, about how many of the 275 people would we expect to be white? How many would we expectto be black?

About 72% of the population is white, so we would expect about 72% of the jurors to be white:0.72× 275 = 198.

Similarly, we would expect about 7% of the jurors to be black, which would correspond to about0.07× 275 = 19.25 black jurors.

GUIDED PRACTICE 6.23

Twelve percent of the population is Hispanic and 9% represent other races. How many of the 275jurors would we expect to be Hispanic or from another race? Answers can be found in Figure 6.6.

The sample proportion represented from each race among the 275 jurors was not a precisematch for any ethnic group. While some sampling variation is expected, we would expect the

230 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

Race White Black Hispanic Other TotalObserved data 205 26 25 19 275Expected counts 198 19.25 33 24.75 275

Figure 6.6: Actual and expected make-up of the jurors.

sample proportions to be fairly similar to the population proportions if there is no bias on juries.We need to test whether the differences are strong enough to provide convincing evidence that thejurors are not a random sample. These ideas can be organized into hypotheses:

H0: The jurors are a random sample, i.e. there is no racial bias in who serves on a jury, and theobserved counts reflect natural sampling fluctuation.

HA: The jurors are not randomly sampled, i.e. there is racial bias in juror selection.

To evaluate these hypotheses, we quantify how different the observed counts are from the expectedcounts. Strong evidence for the alternative hypothesis would come in the form of unusually largedeviations in the groups from what would be expected based on sampling variation alone.

6.3.2 The chi-square test statistic

In previous hypothesis tests, we constructed a test statistic of the following form:

point estimate− null value

SE of point estimate

This construction was based on (1) identifying the difference between a point estimate and anexpected value if the null hypothesis was true, and (2) standardizing that difference using thestandard error of the point estimate. These two ideas will help in the construction of an appropriatetest statistic for count data.

Our strategy will be to first compute the difference between the observed counts and the countswe would expect if the null hypothesis was true, then we will standardize the difference:

Z1 =observed white count− null white count

SE of observed white count

The standard error for the point estimate of the count in binned data is the square root of the countunder the null.32 Therefore:

Z1 =205− 198√

198= 0.50

The fraction is very similar to previous test statistics: first compute a difference, then standardizeit. These computations should also be completed for the black, Hispanic, and other groups:

Black Hispanic Other

Z2 =26− 19.25√

19.25= 1.54 Z3 =

25− 33√33

= −1.39 Z4 =19− 24.75√

24.75= −1.16

We would like to use a single test statistic to determine if these four standardized differences areirregularly far from zero. That is, Z1, Z2, Z3, and Z4 must be combined somehow to help determineif they – as a group – tend to be unusually far from zero. A first thought might be to take theabsolute value of these four standardized differences and add them up:

|Z1|+ |Z2|+ |Z3|+ |Z4| = 4.58

32Using some of the rules learned in earlier chapters, we might think that the standard error would be np(1− p),where n is the sample size and p is the proportion in the population. This would be correct if we were looking onlyat one count. However, we are computing many standardized differences and adding them together. It can be shown– though not here – that the square root of the count is a better way to standardize the count differences.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 231

Indeed, this does give one number summarizing how far the actual counts are from what was ex-pected. However, it is more common to add the squared values:

Z21 + Z2

2 + Z23 + Z2

4 = 5.89

Squaring each standardized difference before adding them together does two things:

• Any standardized difference that is squared will now be positive.

• Differences that already look unusual – e.g. a standardized difference of 2.5 – will becomemuch larger after being squared.

The test statistic X2, which is the sum of the Z2 values, is generally used for these reasons. We canalso write an equation for X2 using the observed counts and null counts:

X2 =(observed count1 − null count1)2

null count1

+ · · ·+ (observed count4 − null count4)2

null count4

The final number X2 summarizes how strongly the observed counts tend to deviate from the nullcounts. In Section 6.3.4, we will see that if the null hypothesis is true, then X2 follows a newdistribution called a chi-square distribution. Using this distribution, we will be able to obtain ap-value to evaluate the hypotheses.

6.3.3 The chi-square distribution and finding areas

The chi-square distribution is sometimes used to characterize data sets and statistics thatare always positive and typically right skewed. Recall a normal distribution had two parameters –mean and standard deviation – that could be used to describe its exact characteristics. The chi-square distribution has just one parameter called degrees of freedom (df), which influences theshape, center, and spread of the distribution.

GUIDED PRACTICE 6.24

Figure 6.7 shows three chi-square distributions.(a) How does the center of the distribution change when the degrees of freedom is larger?(b) What about the variability (spread)?(c) How does the shape change?33

0 5 10 15 20 25

Degrees of Freedom

249

Figure 6.7: Three chi-square distributions with varying degrees of freedom.

33(a) The center becomes larger. If took a careful look, we could see that the mean of each distribution is equalto the distribution’s degrees of freedom. (b) The variability increases as the degrees of freedom increases. (c) Thedistribution is very strongly skewed for df = 2, and then the distributions become more symmetric for the largerdegrees of freedom df = 4 and df = 9. We would see this trend continue if we examined distributions with even morelarger degrees of freedom.

232 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

Figure 6.7 and Guided Practice 6.24 demonstrate three general properties of chi-square distri-butions as the degrees of freedom increases: the distribution becomes more symmetric, the centermoves to the right, and the variability inflates.

Our principal interest in the chi-square distribution is the calculation of p-values, which (aswe have seen before) is related to finding the relevant area in the tail of a distribution. The mostcommon ways to do this are using computer software, using a graphing calculator, or using a table.For folks wanting to use the table option, we provide an outline of how to read the chi-square tablein Appendix C.3, which is also where you may find the table. For the examples below, use yourpreferred approach to confirm you get the same answers.

EXAMPLE 6.25

Figure 6.8(a) shows a chi-square distribution with 3 degrees of freedom and an upper shaded tailstarting at 6.25. Find the shaded area.

Using statistical software or a graphing calculator, we can find that the upper tail area for a chi-square distribution with 3 degrees of freedom (df) and a cutoff of 6.25 is 0.1001. That is, the shadedupper tail of Figure 6.8(a) has area 0.1.

EXAMPLE 6.26

Figure 6.8(b) shows the upper tail of a chi-square distribution with 2 degrees of freedom. The boundfor this upper tail is at 4.3. Find the tail area.

Using software, we can find that the tail area shaded in Figure 6.8(b) to be 0.1165. If using a table,we would only be able to find a range of values for the tail area: between 0.1 and 0.2.

EXAMPLE 6.27

Figure 6.8(c) shows an upper tail for a chi-square distribution with 5 degrees of freedom and a cutoffof 5.1. Find the tail area.

Using software, we would obtain a tail area of 0.4038. If using the table in Appendix C.3, we wouldhave identified that the tail area is larger than 0.3 but not be able to give the precise value.

GUIDED PRACTICE 6.28

Figure 6.8(d) shows a cutoff of 11.7 on a chi-square distribution with 7 degrees of freedom. Find thearea of the upper tail.34

GUIDED PRACTICE 6.29

Figure 6.8(e) shows a cutoff of 10 on a chi-square distribution with 4 degrees of freedom. Find thearea of the upper tail.35

GUIDED PRACTICE 6.30

Figure 6.8(f) shows a cutoff of 9.21 with a chi-square distribution with 3 df. Find the area of theupper tail.36

34 The area is 0.1109. If using a table, we would identify that it falls between 0.1 and 0.2.35Precise value: 0.0404. If using the table: between 0.02 and 0.05.36Precise value: 0.0266. If using the table: between 0.02 and 0.05.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 233

0 5 10 15

(a)

0 5 10 15

(b)

0 5 10 15 20 25

(c)

0 5 10 15 20 25

(d)

0 5 10 15

(e)

0 5 10 15

(f)

Figure 6.8: (a) Chi-square distribution with 3 degrees of freedom, area above6.25 shaded. (b) 2 degrees of freedom, area above 4.3 shaded. (c) 5 degrees offreedom, area above 5.1 shaded. (d) 7 degrees of freedom, area above 11.7 shaded.(e) 4 degrees of freedom, area above 10 shaded. (f) 3 degrees of freedom, areaabove 9.21 shaded.

234 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.3.4 Finding a p-value for a chi-square distribution

In Section 6.3.2, we identified a new test statistic (X2) within the context of assessing whetherthere was evidence of racial bias in how jurors were sampled. The null hypothesis represented theclaim that jurors were randomly sampled and there was no racial bias. The alternative hypothesiswas that there was racial bias in how the jurors were sampled.

We determined that a large X2 value would suggest strong evidence favoring the alternativehypothesis: that there was racial bias. However, we could not quantify what the chance was ofobserving such a large test statistic (X2 = 5.89) if the null hypothesis actually was true. This iswhere the chi-square distribution becomes useful. If the null hypothesis was true and there was noracial bias, then X2 would follow a chi-square distribution, with three degrees of freedom in thiscase. Under certain conditions, the statistic X2 follows a chi-square distribution with k − 1 degreesof freedom, where k is the number of bins.

EXAMPLE 6.31

How many categories were there in the juror example? How many degrees of freedom should beassociated with the chi-square distribution used for X2?

In the jurors example, there were k = 4 categories: white, black, Hispanic, and other. Accordingto the rule above, the test statistic X2 should then follow a chi-square distribution with k − 1 = 3degrees of freedom if H0 is true.

Just like we checked sample size conditions to use a normal distribution in earlier sections, wemust also check a sample size condition to safely apply the chi-square distribution for X2. Eachexpected count must be at least 5. In the juror example, the expected counts were 198, 19.25, 33,and 24.75, all easily above 5, so we can apply the chi-square model to the test statistic, X2 = 5.89.

EXAMPLE 6.32

If the null hypothesis is true, the test statistic X2 = 5.89 would be closely associated with a chi-square distribution with three degrees of freedom. Using this distribution and test statistic, identifythe p-value.

The chi-square distribution and p-value are shown in Figure 6.9. Because larger chi-square valuescorrespond to stronger evidence against the null hypothesis, we shade the upper tail to representthe p-value. Using statistical software (or the table in Appendix C.3), we can determine that thearea is 0.1171. Generally we do not reject the null hypothesis with such a large p-value. In otherwords, the data do not provide convincing evidence of racial bias in the juror selection.

0 5 10 15

Figure 6.9: The p-value for the juror hypothesis test is shaded in the chi-squaredistribution with df = 3.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 235

CHI-SQUARE TEST FOR ONE-WAY TABLE

Suppose we are to evaluate whether there is convincing evidence that a set of observed countsO1, O2, …, Ok in k categories are unusually different from what might be expected under a nullhypothesis. Call the expected counts that are based on the null hypothesis E1, E2, …, Ek. Ifeach expected count is at least 5 and the null hypothesis is true, then the test statistic belowfollows a chi-square distribution with k − 1 degrees of freedom:

X2 =(O1 − E1)2

E1+

(O2 − E2)2

E2+ · · ·+ (Ok − Ek)2

The p-value for this test statistic is found by looking at the upper tail of this chi-square distri-bution. We consider the upper tail because larger values of X2 would provide greater evidenceagainst the null hypothesis.

CONDITIONS FOR THE CHI-SQUARE TEST

There are two conditions that must be checked before performing a chi-square test:

Independence. Each case that contributes a count to the table must be independent of allthe other cases in the table.

Sample size / distribution. Each particular scenario (i.e. cell count) must have at least5 expected cases.

Failing to check conditions may affect the test’s error rates.

When examining a table with just two bins, pick a single bin and use the one-proportionmethods introduced in Section 6.1.

236 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.3.5 Evaluating goodness of fit for a distribution

Section 4.2 would be useful background reading for this example, but it is not a prerequisite.We can apply the chi-square testing framework to the second problem in this section: evaluating

whether a certain statistical model fits a data set. Daily stock returns from the S&P500 for 10 canbe used to assess whether stock activity each day is independent of the stock’s behavior on previousdays. This sounds like a very complex question, and it is, but a chi-square test can be used to studythe problem. We will label each day as Up or Down (D) depending on whether the market was upor down that day. For example, consider the following changes in price, their new labels of up anddown, and then the number of days that must be observed before each Up day:

Change in price 2.52 -1.46 0.51 -4.07 3.36 1.10 -5.46 -1.03 -2.99 1.71Outcome Up D Up D Up Up D D D UpDays to Up 1 – 2 – 2 1 – – – 4

If the days really are independent, then the number of days until a positive trading day should followa geometric distribution. The geometric distribution describes the probability of waiting for the kth

trial to observe the first success. Here each up day (Up) represents a success, and down (D) daysrepresent failures. In the data above, it took only one day until the market was up, so the first waittime was 1 day. It took two more days before we observed our next Up trading day, and two morefor the third Up day. We would like to determine if these counts (1, 2, 2, 1, 4, and so on) followthe geometric distribution. Figure 6.10 shows the number of waiting days for a positive trading dayduring 10 years for the S&P500.

Days 1 2 3 4 5 6 7+ TotalObserved 717 369 155 69 28 14 10 1362

Figure 6.10: Observed distribution of the waiting time until a positive trading dayfor the S&P500.

We consider how many days one must wait until observing an Up day on the S&P500 stockindex. If the stock activity was independent from one day to the next and the probability of apositive trading day was constant, then we would expect this waiting time to follow a geometricdistribution. We can organize this into a hypothesis framework:

H0: The stock market being up or down on a given day is independent from all other days. Wewill consider the number of days that pass until an Up day is observed. Under this hypothesis,the number of days until an Up day should follow a geometric distribution.

HA: The stock market being up or down on a given day is not independent from all other days.Since we know the number of days until an Up day would follow a geometric distribution underthe null, we look for deviations from the geometric distribution, which would support thealternative hypothesis.

There are important implications in our result for stock traders: if information from past tradingdays is useful in telling what will happen today, that information may provide an advantage overother traders.

We consider data for the S&P500 and summarize the waiting times in Figure 6.11 and Fig-ure 6.12. The S&P500 was positive on 54.5% of those days.

Because applying the chi-square framework requires expected counts to be at least 5, we havebinned together all the cases where the waiting time was at least 7 days to ensure each expectedcount is well above this minimum. The actual data, shown in the Observed row in Figure 6.11, canbe compared to the expected counts from the Geometric Model row. The method for computingexpected counts is discussed in Figure 6.11. In general, the expected counts are determined by(1) identifying the null proportion associated with each bin, then (2) multiplying each null proportionby the total count to obtain the expected counts. That is, this strategy identifies what proportionof the total count we would expect to be in each bin.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 237

Days 1 2 3 4 5 6 7+ TotalObserved 717 369 155 69 28 14 10 1362Geometric Model 743 338 154 70 32 14 12 1362

Figure 6.11: Distribution of the waiting time until a positive trading day. Theexpected counts based on the geometric model are shown in the last row. Tofind each expected count, we identify the probability of waiting D days based onthe geometric model (P (D) = (1 − 0.545)D−1(0.545)) and multiply by the totalnumber of streaks, 1362. For example, waiting for three days occurs under thegeometric model about 0.4552× 0.545 = 11.28% of the time, which corresponds to0.1128× 1362 = 154 streaks.

Wait Until Positive Day

1 2 3 4 5 6 7+

200

400

600

800

Fre

quen

ObservedExpected

Figure 6.12: Side-by-side bar plot of the observed and expected counts for eachwaiting time.

238 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

EXAMPLE 6.33

Do you notice any unusually large deviations in the graph? Can you tell if these deviations are dueto chance just by looking?

It is not obvious whether differences in the observed counts and the expected counts from the geo-metric distribution are significantly different. That is, it is not clear whether these deviations mightbe due to chance or whether they are so strong that the data provide convincing evidence againstthe null hypothesis. However, we can perform a chi-square test using the counts in Figure 6.11.

GUIDED PRACTICE 6.34

Figure 6.11 provides a set of count data for waiting times (O1 = 717, O2 = 369, …) and expectedcounts under the geometric distribution (E1 = 743, E2 = 338, …). Compute the chi-square teststatistic, X2.37

GUIDED PRACTICE 6.35

Because the expected counts are all at least 5, we can safely apply the chi-square distribution toX2. However, how many degrees of freedom should we use?38

EXAMPLE 6.36

If the observed counts follow the geometric model, then the chi-square test statistic X2 = 4.61 wouldclosely follow a chi-square distribution with df = 6. Using this information, compute a p-value.

Figure 6.13 shows the chi-square distribution, cutoff, and the shaded p-value. Using software, wecan find the p-value: 0.5951. Ultimately, we do not have sufficient evidence to reject the notion thatthe wait times follow a geometric distribution for the last 10 years of data for the S&P500, i.e. wecannot reject the notion that trading days are independent.

0 5 10 15 20 25

Area representingthe p−value

Figure 6.13: Chi-square distribution with 6 degrees of freedom. The p-value forthe stock analysis is shaded.

EXAMPLE 6.37

In Example 6.36, we did not reject the null hypothesis that the trading days are independent duringthe last 10 of data. Why is this so important?

It may be tempting to think the market is “due” for an Up day if there have been several consecutivedays where it has been down. However, we haven’t found strong evidence that there’s any suchproperty where the market is “due” for a correction. At the very least, the analysis suggests anydependence between days is very weak.

37X2 =(717−743)2

743+

(369−338)2

338+ · · ·+ (10−12)2

12= 4.61

38There are k = 7 groups, so we use df = k − 1 = 6.

6.3. TESTING FOR GOODNESS OF FIT USING CHI-SQUARE 239

Exercises

6.31 True or false, Part I. Determine if the statements below are true or false. For each false statement,suggest an alternative wording to make it a true statement.

(a) The chi-square distribution, just like the normal distribution, has two parameters, mean and standarddeviation.

(b) The chi-square distribution is always right skewed, regardless of the value of the degrees of freedomparameter.

(d) As the degrees of freedom increases, the shape of the chi-square distribution becomes more skewed.

6.32 True or false, Part II. Determine if the statements below are true or false. For each false statement,suggest an alternative wording to make it a true statement.

(a) As the degrees of freedom increases, the mean of the chi-square distribution increases.

(b) If you found χ2 = 10 with df = 5 you would fail to reject H0 at the 5% significance level.

(d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

6.33 Open source textbook. A professor using an open source introductory statistics book predicts that60% of the students will purchase a hard copy of the book, 25% will print it out from the web, and 15%will read it online. At the end of the semester he asks his students to complete a survey where they indicatewhat format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30said they printed it out from the web, and 25 said they read it online.

(a) State the hypotheses for testing if the professor’s predictions were inaccurate.

(b) How many students did the professor expect to buy the book, print the book, and read the bookexclusively online?

(c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verifythey are satisfied.

(d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value.

(e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpretyour conclusion in this context.

6.34 Barking deer. Microhabitat factors associated with forage and bed sites of barking deer in HainanIsland, China were examined. In this region woods make up 4.8% of the land, cultivated grass plot makesup 14.7%, and deciduous forests make up 39.6%. Of the 426 sites where the deer forage, 4 were categorizedas woods, 16 as cultivated grassplot, and 61 as deciduous forests. The table below summarizes these data.39

Woods Cultivated grassplot Deciduous forests Other Total

4 16 61 345 426

(a) Write the hypotheses for testing if barking deer prefer to forage in cer-tain habitats over others.

(b) What type of test can we use to answer this research question?

(d) Do these data provide convincing evidence that barking deer prefer toforage in certain habitats over others? Conduct an appropriate hypoth-esis test to answer this research question. Photo by Shrikant Rao

(http://flic.kr/p/4Xjdkk)

CC BY 2.0 license

39Liwei Teng et al. “Forage and bed sites characteristics of Indian muntjac (Muntiacus muntjak) in Hainan Island,China”. In: Ecological Research 19.6 (2004), pp. 675–681.

240 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.4 Testing for independence in two-way tables

We all buy used products – cars, computers, textbooks, and so on – and we sometimes assumethe sellers of those products will be forthright about any underlying problems with what they’reselling. This is not something we should take for granted. Researchers recruited 219 participants ina study where they would sell a used iPod40 that was known to have frozen twice in the past. Theparticipants were incentivized to get as much money as they could for the iPod since they wouldreceive a 5% cut of the sale on top of $10 for participating. The researchers wanted to understandwhat types of questions would elicit the seller to disclose the freezing issue.

Unbeknownst to the participants who were the sellers in the study, the buyers were collaboratingwith the researchers to evaluate the influence of different questions on the likelihood of getting thesellers to disclose the past issues with the iPod. The scripted buyers started with “Okay, I guess I’msupposed to go first. So you’ve had the iPod for 2 years …” and ended with one of three questions:

• General: What can you tell me about it?

• Positive Assumption: It doesn’t have any problems, does it?

• Negative Assumption: What problems does it have?

The question is the treatment given to the sellers, and the response is whether the question promptedthem to disclose the freezing issue with the iPod. The results are shown in Figure 6.14, and thedata suggest that asking the, What problems does it have?, was the most effective at getting theseller to disclose the past freezing issues. However, you should also be asking yourself: could we seethese results due to chance alone, or is this in fact evidence that some questions are more effectivefor getting at the truth?

General Positive Assumption Negative Assumption TotalDisclose Problem 2 23 36 61Hide Problem 71 50 37 158Total 73 73 73 219

Figure 6.14: Summary of the iPod study, where a question was posed to the studyparticipant who acted

DIFFERENCES OF ONE-WAY TABLES VS TWO-WAY TABLES

A one-way table describes counts for each outcome in a single variable. A two-way tabledescribes counts for combinations of outcomes for two variables. When we consider a two-waytable, we often would like to know, are these variables related in any way? That is, are theydependent (versus independent)?

The hypothesis test for the iPod experiment is really about assessing whether there is statis-tically significant evidence that the success each question had on getting the participant to disclosethe problem with the iPod. In other words, the goal is to check whether the buyer’s question wasindependent of whether the seller disclosed a problem.

40For readers not as old as the authors, an iPod is basically an iPhone without any cellular service, assuming itwas one of the later generations. Earlier generations were more basic.

6.4. TESTING FOR INDEPENDENCE IN TWO-WAY TABLES 241

6.4.1 Expected counts in two-way tables

Like with one-way tables, we will need to compute estimated counts for each cell in a two-way table.

EXAMPLE 6.38

From the experiment, we can compute the proportion of all sellers who disclosed the freezing problemas 61/219 = 0.2785. If there really is no difference among the questions and 27.85% of sellers weregoing to disclose the freezing problem no matter the question that was put to them, how many ofthe 73 people in the General group would we have expected to disclose the freezing problem?

We would predict that 0.2785×73 = 20.33 sellers would disclose the problem. Obviously we observedfewer than this, though it is not yet clear if that is due to chance variation or whether that is becausethe questions vary in how effective they are at getting to the truth.

GUIDED PRACTICE 6.39

If the questions were actually equally effective, meaning about 27.85% of respondents would disclosethe freezing issue regardless of what question they were asked, about how many sellers would weexpect to hide the freezing problem from the Positive Assumption group?41

We can compute the expected number of sellers who we would expect to disclose or hide thefreezing issue for all groups, if the questions had no impact on what they disclosed, using the samestrategy employed in Example 6.38 and Guided Practice 6.39. These expected counts were used toconstruct Figure 6.15, which is the same as Figure 6.14, except now the expected counts have beenadded in parentheses.

General Positive Assumption Negative Assumption TotalDisclose Problem 2 (20.33) 23 (20.33) 36 (20.33) 61Hide Problem 71 (52.67) 50 (52.67) 37 (52.67) 158Total 73 73 73 219

Figure 6.15: The observed counts and the (expected counts).

The examples and exercises above provided some help in computing expected counts. In general,expected counts for a two-way table may be computed using the row totals, column totals, and thetable total. For instance, if there was no difference between the groups, then about 27.85% of eachcolumn should be in the first row:

0.2785× (column 1 total) = 20.33

0.2785× (column 2 total) = 20.33

0.2785× (column 3 total) = 20.33

Looking back to how 0.2785 was computed – as the fraction of sellers who disclosed the freezingissue (158/219) – these three expected counts could have been computed as(

row 1 total

table total

)(column 1 total) = 20.33(

row 1 total

table total

)(column 2 total) = 20.33(

row 1 total

table total

)(column 3 total) = 20.33

This leads us to a general formula for computing expected counts in a two-way table when we wouldlike to test whether there is strong evidence of an association between the column variable and rowvariable.

41We would expect (1 − 0.2785) × 73 = 52.67. It is okay that this result, like the result from Example 6.38, is afraction.

242 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

COMPUTING EXPECTED COUNTS IN A TWO-WAY TABLE

To identify the expected count for the ith row and jth column, compute

Expected Countrow i, col j =(row i total)× (column j total)

table total

6.4.2 The chi-square test for two-way tables

The chi-square test statistic for a two-way table is found the same way it is found for a one-waytable. For each table count, compute

General formula(observed count − expected count)2

expected count

Row 1, Col 1(2− 20.33)2

20.33= 16.53

Row 1, Col 2(23− 20.33)2

20.33= 0.35

……

Row 2, Col 3(37− 52.67)2

52.67= 4.66

Adding the computed value for each cell gives the chi-square test statistic X2:

X2 = 16.53 + 0.35 + · · ·+ 4.66 = 40.13

Just like before, this test statistic follows a chi-square distribution. However, the degrees of freedomare computed a little differently for a two-way table.42 For two way tables, the degrees of freedomis equal to

df = (number of rows minus 1)× (number of columns minus 1)

In our example, the degrees of freedom parameter is

df = (2− 1)× (3− 1) = 2

If the null hypothesis is true (i.e. the questions had no impact on the sellers in the experiment),then the test statistic X2 = 40.13 closely follows a chi-square distribution with 2 degrees of freedom.Using this information, we can compute the p-value for the test, which is depicted in Figure 6.16.

COMPUTING DEGREES OF FREEDOM FOR A TWO-WAY TABLE

When applying the chi-square test to a two-way table, we use

df = (R− 1)× (C − 1)

where R is the number of rows in the table and C is the number of columns.

When analyzing 2-by-2 contingency tables, one guideline is to use the two-proportion methodsintroduced in Section 6.2.

42Recall: in the one-way table, the degrees of freedom was the number of cells minus 1.

6.4. TESTING FOR INDEPENDENCE IN TWO-WAY TABLES 243

0 10 20 30 40 50

Tail area (1 / 500 million)is too small to see

Figure 6.16: Visualization of the p-value for X2 = 40.13 when df = 2.

EXAMPLE 6.40

Find the p-value and draw a conclusion about whether the question affects the sellers likelihood ofreporting the freezing problem.

Using a computer, we can compute a very precise value for the tail area above X2 = 40.13 for achi-square distribution with 2 degrees of freedom: 0.000000002. (If using the table in Appendix C.3,we would identify the p-value is smaller than 0.001.) Using a significance level of α = 0.05, the nullhypothesis is rejected since the p-value is smaller. That is, the data provide convincing evidence thatthe question asked did affect a seller’s likelihood to tell the truth about problems with the iPod.

EXAMPLE 6.41

Figure 6.17 summarizes the results of an experiment evaluating three treatments for Type 2 Diabetesin patients aged 10-17 who were being treated with metformin. The three treatments consideredwere continued treatment with metformin (met), treatment with metformin combined with rosigli-tazone (rosi), or a lifestyle intervention program. Each patient had a primary outcome, which waseither lacked glycemic control (failure) or did not lack that control (success). What are appropriatehypotheses for this test?

H0: There is no difference in the effectiveness of the three treatments.

HA: There is some difference in effectiveness between the three treatments, e.g. perhaps the rosi

treatment performed better than lifestyle.

Failure Success Totallifestyle 109 125 234met 120 112 232rosi 90 143 233Total 319 380 699

Figure 6.17: Results for the Type 2 Diabetes study.

244 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

GUIDED PRACTICE 6.42

A chi-square test for a two-way table may be used to test the hypotheses in Example 6.41. As afirst step, compute the expected values for each of the six table cells.43

GUIDED PRACTICE 6.43

Compute the chi-square test statistic for the data in Figure 6.17.44

GUIDED PRACTICE 6.44

Because there are 3 rows and 2 columns, the degrees of freedom for the test is df = (3−1)×(2−1) = 2.Use X2 = 8.16, df = 2, evaluate whether to reject the null hypothesis using a significance levelof 0.05.45

43The expected count for row one / column one is found by multiplying the row one total (234) and column onetotal (319), then dividing by the table total (699): 234×319

699= 106.8. Similarly for the second column and the first

row: 234×380699

= 127.2. Row 2: 105.9 and 126.1. Row 3: 106.3 and 126.7.

44For each cell, compute(obs−exp)2

exp. For instance, the first row and first column:

(109−106.8)2

106.8= 0.05. Adding the

results of each cell gives the chi-square test statistic: X2 = 0.05 + · · ·+ 2.11 = 8.16.45 If using a computer, we can identify the p-value as 0.017. That is, we reject the null hypothesis because the

p-value is less than 0.05, and we conclude that at least one of the treatments is more or less effective than the othersat treating Type 2 Diabetes for glycemic control.

6.4. TESTING FOR INDEPENDENCE IN TWO-WAY TABLES 245

Exercises

6.35 Quitters. Does being part of a support group affect the ability of people to quit smoking? A countyhealth department enrolled 300 smokers in a randomized experiment. 150 participants were assigned to agroup that used a nicotine patch and met weekly with a support group; the other 150 received the patchand did not meet with a support group. At the end of the study, 40 of the participants in the patch plussupport group had quit smoking while only 30 smokers had quit in the other group.

(a) Create a two-way table presenting the results of this study.

(b) Answer each of the following questions under the null hypothesis that being part of a support group doesnot affect the ability of people to quit smoking, and indicate whether the expected values are higher orlower than the observed values.

i. How many subjects in the “patch + support” group would you expect to quit?

ii. How many subjects in the “patch only” group would you expect to not quit?

6.36 Full body scan, Part II. The table below summarizes a data set we first encountered in Exercise 6.26regarding views on full-body scans and political affiliation. The differences in each political group may bedue to chance. Complete the following computations under the null hypothesis of independence between anindividual’s party affiliation and his support of full-body scans. It may be useful to first add on an extracolumn for row totals before proceeding with the computations.

Party AffiliationRepublican Democrat Independent

AnswerShould 264 299 351Should not 38 55 77Don’t know/No answer 16 15 22Total 318 369 450

(a) How many Republicans would you expect to not support the use of full-body scans?

(b) How many Democrats would you expect to support the use of full- body scans?

6.37 Offshore drilling, Part III. The table below summarizes a data set we first encountered in Exercise 6.23that examines the responses of a random sample of college graduates and non-graduates on the topic of oildrilling. Complete a chi-square test for these data to check whether there is a statistically significantdifference in responses from college graduates and non-graduates.

College GradYes No

Support 154 132Oppose 180 126Do not know 104 131Total 438 389

6.38 Parasitic worm. Lymphatic filariasis is a disease caused by a parasitic worm. Complications of thedisease can lead to extreme swelling and other complications. Here we consider results from a randomizedexperiment that compared three different drug treatment options to clear people of the this parasite, whichpeople are working to eliminate entirely. The results for the second year of the study are given below:46

Clear at Year 2 Not Clear at Year 2

Three drugs 52 2Two drugs 31 24Two drugs annually 42 14

(a) Set up hypotheses for evaluating whether there is any difference in the performance of the treatments,and also check conditions.

(b) Statistical software was used to run a chi-square test, which output:

X2 = 23.7 df = 2 p-value = 7.2e-6

Use these results to evaluate the hypotheses from part (a), and provide a conclusion in the context ofthe problem.

46Christopher King et al. “A Trial of a Triple-Drug Treatment for Lymphatic Filariasis”. In: New England Journalof Medicine 379 (2018), pp. 1801–1810.

246 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

Chapter exercises

6.39 Active learning. A teacher wanting to increase the active learning component of her course is con-cerned about student reactions to changes she is planning to make. She conducts a survey in her class,asking students whether they believe more active learning in the classroom (hands on exercises) instead oftraditional lecture will helps improve their learning. She does this at the beginning and end of the semesterand wants to evaluate whether students’ opinions have changed over the semester. Can she used the methodswe learned in this chapter for this analysis? Explain your reasoning.

6.40 Website experiment. The OpenIntro website occasionally experiments with design and link place-ment. We conducted one experiment testing three different placements of a download link for this textbookon the book’s main page to see which location, if any, led to the most downloads. The number of site visitorsincluded in the experiment was 701 and is captured in one of the response combinations in the followingtable:

Download No Download

Position 1 13.8% 18.3%Position 2 14.6% 18.5%Position 3 12.1% 22.7%

(a) Calculate the actual number of site visitors in each of the six response categories.

(b) Each individual in the experiment had an equal chance of being in any of the three experiment groups.However, we see that there are slightly different totals for the groups. Is there any evidence that thegroups were actually imbalanced? Make sure to clearly state hypotheses, check conditions, calculate theappropriate test statistic and the p-value, and make your conclusion in context of the data.

(c) Complete an appropriate hypothesis test to check whether there is evidence that there is a higher rateof site visitors clicking on the textbook link in any of the three groups.

6.41 Shipping holiday gifts. A local news survey asked 500 randomly sampled Los Angeles residentswhich shipping carrier they prefer to use for shipping holiday gifts. The table below shows the distributionof responses by age group as well as the expected counts for each cell (shown in parentheses).

Age18-34 35-54 55+ Total

Shipping Method

USPS 72 (81) 97 (102) 76 (62) 245UPS 52 (53) 76 (68) 34 (41) 162FedEx 31 (21) 24 (27) 9 (16) 64Something else 7 (5) 6 (7) 3 (4) 16Not sure 3 (5) 6 (5) 4 (3) 13Total 165 209 126 500

(a) State the null and alternative hypotheses for testing for independence of age and preferred shippingmethod for holiday gifts among Los Angeles residents.

(b) Are the conditions for inference using a chi-square test satisfied?

6.42 The Civil War. A national survey conducted among a simple random sample of 1,507 adults showsthat 56% of Americans think the Civil War is still relevant to American politics and political life.47

(a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of theAmericans think the Civil War is still relevant.

(b) Interpret the p-value in this context.

(c) Calculate a 90% confidence interval for the proportion of Americans who think the Civil War is stillrelevant. Interpret the interval in this context, and comment on whether or not the confidence intervalagrees with the conclusion of the hypothesis test.

47Pew Research Center Publications, Civil War at 150: Still Relevant, Still Divisive, data collected between March30 – April 3, 2011.

6.4. TESTING FOR INDEPENDENCE IN TWO-WAY TABLES 247

6.43 College smokers. We are interested in estimating the proportion of students at a university whosmoke. Out of a random sample of 200 students from this university, 40 students smoke.

(a) Calculate a 95% confidence interval for the proportion of students at this university who smoke, andinterpret this interval in context. (Reminder: Check conditions.)

(b) If we wanted the margin of error to be no larger than 2% at a 95% confidence level for the proportionof students who smoke, how big of a sample would we need?

6.44 Acetaminophen and liver damage. It is believed that large doses of acetaminophen (the activeingredient in over the counter pain relievers like Tylenol) may cause damage to the liver. A researcherwants to conduct a study to estimate the proportion of acetaminophen users who have liver damage. Forparticipating in this study, he will pay each subject $20 and provide a free medical consultation if the patienthas liver damage.

(a) If he wants to limit the margin of error of his 98% confidence interval to 2%, what is the minimumamount of money he needs to set aside to pay his subjects?

(b) The amount you calculated in part (a) is substantially over his budget so he decides to use fewer subjects.How will this affect the width of his confidence interval?

6.45 Life after college. We are interested in estimating the proportion of graduates at a mid-sized uni-versity who found a job within one year of completing their undergraduate degree. Suppose we conduct asurvey and find out that 348 of the 400 randomly sampled graduates found jobs. The graduating class underconsideration included over 4500 students.

(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?

(b) Check if the conditions for constructing a confidence interval based on these data are met.

(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year ofcompleting their undergraduate degree at this university, and interpret it in the context of the data.

(d) What does “95% confidence” mean?

(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of thedata.

(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider? Explain.

6.46 Diabetes and unemployment. A Gallup poll surveyed Americans about their employment statusand whether or not they have diabetes. The survey results indicate that 1.5% of the 47,774 employed (fullor part time) and 2.5% of the 5,855 unemployed 18-29 year olds have diabetes.48

(a) Create a two-way table presenting the results of this study.

(b) State appropriate hypotheses to test for difference in proportions of diabetes between employed andunemployed Americans.

(c) The sample difference is about 1%. If we completed the hypothesis test, we would find that the p-valueis very small (about 0), meaning the difference is statistically significant. Use this result to explain thedifference between statistically significant and practically significant findings.

6.47 Rock-paper-scissors. Rock-paper-scissors is a hand game played by two or more people whereplayers choose to sign either rock, paper, or scissors with their hands. For your statistics class project, youwant to evaluate whether players choose between these three options randomly, or if certain options arefavored above others. You ask two friends to play rock-paper-scissors and count the times each option isplayed. The following table summarizes the data:

Rock Paper Scissors

43 21 35

Use these data to evaluate whether players choose between these three options randomly, or if certain optionsare favored above others. Make sure to clearly outline each step of your analysis, and interpret your resultsin context of the data and the research question.

48Gallup Wellbeing, Employed Americans in Better Health Than the Unemployed, data collected Jan. 2, 2011 -May 21, 2012.

248 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA

6.48 2010 Healthcare Law. On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that46% of 1,012 Americans agree with this decision. At a 95% confidence level, this sample has a 3% margin oferror. Based on this information, determine if the following statements are true or false, and explain yourreasoning.49

(a) We are 95% confident that between 43% and 49% of Americans in this sample support the decision ofthe U.S. Supreme Court on the 2010 healthcare law.

(b) We are 95% confident that between 43% and 49% of Americans support the decision of the U.S. SupremeCourt on the 2010 healthcare law.

(c) If we considered many random samples of 1,012 Americans, and we calculated the sample proportionsof those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will bebetween 43% and 49%.

(d) The margin of error at a 90% confidence level would be higher than 3%.

6.49 Browsing on the mobile device. A survey of 2,254 American adults indicates that 17% of cell phoneowners browse the internet exclusively on their phone rather than a computer or other device.50

(a) According to an online article, a report from a mobile research company indicates that 38 percent ofChinese mobile web users only access the internet through their cell phones.51 Conduct a hypothesistest to determine if these data provide strong evidence that the proportion of Americans who only usetheir cell phones to access the internet is different than the Chinese proportion of 38%.

(b) Interpret the p-value in this context.

(c) Calculate a 95% confidence interval for the proportion of Americans who access the internet on theircell phones, and interpret the interval in this context.

6.50 Coffee and Depression. Researchers conducted a study investigating the relationship between caf-feinated coffee consumption and risk of depression in women. They collected data on 50,739 women free ofdepression symptoms at the start of the study in the year 1996, and these women were followed through 2006.The researchers used questionnaires to collect data on caffeinated coffee consumption, asked each individualabout physician- diagnosed depression, and also asked about the use of antidepressants. The table belowshows the distribution of incidences of depression by amount of caffeinated coffee consumption.52

Caffeinated coffee consumption≤ 1 2-6 1 2-3 ≥ 4

cup/week cups/week cup/day cups/day cups/day Total

Clinical Yes 670 373 905 564 95 2,607depression No 11,545 6,244 16,329 11,726 2,288 48,132

Total 12,215 6,617 17,234 12,290 2,383 50,739

(a) What type of test is appropriate for evaluating if there is an association between coffee intake anddepression?

(b) Write the hypotheses for the test you identified in part (a).

(d) Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the teststatistic, i.e. (Observed− Expected)2/Expected.

(e) The test statistic is χ2 = 20.93. What is the p-value?

(f) What is the conclusion of the hypothesis test?

(g) One of the authors of this study was quoted on the NYTimes as saying it was “too early to recommendthat women load up on extra coffee” based on just this study.53 Do you agree with this statement?Explain your reasoning.

49Gallup, Americans Issue Split Decision on Healthcare Ruling, data collected June 28, 2012.50Pew Internet, Cell Internet Use 2012, data collected between March 15 – April 13, 2012.51S. Chang. “The Chinese Love to Use Feature Phone to Access the Internet”. In: M.I.C Gadget (2012).52M. Lucas et al. “Coffee, caffeine, and risk of depression among women”. In: Archives of internal medicine 171.17

(2011), p. 1571.53A. O’Connor. “Coffee Drinking Linked to Less Depression in Women”. In: New York Times (2011).

249

Chapter 7Inference for numerical data

7.1 One-sample means with the t-distribution

7.2 Paired data

7.3 Difference of two means

7.4 Power calculations for a difference of means

7.5 Comparing many means with ANOVA

250

Chapter 5 introduced a framework for statistical inference based on

confidence intervals and hypotheses using the normal distribution for

sample proportions. In this chapter, we encounter several new point

estimates and a couple new distributions. In each case, the inference

ideas remain the same: determine which point estimate or test statistic

is useful, identify an appropriate distribution for the point estimate or

test statistic, and apply the ideas of inference.

For videos, slides, and other resources, please visit

www.openintro.org/os

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 251

7.1 One-sample means with the ttt-distribution

Similar to how we can model the behavior of the sample proportion p̂ using a normal distribution,the sample mean x̄ can also be modeled using a normal distribution when certain conditions aremet. However, we’ll soon learn that a new distribution, called the t-distribution, tends to be moreuseful when working with the sample mean. We’ll first learn about this new distribution, then we’lluse it to construct confidence intervals and conduct hypothesis tests for the mean.

7.1.1 The sampling distribution of x̄̄x̄x

The sample mean tends to follow a normal distribution centered at the population mean, µ,when certain conditions are met. Additionally, we can compute a standard error for the samplemean using the population standard deviation σ and the sample size n.

CENTRAL LIMIT THEOREM FOR THE SAMPLE MEAN

When we collect a sufficiently large sample of n independent observations from a populationwith mean µ and standard deviation σ, the sampling distribution of x̄ will be nearly normalwith

Mean = µ Standard Error (SE) =σ√n

Before diving into confidence intervals and hypothesis tests using x̄, we first need to cover two topics:

• When we modeled p̂ using the normal distribution, certain conditions had to be satisfied.The conditions for working with x̄ are a little more complex, and we’ll spend Section 7.1.2discussing how to check conditions for inference.

• The standard error is dependent on the population standard deviation, σ. However, we rarelyknow σ, and instead we must estimate it. Because this estimation is itself imperfect, we use anew distribution called the t-distribution to fix this problem, which we discuss in Section 7.1.3.

7.1.2 Evaluating the two conditions required for modeling x̄̄x̄x

Two conditions are required to apply the Central Limit Theorem for a sample mean x̄:

Independence. The sample observations must be independent, The most common way to satisfythis condition is when the sample is a simple random sample from the population. If thedata come from a random process, analogous to rolling a die, this would also satisfy theindependence condition.

Normality. When a sample is small, we also require that the sample observations come from anormally distributed population. We can relax this condition more and more for larger andlarger sample sizes. This condition is obviously vague, making it difficult to evaluate, so nextwe introduce a couple rules of thumb to make checking this condition easier.

RULES OF THUMB: HOW TO PERFORM THE NORMALITY CHECK

There is no perfect way to check the normality condition, so instead we use two rules of thumb:

n < 30: If the sample size n is less than 30 and there are no clear outliers in the data, then wetypically assume the data come from a nearly normal distribution to satisfy the condition.

n ≥ 30: If the sample size n is at least 30 and there are no particularly extreme outliers, thenwe typically assume the sampling distribution of x̄ is nearly normal, even if the underlyingdistribution of individual observations is not.

252 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

In this first course in statistics, you aren’t expected to develop perfect judgement on the nor-mality condition. However, you are expected to be able to handle clear cut cases based on the rulesof thumb.1

EXAMPLE 7.1

Consider the following two plots that come from simple random samples from different populations.Their sample sizes are n1 = 15 and n2 = 50.

Sample 1 Observations (n = 15)

0 2 4 6

Fre

quen

Sample 2 Observations (n = 50)

0 10 20

Fre

quen

cyAre the independence and normality conditions met in each case?

Each samples is from a simple random sample of its respective population, so the independencecondition is satisfied. Let’s next check the normality condition for each using the rule of thumb.

The first sample has fewer than 30 observations, so we are watching for any clear outliers. None arepresent; while there is a small gap in the histogram on the right, this gap is small and 20% of theobservations in this small sample are represented in that far right bar of the histogram, so we canhardly call these clear outliers. With no clear outliers, the normality condition is reasonably met.

The second sample has a sample size greater than 30 and includes an outlier that appears to beroughly 5 times further from the center of the distribution than the next furthest observation. Thisis an example of a particularly extreme outlier, so the normality condition would not be satisfied.

In practice, it’s typical to also do a mental check to evaluate whether we have reason to believethe underlying population would have moderate skew (if n < 30) or have particularly extreme outliers(n ≥ 30) beyond what we observe in the data. For example, consider the number of followers for eachindividual account on Twitter, and then imagine this distribution. The large majority of accountshave built up a couple thousand followers or fewer, while a relatively tiny fraction have amassedtens of millions of followers, meaning the distribution is extremely skewed. When we know the datacome from such an extremely skewed distribution, it takes some effort to understand what samplesize is large enough for the normality condition to be satisfied.

7.1.3 Introducing the ttt-distribution

In practice, we cannot directly calculate the standard error for x̄ since we do not know thepopulation standard deviation, σ. We encountered a similar issue when computing the standarderror for a sample proportion, which relied on the population proportion, p. Our solution in theproportion context was to use sample value in place of the population value when computing thestandard error. We’ll employ a similar strategy for computing the standard error of x̄, using thesample standard deviation s in place of σ:

SE =σ√n≈ s√

This strategy tends to work well when we have a lot of data and can estimate σ using s accurately.However, the estimate is less precise with smaller samples, and this leads to problems when usingthe normal distribution to model x̄.

1More nuanced guidelines would consider further relaxing the particularly extreme outlier check when the samplesize is very large. However, we’ll leave further discussion here to a future course.

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 253

We’ll find it useful to use a new distribution for inference calculations called the ttt-distribution.A t-distribution, shown as a solid line in Figure 7.1, has a bell shape. However, its tails are thickerthan the normal distribution’s, meaning observations are more likely to fall beyond two standard de-viations from the mean than under the normal distribution. The extra thick tails of the t-distributionare exactly the correction needed to resolve the problem of using s in place of σ in the SE calculation.

−4 −2 0 2 4

Normal

t−distribution

Figure 7.1: Comparison of a t-distribution and a normal distribution.

The t-distribution is always centered at zero and has a single parameter: degrees of freedom.The degrees of freedom (dfdfdf) describes the precise form of the bell-shaped t-distribution. Severalt-distributions are shown in Figure 7.2 in comparison to the normal distribution.

In general, we’ll use a t-distribution with df = n−1 to model the sample mean when the samplesize is n. That is, when we have more observations, the degrees of freedom will be larger and thet-distribution will look more like the standard normal distribution; when the degrees of freedom isabout 30 or more, the t-distribution is nearly indistinguishable from the normal distribution.

−4 −2 0 2 4

normalt, df = 8t, df = 4t, df = 2t, df = 1

Figure 7.2: The larger the degrees of freedom, the more closely the t-distributionresembles the standard normal distribution.

DEGREES OF FREEDOM (dfdfdf )

The degrees of freedom describes the shape of the t-distribution. The larger the degrees offreedom, the more closely the distribution approximates the normal model.

When modeling x̄ using the t-distribution, use df = n− 1.

The t-distribution allows us greater flexibility than the normal distribution when analyzingnumerical data. In practice, it’s common to use statistical software, such as R, Python, or SASfor these analyses. Alternatively, a graphing calculator or a ttt-table may be used; the t-table issimilar to the normal distribution table, and it may be found in Appendix C.2, which includesusage instructions and examples for those who wish to use this option. No matter the approach youchoose, apply your method using the examples below to confirm your working understanding of thet-distribution.

254 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

−4 −2 0 2 4

Figure 7.3: The t-distribution with 18 degrees of freedom. The area below -2.10has been shaded.

−4 −2 0 2 4 −4 −2 0 2 4

Figure 7.4: Left: The t-distribution with 20 degrees of freedom, with the areaabove 1.65 shaded. Right: The t-distribution with 2 degrees of freedom, with thearea further than 3 units from 0 shaded.

EXAMPLE 7.2

What proportion of the t-distribution with 18 degrees of freedom falls below -2.10?

Just like a normal probability problem, we first draw the picture in Figure 7.3 and shade the areabelow -2.10. Using statistical software, we can obtain a precise value: 0.0250.

EXAMPLE 7.3

A t-distribution with 20 degrees of freedom is shown in the left panel of Figure 7.4. Estimate theproportion of the distribution falling above 1.65.

With a normal distribution, this would correspond to about 0.05, so we should expect the t-distribution to give us a value in this neighborhood. Using statistical software: 0.0573.

EXAMPLE 7.4

A t-distribution with 2 degrees of freedom is shown in the right panel of Figure 7.4. Estimate theproportion of the distribution falling more than 3 units from the mean (above or below).

With so few degrees of freedom, the t-distribution will give a more notably different value than thenormal distribution. Under a normal distribution, the area would be about 0.003 using the 68-95-99.7 rule. For a t-distribution with df = 2, the area in both tails beyond 3 units totals 0.0955. Thisarea is dramatically different than what we obtain from the normal distribution.

GUIDED PRACTICE 7.5

What proportion of the t-distribution with 19 degrees of freedom falls above -1.79 units? Use yourpreferred method for finding tail areas.2

2We want to find the shaded area above -1.79 (we leave the picture to you). The lower tail area has an area of0.0447, so the upper area would have an area of 1− 0.0447 = 0.9553.

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 255

7.1.4 One sample ttt-confidence intervals

Let’s get our first taste of applying the t-distribution in the context of an example about themercury content of dolphin muscle. Elevated mercury concentrations are an important problem forboth dolphins and other animals, like humans, who occasionally eat them.

Figure 7.5: A Risso’s dolphin.—————————–Photo by Mike Baird (www.bairdphotos.com). CC BY 2.0 license.

We will identify a confidence interval for the average mercury content in dolphin muscle using asample of 19 Risso’s dolphins from the Taiji area in Japan. The data are summarized in Figure 7.6.The minimum and maximum observed values can be used to evaluate whether or not there are clearoutliers.

n x̄ s minimum maximum19 4.4 2.3 1.7 9.2

Figure 7.6: Summary of mercury content in the muscle of 19 Risso’s dolphins fromthe Taiji area. Measurements are in micrograms of mercury per wet gram of muscle(µg/wet g).

EXAMPLE 7.6

Are the independence and normality conditions satisfied for this data set?

The observations are a simple random sample, therefore independence is reasonable. The summarystatistics in Figure 7.6 do not suggest any clear outliers, since all observations are within 2.5 standarddeviations of the mean. Based on this evidence, the normality condition seems reasonable.

In the normal model, we used z? and the standard error to determine the width of a confidenceinterval. We revise the confidence interval formula slightly when using the t-distribution:

point estimate ± t?df × SE → x̄ ± t?df ×s√n

EXAMPLE 7.7

Using the summary statistics in Figure 7.6, compute the standard error for the average mercurycontent in the n = 19 dolphins.

We plug in s and n into the formula: SE = s/√n = 2.3/

√19 = 0.528.

256 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

The value t?df is a cutoff we obtain based on the confidence level and the t-distribution with dfdegrees of freedom. That cutoff is found in the same way as with a normal distribution: we findt?df such that the fraction of the t-distribution with df degrees of freedom within a distance t?df of 0matches the confidence level of interest.

EXAMPLE 7.8

When n = 19, what is the appropriate degrees of freedom? Find t?df for this degrees of freedom andthe confidence level of 95%

The degrees of freedom is easy to calculate: df = n− 1 = 18.

Using statistical software, we find the cutoff where the upper tail is equal to 2.5%: t?18 = 2.10. Thearea below -2.10 will also be equal to 2.5%. That is, 95% of the t-distribution with df = 18 lieswithin 2.10 units of 0.

EXAMPLE 7.9

Compute and interpret the 95% confidence interval for the average mercury content in Risso’sdolphins.

We can construct the confidence interval as

x̄ ± t?18 × SE → 4.4 ± 2.10× 0.528 → (3.29, 5.51)

We are 95% confident the average mercury content of muscles in Risso’s dolphins is between 3.29and 5.51 µg/wet gram, which is considered extremely high.

FINDING A ttt-CONFIDENCE INTERVAL FOR THE MEAN

Based on a sample of n independent and nearly normal observations, a confidence interval forthe population mean is

point estimate ± t?df × SE → x̄ ± t?df ×s√n

where x̄ is the sample mean, t?df corresponds to the confidence level and degrees of freedom df ,and SE is the standard error as estimated by the sample.

GUIDED PRACTICE 7.10

The FDA’s webpage provides some data on mercury content of fish. Based on a sample of 15 croakerwhite fish (Pacific), a sample mean and standard deviation were computed as 0.287 and 0.069 ppm(parts per million), respectively. The 15 observations ranged from 0.18 to 0.41 ppm. We will assumethese observations are independent. Based on the summary statistics of the data, do you have anyobjections to the normality condition of the individual observations?3

EXAMPLE 7.11

Estimate the standard error of x̄ = 0.287 ppm using the data summaries in Guided Practice 7.10.If we are to use the t-distribution to create a 90% confidence interval for the actual mean of themercury content, identify the degrees of freedom and t?df .

The standard error: SE = 0.069√15

= 0.0178.

Degrees of freedom: df = n− 1 = 14.

Since the goal is a 90% confidence interval, we choose t?14 so that the two-tail area is 0.1: t?14 = 1.76.

3The sample size is under 30, so we check for obvious outliers: since all observations are within 2 standarddeviations of the mean, there are no such clear outliers.

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 257

CONFIDENCE INTERVAL FOR A SINGLE MEAN

Once you’ve determined a one-mean confidence interval would be helpful for an application,there are four steps to constructing the interval:

Prepare. Identify x̄, s, n, and determine what confidence level you wish to use.

Check. Verify the conditions to ensure x̄ is nearly normal.

Calculate. If the conditions hold, compute SE, find t?df , and construct the interval.

Conclude. Interpret the confidence interval in the context of the problem.

GUIDED PRACTICE 7.12

Using the information and results of Guided Practice 7.10 and Example 7.11, compute a 90% con-fidence interval for the average mercury content of croaker white fish (Pacific).4

GUIDED PRACTICE 7.13

The 90% confidence interval from Guided Practice 7.12 is 0.256 ppm to 0.318 ppm. Can we say that90% of croaker white fish (Pacific) have mercury levels between 0.256 and 0.318 ppm?5

7.1.5 One sample ttt-tests

Is the typical US runner getting faster or slower over time? We consider this question in the contextof the Cherry Blossom Race, which is a 10-mile race in Washington, DC each spring.

The average time for all runners who finished the Cherry Blossom Race in 2006 was 93.29minutes (93 minutes and about 17 seconds). We want to determine using data from 100 participantsin the 2017 Cherry Blossom Race whether runners in this race are getting faster or slower, versusthe other possibility that there has been no change.

GUIDED PRACTICE 7.14

What are appropriate hypotheses for this context?6

GUIDED PRACTICE 7.15

The data come from a simple random sample of all participants, so the observations are independent.However, should we be worried about the normality condition? See Figure 7.7 for a histogram ofthe differences and evaluate if we can move forward.7

When completing a hypothesis test for the one-sample mean, the process is nearly identical tocompleting a hypothesis test for a single proportion. First, we find the Z-score using the observedvalue, null value, and standard error; however, we call it a T-score since we use a t-distribution forcalculating the tail area. Then we finding the p-value using the same ideas we used previously: findthe one-tail area under the sampling distribution, and double it.

4 x̄ ± t?14 × SE → 0.287 ± 1.76× 0.0178 → (0.256, 0.318). We are 90% confident that the average mercurycontent of croaker white fish (Pacific) is between 0.256 and 0.318 ppm.

5 No, a confidence interval only provides a range of plausible values for a population parameter, in this case thepopulation mean. It does not describe what we might observe for individual observations.

6H0: The average 10-mile run time was the same for 2006 and 2017. µ = 93.29 minutes. HA: The average 10-milerun time for 2017 was different than that of 2006. µ 6= 93.29 minutes.

7With a sample of 100, we should only be concerned if there is are particularly extreme outliers. The histogramof the data doesn’t show any outliers of concern (and arguably, no outliers at all).

258 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

Time (Minutes)

Fre

quen

60 80 100 120 140

Figure 7.7: A histogram of time for the sample Cherry Blossom Race data.

EXAMPLE 7.16

With both the independence and normality conditions satisfied, we can proceed with a hypothesistest using the t-distribution. The sample mean and sample standard deviation of the sample of 100runners from the 2017 Cherry Blossom Race are 97.32 and 16.98 minutes, respectively. Recall thatthe sample size is 100 and the average run time in 2006 was 93.29 minutes. Find the test statisticand p-value. What is your conclusion?

To find the test statistic (T-score), we first must determine the standard error:

SE = 16.98/√

100 = 1.70

Now we can compute the T-score using the sample mean (97.32), null value (93.29), and SE:

T =97.32− 93.29

1.70= 2.37

For df = 100 − 1 = 99, we can determine using statistical software (or a t-table) that the one-tailarea is 0.01, which we double to get the p-value: 0.02.

Because the p-value is smaller than 0.05, we reject the null hypothesis. That is, the data providestrong evidence that the average run time for the Cherry Blossom Run in 2017 is different thanthe 2006 average. Since the observed value is above the null value and we have rejected the nullhypothesis, we would conclude that runners in the race were slower on average in 2017 than in 2006.

HYPOTHESIS TESTING FOR A SINGLE MEAN

Once you’ve determined a one-mean hypothesis test is the correct procedure, there are foursteps to completing the test:

Prepare. Identify the parameter of interest, list out hypotheses, identify the significance level,and identify x̄, s, and n.

Check. Verify conditions to ensure x̄ is nearly normal.

Calculate. If the conditions hold, compute SE, compute the T-score, and identify the p-value.

Conclude. Evaluate the hypothesis test by comparing the p-value to α, and provide a conclu-sion in the context of the problem.

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 259

Exercises

7.1 Identify the critical t. An independent random sample is selected from an approximately normalpopulation with unknown standard deviation. Find the degrees of freedom and the critical t-value (t?) forthe given sample size and confidence level.

(a) n = 6, CL = 90%

(b) n = 21, CL = 98%

(d) n = 12, CL = 99%

7.2 t-distribution. The figure on the right shows three unimodal and symmetric curves: the standardnormal (z) distribution, the t-distribution with 5 degrees of freedom, and the t-distribution with 1 degree offreedom. Determine which is which, and explain your reasoning.

−4 −2 0 2 4

soliddasheddotted

7.3 Find the p-value, Part I. An independent random sample is selected from an approximately normalpopulation with an unknown standard deviation. Find the p-value for the given sample size and test statistic.Also determine if the null hypothesis would be rejected at α = 0.05.

(a) n = 11, T = 1.91

(b) n = 17, T = −3.45

(d) n = 28, T = 2.13

7.4 Find the p-value, Part II. An independent random sample is selected from an approximately normalpopulation with an unknown standard deviation. Find the p-value for the given sample size and test statistic.Also determine if the null hypothesis would be rejected at α = 0.01.

(a) n = 26, T = 2.485

(b) n = 18, T = 0.5

7.5 Working backwards, Part I. A 95% confidence interval for a population mean, µ, is given as (18.985,21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate thesample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Usethe t-distribution in any calculations.

7.6 Working backwards, Part II. A 90% confidence interval for a population mean is (65, 77). Thepopulation distribution is approximately normal and the population standard deviation is unknown. Thisconfidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, themargin of error, and the sample standard deviation.

260 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.7 Sleep habits of New Yorkers. New York is known as “the city that never sleeps”. A random sampleof 25 New Yorkers were asked how much sleep they get per night. Statistical summaries of these data areshown below. The point estimate suggests New Yorkers sleep less than 8 hours a night on average. Is theresult statistically significant?

n x̄ s min max

25 7.73 0.77 6.17 9.78

(a) Write the hypotheses in symbols and in words.

(b) Check conditions, then calculate the test statistic, T , and the associated degrees of freedom.

(d) What is the conclusion of the hypothesis test?

(e) If you were to construct a 90% confidence interval that corresponded to this hypothesis test, would youexpect 8 hours to be in the interval?

7.8 Heights of adults. Researchers studying anthropometry collected body girth measurements and skele-tal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals.The histogram below shows the sample distribution of heights in centimeters.8

Height150 160 170 180 190 200

100

Min 147.2Q1 163.8Median 170.3Mean 171.1SD 9.4Q3 177.8Max 198.1

(a) What is the point estimate for the average height of active individuals? What about the median?

(b) What is the point estimate for the standard deviation of the heights of active individuals? What aboutthe IQR?

(c) Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm(155cm) considered unusually short? Explain your reasoning.

(d) The researchers take another random sample of physically active individuals. Would you expect the meanand the standard deviation of this new sample to be the ones given above? Explain your reasoning.

(e) The sample means obtained are point estimates for the mean height of all active individuals, if thesample of individuals is equivalent to a simple random sample. What measure do we use to quantify thevariability of such an estimate? Compute this quantity using the data from the original sample underthe condition that the data are a simple random sample.

7.9 Find the mean. You are given the following hypotheses:

H0 : µ = 60

HA : µ 6= 60

We know that the sample standard deviation is 8 and the sample size is 20. For what sample mean wouldthe p-value be equal to 0.05? Assume that all conditions necessary for inference are satisfied.

8G. Heinz et al. “Exploring relationships in body dimensions”. In: Journal of Statistics Education 11.2 (2003).

7.1. ONE-SAMPLE MEANS WITH THE T -DISTRIBUTION 261

7.10 t? vs. z?. For a given confidence level, t?df is larger than z?. Explain how t∗df being slightly largerthan z∗ affects the width of the confidence interval.

7.11 Play the piano. Georgianna claims that in a small city renowned for its music school, the averagechild takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with amean of 4.6 years of piano lessons and a standard deviation of 2.2 years.

(a) Evaluate Georgianna’s claim (or that the opposite might be true) using a hypothesis test.

(b) Construct a 95% confidence interval for the number of years students in this city take piano lessons, andinterpret it in context of the data.

7.12 Auto exhaust and lead exposure. Researchers interested in lead exposure due to car exhaust sampledthe blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while workingtraffic enforcement in a primarily urban environment. The blood samples of these officers had an averagelead concentration of 124.32 µg/l and a SD of 37.74 µg/l; a previous study of individuals from a nearbysuburb, with no history of exposure, found an average blood level concentration of 35 µg/l.9

(a) Write down the hypotheses that would be appropriate for testing if the police officers appear to havebeen exposed to a different concentration of lead.

(b) Explicitly state and check all conditions necessary for inference on these data.

(c) Regardless of your answers in part (b), test the hypothesis that the downtown police officers have ahigher lead exposure than the group in the previous study. Interpret your results in context.

7.13 Car insurance savings. A market researcher wants to evaluate car insurance savings at a competingcompany. Based on past studies he is assuming that the standard deviation of savings is $100. He wantsto collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. Howlarge of a sample should he collect?

7.14 SAT scores. The standard deviation of SAT scores for students at a particular Ivy League college is250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students atthis college as part of a class project. They want their margin of error to be no more than 25 points.

(a) Raina wants to use a 90% confidence interval. How large a sample should she collect?

(b) Luke wants to use a 99% confidence interval. Without calculating the actual sample size, determinewhether his sample should be larger or smaller than Raina’s, and explain your reasoning.

9WI Mortada et al. “Study of lead exposure from automobile exhaust as a risk for nephrotoxicity among trafficpolicemen.” In: American journal of nephrology 21.4 (2000), pp. 274–279.

262 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.2 Paired data

In an earlier edition of this textbook, we found that Amazon prices were, on average, lower thanthose of the UCLA Bookstore for UCLA courses in 2010. It’s been several years, and many storeshave adapted to the online market, so we wondered, how is the UCLA Bookstore doing today?

We sampled 201 UCLA courses. Of those, 68 required books could be found on Amazon.A portion of the data set from these courses is shown in Figure 7.8, where prices are in US dollars.

subject course number bookstore amazon price difference1 American Indian Studies M10 47.97 47.45 0.522 Anthropology 2 14.26 13.55 0.713 Arts and Architecture 10 13.50 12.53 0.97…

……

…68 Jewish Studies M10 35.96 32.40 3.56

Figure 7.8: Four cases of the textbooks data set.

7.2.1 Paired observations

Each textbook has two corresponding prices in the data set: one for the UCLA Bookstore andone for Amazon. When two sets of observations have this special correspondence, they are said tobe paired.

PAIRED DATA

Two sets of observations are paired if each observation in one set has a special correspondenceor connection with exactly one observation in the other data set.

To analyze paired data, it is often useful to look at the difference in outcomes of each pair ofobservations. In the textbook data, we look at the differences in prices, which is represented as theprice difference variable in the data set. Here the differences are taken as

UCLA Bookstore price−Amazon price

It is important that we always subtract using a consistent order; here Amazon prices are alwayssubtracted from UCLA prices. The first difference shown in Figure 7.8 is computed as 47.97 −47.45 = 0.52. Similarly, the second difference is computed as 14.26 − 13.55 = 0.71, and the thirdis 13.50 − 12.53 = 0.97. A histogram of the differences is shown in Figure 7.9. Using differencesbetween paired observations is a common and useful way to analyze paired data.

UCLA Bookstore Price − Amazon Price (USD)

−$20 $0 $20 $40 $60 $80

Fre

quen

Figure 7.9: Histogram of the difference in price for each book sampled.

7.2. PAIRED DATA 263

7.2.2 Inference for paired data

To analyze a paired data set, we simply analyze the differences. We can use the same t-distribution techniques we applied in Section 7.1.

ndiff

x̄diff

sdiff

68 3.58 13.42

Figure 7.10: Summary statistics for the 68 price differences.

EXAMPLE 7.17

Set up a hypothesis test to determine whether, on average, there is a difference between Amazon’sprice for a book and the UCLA bookstore’s price. Also, check the conditions for whether we canmove forward with the test using the t-distribution.

We are considering two scenarios: there is no difference or there is some difference in average prices.

H0: µdiff = 0. There is no difference in the average textbook price.

HA: µdiff 6= 0. There is a difference in average prices.

Next, we check the independence and normality conditions. The observations are based on a simplerandom sample, so independence is reasonable. While there are some outliers, n = 68 and noneof the outliers are particularly extreme, so the normality of x̄ is satisfied. With these conditionssatisfied, we can move forward with the t-distribution.

EXAMPLE 7.18

Complete the hypothesis test started in Example 7.17.

To compute the test compute the standard error associated with x̄diff using the standard deviationof the differences (s

diff= 13.42) and the number of differences (n

diff= 68):

SEx̄diff=

sdiff√ndiff

=13.42√

68= 1.63

The test statistic is the T-score of x̄diff under the null condition that the actual mean difference is 0:

T =x̄diff − 0

SEx̄diff

=3.58− 0

1.63= 2.20

To visualize the p-value, the sampling distribution of x̄diff is drawn as though H0 is true, and thep-value is represented by the two shaded tails:

µ0 = 0 xdiff = 2.98

The degrees of freedom is df = 68 − 1 = 67. Using statistical software, we find the one-tail area of0.0156. Doubling this area gives the p-value: 0.0312.

Because the p-value is less than 0.05, we reject the null hypothesis. Amazon prices are, on average,lower than the UCLA Bookstore prices for UCLA courses.

264 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

GUIDED PRACTICE 7.19

Create a 95% confidence interval for the average price difference between books at the UCLA book-store and books on Amazon.10

GUIDED PRACTICE 7.20

We have strong evidence that Amazon is, on average, less expensive. How should this conclusionaffect UCLA student buying habits? Should UCLA students always buy their books on Amazon?11

10Conditions have already verified and the standard error computed in Example 7.17. To find the interval, identifyt?67 using statistical software or the t-table (t?67 = 2.00), and plug it, the point estimate, and the standard error intothe confidence interval formula:

point estimate ± z? × SE → 3.58 ± 2.00× 1.63 → (0.32, 6.84)

We are 95% confident that Amazon is, on average, between $0.32 and $6.84 less expensive than the UCLA Bookstorefor UCLA course books.

11The average price difference is only mildly useful for this question. Examine the distribution shown in Figure 7.9.There are certainly a handful of cases where Amazon prices are far below the UCLA Bookstore’s, which suggests itis worth checking Amazon (and probably other online sites) before purchasing. However, in many cases the Amazonprice is above what the UCLA Bookstore charges, and most of the time the price isn’t that different. Ultimately, ifgetting a book immediately from the bookstore is notably more convenient, e.g. to get started on reading or homework,it’s likely a good idea to go with the UCLA Bookstore unless the price difference on a specific book happens to bequite large.

For reference, this is a very different result from what we (the authors) had seen in a similar data set from 2010. Atthat time, Amazon prices were almost uniformly lower than those of the UCLA Bookstore’s and by a large margin,making the case to use Amazon over the UCLA Bookstore quite compelling at that time. Now we frequently checkmultiple websites to find the best price.

7.2. PAIRED DATA 265

Exercises

7.15 Air quality. Air quality measurements were collected in a random sample of 25 country capitals in2013, and then again in the same cities in 2014. We would like to use these data to compare average airquality between the two years. Should we use a paired or non-paired test? Explain your reasoning.

7.16 True / False: paired. Determine if the following statements are true or false. If false, explain.

(a) In a paired analysis we first take the difference of each pair of observations, and then we do inferenceon these differences.

(b) Two data sets of different sizes cannot be analyzed as paired data.

(c) Consider two sets of data that are paired with each other. Each observation in one data set has a naturalcorrespondence with exactly one observation from the other data set.

(d) Consider two sets of data that are paired with each other. Each observation in one data set is subtractedfrom the average of the other data set’s observations.

7.17 Paired or not? Part I. In each of the following scenarios, determine if the data are paired.

(a) Compare pre- (beginning of semester) and post-test (end of semester) scores of students.

(b) Assess gender-related salary gap by comparing salaries of randomly sampled men and women.

(c) Compare artery thicknesses at the beginning of a study and after 2 years of taking Vitamin E for thesame group of patients.

(d) Assess effectiveness of a diet regimen by comparing the before and after weights of subjects.

7.18 Paired or not? Part II. In each of the following scenarios, determine if the data are paired.

(a) We would like to know if Intel’s stock and Southwest Airlines’ stock have similar rates of return. To findout, we take a random sample of 50 days, and record Intel’s and Southwest’s stock on those same days.

(b) We randomly sample 50 items from Target stores and note the price for each. Then we visit Walmartand collect the price for each of those same 50 items.

(c) A school board would like to determine whether there is a difference in average SAT scores for studentsat one high school versus another high school in the district. To check, they take a simple randomsample of 100 students from each high school.

7.19 Global warming, Part I. Let’s consider a limited set of climate data, examining temperature differencesin 1948 vs 2018. We sampled 197 locations from the National Oceanic and Atmospheric Administration’s(NOAA) historical data, where the data was available for both years of interest. We want to know: werethere more days with temperatures exceeding 90°F in 2018 or in 1948?12 The difference in number of daysexceeding 90°F (number of days in 2018 – number of days in 1948) was calculated for each of the 197 locations.The average of these differences was 2.9 days with a standard deviation of 17.2 days. We are interested indetermining whether these data provide strong evidence that there were more days in 2018 that exceeded90°F from NOAA’s weather stations.

(a) Is there a relationship between the observations collected in 1948and 2018? Or are the observations in the two groups independent?Explain.

(b) Write hypotheses for this research in symbols and in words.

(d) Calculate the test statistic and find the p-value.

(e) Use α = 0.05 to evaluate the test, and interpret your conclusionin context.

(f) What type of error might we have made? Explain in context whatthe error means.

(g) Based on the results of this hypothesis test, would you expect aconfidence interval for the average difference between the numberof days exceeding 90°F from 1948 and 2018 to include 0? Explainyour reasoning.

Differences in Number of Days

−60 −40 −20 0 20 40 60

12NOAA, www.ncdc.noaa.gov/cdo-web/datasets, April 24, 2019.

266 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.20 High School and Beyond, Part I. The National Center of Education Statistics conducted a survey ofhigh school seniors, collecting test data on reading, writing, and several other subjects. Here we examine asimple random sample of 200 students from this survey. Side-by-side box plots of reading and writing scoresas well as a histogram of the differences in scores are shown below.

scor

read write

Differences in scores (read − write)−20 −10 0 10 20

(a) Is there a clear difference in the average reading and writing scores?

(b) Are the reading and writing scores of each student independent of each other?

(c) Create hypotheses appropriate for the following research question: is there an evident difference in theaverage scores of students in the reading and writing exam?

(d) Check the conditions required to complete this test.

(e) The average observed difference in scores is x̄read−write = −0.545, and the standard deviation of thedifferences is 8.887 points. Do these data provide convincing evidence of a difference between the averagescores on the two exams?

(f) What type of error might we have made? Explain what the error means in the context of the application.

(g) Based on the results of this hypothesis test, would you expect a confidence interval for the averagedifference between the reading and writing scores to include 0? Explain your reasoning.

7.21 Global warming, Part II. We considered the change in the number of days exceeding 90°F from 1948and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean andstandard deviation of the reported differences are 2.9 days and 17.2 days.

(a) Calculate a 90% confidence interval for the average difference between number of days exceeding 90°Fbetween 1948 and 2018. We’ve already checked the conditions for you.

(b) Interpret the interval in context.

(c) Does the confidence interval provide convincing evidence that there were more days exceeding 90°F in2018 than in 1948 at NOAA stations? Explain.

7.22 High school and beyond, Part II. We considered the differences between the reading and writingscores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20.The mean and standard deviation of the differences are x̄read−write = −0.545 and 8.887 points.

(a) Calculate a 95% confidence interval for the average difference between the reading and writing scores ofall students.

(b) Interpret this interval in context.

(c) Does the confidence interval provide convincing evidence that there is a real difference in the averagescores? Explain.

7.3. DIFFERENCE OF TWO MEANS 267

7.3 Difference of two means

In this section we consider a difference in two population means, µ1 − µ2, under the condition thatthe data are not paired. Just as with a single sample, we identify conditions to ensure we can usethe t-distribution with a point estimate of the difference, x̄1− x̄2, and a new standard error formula.Other than these two differences, the details are almost identical to the one-mean procedures.

We apply these methods in three contexts: determining whether stem cells can improve heartfunction, exploring the relationship between pregnant womens’ smoking habits and birth weightsof newborns, and exploring whether there is statistically significant evidence that one variation ofan exam is harder than another variation. This section is motivated by questions like “Is thereconvincing evidence that newborns from mothers who smoke have a different average birth weightthan newborns from mothers who don’t smoke?”

7.3.1 Confidence interval for a difference of means

Does treatment using embryonic stem cells (ESCs) help improve heart function following aheart attack? Figure 7.11 contains summary statistics for an experiment to test ESCs in sheepthat had a heart attack. Each of these sheep was randomly assigned to the ESC or control group,and the change in their hearts’ pumping capacity was measured in the study. Figure 7.12 provideshistograms of the two data sets. A positive value corresponds to increased pumping capacity, whichgenerally suggests a stronger recovery. Our goal will be to identify a 95% confidence interval for theeffect of ESCs on the change in heart pumping capacity relative to the control group.

n x̄ sESCs 9 3.50 5.17control 9 -4.33 2.76

Figure 7.11: Summary statistics of the embryonic stem cell study.

The point estimate of the difference in the heart pumping variable is straightforward to find:it is the difference in the sample means.

x̄esc − x̄control = 3.50− (−4.33) = 7.83

For the question of whether we can model this difference using a t-distribution, we’ll need to checknew conditions. Like the 2-proportion cases, we will require a more robust version of independenceso we are confident the two groups are also independent. Secondly, we also check for normality ineach group separately, which in practice is a check for outliers.

USING THE ttt-DISTRIBUTION FOR A DIFFERENCE IN MEANS

The t-distribution can be used for inference when working with the standardized difference oftwo means if

• Independence, extended. The data are independent within and between the two groups,e.g. the data come from independent random samples or from a randomized experiment.

• Normality. We check the outliers rules of thumb for each group separately.

The standard error may be computed as

SE =

√σ2

n1+σ2

The official formula for the degrees of freedom is quite complex and is generally computed usingsoftware, so instead you may use the smaller of n1 − 1 and n2 − 1 for the degrees of freedom ifsoftware isn’t readily available.

268 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.21

Can the t-distribution be used to make inference using the point estimate, x̄esc − x̄control = 7.83?

First, we check for independence. Because the sheep were randomized into the groups, independencewithin and between groups is satisfied.

Figure 7.12 does not reveal any clear outliers in either group. (The ESC group does look a bit morevariability, but this is not the same as having clear outliers.)

With both conditions met, we can use the t-distribution to model the difference of sample means.

freq

uenc

−10% −5% 0% 5% 10% 15%

Embryonic stem cell transplant

Change in heart pumping function

freq

uenc

−10% −5% 0% 5% 10% 15%

Control (no treatment)

Change in heart pumping function

Figure 7.12: Histograms for both the embryonic stem cell and control group.

As with the one-sample case, we always compute the standard error using sample standarddeviations rather than population standard deviations:

SE =

√s2esc

nesc+s2control

ncontrol=

√5.172

2.762

9= 1.95

Generally, we use statistical software to find the appropriate degrees of freedom, or if software isn’tavailable, we can use the smaller of n1 − 1 and n2 − 1 for the degrees of freedom, e.g. if using at-table to find tail areas. For transparency in the Examples and Guided Practice, we’ll use the latterapproach for finding df ; in the case of the ESC example, this means we’ll use df = 8.

7.3. DIFFERENCE OF TWO MEANS 269

EXAMPLE 7.22

Calculate a 95% confidence interval for the effect of ESCs on the change in heart pumping capacityof sheep after they’ve suffered a heart attack.

We will use the sample difference and the standard error that we computed earlier calculations:

x̄esc − x̄control = 7.83 SE =

√5.172

2.762

9= 1.95

Using df = 8, we can identify the critical value of t?8 = 2.31 for a 95% confidence interval. Finally,we can enter the values into the confidence interval formula:

point estimate ± t? × SE → 7.83 ± 2.31× 1.95 → (3.32, 12.34)

We are 95% confident that embryonic stem cells improve the heart’s pumping function in sheep thathave suffered a heart attack by 3.32% to 12.34%.

As with past statistical inference applications, there is a well-trodden procedure.

Prepare. Retrieve critical contextual information, and if appropriate, set up hypotheses.

Check. Ensure the required conditions are reasonably satisfied.

Calculate. Find the standard error, and then construct a confidence interval, or if conducting ahypothesis test, find a test statistic and p-value.

Conclude. Interpret the results in the context of the application.

The details change a little from one setting to the next, but this general approach remain the same.

7.3.2 Hypothesis tests for the difference of two means

A data set called ncbirths represents a random sample of 150 cases of mothers and their new-borns in North Carolina over a year. Four cases from this data set are represented in Figure 7.13. Weare particularly interested in two variables: weight and smoke. The weight variable represents theweights of the newborns and the smoke variable describes which mothers smoked during pregnancy.We would like to know, is there convincing evidence that newborns from mothers who smoke havea different average birth weight than newborns from mothers who don’t smoke? We will use theNorth Carolina sample to try to answer this question. The smoking group includes 50 cases and thenonsmoking group contains 100 cases.

fage mage weeks weight sex smoke1 NA 13 37 5.00 female nonsmoker2 NA 14 36 5.88 female nonsmoker3 19 15 41 8.13 male smoker…

……

…150 45 50 36 9.25 female nonsmoker

Figure 7.13: Four cases from the ncbirths data set. The value “NA”, shown forthe first two entries of the first variable, indicates that piece of data is missing.

270 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.23

Set up appropriate hypotheses to evaluate whether there is a relationship between a mother smokingand average birth weight.

The null hypothesis represents the case of no difference between the groups.

H0: There is no difference in average birth weight for newborns from mothers who did and did notsmoke. In statistical notation: µn−µs = 0, where µn represents non-smoking mothers and µsrepresents mothers who smoked.

HA: There is some difference in average newborn weights from mothers who did and did not smoke(µn − µs 6= 0).

We check the two conditions necessary to model the difference in sample means using thet-distribution.

• Because the data come from a simple random sample, the observations are independent, bothwithin and between samples.

• With both data sets over 30 observations, we inspect the data in Figure 7.14 for any particularlyextreme outliers and find none.

Since both conditions are satisfied, the difference in sample means may be modeled using a t-distribution.

Mothers Who Smoked

Newborn Weights (lbs)0 2 4 6 8 10

Mothers Who Did Not Smoke

Newborn Weights (lbs)

freq

uenc

0 2 4 6 8 10

Figure 7.14: The top panel represents birth weights for infants whose motherssmoked. The bottom panel represents the birth weights for infants whose motherswho did not smoke.

GUIDED PRACTICE 7.24

The summary statistics in Figure 7.15 may be useful for this Guided Practice.13

(a) What is the point estimate of the population difference, µn − µs?(b) Compute the standard error of the point estimate from part (a).

smoker nonsmoker

mean 6.78 7.18st. dev. 1.43 1.60samp. size 50 100

Figure 7.15: Summary statistics for the ncbirths data set.

13(a) The difference in sample means is an appropriate point estimate: x̄n − x̄s = 0.40. (b) The standard error ofthe estimate can be calculated using the standard error formula:

SE =

√σ2n

nn+σ2s

ns≈

√s2nnn

+s2sns

√1.602

100+

1.432

50= 0.26

7.3. DIFFERENCE OF TWO MEANS 271

EXAMPLE 7.25

Complete the hypothesis test started in Example 7.23 and Guided Practice 7.24. Use a significancelevel of α = 0.05. For reference, x̄n − x̄s = 0.40, SE = 0.26, and the sample sizes were nn = 100and ns = 50.

We can find the test statistic for this test using the values from Guided Practice 7.24:

T =0.40− 0

0.26= 1.54

The p-value is represented by the two shaded tails in the following plot:

µn − µs = 0 obs. diff

We find the single tail area using software (or the t-table in Appendix C.2). We’ll use the smaller ofnn−1 = 99 and ns−1 = 49 as the degrees of freedom: df = 49. The one tail area is 0.065; doublingthis value gives the two-tail area and p-value, 0.135.

The p-value is larger than the significance value, 0.05, so we do not reject the null hypothesis. Thereis insufficient evidence to say there is a difference in average birth weight of newborns from NorthCarolina mothers who did smoke during pregnancy and newborns from North Carolina mothers whodid not smoke during pregnancy.

GUIDED PRACTICE 7.26

We’ve seen much research suggesting smoking is harmful during pregnancy, so how could we fail toreject the null hypothesis in Example 7.25? 14

GUIDED PRACTICE 7.27

If we made a Type 2 Error and there is a difference, what could we have done differently in datacollection to be more likely to detect the difference?15

Public service announcement: while we have used this relatively small data set as an example,larger data sets show that women who smoke tend to have smaller newborns. In fact, some in thetobacco industry actually had the audacity to tout that as a benefit of smoking:

It’s true. The babies born from women who smoke are smaller, but they’re just as healthyas the babies born from women who do not smoke. And some women would prefer havingsmaller babies.

– Joseph Cullman, Philip Morris’ Chairman of the Board…on CBS’ Face the Nation, Jan 3, 1971

Fact check: the babies from women who smoke are not actually as healthy as the babies from womenwho do not smoke.16

14It is possible that there is a difference but we did not detect it. If there is a difference, we made a Type 2 Error.15We could have collected more data. If the sample sizes are larger, we tend to have a better shot at finding a

difference if one exists. In fact, this is exactly what we would find if we examined a larger data set!16You can watch an episode of John Oliver on Last Week Tonight to explore the present day offenses of the tobacco

industry. Please be aware that there is some adult language: youtu.be/6UsHHOCH4q8.

272 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.3.3 Case study: two versions of a course exam

An instructor decided to run two slight variations of the same exam. Prior to passing out theexams, she shuffled the exams together to ensure each student received a random version. Summarystatistics for how students performed on these two exams are shown in Figure 7.16. Anticipatingcomplaints from students who took Version B, she would like to evaluate whether the differenceobserved in the groups is so large that it provides convincing evidence that Version B was moredifficult (on average) than Version A.

Version n x̄ s min maxA 30 79.4 14 45 100B 27 74.1 20 32 100

Figure 7.16: Summary statistics of scores for each exam version.

GUIDED PRACTICE 7.28

Construct hypotheses to evaluate whether the observed difference in sample means, x̄A − x̄B = 5.3,is due to chance. We will later evaluate these hypotheses using α = 0.01.17

GUIDED PRACTICE 7.29

To evaluate the hypotheses in Guided Practice 7.28 using the t-distribution, we must first verifyconditions.18

(a) Does it seem reasonable that the scores are independent?

(b) Any concerns about outliers?

After verifying the conditions for each sample and confirming the samples are independent ofeach other, we are ready to conduct the test using the t-distribution. In this case, we are estimatingthe true difference in average test scores using the sample data, so the point estimate is x̄A−x̄B = 5.3.The standard error of the estimate can be calculated as

SE =

√s2A

nA+s2B

nB=

√142

30+

202

27= 4.62

Finally, we construct the test statistic:

T =point estimate− null value

SE=

(79.4− 74.1)− 0

4.62= 1.15

If we have a computer handy, we can identify the degrees of freedom as 45.97. Otherwise we use thesmaller of n1 − 1 and n2 − 1: df = 26.

17H0: the exams are equally difficult, on average. µA − µB = 0. HA: one exam was more difficult than the other,on average. µA − µB 6= 0.

18(a) Since the exams were shuffled, the “treatment” in this case was randomly assigned, so independence withinand between groups is satisfied. (b) The summary statistics suggest the data are roughly symmetric about the mean,and the min/max values don’t suggest any outliers of concern.

7.3. DIFFERENCE OF TWO MEANS 273

−3 −2 −1 0 1 2 3

T = 1.15

Figure 7.17: The t-distribution with 26 degrees of freedom and the p-value fromexam example represented as the shaded areas.

EXAMPLE 7.30

Identify the p-value depicted in Figure 7.17 using df = 26, and provide a conclusion in the contextof the case study.

Using software, we can find the one-tail area (0.13) and then double this value to get the two-tailarea, which is the p-value: 0.26. (Alternatively, we could use the t-table in Appendix C.2.)

In Guided Practice 7.28, we specified that we would use α = 0.01. Since the p-value is larger thanα, we do not reject the null hypothesis. That is, the data do not convincingly show that one examversion is more difficult than the other, and the teacher should not be convinced that she shouldadd points to the Version B exam scores.

7.3.4 Pooled standard deviation estimate (special topic)

Occasionally, two populations will have standard deviations that are so similar that they canbe treated as identical. For example, historical data or a well-understood biological mechanism mayjustify this strong assumption. In such cases, we can make the t-distribution approach slightly moreprecise by using a pooled standard deviation.

The pooled standard deviation of two groups is a way to use data from both samples tobetter estimate the standard deviation and standard error. If s1 and s2 are the standard deviationsof groups 1 and 2 and there are very good reasons to believe that the population standard deviationsare equal, then we can obtain an improved estimate of the group variances by pooling their data:

s2pooled =

s21 × (n1 − 1) + s2

2 × (n2 − 1)

n1 + n2 − 2

where n1 and n2 are the sample sizes, as before. To use this new statistic, we substitute s2pooled in

place of s21 and s2

2 in the standard error formula, and we use an updated formula for the degrees offreedom:

df = n1 + n2 − 2

The benefits of pooling the standard deviation are realized through obtaining a better estimateof the standard deviation for each group and using a larger degrees of freedom parameter for thet-distribution. Both of these changes may permit a more accurate model of the sampling distributionof x̄1 − x̄2, if the standard deviations of the two groups are indeed equal.

POOL STANDARD DEVIATIONS ONLY AFTER CAREFUL CONSIDERATION

A pooled standard deviation is only appropriate when background research indicates the pop-ulation standard deviations are nearly equal. When the sample size is large and the conditionmay be adequately checked with data, the benefits of pooling the standard deviations greatlydiminishes.

274 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

Exercises

7.23 Friday the 13th, Part I. In the early 1990’s, researchers in the UK collected data on traffic flow, numberof shoppers, and traffic accident related emergency room admissions on Friday the 13th and the previousFriday, Friday the 6th. The histograms below show the distribution of number of cars passing by a specificintersection on Friday the 6th and Friday the 13th for many such date pairs. Also given are some samplestatistics, where the difference is the number of cars on the 6th minus the number of cars on the 13th.19

Friday the 6th

120000 130000 140000

Friday the 13th

120000 130000 140000

Difference

0 2000 4000

6th 13th Diff.

x̄ 128,385 126,550 1,835s 7,259 7,664 1,176n 10 10 10

(a) Are there any underlying structures in these data that should be considered in an analysis? Explain.

(b) What are the hypotheses for evaluating whether the number of people out on Friday the 6th is differentthan the number out on Friday the 13th?

(d) Calculate the test statistic and the p-value.

(e) What is the conclusion of the hypothesis test?

(f) Interpret the p-value in this context.

(g) What type of error might have been made in the conclusion of your test? Explain.

7.24 Diamonds, Part I. Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity,color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is notsmooth. For example, the difference between the size of a 0.99 carat diamond and a 1 carat diamond isundetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than theprice of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat,each sample of size 23, and compare the average prices of the diamonds. In order to be able to compareequivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, fora 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by 100. Thedistributions and some sample statistics are shown below.20

Conduct a hypothesis test to evaluate if there is a differencebetween the average standardized prices of 0.99 and 1 caratdiamonds. Make sure to state your hypotheses clearly, checkrelevant conditions, and interpret your results in context ofthe data.

0.99 carats 1 carat

Mean $44.51 $56.81SD $13.32 $16.13n 23 23

Poi

nt p

rice

(in d

olla

rs)

0.99 carats 1 carat

19T.J. Scanlon et al. “Is Friday the 13th Bad For Your Health?” In: BMJ 307 (1993), pp. 1584–1586.20H. Wickham. ggplot2: elegant graphics for data analysis. Springer New York, 2009.

7.3. DIFFERENCE OF TWO MEANS 275

7.25 Friday the 13th, Part II. The Friday the 13th study reported in Exercise 7.23 also provides data ontraffic accident related emergency room admissions. The distributions of these counts from Friday the 6th

and Friday the 13th are shown below for six such paired dates along with summary statistics. You mayassume that conditions for inference are met.

Friday the 6th5 10

Friday the 13th5 10

Difference−5 0

6th 13th diff

Mean 7.5 10.83 -3.33SD 3.33 3.6 3.01n 6 6 6

(a) Conduct a hypothesis test to evaluate if there is a difference between the average numbers of trafficaccident related emergency room admissions between Friday the 6th and Friday the 13th.

(b) Calculate a 95% confidence interval for the difference between the average numbers of traffic accidentrelated emergency room admissions between Friday the 6th and Friday the 13th.

(c) The conclusion of the original study states, “Friday 13th is unlucky for some. The risk of hospitaladmission as a result of a transport accident may be increased by as much as 52%. Staying at home isrecommended.” Do you agree with this statement? Explain your reasoning.

7.26 Diamonds, Part II. In Exercise 7.24, we discussed diamond prices (standardized by weight) for dia-monds with weights 0. 99 carats and 1 carat. See the table for summary statistics, and then construct a 95%confidence interval for the average difference between the standardized prices of 0.99 and 1 carat diamonds.You may assume the conditions for inference are met.

0.99 carats 1 carat

Mean $44.51 $56.81SD $13.32 $16.13n 23 23

7.27 Chicken diet and weight, Part I. Chicken farming is a multi-billion dollar industry, and any methodsthat increase the growth rate of young chicks can reduce consumer costs while increasing company profits,possibly by millions of dollars. An experiment was conducted to measure and compare the effectiveness ofvarious feed supplements on the growth rate of chickens. Newly hatched chicks were randomly allocated intosix groups, and each group was given a different feed supplement. Below are some summary statistics fromthis data set along with box plots showing the distribution of weights by feed type.21

Wei

ght (

in g

ram

casein horsebean linseed meatmeal soybean sunflower100

150

200

250

300

350

400●

●

Mean SD ncasein 323.58 64.43 12horsebean 160.20 38.63 10linseed 218.75 52.24 12meatmeal 276.91 64.90 11soybean 246.43 54.13 14sunflower 328.92 48.84 12

(a) Describe the distributions of weights of chickens that were fed linseed and horsebean.

(b) Do these data provide strong evidence that the average weights of chickens that were fed linseed andhorsebean are different? Use a 5% significance level.

(d) Would your conclusion change if we used α = 0.01?

21Chicken Weights by Feed Type, from the datasets package in R..

276 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.28 Fuel efficiency of manual and automatic cars, Part I. Each year the US Environmental ProtectionAgency (EPA) releases fuel economy data on cars manufactured in that year. Below are summary statisticson fuel efficiency (in miles/gallon) from random samples of cars with manual and automatic transmissions.Do these data provide strong evidence of a difference between the average fuel efficiency of cars with manualand automatic transmissions in terms of their average city mileage? Assume that conditions for inferenceare satisfied.22

City MPG

Automatic ManualMean 16.12 19.85SD 3.58 4.51n 26 26

City MPG

automatic manual

7.29 Chicken diet and weight, Part II. Casein is a common weight gain supplement for humans. Does ithave an effect on chickens? Using data provided in Exercise 7.27, test the hypothesis that the average weightof chickens that were fed casein is different than the average weight of chickens that were fed soybean. Ifyour hypothesis test yields a statistically significant result, discuss whether or not the higher average weightof chickens can be attributed to the casein diet. Assume that conditions for inference are satisfied.

7.30 Fuel efficiency of manual and automatic cars, Part II. The table provides summary statistics onhighway fuel economy of the same 52 cars from Exercise 7.28. Use these statistics to calculate a 98%confidence interval for the difference between average highway mileage of manual and automatic cars, andinterpret this interval in the context of the data.23

Hwy MPG

Automatic ManualMean 22.92 27.88SD 5.29 5.01n 26 26

Hwy MPG

automatic manual

22U.S. Department of Energy, Fuel Economy Data, 2012 Datafile.23U.S. Department of Energy, Fuel Economy Data, 2012 Datafile.

7.3. DIFFERENCE OF TWO MEANS 277

7.31 Prison isolation experiment, Part I. Subjects from Central Prison in Raleigh, NC, volunteered foran experiment involving an “isolation” experience. The goal of the experiment was to find a treatmentthat reduces subjects’ psychopathic deviant T scores. This score measures a person’s need for control ortheir rebellion against control, and it is part of a commonly used mental health test called the MinnesotaMultiphasic Personality Inventory (MMPI) test. The experiment had three treatment groups:

(1) Four hours of sensory restriction plus a 15 minute “therapeutic” tape advising that professional help isavailable.

(2) Four hours of sensory restriction plus a 15 minute “emotionally neutral” tape on training hunting dogs.

(3) Four hours of sensory restriction but no taped message.

Forty-two subjects were randomly assigned to these treatment groups, and an MMPI test was administeredbefore and after the treatment. Distributions of the differences between pre and post treatment scores (pre- post) are shown below, along with some sample statistics. Use this information to independently test theeffectiveness of each treatment. Make sure to clearly state your hypotheses, check conditions, and interpretresults in the context of the data.24

Treatment 1

0 20 40

Treatment 2

−20 −10 0 10 20

Treatment 3

−20 −10 0

Tr 1 Tr 2 Tr 3

Mean 6.21 2.86 -3.21SD 12.3 7.94 8.57n 14 14 14

7.32 True / False: comparing means. Determine if the following statements are true or false, and explainyour reasoning for statements you identify as false.

(a) When comparing means of two samples where n1 = 20 and n2 = 40, we can use the normal model forthe difference in means since n2 ≥ 30.

(b) As the degrees of freedom increases, the t-distribution approaches normality.

(c) We use a pooled standard error for calculating the standard error of the difference between means whensample sizes of groups are equal to each other.

24Prison isolation experiment, stat.duke.edu/resources/datasets/prison-isolation.

278 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.4 Power calculations for a difference of means

Often times in experiment planning, there are two competing considerations:

• We want to collect enough data that we can detect important effects.

• Collecting data can be expensive, and in experiments involving people, there may be some riskto patients.

In this section, we focus on the context of a clinical trial, which is a health-related experiment wherethe subject are people, and we will determine an appropriate sample size where we can be 80% surethat we would detect any practically important effects.25

7.4.1 Going through the motions of a test

We’re going to go through the motions of a hypothesis test. This will help us frame ourcalculations for determining an appropriate sample size for the study.

EXAMPLE 7.31

Suppose a pharmaceutical company has developed a new drug for lowering blood pressure, and theyare preparing a clinical trial (experiment) to test the drug’s effectiveness. They recruit people whoare taking a particular standard blood pressure medication. People in the control group will continueto take their current medication through generic-looking pills to ensure blinding. Write down thehypotheses for a two-sided hypothesis test in this context.

Generally, clinical trials use a two-sided alternative hypothesis, so below are suitable hypotheses forthis context:

H0: The new drug performs exactly as well as the standard medication.µtrmt − µctrl = 0.

HA: The new drug’s performance differs from the standard medication.µtrmt − µctrl 6= 0.

EXAMPLE 7.32

The researchers would like to run the clinical trial on patients with systolic blood pressures between140 and 180 mmHg. Suppose previously published studies suggest that the standard deviation of thepatients’ blood pressures will be about 12 mmHg and the distribution of patient blood pressures willbe approximately symmetric.26 If we had 100 patients per group, what would be the approximatestandard error for x̄trmt − x̄ctrl?

The standard error is calculated as follows:

SEx̄trmt−x̄ctrl =

√s2trmt

ntrmt+s2ctrl

nctrl=

√122

100+

122

100= 1.70

This may be an imperfect estimate of SEx̄trmt−x̄ctrl , since the standard deviation estimate we usedmay not be perfectly correct for this group of patients. However, it is sufficient for our purposes.

25Even though we don’t cover it explicitly, similar sample size planning is also helpful for observational studies.26In this particular study, we’d generally measure each patient’s blood pressure at the beginning and end of the

study, and then the outcome measurement for the study would be the average change in blood pressure. That is, bothµtrmt and µctrl would represent average differences. This is what you might think of as a 2-sample paired testingstructure, and we’d analyze it exactly just like a hypothesis test for a difference in the average change for patients.In the calculations we perform here, we’ll suppose that 12 mmHg is the predicted standard deviation of a patient’sblood pressure difference over the course of the study.

7.4. POWER CALCULATIONS FOR A DIFFERENCE OF MEANS 279

EXAMPLE 7.33

What does the null distribution of x̄trmt − x̄ctrl look like?

The degrees of freedom are greater than 30, so the distribution of x̄trmt− x̄ctrl will be approximatelynormal. The standard deviation of this distribution (the standard error) would be about 1.70, andunder the null hypothesis, its mean would be 0.

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distribution

EXAMPLE 7.34

For what values of x̄trmt − x̄ctrl would we reject the null hypothesis?

For α = 0.05, we would reject H0 if the difference is in the lower 2.5% or upper 2.5% tail:

Lower 2.5%: For the normal model, this is 1.96 standard errors below 0, so any difference smallerthan −1.96× 1.70 = −3.332 mmHg.

Upper 2.5%: For the normal model, this is 1.96 standard errors above 0, so any difference largerthan 1.96× 1.70 = 3.332 mmHg.

The boundaries of these rejection regions are shown below:

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distribution

Reject H0Do not

reject H0Reject H0

Next, we’ll perform some hypothetical calculations to determine the probability we reject thenull hypothesis, if the alternative hypothesis were actually true.

7.4.2 Computing the power for a 2-sample test

When planning a study, we want to know how likely we are to detect an effect we care about.In other words, if there is a real effect, and that effect is large enough that it has practical value,then what’s the probability that we detect that effect? This probability is called the power, andwe can compute it for different sample sizes or for different effect sizes.

We first determine what is a practically significant result. Suppose that the company researcherscare about finding any effect on blood pressure that is 3 mmHg or larger vs the standard medication.Here, 3 mmHg is the minimum effect size of interest, and we want to know how likely we are todetect this size of an effect in the study.

280 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.35

Suppose we decided to move forward with 100 patients per treatment group and the new drugreduces blood pressure by an additional 3 mmHg relative to the standard medication. What is theprobability that we detect a drop?

Before we even do any calculations, notice that if x̄trmt− x̄ctrl = −3 mmHg, there wouldn’t even besufficient evidence to reject H0. That’s not a good sign.

To calculate the probability that we will reject H0, we need to determine a few things:

• The sampling distribution for x̄trmt − x̄ctrl when the true difference is -3 mmHg. This is thesame as the null distribution, except it is shifted to the left by 3:

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distributionDistribution withµtrmt − µctrl = −3

• The rejection regions, which are outside of the dotted lines above.

• The fraction of the distribution that falls in the rejection region.

In short, we need to calculate the probability that x < −3.332 for a normal distribution with mean-3 and standard deviation 1.7. To do so, we first shade the area we want to calculate:

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distributionDistribution withµtrmt − µctrl = −3

We’ll use a normal approximation, which is good approximation when the degrees of freedom isabout 30 or more. We’ll start by calculating the Z-score and find the tail area using either statisticalsoftware or the probability table:

Z =−3.332− (−3)

1.7= −0.20 → 0.42

The power for the test is about 42% when µtrmt − µctrl = −3 and each group has a sample sizeof 100.

In Example 7.35, we ignored the upper rejection region in the calculation, which was in theopposite direction of the hypothetical truth, i.e. -3. The reasoning? There wouldn’t be any value inrejecting the null hypothesis and concluding there was an increase when in fact there was a decrease.

We’ve also used a normal distribution instead of the t-distribution. This is a convenience, andif the sample size is too small, we’d need to revert back to using the t-distribution. We’ll discussthis a bit further at the end of this section.

7.4. POWER CALCULATIONS FOR A DIFFERENCE OF MEANS 281

7.4.3 Determining a proper sample size

In the last example, we found that if we have a sample size of 100 in each group, we canonly detect an effect size of 3 mmHg with a probability of about 0.42. Suppose the researchersmoved forward and only used 100 patients per group, and the data did not support the alternativehypothesis, i.e. the researchers did not reject H0. This is a very bad situation to be in for a fewreasons:

• In the back of the researchers’ minds, they’d all be wondering, maybe there is a real andmeaningful difference, but we weren’t able to detect it with such a small sample.

• The company probably invested hundreds of millions of dollars in developing the new drug, sonow they are left with great uncertainty about its potential since the experiment didn’t havea great shot at detecting effects that could still be important.

• Patients were subjected to the drug, and we can’t even say with much certainty that the drugdoesn’t help (or harm) patients.

• Another clinical trial may need to be run to get a more conclusive answer as to whether thedrug does hold any practical value, and conducting a second clinical trial may take years andmany millions of dollars.

We want to avoid this situation, so we need to determine an appropriate sample size to ensure we canbe pretty confident that we’ll detect any effects that are practically important. As mentioned earlier,a change of 3 mmHg was deemed to be the minimum difference that was practically important. As afirst step, we could calculate power for several different sample sizes. For instance, let’s try 500patients per group.

GUIDED PRACTICE 7.36

Calculate the power to detect a change of -3 mmHg when using a sample size of 500 per group.27

(a) Determine the standard error (recall that the standard deviation for patients was expected tobe about 12 mmHg).

(b) Identify the null distribution and rejection regions.

(d) Compute the probability we reject the null hypothesis.

The researchers decided 3 mmHg was the minimum difference that was practically important,and with a sample size of 500, we can be very certain (97.7% or better) that we will detect any suchdifference. We now have moved to another extreme where we are exposing an unnecessary numberof patients to the new drug in the clinical trial. Not only is this ethically questionable, but it wouldalso cost a lot more money than is necessary to be quite sure we’d detect any important effects.

The most common practice is to identify the sample size where the power is around 80%, andsometimes 90%. Other values may be reasonable for a specific context, but 80% and 90% are mostcommonly targeted as a good balance between high power and not exposing too many patients to anew treatment (or wasting too much money).

We could compute the power of the test at several other possible sample sizes until we find onethat’s close to 80%, but there’s a better way. We should solve the problem backwards.

27(a) The standard error is given as SE =√

122

500+ 122

500= 0.76.

(b) & (c) The null distribution, rejection boundaries, and alternative distribution are shown below:

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distributionDistribution withµtrmt − µctrl = −3

The rejection regions are the areas on the outside of the two dotted lines and are at ±0.76× 1.96 = ±1.49.(d) The area of the alternative distribution where µtrmt − µctrl = −3 has been shaded. We compute the Z-score and

find the tail area: Z =−1.49−(−3)

0.76= 1.99 → 0.977. With 500 patients per group, we would be about 97.7% sure

(or more) that we’d detect any effects that are at least 3 mmHg in size.

282 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.37

What sample size will lead to a power of 80%?

We’ll assume we have a large enough sample that the normal distribution is a good approximationfor the test statistic, since the normal distribution and the t-distribution look almost identical whenthe degrees of freedom are moderately large (e.g. df ≥ 30). If that doesn’t turn out to be true, thenwe’d need to make a correction.

We start by identifying the Z-score that would give us a lower tail of 80%. For a moderately largesample size per group, the Z-score for a lower tail of 80% would be about Z = 0.84.

−9 −6 −3 0 3 6 9xtrmt − xctrl

Null distributionDistribution withµtrmt − µctrl = −3

0.84 SE 1.96 SE

Additionally, the rejection region extends 1.96 × SE from the center of the null distribution forα = 0.05. This allows us to calculate the target distance between the center of the null and alternativedistributions in terms of the standard error:

0.84× SE + 1.96× SE = 2.8× SE

In our example, we want the distance between the null and alternative distributions’ centers to equalthe minimum effect size of interest, 3 mmHg, which allows us to set up an equation between thisdifference and the standard error:

3 = 2.8× SE

3 = 2.8×√

122

n =2.82

32×(122 + 122

)= 250.88

We should target 251 patients per group in order to achieve 80% power at the 0.05 significance levelfor this context.

The standard error difference of 2.8 × SE is specific to a context where the targeted power is80% and the significance level is α = 0.05. If the targeted power is 90% or if we use a differentsignificance level, then we’ll use something a little different than 2.8× SE.

Had the suggested sample size been relatively small – roughly 30 or smaller – it would havebeen a good idea to rework the calculations using the degrees of fredom for the smaller sample sizeunder that initial sample size. That is, we would have revised the 0.84 and 1.96 values based ondegrees of freedom implied by the initial sample size. The revised sample size target would generallyhave then been a little larger.

7.4. POWER CALCULATIONS FOR A DIFFERENCE OF MEANS 283

GUIDED PRACTICE 7.38

Suppose the targeted power was 90% and we were using α = 0.01. How many standard errors shouldseparate the centers of the null and alternative distribution, where the alternative distribution iscentered at the minimum effect size of interest?28

GUIDED PRACTICE 7.39

What are some considerations that are important in determining what the power should be for anexperiment?29

Figure 7.18 shows the power for sample sizes from 20 patients to 5,000 patients when α = 0.05and the true difference is -3. This curve was constructed by writing a program to compute the powerfor many different sample sizes.

Sample Size Per Group

Pow

20 50 100 200 500 1000 2000 5000

0.0

0.2

0.4

0.6

0.8

1.0

Figure 7.18: The curve shows the power for different sample sizes in the contextof the blood pressure example when the true difference is -3. Having more thanabout 250 to 350 observations doesn’t provide much additional value in detectingan effect when α = 0.05.

Power calculations for expensive or risky experiments are critical. However, what about experi-ments that are inexpensive and where the ethical considerations are minimal? For example, if we aredoing final testing on a new feature on a popular website, how would our sample size considerationschange? As before, we’d want to make sure the sample is big enough. However, suppose the featurehas undergone some testing and is known to perform well (e.g. the website’s users seem to enjoy thefeature). Then it may be reasonable to run a larger experiment if there’s value from having a moreprecise estimate of the feature’s effect, such as helping guide the development of the next usefulfeature.

28First, find the Z-score such that 90% of the distribution is below it: Z = 1.28. Next, find the cutoffs for therejection regions: ±2.58. Then the difference in centers should be about 1.28× SE + 2.58× SE = 3.86× SE.

29Answers will vary, but here are a few important considerations:

• Whether there is any risk to patients in the study.

• The cost of enrolling more patients.

• The potential downside of not detecting an effect of interest.

284 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

Exercises

7.33 Increasing corn yield. A large farm wants to try out a new type of fertilizer to evaluate whetherit will improve the farm’s corn production. The land is broken into plots that produce an average of 1,215pounds of corn with a standard deviation of 94 pounds per plot. The owner is interested in detecting anyaverage difference of at least 40 pounds per plot. How many plots of land would be needed for the experimentif the desired power level is 90%? Assume each plot of land gets treated with either the current fertilizer orthe new fertilizer.

7.34 Email outreach efforts. A medical research group is recruiting people to complete short surveysabout their medical history. For example, one survey asks for information on a person’s family history inregards to cancer. Another survey asks about what topics were discussed during the person’s last visit to ahospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviationof the number of surveys is about 2.2. The research group wants to try a new interface that they think willencourage new enrollees to complete more surveys, where they will randomize each enrollee to either get thenew interface or the current interface. How many new enrollees do they need for each interface to detect aneffect size of 0.5 surveys per enrollee, if the desired power level is 80%?

7.5. COMPARING MANY MEANS WITH ANOVA 285

7.5 Comparing many means with ANOVA

Sometimes we want to compare means across many groups. We might initially think to do pairwisecomparisons. For example, if there were three groups, we might be tempted to compare the firstmean with the second, then with the third, and then finally compare the second and third meansfor a total of three comparisons. However, this strategy can be treacherous. If we have many groupsand do many comparisons, it is likely that we will eventually find a difference just by chance, evenif there is no difference in the populations. Instead, we should apply a holistic test to check whetherthere is evidence that at least one pair groups are in fact different, and this is where ANOVA savesthe day.

7.5.1 Core ideas of ANOVA

In this section, we will learn a new method called analysis of variance (ANOVA) and anew test statistic called F . ANOVA uses a single hypothesis test to check whether the means acrossmany groups are equal:

H0: The mean outcome is the same across all groups. In statistical notation, µ1 = µ2 = · · · = µkwhere µi represents the mean of the outcome for observations in category i.

HA: At least one mean is different.

Generally we must check three conditions on the data before performing ANOVA:

• the observations are independent within and across groups,

• the data within each group are nearly normal, and

• the variability across the groups is about equal.

When these three conditions are met, we may perform an ANOVA to determine whether the dataprovide strong evidence against the null hypothesis that all the µi are equal.

EXAMPLE 7.40

College departments commonly run multiple lectures of the same introductory course each semesterbecause of high demand. Consider a statistics department that runs three lectures of an introductorystatistics course. We might like to determine whether there are statistically significant differences infirst exam scores in these three classes (A, B, and C). Describe appropriate hypotheses to determinewhether there are any differences between the three classes.

The hypotheses may be written in the following form:

H0: The average score is identical in all lectures. Any observed difference is due to chance. Nota-tionally, we write µA = µB = µC .

HA: The average score varies by class. We would reject the null hypothesis in favor of the alternativehypothesis if there were larger differences among the class averages than what we might expectfrom chance alone.

Strong evidence favoring the alternative hypothesis in ANOVA is described by unusually largedifferences among the group means. We will soon learn that assessing the variability of the groupmeans relative to the variability among individual observations within each group is key to ANOVA’ssuccess.

286 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.41

Examine Figure 7.19. Compare groups I, II, and III. Can you visually determine if the differences inthe group centers is due to chance or not? Now compare groups IV, V, and VI. Do these differencesappear to be due to chance?

Any real difference in the means of groups I, II, and III is difficult to discern, because the data withineach group are very volatile relative to any differences in the average outcome. On the other hand, itappears there are differences in the centers of groups IV, V, and VI. For instance, group V appearsto have a higher mean than that of the other two groups. Investigating groups IV, V, and VI, wesee the differences in the groups’ centers are noticeable because those differences are large relativeto the variability in the individual observations within each group.

Out

com

−1

IV V VI

Figure 7.19: Side-by-side dot plot for the outcomes for six groups.

7.5.2 Is batting performance related to player position in MLB?

We would like to discern whether there are real differences between the batting performance ofbaseball players according to their position: outfielder (OF), infielder (IF), and catcher (C). We willuse a data set called bat18, which includes batting records of 429 Major League Baseball (MLB)players from the 2018 season who had at least 100 at bats. Six of the 429 cases represented inbat18 are shown in Figure 7.20, and descriptions for each variable are provided in Figure 7.21. Themeasure we will use for the player batting performance (the outcome variable) is on-base percentage(OBP). The on-base percentage roughly represents the fraction of the time a player successfully getson base or hits a home run.

name team position AB H HR RBI AVG OBP1 Abreu, J CWS IF 499 132 22 78 0.265 0.3252 Acuna Jr., R ATL OF 433 127 26 64 0.293 0.3663 Adames, W TB IF 288 80 10 34 0.278 0.348…

……

…427 Zimmerman, R WSH IF 288 76 13 51 0.264 0.337428 Zobrist, B CHC IF 455 139 9 58 0.305 0.378429 Zunino, M SEA C 373 75 20 44 0.201 0.259

Figure 7.20: Six cases from the bat18 data matrix.

7.5. COMPARING MANY MEANS WITH ANOVA 287

variable description

name Player nameteam The abbreviated name of the player’s teamposition The player’s primary field position (OF, IF, C)AB Number of opportunities at batH Number of hitsHR Number of home runsRBI Number of runs batted inAVG Batting average, which is equal to H/ABOBP On-base percentage, which is roughly equal to the fraction

of times a player gets on base or hits a home run

Figure 7.21: Variables and their descriptions for the bat18 data set.

GUIDED PRACTICE 7.42

The null hypothesis under consideration is the following: µOF = µIF = µC. Write the null andcorresponding alternative hypotheses in plain language.30

EXAMPLE 7.43

The player positions have been divided into four groups: outfield (OF), infield (IF), and catcher (C).What would be an appropriate point estimate of the on-base percentage by outfielders, µOF?

A good estimate of the on-base percentage by outfielders would be the sample average of OBP forjust those players whose position is outfield: x̄OF = 0.320.

Figure 7.22 provides summary statistics for each group. A side-by-side box plot for the on-base percentage is shown in Figure 7.23. Notice that the variability appears to be approximatelyconstant across groups; nearly constant variance across groups is an important assumption thatmust be satisfied before we consider the ANOVA approach.

OF IF C

Sample size (ni) 160 205 64Sample mean (x̄i) 0.320 0.318 0.302Sample SD (si) 0.043 0.038 0.038

Figure 7.22: Summary statistics of on-base percentage, split by player position.

On−

Bas

e P

erce

ntag

PositionOF IF C

0.20

0.25

0.30

0.35

0.40

0.45

Figure 7.23: Side-by-side box plot of the on-base percentage for 429 players acrossfour groups. There is one prominent outlier visible in the infield group, but with154 observations in the infield group, this outlier is not a concern.

30H0: The average on-base percentage is equal across the four positions. HA: The average on-base percentagevaries across some (or all) groups.

288 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

EXAMPLE 7.44

The largest difference between the sample means is between the catcher and the outfielder positions.Consider again the original hypotheses:

H0: µOF = µIF = µC

HA: The average on-base percentage (µi) varies across some (or all) groups.

Why might it be inappropriate to run the test by simply estimating whether the difference of µCand µOF is statistically significant at a 0.05 significance level?

The primary issue here is that we are inspecting the data before picking the groups that will becompared. It is inappropriate to examine all data by eye (informal testing) and only afterwardsdecide which parts to formally test. This is called data snooping or data fishing. Naturally,we would pick the groups with the large differences for the formal test, and this would leading toan inflation in the Type 1 Error rate. To understand this better, let’s consider a slightly differentproblem.

Suppose we are to measure the aptitude for students in 20 classes in a large elementary school atthe beginning of the year. In this school, all students are randomly assigned to classrooms, so anydifferences we observe between the classes at the start of the year are completely due to chance.However, with so many groups, we will probably observe a few groups that look rather different fromeach other. If we select only these classes that look so different and then perform a formal test, wewill probably make the wrong conclusion that the assignment wasn’t random. While we might onlyformally test differences for a few pairs of classes, we informally evaluated the other classes by eyebefore choosing the most extreme cases for a comparison.

For additional information on the ideas expressed in Example 7.44, we recommend readingabout the prosecutor’s fallacy.31

In the next section we will learn how to use the F statistic and ANOVA to test whether observeddifferences in sample means could have happened just by chance even if there was no difference inthe respective population means.

31See, for example, andrewgelman.com/2007/05/18/the prosecutors.

7.5. COMPARING MANY MEANS WITH ANOVA 289

7.5.3 Analysis of variance (ANOVA) and the FFF -test

The method of analysis of variance in this context focuses on answering one question: is thevariability in the sample means so large that it seems unlikely to be from chance alone? This questionis different from earlier testing procedures since we will simultaneously consider many groups, andevaluate whether their sample means differ more than we would expect from natural variation.We call this variability the mean square between groups (MSG), and it has an associateddegrees of freedom, dfG = k − 1 when there are k groups. The MSG can be thought of as a scaledvariance formula for means. If the null hypothesis is true, any variation in the sample means is dueto chance and shouldn’t be too large. Details of MSG calculations are provided in the footnote.32

However, we typically use software for these computations.The mean square between the groups is, on its own, quite useless in a hypothesis test. We need

a benchmark value for how much variability should be expected among the sample means if thenull hypothesis is true. To this end, we compute a pooled variance estimate, often abbreviated asthe mean square error (MSE), which has an associated degrees of freedom value dfE = n − k.It is helpful to think of MSE as a measure of the variability within the groups. Details of thecomputations of the MSE and a link to an extra online section for ANOVA calculations are providedin the footnote33 for interested readers.

When the null hypothesis is true, any differences among the sample means are only due tochance, and the MSG and MSE should be about equal. As a test statistic for ANOVA, we examinethe fraction of MSG and MSE:

F =MSG

MSE

The MSG represents a measure of the between-group variability, and MSE measures the variabilitywithin each of the groups.

GUIDED PRACTICE 7.45

For the baseball data, MSG = 0.00803 and MSE = 0.00158. Identify the degrees of freedomassociated with MSG and MSE and verify the F statistic is approximately 5.077.34

We can use the F statistic to evaluate the hypotheses in what is called an FFF -test. A p-valuecan be computed from the F statistic using an F distribution, which has two associated parameters:df1 and df2. For the F statistic in ANOVA, df1 = dfG and df2 = dfE . An F distribution with 2 and426 degrees of freedom, corresponding to the F statistic for the baseball hypothesis test, is shownin Figure 7.24.

32Let x̄ represent the mean of outcomes across all groups. Then the mean square between groups is computed as

MSG =1

dfGSSG =

k − 1

k∑i=1

ni (x̄i − x̄)2

where SSG is called the sum of squares between groups and ni is the sample size of group i.33Let x̄ represent the mean of outcomes across all groups. Then the sum of squares total (SST ) is computed as

SST =n∑i=1

(xi − x̄)2

where the sum is over all observations in the data set. Then we compute the sum of squared errors (SSE) in oneof two equivalent ways:

SSE = SST − SSG

= (n1 − 1)s21 + (n2 − 1)s22 + · · ·+ (nk − 1)s2k

where s2i is the sample variance (square of the standard deviation) of the residuals in group i. Then the MSE is the

standardized form of SSE: MSE = 1dfE

SSE.

For additional details on ANOVA calculations, see www.openintro.org/d?file=stat extra anova calculations34There are k = 3 groups, so dfG = k − 1 = 2. There are n = n1 + n2 + n3 = 429 total observations, so dfE =

n−k = 426. Then the F statistic is computed as the ratio of MSG and MSE: F = MSGMSE

= 0.008030.00158

= 5.082 ≈ 5.077.(F = 5.077 was computed by using values for MSG and MSE that were not rounded.)

290 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

0 2 4 6 8

Small tail area

Figure 7.24: An F distribution with df1 = 3 and df2 = 323.

The larger the observed variability in the sample means (MSG) relative to the within-groupobservations (MSE), the larger F will be and the stronger the evidence against the null hypothesis.Because larger values of F represent stronger evidence against the null hypothesis, we use the uppertail of the distribution to compute a p-value.

THE FFF STATISTIC AND THE FFF -TEST

Analysis of variance (ANOVA) is used to test whether the mean outcome differs across 2 ormore groups. ANOVA uses a test statistic F , which represents a standardized ratio of variabilityin the sample means relative to the variability within the groups. If H0 is true and the modelconditions are satisfied, the statistic F follows an F distribution with parameters df1 = k − 1and df2 = n− k. The upper tail of the F distribution is used to represent the p-value.

EXAMPLE 7.46

The p-value corresponding to the shaded area in Figure 7.24 is equal to about 0.0066. Does thisprovide strong evidence against the null hypothesis?

The p-value is smaller than 0.05, indicating the evidence is strong enough to reject the null hypothesisat a significance level of 0.05. That is, the data provide strong evidence that the average on-basepercentage varies by player’s primary field position.

7.5.4 Reading an ANOVA table from software

The calculations required to perform an ANOVA by hand are tedious and prone to humanerror. For these reasons, it is common to use statistical software to calculate the F statistic andp-value.

An ANOVA can be summarized in a table very similar to that of a regression summary, whichwe will see in Chapters 8 and 9. Figure 7.25 shows an ANOVA summary to test whether the mean ofon-base percentage varies by player positions in the MLB. Many of these values should look familiar;in particular, the F -test statistic and p-value can be retrieved from the last two columns.

Df Sum Sq Mean Sq F value Pr(>F)position 2 0.0161 0.0080 5.0766 0.0066Residuals 426 0.6740 0.0016

spooled = 0.040 on df = 423

Figure 7.25: ANOVA summary for testing whether the average on-base percentagediffers across player positions.

7.5. COMPARING MANY MEANS WITH ANOVA 291

7.5.5 Graphical diagnostics for an ANOVA analysis

There are three conditions we must check for an ANOVA analysis: all observations must beindependent, the data in each group must be nearly normal, and the variance within each groupmust be approximately equal.

Independence. If the data are a simple random sample, this condition is satisfied. For processesand experiments, carefully consider whether the data may be independent (e.g. no pairing).For example, in the MLB data, the data were not sampled. However, there are not obviousreasons why independence would not hold for most or all observations.

Approximately normal. As with one- and two-sample testing for means, the normality assump-tion is especially important when the sample size is quite small when it is ironically difficultto check for non-normality. A histogram of the observations from each group is shown in Fig-ure 7.26. Since each of the groups we’re considering have relatively large sample sizes, whatwe’re looking for are major outliers. None are apparent, so this conditions is reasonably met.

Outfielders

On−Base Percentage

Fre

quen

0.2 0.3 0.4

In−fielders

On−Base Percentage

Fre

quen

0.2 0.3 0.4

50Catchers

On−Base Percentage

Fre

quen

cy0.2 0.3 0.4

02468

101214

Figure 7.26: Histograms of OBP for each field position.

Constant variance. The last assumption is that the variance in the groups is about equal fromone group to the next. This assumption can be checked by examining a side-by-side box plotof the outcomes across the groups, as in Figure 7.23 on page 287. In this case, the variabilityis similar in the four groups but not identical. We see in Table 7.22 on page 287 that thestandard deviation doesn’t vary much from one group to the next.

DIAGNOSTICS FOR AN ANOVA ANALYSIS

Independence is always important to an ANOVA analysis. The normality condition is veryimportant when the sample sizes for each group are relatively small. The constant variancecondition is especially important when the sample sizes differ between groups.

292 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.5.6 Multiple comparisons and controlling Type 1 Error rate

When we reject the null hypothesis in an ANOVA analysis, we might wonder, which of thesegroups have different means? To answer this question, we compare the means of each possible pairof groups. For instance, if there are three groups and there is strong evidence that there are somedifferences in the group means, there are three comparisons to make: group 1 to group 2, group 1to group 3, and group 2 to group 3. These comparisons can be accomplished using a two-samplet-test, but we use a modified significance level and a pooled estimate of the standard deviation acrossgroups. Usually this pooled standard deviation can be found in the ANOVA table, e.g. along thebottom of Figure 7.25.

EXAMPLE 7.47

Example 7.40 on page 285 discussed three statistics lectures, all taught during the same semester.Figure 7.27 shows summary statistics for these three courses, and a side-by-side box plot of the datais shown in Figure 7.28. We would like to conduct an ANOVA for these data. Do you see anydeviations from the three conditions for ANOVA?

In this case (like many others) it is difficult to check independence in a rigorous way. Instead,the best we can do is use common sense to consider reasons the assumption of independence maynot hold. For instance, the independence assumption may not be reasonable if there is a starteaching assistant that only half of the students may access; such a scenario would divide a classinto two subgroups. No such situations were evident for these particular data, and we believe thatindependence is acceptable.

The distributions in the side-by-side box plot appear to be roughly symmetric and show no noticeableoutliers.

The box plots show approximately equal variability, which can be verified in Figure 7.27, supportingthe constant variance assumption.

Class i A B Cni 58 55 51x̄i 75.1 72.0 78.9si 13.9 13.8 13.1

Figure 7.27: Summary statistics for the first midterm scores in three differentlectures of the same course.

Lecture

Mid

term

Sco

res

A B C

100

Figure 7.28: Side-by-side box plot for the first midterm scores in three differentlectures of the same course.

7.5. COMPARING MANY MEANS WITH ANOVA 293

GUIDED PRACTICE 7.48

ANOVA was conducted for the midterm data, and summary results are shown in Figure 7.29. Whatshould we conclude?35

Df Sum Sq Mean Sq F value Pr(>F)lecture 2 1290.11 645.06 3.48 0.0330Residuals 161 29810.13 185.16

spooled = 13.61 on df = 161

Figure 7.29: ANOVA summary table for the midterm data.

There is strong evidence that the different means in each of the three classes is not simply dueto chance. We might wonder, which of the classes are actually different? As discussed in earlierchapters, a two-sample t-test could be used to test for differences in each possible pair of groups.However, one pitfall was discussed in Example 7.44 on page 288: when we run so many tests, theType 1 Error rate increases. This issue is resolved by using a modified significance level.

MULTIPLE COMPARISONS AND THE BONFERRONI CORRECTION FOR ααα

The scenario of testing many pairs of groups is called multiple comparisons. The Bonferronicorrection suggests that a more stringent significance level is more appropriate for these tests:

α? = α/K

where K is the number of comparisons being considered (formally or informally). If there are

k groups, then usually all possible pairs are compared and K = k(k−1)2 .

EXAMPLE 7.49

In Guided Practice 7.48, you found strong evidence of differences in the average midterm gradesbetween the three lectures. Complete the three possible pairwise comparisons using the Bonferronicorrection and report any differences.

We use a modified significance level of α? = 0.05/3 = 0.0167. Additionally, we use the pooledestimate of the standard deviation: spooled = 13.61 on df = 161, which is provided in the ANOVAsummary table.

Lecture A versus Lecture B: The estimated difference and standard error are, respectively,

x̄A − x̄B = 75.1− 72 = 3.1 SE =

√13.612

58+

13.612

55= 2.56

(See Section 7.3.4 on page 273 for additional details.) This results in a T-score of 1.21 on df = 161(we use the df associated with spooled). Statistical software was used to precisely identify the two-sided p-value since the modified significance level of 0.0167 is not found in the t-table. The p-value(0.228) is larger than α∗ = 0.0167, so there is not strong evidence of a difference in the means oflectures A and B.

Lecture A versus Lecture C: The estimated difference and standard error are 3.8 and 2.61, respec-tively. This results in a T score of 1.46 on df = 161 and a two-sided p-value of 0.1462. This p-valueis larger than α∗, so there is not strong evidence of a difference in the means of lectures A and C.

Lecture B versus Lecture C: The estimated difference and standard error are 6.9 and 2.65, respec-tively. This results in a T score of 2.60 on df = 161 and a two-sided p-value of 0.0102. This p-valueis smaller than α∗. Here we find strong evidence of a difference in the means of lectures B and C.

35The p-value of the test is 0.0330, less than the default significance level of 0.05. Therefore, we reject the nullhypothesis and conclude that the difference in the average midterm scores are not due to chance.

294 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

We might summarize the findings of the analysis from Example 7.49 using the following notation:

µA?= µB µA

?= µC µB 6= µC

The midterm mean in lecture A is not statistically distinguishable from those of lectures B or C.However, there is strong evidence that lectures B and C are different. In the first two pairwisecomparisons, we did not have sufficient evidence to reject the null hypothesis. Recall that failing toreject H0 does not imply H0 is true.

REJECTH0H0H0 WITH ANOVA BUT FIND NO DIFFERENCES IN GROUP MEANS

It is possible to reject the null hypothesis using ANOVA and then to not subsequently identifydifferences in the pairwise comparisons. However, this does not invalidate the ANOVA conclu-sion. It only means we have not been able to successfully identify which specific groups differin their means.

The ANOVA procedure examines the big picture: it considers all groups simultaneously todecipher whether there is evidence that some difference exists. Even if the test indicates that thereis strong evidence of differences in group means, identifying with high confidence a specific differenceas statistically significant is more difficult.

Consider the following analogy: we observe a Wall Street firm that makes large quantities ofmoney based on predicting mergers. Mergers are generally difficult to predict, and if the predictionsuccess rate is extremely high, that may be considered sufficiently strong evidence to warrant inves-tigation by the Securities and Exchange Commission (SEC). While the SEC may be quite certainthat there is insider trading taking place at the firm, the evidence against any single trader may notbe very strong. It is only when the SEC considers all the data that they identify the pattern. Thisis effectively the strategy of ANOVA: stand back and consider all the groups simultaneously.

7.5. COMPARING MANY MEANS WITH ANOVA 295

Exercises

7.35 Fill in the blank. When doing an ANOVA, you observe large differences in means between groups.Within the ANOVA framework, this would most likely be interpreted as evidence strongly favoring the

hypothesis.

7.36 Which test? We would like to test if students who are in the social sciences, natural sciences, artsand humanities, and other fields spend the same amount of time studying for this course. What type of testshould we use? Explain your reasoning.

7.37 Chicken diet and weight, Part III. In Exercises 7.27 and 7.29 we compared the effects of two typesof feed at a time. A better analysis would first consider all feed types at once: casein, horsebean, linseed,meat meal, soybean, and sunflower. The ANOVA output below can be used to test for differences betweenthe average weights of chicks on different diets.

Df Sum Sq Mean Sq F value Pr(>F)

feed 5 231,129.16 46,225.83 15.36 0.0000Residuals 65 195,556.02 3,008.55

Conduct a hypothesis test to determine if these data provide convincing evidence that the average weightof chicks varies across some (or all) groups. Make sure to check relevant conditions. Figures and summarystatistics are shown below.

Wei

ght (

in g

ram

casein horsebean linseed meatmeal soybean sunflower100

150

200

250

300

350

400●

●

Mean SD ncasein 323.58 64.43 12horsebean 160.20 38.63 10linseed 218.75 52.24 12meatmeal 276.91 64.90 11soybean 246.43 54.13 14sunflower 328.92 48.84 12

7.38 Teaching descriptive statistics. A study compared five different methods for teaching descriptivestatistics. The five methods were traditional lecture and discussion, programmed textbook instruction,programmed text with lectures, computer instruction, and computer instruction with lectures. 45 studentswere randomly assigned, 9 to each method. After completing the course, students took a 1-hour exam.

(a) What are the hypotheses for evaluating if the average test scores are different for the different teachingmethods?

(b) What are the degrees of freedom associated with the F -test for evaluating these hypotheses?

296 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.39 Coffee, depression, and physical activity. Caffeine is the world’s most widely used stimulant, withapproximately 80% consumed in the form of coffee. Participants in a study investigating the relationshipbetween coffee consumption and exercise were asked to report the number of hours they spent per week onmoderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on thesedata the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value alwaysgreater than 0. The table below gives summary statistics of MET for women in this study based on theamount of coffee consumed.36

Caffeinated coffee consumption≤ 1 cup/week 2-6 cups/week 1 cup/day 2-3 cups/day ≥ 4 cups/day Total

Mean 18.7 19.6 19.3 18.9 17.5SD 21.1 25.5 22.5 22.0 22.0n 12,215 6,617 17,234 12,290 2,383 50,739

(a) Write the hypotheses for evaluating if the average physical activity level varies among the different levelsof coffee consumption.

(b) Check conditions and describe any assumptions you must make to proceed with the test.

Df Sum Sq Mean Sq F value Pr(>F)

coffee XXXXX XXXXX XXXXX XXXXX 0.0003

Residuals XXXXX 25,564,819 XXXXX

Total XXXXX 25,575,327

(d) What is the conclusion of the test?

7.40 Student performance across discussion sections. A professor who teaches a large introductorystatistics class (197 students) with eight discussion sections would like to test if student performance differsby discussion section, where each discussion section has a different teaching assistant. The summary tablebelow shows the average final exam score for each discussion section as well as the standard deviation ofscores and the number of students in each section.

Sec 1 Sec 2 Sec 3 Sec 4 Sec 5 Sec 6 Sec 7 Sec 8

ni 33 19 10 29 33 10 32 31x̄i 92.94 91.11 91.80 92.45 89.30 88.30 90.12 93.35si 4.21 5.58 3.43 5.92 9.32 7.27 6.93 4.57

The ANOVA output below can be used to test for differences between the average scores from the differentdiscussion sections.

Df Sum Sq Mean Sq F value Pr(>F)

section 7 525.01 75.00 1.87 0.0767Residuals 189 7584.11 40.13

Conduct a hypothesis test to determine if these data provide convincing evidence that the average score variesacross some (or all) groups. Check conditions and describe any assumptions you must make to proceed withthe test.

36M. Lucas et al. “Coffee, caffeine, and risk of depression among women”. In: Archives of internal medicine 171.17(2011), p. 1571.

7.5. COMPARING MANY MEANS WITH ANOVA 297

7.41 GPA and major. Undergraduate students taking an introductory statistics course at Duke Universityconducted a survey about GPA and major. The side-by-side box plots show the distribution of GPA amongthree groups of majors. Also provided is the ANOVA output.

GPA

●

Arts and Humanities Natural Sciences Social Sciences

2.5

3.0

3.5

4.0

Df Sum Sq Mean Sq F value Pr(>F)

major 2 0.03 0.015 0.185 0.8313Residuals 195 15.77 0.081

(a) Write the hypotheses for testing for a difference between average GPA across majors.

(b) What is the conclusion of the hypothesis test?

7.42 Work hours and education. The General Social Survey collects data on demographics, education,and work, among many other characteristics of US residents.37 Using ANOVA, we can consider educa-tional attainment levels for all 1,172 respondents at once. Below are the distributions of hours worked byeducational attainment and relevant summary statistics that will be helpful in carrying out this analysis.

Educational attainmentLess than HS HS Jr Coll Bachelor’s Graduate Total

Mean 38.67 39.6 41.39 42.55 40.85 40.45SD 15.81 14.97 18.1 13.62 15.51 15.17n 121 546 97 253 155 1,172

Hou

rs w

orke

d pe

r w

eek

Less than HS HS Jr Coll Bachelor's Graduate

(a) Write hypotheses for evaluating whether the average number of hours worked varies across the fivegroups.

(b) Check conditions and describe any assumptions you must make to proceed with the test.

Df Sum Sq Mean Sq F-value Pr(>F)

degree XXXXX XXXXX 501.54 XXXXX 0.0682

Residuals XXXXX 267,382 XXXXX

Total XXXXX XXXXX

(d) What is the conclusion of the test?

7.43 True / False: ANOVA, Part I. Determine if the following statements are true or false in ANOVA, andexplain your reasoning for statements you identify as false.

(a) As the number of groups increases, the modified significance level for pairwise tests increases as well.

(b) As the total sample size increases, the degrees of freedom for the residuals increases as well.

(c) The constant variance condition can be somewhat relaxed when the sample sizes are relatively consistentacross groups.

(d) The independence assumption can be relaxed when the total sample size is large.

37National Opinion Research Center, General Social Survey, 2018.

298 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.44 Child care hours. The China Health and Nutrition Survey aims to examine the effects of the health,nutrition, and family planning policies and programs implemented by national and local governments.38 It,for example, collects information on number of hours Chinese parents spend taking care of their childrenunder age 6. The side-by-side box plots below show the distribution of this variable by educational attainmentof the parent. Also provided below is the ANOVA output for comparing average hours across educationalattainment categories.

Chi

ld c

are

hour

Primary school Lower middle school Upper middle school Technical or vocational College0

100

150

Df Sum Sq Mean Sq F value Pr(>F)

education 4 4142.09 1035.52 1.26 0.2846Residuals 794 653047.83 822.48

(a) Write the hypotheses for testing for a difference between the average number of hours spent on childcare across educational attainment levels.

(b) What is the conclusion of the hypothesis test?

7.45 Prison isolation experiment, Part II. Exercise 7.31 introduced an experiment that was conductedwith the goal of identifying a treatment that reduces subjects’ psychopathic deviant T scores, where thisscore measures a person’s need for control or his rebellion against control. In Exercise 7.31 you evaluated thesuccess of each treatment individually. An alternative analysis involves comparing the success of treatments.The relevant ANOVA output is given below.

Df Sum Sq Mean Sq F value Pr(>F)

treatment 2 639.48 319.74 3.33 0.0461Residuals 39 3740.43 95.91

spooled = 9.793 on df = 39

(a) What are the hypotheses?

(b) What is the conclusion of the test? Use a 5% significance level.

(c) If in part (b) you determined that the test is significant, conduct pairwise tests to determine whichgroups are different from each other. If you did not reject the null hypothesis in part (b), recheck youranswer. Summary statistics for each group are provided below

Tr 1 Tr 2 Tr 3

Mean 6.21 2.86 -3.21SD 12.3 7.94 8.57n 14 14 14

7.46 True / False: ANOVA, Part II. Determine if the following statements are true or false, and explainyour reasoning for statements you identify as false.

If the null hypothesis that the means of four groups are all the same is rejected using ANOVA at a 5%significance level, then …

(a) we can then conclude that all the means are different from one another.

(b) the standardized variability between groups is higher than the standardized variability within groups.

(d) the appropriate α to be used in pairwise comparisons is 0.05 / 4 = 0.0125 since there are four groups.

38UNC Carolina Population Center, China Health and Nutrition Survey, 2006.

7.5. COMPARING MANY MEANS WITH ANOVA 299

Chapter exercises

7.47 Gaming and distracted eating, Part I. A group of researchers are interested in the possible effects ofdistracting stimuli during eating, such as an increase or decrease in the amount of food consumption. To testthis hypothesis, they monitored food intake for a group of 44 patients who were randomized into two equalgroups. The treatment group ate lunch while playing solitaire, and the control group ate lunch without anyadded distractions. Patients in the treatment group ate 52.1 grams of biscuits, with a standard deviation of45.1 grams, and patients in the control group ate 27.1 grams of biscuits, with a standard deviation of 26.4grams. Do these data provide convincing evidence that the average food intake (measured in amount ofbiscuits consumed) is different for the patients in the treatment group? Assume that conditions for inferenceare satisfied.39

7.48 Gaming and distracted eating, Part II. The researchers from Exercise 7.47 also investigated the effectsof being distracted by a game on how much people eat. The 22 patients in the treatment group who atetheir lunch while playing solitaire were asked to do a serial-order recall of the food lunch items they ate.The average number of items recalled by the patients in this group was 4. 9, with a standard deviation of1.8. The average number of items recalled by the patients in the control group (no distraction) was 6.1, witha standard deviation of 1.8. Do these data provide strong evidence that the average number of food itemsrecalled by the patients in the treatment and control groups are different?

7.49 Sample size and pairing. Determine if the following statement is true or false, and if false, explainyour reasoning: If comparing means of two groups with equal sample sizes, always use a paired test.

7.50 College credits. A college counselor is interested in estimating how many credits a student typicallyenrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar’sdatabase of students. The histogram below shows the distribution of the number of credits taken by thesestudents. Sample statistics for this distribution are also provided.

Number of credits

Fre

quen

8 10 12 14 16 18

20 Min 8Q1 13Median 14Mean 13.65SD 1.91Q3 15Max 18

(a) What is the point estimate for the average number of credits taken per semester by students at thiscollege? What about the median?

(b) What is the point estimate for the standard deviation of the number of credits taken per semester bystudents at this college? What about the IQR?

(d) The college counselor takes another random sample of 100 students and this time finds a sample meanof 14.02 units. Should she be surprised that this sample statistic is slightly different than the one fromthe original sample? Explain your reasoning.

(e) The sample means given above are point estimates for the mean number of credits taken by all studentsat that college. What measures do we use to quantify the variability of this estimate? Compute thisquantity using the data from the original sample.

39R.E. Oldham-Cooper et al. “Playing a computer game during lunch affects fullness, memory for lunch, and latersnack intake”. In: The American Journal of Clinical Nutrition 93.2 (2011), p. 308.

300 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.51 Hen eggs. The distribution of the number of eggs laid by a certain species of hen during their breedingperiod has a mean of 35 eggs with a standard deviation of 18.2. Suppose a group of researchers randomlysamples 45 hens of this species, counts the number of eggs laid during their breeding period, and recordsthe sample mean. They repeat this 1,000 times, and build a distribution of sample means.

(a) What is this distribution called?

(b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explainyour reasoning.

(d) Suppose the researchers’ budget is reduced and they are only able to collect random samples of 10 hens.The sample mean of the number of eggs is recorded, and we repeat this 1,000 times, and build a newdistribution of sample means. How will the variability of this new distribution compare to the variabilityof the original distribution?

7.52 Forest management. Forest rangers wanted to better understand the rate of growth for younger treesin the park. They took measurements of a random sample of 50 young trees in 2009 and again measuredthose same trees in 2019. The data below summarize their measurements, where the heights are in feet:

2009 2019 Differences

x̄ 12.0 24.5 12.5s 3.5 9.5 7.2n 50 50 50

Construct a 99% confidence interval for the average growth of (what had been) younger trees in the parkover 2009-2019.

7.53 Experiment resizing. At a startup company running a new weather app, an engineering teamgenerally runs experiments where a random sample of 1% of the app’s visitors in the control group andanother 1% were in the treatment group to test each new feature. The team’s core goal is to increase ametric called daily visitors, which is essentially the number of visitors to the app each day. They trackthis metric in each experiment arm and as their core experiment metric. In their most recent experiment,the team tested including a new animation when the app started, and the number of daily visitors in thisexperiment stabilized at +1.2% with a 95% confidence interval of (-0.2%, +2.6%). This means if this newapp start animation was launched, the team thinks they might lose as many as 0.2% of daily visitors orgain as many as 2.6% more daily visitors. Suppose you are consulting as the team’s data scientist, and afterdiscussing with the team, you and they agree that they should run another experiment that is bigger. Youalso agree that this new experiment should be able to detect a gain in the daily visitors metric of 1.0% ormore with 80% power. Now they turn to you and ask, “How big of an experiment do we need to run toensure we can detect this effect?”

(a) How small must the standard error be if the team is to be able to detect an effect of 1.0% with 80%power and a significance level of α = 0.05? You may safely assume the percent change in daily visitorsmetric follows a normal distribution.

(b) Consider the first experiment, where the point estimate was +1.2% and the 95% confidence interval was(-0.2%, +2.6%). If that point estimate followed a normal distribution, what was the standard error ofthe estimate?

(c) The ratio of the standard error from part (a) vs the standard error from part (b) should be 2.03. Howmuch bigger of an experiment is needed to shrink a standard error by a factor of 2.03?

(d) Using your answer from part (c) and that the original experiment was a 1% vs 1% experiment torecommend an experiment size to the team.

7.5. COMPARING MANY MEANS WITH ANOVA 301

7.54 Torque on a rusty bolt. Project Farm is a YouTube channel that routinely compares differentproducts. In one episode, the channel evaluated different options for loosening rusty bolts.40 Eight optionswere evaluated, including a control group where no treatment was given (“none” in the graph), to determinewhich was most effective. For all treatments, there were four bolts tested, except for a treatment of heatwith a blow torch, where only two data points were collected. The results are shown in the figure below:

Torque Required to Loosen Rusty Bolt, in Foot−Pounds

90 100 110 120 130 140

none

Acetone/ATF

AeroKroil

Liquid Wrench

PB Blaster

Royal Purple

WD−40

Heat

(a) Do you think it is reasonable to apply ANOVA in this case?

(b) Regardless of your answer in part (a), describe hypotheses for ANOVA in this context, and use the tablebelow to carry out the test. Give your conclusion in the context of the data.

Df Sum Sq Mean Sq F value Pr(>F)

treatment 7 3603.43 514.78 4.03 0.0056Residuals 22 2812.80 127.85

(c) The table below are p-values for pairwise t-tests comparing each of the different groups. These p-valueshave not been corrected for multiple comparisons. Which pair of groups appears most likely to representa difference?

AeroKroil Heat Liquid Wrench none PB Blaster Royal Purple WD-40Acetone/ATF 0.2026 0.0308 0.0476 0.1542 0.3294 0.5222 0.3744AeroKroil 0.0027 0.0025 0.8723 0.7551 0.5143 0.6883Heat 0.5580 0.0020 0.0050 0.0096 0.0059Liquid Wrench 0.0017 0.0053 0.0117 0.0065none 0.6371 0.4180 0.5751PB Blaster 0.7318 0.9286Royal Purple 0.8000

(d) There are 28 p-values shown in the table in part (c). Determine if any of them are statistically significantafter correcting for multiple comparisons. If so, which one(s)? Explain your answer.

7.55 Exclusive relationships. A survey conducted on a reasonably random sample of 203 undergraduatesasked, among many other questions, about the number of exclusive relationships these students have beenin. The histogram below shows the distribution of the data from this sample. The sample average is 3.2with a standard deviation of 1.97.

Number of exclusive relationships0 2 4 6 8 10

100

Estimate the average number of exclusive relationships Duke students have been in using a 90% confidenceinterval and interpret this interval in context. Check any conditions required for inference, and note anyassumptions you must make as you proceed with your calculations and conclusions.

40Project Farm on YouTube, youtu.be/xUEob2oAKVs, April 16, 2018.

302 CHAPTER 7. INFERENCE FOR NUMERICAL DATA

7.56 Age at first marriage, Part I. The National Survey of Family Growth conducted by the Centersfor Disease Control gathers information on family life, marriage and divorce, pregnancy, infertility, use ofcontraception, and men’s and women’s health. One of the variables collected on this survey is the age atfirst marriage. The histogram below shows the distribution of ages at first marriage of 5,534 randomlysampled women between 2006 and 2010. The average age at first marriage among these women is 23.44 witha standard deviation of 4.72.41

Age at first marriage10 15 20 25 30 35 40 45

200

400

600

800

1000

Estimate the average age at first marriage of women using a 95% confidence interval, and interpret thisinterval in context. Discuss any relevant assumptions.

7.57 Online communication. A study suggests that the average college student spends 10 hours per weekcommunicating with others online. You believe that this is an underestimate and decide to collect your ownsample for a hypothesis test. You randomly sample 60 students from your dorm and find that on averagethey spent 13.5 hours a week communicating with others online. A friend of yours, who offers to help youwith the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see.

H0 : x̄ < 10 hours

HA : x̄ > 13.5 hours

7.58 Age at first marriage, Part II. Exercise 7.56 presents the results of a 2006 – 2010 survey showing thatthe average age of women at first marriage is 23.44. Suppose a social scientist thinks this value has changedsince the survey was taken. Below is how she set up her hypotheses. Indicate any errors you see.

H0 : x̄ 6= 23.44 years old

HA : x̄ = 23.44 years old

41Centers for Disease Control and Prevention, National Survey of Family Growth, 2010.

303

Chapter 8Introduction to linearregression

8.1 Fitting a line, residuals, and correlation

8.2 Least squares regression

8.3 Types of outliers in linear regression

8.4 Inference for linear regression

304

Linear regression is a very powerful statistical technique. Many people

have some familiarity with regression just from reading the news, where

straight lines are overlaid on scatterplots. Linear models can be used for

prediction or to evaluate whether there is a linear relationship between

two numerical variables.

For videos, slides, and other resources, please visit

www.openintro.org/os

8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 305

8.1 Fitting a line, residuals, and correlation

It’s helpful to think deeply about the line fitting process. In this section, we define the formof a linear model, explore criteria for what makes a good fit, and introduce a new statistic calledcorrelation.

8.1.1 Fitting a line to data

Figure 8.1 shows two variables whose relationship can be modeled perfectly with a straight line.The equation for the line is

y = 5 + 64.96x

Consider what a perfect linear relationship means: we know the exact value of y just by knowing thevalue of x. This is unrealistic in almost any natural process. For example, if we took family income(x), this value would provide some useful information about how much financial support a collegemay offer a prospective student (y). However, the prediction would be far from perfect, since otherfactors play a role in financial support beyond a family’s finances.

●

●●

●

Number of Target Corporation Stocks to Purchase

0 10 20 30

$1k

$2k

Tota

l Cos

t of t

he S

hare

Pur

chas

Figure 8.1: Requests from twelve separate buyers were simultaneously placed witha trading company to purchase Target Corporation stock (ticker TGT, December28th, 2018), and the total cost of the shares were reported. Because the cost iscomputed using a linear formula, the linear fit is perfect.

Linear regression is the statistical method for fitting a line to data where the relationshipbetween two variables, x and y, can be modeled by a straight line with some error:

y = β0 + β1x+ ε

The values β0 and β1 represent the model’s parameters (β is the Greek letter beta), and the error isrepresented by ε (the Greek letter epsilon). The parameters are estimated using data, and we writetheir point estimates as b0 and b1. When we use x to predict y, we usually call x the explanatory orpredictor variable, and we call y the response; we also often drop the ε term when writing downthe model since our main focus is often on the prediction of the average outcome.

It is rare for all of the data to fall perfectly on a straight line. Instead, it’s more common fordata to appear as a cloud of points, such as those examples shown in Figure 8.2. In each case, thedata fall around a straight line, even if none of the observations fall exactly on the line. The firstplot shows a relatively strong downward linear trend, where the remaining variability in the dataaround the line is minor relative to the strength of the relationship between x and y. The secondplot shows an upward trend that, while evident, is not as strong as the first. The last plot shows a

306 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

−50 0 50

−50

500 1000 1500

10000

20000

0 20 40

−200

200

400

Figure 8.2: Three data sets where a linear model may be useful even though thedata do not all fall exactly on the line.

●

●●

● ● ● ● ●●

●

Angle of Incline (Degrees)

Dis

tanc

e Tr

avel

ed (

0 30 60 90

Best fitting straight line is flat (!)

Figure 8.3: A linear model is not useful in this nonlinear case. These data are froman introductory physics experiment.

very weak downward trend in the data, so slight we can hardly notice it. In each of these examples,we will have some uncertainty regarding our estimates of the model parameters, β0 and β1. Forinstance, we might wonder, should we move the line up or down a little, or should we tilt it moreor less? As we move forward in this chapter, we will learn about criteria for line-fitting, and we willalso learn about the uncertainty associated with estimates of model parameters.

There are also cases where fitting a straight line to the data, even if there is a clear relationshipbetween the variables, is not helpful. One such case is shown in Figure 8.3 where there is a very clearrelationship between the variables even though the trend is not linear. We discuss nonlinear trendsin this chapter and the next, but details of fitting nonlinear models are saved for a later course.

8.1.2 Using linear regression to predict possum head lengths

Brushtail possums are a marsupial that lives in Australia, and a photo of one is shown inFigure 8.4. Researchers captured 104 of these animals and took body measurements before releasingthe animals back into the wild. We consider two of these measurements: the total length of eachpossum, from head to tail, and the length of each possum’s head.

Figure 8.5 shows a scatterplot for the head length and total length of the possums. Eachpoint represents a single possum from the data. The head and total length variables are associated:possums with an above average total length also tend to have above average head lengths. While therelationship is not perfectly linear, it could be helpful to partially explain the connection betweenthese variables with a straight line.

8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 307

Figure 8.4: The common brushtail possum of Australia.—————————–Photo by Greg Schechter (https://flic.kr/p/9BAFbR). CC BY 2.0 license.

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d Le

ngth

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Figure 8.5: A scatterplot showing head length against total length for 104 brushtailpossums. A point representing a possum with head length 94.1mm and total length89cm is highlighted.

We want to describe the relationship between the head length and total length variables in thepossum data set using a line. In this example, we will use the total length as the predictor variable,x, to predict a possum’s head length, y. We could fit the linear relationship by eye, as in Figure 8.6.The equation for this line is

ŷ = 41 + 0.59x

A “hat” on y is used to signify that this is an estimate. We can use this line to discuss propertiesof possums. For instance, the equation predicts a possum with a total length of 80 cm will have ahead length of

ŷ = 41 + 0.59× 80

= 88.2

The estimate may be viewed as an average: the equation predicts that possums with a total lengthof 80 cm will have an average head length of 88.2 mm. Absent further information about an 80 cmpossum, the prediction for head length that uses the average is a reasonable estimate.

308 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

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d Le

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Figure 8.6: A reasonable linear model was fit to represent the relationship betweenhead length and total length.

EXAMPLE 8.1

What other variables might help us predict the head length of a possum besides its length?

Perhaps the relationship would be a little different for male possums than female possums, or perhapsit would differ for possums from one region of Australia versus another region. In Chapter 9, we’lllearn about how we can include more than one predictor. Before we get there, we first need to betterunderstand how to best build a simple linear model with one predictor.

8.1.3 Residuals

Residuals are the leftover variation in the data after accounting for the model fit:

Data = Fit + Residual

Each observation will have a residual, and three of the residuals for the linear model we fit for thepossum data is shown in Figure 8.6. If an observation is above the regression line, then its residual,the vertical distance from the observation to the line, is positive. Observations below the line havenegative residuals. One goal in picking the right linear model is for these residuals to be as small aspossible.

Let’s look closer at the three residuals featured in Figure 8.6. The observation marked by an“×” has a small, negative residual of about -1; the observation marked by “+” has a large residualof about +7; and the observation marked by “4” has a moderate residual of about -4. The size ofa residual is usually discussed in terms of its absolute value. For example, the residual for “4” islarger than that of “×” because | − 4| is larger than | − 1|.

RESIDUAL: DIFFERENCE BETWEEN OBSERVED AND EXPECTED

The residual of the ith observation (xi, yi) is the difference of the observed response (yi) andthe response we would predict based on the model fit (ŷi):

ei = yi − ŷi

We typically identify ŷi by plugging xi into the model.

8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 309

EXAMPLE 8.2

The linear fit shown in Figure 8.6 is given as ŷ = 41 + 0.59x. Based on this line, formally computethe residual of the observation (77.0, 85.3). This observation is denoted by “×” in Figure 8.6. Checkit against the earlier visual estimate, -1.

We first compute the predicted value of point “×” based on the model:

ŷ× = 41 + 0.59x× = 41 + 0.59× 77.0 = 86.4

Next we compute the difference of the actual head length and the predicted head length:

e× = y× − ŷ× = 85.3− 86.4 = −1.1

The model’s error is e× = −1.1mm, which is very close to the visual estimate of -1mm. The negativeresidual indicates that the linear model overpredicted head length for this particular possum.

GUIDED PRACTICE 8.3

If a model underestimates an observation, will the residual be positive or negative? What about ifit overestimates the observation?1

GUIDED PRACTICE 8.4

Compute the residuals for the “+” observation (85.0, 98.6) and the “4” observation (95.5, 94.0) inthe figure using the linear relationship ŷ = 41 + 0.59x.2

Residuals are helpful in evaluating how well a linear model fits a data set. We often displaythem in a residual plot such as the one shown in Figure 8.7 for the regression line in Figure 8.6.The residuals are plotted at their original horizontal locations but with the vertical coordinate asthe residual. For instance, the point (85.0, 98.6)+ had a residual of 7.45, so in the residual plot itis placed at (85.0, 7.45). Creating a residual plot is sort of like tipping the scatterplot over so theregression line is horizontal.

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−5

Total Length (cm)

Res

idua

Figure 8.7: Residual plot for the model in Figure 8.6.

1If a model underestimates an observation, then the model estimate is below the actual. The residual, which isthe actual observation value minus the model estimate, must then be positive. The opposite is true when the modeloverestimates the observation: the residual is negative.

2(+) First compute the predicted value based on the model:

ŷ+ = 41 + 0.59x+ = 41 + 0.59× 85.0 = 91.15

Then the residual is given by

e+ = y+ − ŷ+ = 98.6− 91.15 = 7.45

This was close to the earlier estimate of 7.(4) ŷ4 = 41 + 0.59×4 = 97.3. e4 = y4 − ŷ4 = −3.3, close to the estimate of -4.

310 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

x x

yg$residuals

Figure 8.8: Sample data with their best fitting lines (top row) and their corre-sponding residual plots (bottom row).

EXAMPLE 8.5

One purpose of residual plots is to identify characteristics or patterns still apparent in data afterfitting a model. Figure 8.8 shows three scatterplots with linear models in the first row and residualplots in the second row. Can you identify any patterns remaining in the residuals?

In the first data set (first column), the residuals show no obvious patterns. The residuals appear tobe scattered randomly around the dashed line that represents 0.

The second data set shows a pattern in the residuals. There is some curvature in the scatterplot,which is more obvious in the residual plot. We should not use a straight line to model these data.Instead, a more advanced technique should be used.

The last plot shows very little upwards trend, and the residuals also show no obvious patterns. It isreasonable to try to fit a linear model to the data. However, it is unclear whether there is statisticallysignificant evidence that the slope parameter is different from zero. The point estimate of the slopeparameter, labeled b1, is not zero, but we might wonder if this could just be due to chance. We willaddress this sort of scenario in Section 8.4.

8.1.4 Describing linear relationships with correlation

We’ve seen plots with strong linear relationships and others with very weak linear relationships. Itwould be useful if we could quantify the strength of these linear relationships with a statistic.

CORRELATION: STRENGTH OF A LINEAR RELATIONSHIP

Correlation, which always takes values between -1 and 1, describes the strength of the linearrelationship between two variables. We denote the correlation by R.

We can compute the correlation using a formula, just as we did with the sample mean and stan-dard deviation. This formula is rather complex,3 and like with other statistics, we generally perform

3Formally, we can compute the correlation for observations (x1, y1), (x2, y2), …, (xn, yn) using the formula

R =1

n− 1

n∑i=1

xi − x̄sx

yi − ȳsy

where x̄, ȳ, sx, and sy are the sample means and standard deviations for each variable.

8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 311

R = 0.33y

R = 0.69

R = 0.98

R = 1.00

R = 0.08

R = −0.64

R = −0.92

R = −1.00

Figure 8.9: Sample scatterplots and their correlations. The first row shows vari-ables with a positive relationship, represented by the trend up and to the right.The second row shows variables with a negative trend, where a large value in onevariable is associated with a low value in the other.

the calculations on a computer or calculator. Figure 8.9 shows eight plots and their correspondingcorrelations. Only when the relationship is perfectly linear is the correlation either -1 or 1. If therelationship is strong and positive, the correlation will be near +1. If it is strong and negative, itwill be near -1. If there is no apparent linear relationship between the variables, then the correlationwill be near zero.

The correlation is intended to quantify the strength of a linear trend. Nonlinear trends, evenwhen strong, sometimes produce correlations that do not reflect the strength of the relationship; seethree such examples in Figure 8.10.

R = −0.23

R = 0.31

R = 0.50

Figure 8.10: Sample scatterplots and their correlations. In each case, there is astrong relationship between the variables. However, because the relationship isnonlinear, the correlation is relatively weak.

GUIDED PRACTICE 8.6

No straight line is a good fit for the data sets represented in Figure 8.10. Try drawing nonlinearcurves on each plot. Once you create a curve for each, describe what is important in your fit.4

4We’ll leave it to you to draw the lines. In general, the lines you draw should be close to most points and reflectoverall trends in the data.

312 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

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8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 313

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8.5 Exams and grades. The two scatterplots below show the relationship between final and mid-semesterexam grades recorded during several years for a Statistics course at a university.

(a) Based on these graphs, which of the two exams has the strongest correlation with the final exam grade?Explain.

(b) Can you think of a reason why the correlation between the exam you chose in part (a) and the finalexam is higher?

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314 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.6 Husbands and wives, Part I. The Great Britain Office of Population Census and Surveys once collecteddata on a random sample of 170 married couples in Britain, recording the age (in years) and heights(converted here to inches) of the husbands and wives.5 The scatterplot on the left shows the wife’s ageplotted against her husband’s age, and the plot on the right shows wife’s height plotted against husband’sheight.

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Wife

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ge (

in y

ears

)

20 40 60

Husband's height (in inches)

Wife

's h

eigh

t (in

inch

es)

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(a) Describe the relationship between husbands’ and wives’ ages.

(b) Describe the relationship between husbands’ and wives’ heights.

(d) Data on heights were originally collected in centimeters, and then converted to inches. Does this con-version affect the correlation between husbands’ and wives’ heights?

8.7 Match the correlation, Part I. Match each correlation to the corresponding scatterplot.

(a) R = −0.7

(b) R = 0.45

(d) R = 0.92

(1) (2) (3) (4)

8.8 Match the correlation, Part II. Match each correlation to the corresponding scatterplot.

(a) R = 0.49

(b) R = −0.48

(d) R = −0.85

(1) (2) (3) (4)

8.9 Speed and height. 1,302 UCLA students were asked to fill out a survey where they were askedabout their height, fastest speed they have ever driven, and gender. The scatterplot on the left displays therelationship between height and fastest speed, and the scatterplot on the right displays the breakdown bygender in this relationship.

Height (in inches)

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5D.J. Hand. A handbook of small data sets. Chapman & Hall/CRC, 1994.

8.1. FITTING A LINE, RESIDUALS, AND CORRELATION 315

8.10 Guess the correlation. Eduardo and Rosie are both collecting data on number of rainy days in ayear and the total rainfall for the year. Eduardo records rainfall in inches and Rosie in centimeters. Howwill their correlation coefficients compare?

8.11 The Coast Starlight, Part I. The Coast Starlight Amtrak train runs from Seattle to Los Angeles. Thescatterplot below displays the distance between each stop (in miles) and the amount of time it takes to travelfrom one stop to another (in minutes).

(a) Describe the relationship betweendistance and travel time.

(b) How would the relationship change iftravel time was instead measured inhours, and distance was insteadmeasured in kilometers?

(c) Correlation between travel time (inmiles) and distance (in minutes) isr = 0.636. What is the correlationbetween travel time (in kilometers)and distance (in hours)?

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8.12 Crawling babies, Part I. A study conducted at the University of Denver investigated whether babiestake longer to learn to crawl in cold months, when they are often bundled in clothes that restrict theirmovement, than in warmer months.6 Infants born during the study year were split into twelve groups, onefor each birth month. We consider the average crawling age of babies in each group against the average tem-perature when the babies are six months old (that’s when babies often begin trying to crawl). Temperatureis measured in degrees Fahrenheit (◦F) and age is measured in weeks.

(a) Describe the relationship betweentemperature and crawling age.

(b) How would the relationship change iftemperature was measured in degreesCelsius (◦C) and age was measured inmonths?

(c) The correlation between temperaturein ◦F and age in weeks was r = −0.70.If we converted the temperature to ◦Cand age to months, what would thecorrelation be?

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wlin

g ag

e (w

eeks

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6J.B. Benson. “Season of birth and onset of locomotion: Theoretical and methodological implications”. In: Infantbehavior and development 16.1 (1993), pp. 69–81. issn: 0163-6383.

316 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.13 Body measurements, Part I. Researchers studying anthropometry collected body girth measurementsand skeletal diameter measurements, as well as age, weight, height and gender for 507 physically activeindividuals.7 The scatterplot below shows the relationship between height and shoulder girth (over deltoidmuscles), both measured in centimeters.

(a) Describe the relationship betweenshoulder girth and height.

(b) How would the relationship change ifshoulder girth was measured in incheswhile the units of height remained incentimeters?

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180

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eigh

t (cm

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8.14 Body measurements, Part II. The scatterplot below shows the relationship between weight measuredin kilograms and hip girth measured in centimeters from the data described in Exercise 8.13.

(a) Describe the relationship between hipgirth and weight.

(b) How would the relationship change ifweight was measured in pounds whilethe units for hip girth remained incentimeters?

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ght (

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8.15 Correlation, Part I. What would be the correlation between the ages of husbands and wives if menalways married woman who were

(a) 3 years younger than themselves?

(b) 2 years older than themselves?

8.16 Correlation, Part II. What would be the correlation between the annual salaries of males and femalesat a company if for a certain type of position men always made

(a) $5,000 more than women?

(b) 25% more than women?

7G. Heinz et al. “Exploring relationships in body dimensions”. In: Journal of Statistics Education 11.2 (2003).

8.2. LEAST SQUARES REGRESSION 317

8.2 Least squares regression

Fitting linear models by eye is open to criticism since it is based on an individual’s preference.In this section, we use least squares regression as a more rigorous approach.

8.2.1 Gift aid for freshman at Elmhurst College

This section considers family income and gift aid data from a random sample of fifty studentsin the freshman class of Elmhurst College in Illinois. Gift aid is financial aid that does not need tobe paid back, as opposed to a loan. A scatterplot of the data is shown in Figure 8.11 along with twolinear fits. The lines follow a negative trend in the data; students who have higher family incomestended to have lower gift aid from the university.

Family Income

$0 $50k $100k $150k $200k $250k

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$20k

$30k

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Aid

Fro

m U

nive

rsity

Figure 8.11: Gift aid and family income for a random sample of 50 freshmanstudents from Elmhurst College. Two lines are fit to the data, the solid line beingthe least squares line.

GUIDED PRACTICE 8.7

Is the correlation positive or negative in Figure 8.11?8

8.2.2 An objective measure for finding the best line

We begin by thinking about what we mean by “best”. Mathematically, we want a line thathas small residuals. The first option that may come to mind is to minimize the sum of the residualmagnitudes:

|e1|+ |e2|+ · · ·+ |en|

which we could accomplish with a computer program. The resulting dashed line shown in Figure 8.11demonstrates this fit can be quite reasonable. However, a more common practice is to choose theline that minimizes the sum of the squared residuals:

e21 + e2

2 + · · ·+ e2n

8Larger family incomes are associated with lower amounts of aid, so the correlation will be negative. Using acomputer, the correlation can be computed: -0.499.

318 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

The line that minimizes this least squares criterion is represented as the solid line in Fig-ure 8.11. This is commonly called the least squares line. The following are three possible reasonsto choose this option instead of trying to minimize the sum of residual magnitudes without anysquaring:

1. It is the most commonly used method.

2. Computing the least squares line is widely supported in statistical software.

3. In many applications, a residual twice as large as another residual is more than twice as bad.For example, being off by 4 is usually more than twice as bad as being off by 2. Squaring theresiduals accounts for this discrepancy.

The first two reasons are largely for tradition and convenience; the last reason explains why the leastsquares criterion is typically most helpful.9

8.2.3 Conditions for the least squares line

When fitting a least squares line, we generally require

Linearity. The data should show a linear trend. If there is a nonlinear trend (e.g. left panel ofFigure 8.12), an advanced regression method from another book or later course should beapplied.

Nearly normal residuals. Generally, the residuals must be nearly normal. When this conditionis found to be unreasonable, it is usually because of outliers or concerns about influentialpoints, which we’ll talk about more in Sections 8.3. An example of a residual that would bea potentially concern is shown in Figure 8.12, where one observation is clearly much furtherfrom the regression line than the others.

Constant variability. The variability of points around the least squares line remains roughly con-stant. An example of non-constant variability is shown in the third panel of Figure 8.12, whichrepresents the most common pattern observed when this condition fails: the variability of y islarger when x is larger.

Independent observations. Be cautious about applying regression to time series data, whichare sequential observations in time such as a stock price each day. Such data may have anunderlying structure that should be considered in a model and analysis. An example of adata set where successive observations are not independent is shown in the fourth panel ofFigure 8.12. There are also other instances where correlations within the data are important,which is further discussed in Chapter 9.

GUIDED PRACTICE 8.8

Should we have concerns about applying least squares regression to the Elmhurst data in Fig-ure 8.11?10

9There are applications where the sum of residual magnitudes may be more useful, and there are plenty of othercriteria we might consider. However, this book only applies the least squares criterion.

10The trend appears to be linear, the data fall around the line with no obvious outliers, the variance is roughlyconstant. These are also not time series observations. Least squares regression can be applied to these data.

8.2. LEAST SQUARES REGRESSION 319

x x

yg$

resi

dual

yg$

resi

dual

yg$

resi

dual

Figure 8.12: Four examples showing when the methods in this chapter are insuf-ficient to apply to the data. First panel: linearity fails. Second panel: there areoutliers, most especially one point that is very far away from the line. Third panel:the variability of the errors is related to the value of x. Fourth panel: a time seriesdata set is shown, where successive observations are highly correlated.

8.2.4 Finding the least squares line

For the Elmhurst data, we could write the equation of the least squares regression line as

âid = β0 + β1 × family income

Here the equation is set up to predict gift aid based on a student’s family income, which would beuseful to students considering Elmhurst. These two values, β0 and β1, are the parameters of theregression line.

As in Chapters 5, 6, and 7, the parameters are estimated using observed data. In practice, thisestimation is done using a computer in the same way that other estimates, like a sample mean, canbe estimated using a computer or calculator. However, we can also find the parameter estimates byapplying two properties of the least squares line:

• The slope of the least squares line can be estimated by

b1 =sysxR

where R is the correlation between the two variables, and sx and sy are the sample standarddeviations of the explanatory variable and response, respectively.

• If x̄ is the sample mean of the explanatory variable and ȳ is the sample mean of the verticalvariable, then the point (x̄, ȳ) is on the least squares line.

Figure 8.13 shows the sample means for the family income and gift aid as $101,780 and $19,940,respectively. We could plot the point (101.8, 19.94) on Figure 8.11 on page 317 to verify it fallson the least squares line (the solid line).

Next, we formally find the point estimates b0 and b1 of the parameters β0 and β1.

Family Income (x) Gift Aid (y)

mean x̄ = $101,780 ȳ = $19,940sd sx = $63,200 sy = $5,460

R = −0.499

Figure 8.13: Summary statistics for family income and gift aid.

320 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

GUIDED PRACTICE 8.9

Using the summary statistics in Figure 8.13, compute the slope for the regression line of gift aidagainst family income.11

You might recall the point-slope form of a line from math class, which we can use to find themodel fit, including the estimate of b0. Given the slope of a line and a point on the line, (x0, y0),the equation for the line can be written as

y − y0 = slope× (x− x0)

IDENTIFYING THE LEAST SQUARES LINE FROM SUMMARY STATISTICS

To identify the least squares line from summary statistics:

• Estimate the slope parameter, b1 = (sy/sx)R.

• Noting that the point (x̄, ȳ) is on the least squares line, use x0 = x̄ and y0 = ȳ with thepoint-slope equation: y − ȳ = b1(x− x̄).

• Simplify the equation, which would reveal that b0 = ȳ − b1x̄.

EXAMPLE 8.10

Using the point (101780, 19940) from the sample means and the slope estimate b1 = −0.0431 fromGuided Practice 8.9, find the least-squares line for predicting aid based on family income.

Apply the point-slope equation using (101.8, 19.94) and the slope b1 = −0.0431:

y − y0 = b1(x− x0)

y − 19,940 = −0.0431(x− 101,780)

Expanding the right side and then adding 19,940 to each side, the equation simplifies:

âid = 24,320− 0.0431× family income

Here we have replaced y with âid and x with family income to put the equation in context. Thefinal equation should always include a “hat” on the variable being predicted, whether it is a generic“y” or a named variable like “aid”.

A computer is usually used to compute the least squares line, and a summary table generatedusing software for the Elmhurst regression line is shown in Figure 8.14. The first column of numbersprovides estimates for b0 and b1, respectively. These results match those from Example 8.10.

Estimate Std. Error t value Pr(>|t|)(Intercept) 24319.3 1291.5 18.83 <0.0001family income -0.0431 0.0108 -3.98 0.0002

Figure 8.14: Summary of least squares fit for the Elmhurst data. Compare theparameter estimates in the first column to the results of Example 8.10.

11Compute the slope using the summary statistics from Figure 8.13:

b1 =sy

sxR =

5,460

63,200(−0.499) = −0.0431

8.2. LEAST SQUARES REGRESSION 321

EXAMPLE 8.11

Examine the second, third, and fourth columns in Figure 8.14. Can you guess what they represent?(If you have not reviewed any inference chapter yet, skip this example.)

We’ll describe the meaning of the columns using the second row, which corresponds to β1. The firstcolumn provides the point estimate for β1, as we calculated in an earlier example: b1 = −0.0431.The second column is a standard error for this point estimate: SEb1 = 0.0108. The third columnis a t-test statistic for the null hypothesis that β1 = 0: T = −3.98. The last column is the p-valuefor the t-test statistic for the null hypothesis β1 = 0 and a two-sided alternative hypothesis: 0.0002.We will get into more of these details in Section 8.4.

EXAMPLE 8.12

Suppose a high school senior is considering Elmhurst College. Can she simply use the linear equationthat we have estimated to calculate her financial aid from the university?

She may use it as an estimate, though some qualifiers on this approach are important. First, the dataall come from one freshman class, and the way aid is determined by the university may change fromyear to year. Second, the equation will provide an imperfect estimate. While the linear equation isgood at capturing the trend in the data, no individual student’s aid will be perfectly predicted.

8.2.5 Interpreting regression model parameter estimates

Interpreting parameters in a regression model is often one of the most important steps in the analysis.

EXAMPLE 8.13

The intercept and slope estimates for the Elmhurst data are b0 = 24,319 and b1 = −0.0431. Whatdo these numbers really mean?

Interpreting the slope parameter is helpful in almost any application. For each additional $1,000 offamily income, we would expect a student to receive a net difference of $1,000×(−0.0431) = −$43.10in aid on average, i.e. $43.10 less. Note that a higher family income corresponds to less aid becausethe coefficient of family income is negative in the model. We must be cautious in this interpretation:while there is a real association, we cannot interpret a causal connection between the variablesbecause these data are observational. That is, increasing a student’s family income may not causethe student’s aid to drop. (It would be reasonable to contact the college and ask if the relationshipis causal, i.e. if Elmhurst College’s aid decisions are partially based on students’ family income.)

The estimated intercept b0 = 24,319 describes the average aid if a student’s family had no income.The meaning of the intercept is relevant to this application since the family income for some studentsat Elmhurst is $0. In other applications, the intercept may have little or no practical value if thereare no observations where x is near zero.

INTERPRETING PARAMETERS ESTIMATED BY LEAST SQUARES

The slope describes the estimated difference in the y variable if the explanatory variable x for acase happened to be one unit larger. The intercept describes the average outcome of y if x = 0and the linear model is valid all the way to x = 0, which in many applications is not the case.

322 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.2.6 Extrapolation is treacherous

When those blizzards hit the East Coast this winter, it proved to my satisfaction that global warming

was a fraud. That snow was freezing cold. But in an alarming trend, temperatures this spring have risen.

Consider this: On February 6th it was 10 degrees. Today it hit almost 80. At this rate, by August it will be

220 degrees. So clearly folks the climate debate rages on.

Stephen ColbertApril 6th, 201012

Linear models can be used to approximate the relationship between two variables. However,these models have real limitations. Linear regression is simply a modeling framework. The truth isalmost always much more complex than our simple line. For example, we do not know how the dataoutside of our limited window will behave.

EXAMPLE 8.14

Use the model âid = 24,319−0.0431×family income to estimate the aid of another freshman studentwhose family had income of $1 million.

We want to calculate the aid for family income = 1,000,000:

24,319− 0.0431× family income = 24,319− 0.0431× 1,000,000 = −18,781

The model predicts this student will have -$18,781 in aid (!). However, Elmhurst College does notoffer negative aid where they select some students to pay extra on top of tuition to attend.

Applying a model estimate to values outside of the realm of the original data is called extrap-olation. Generally, a linear model is only an approximation of the real relationship between twovariables. If we extrapolate, we are making an unreliable bet that the approximate linear relationshipwill be valid in places where it has not been analyzed.

8.2.7 Using R2 to describe the strength of a fit

We evaluated the strength of the linear relationship between two variables earlier using thecorrelation, R. However, it is more common to explain the strength of a linear fit using R2, calledR-squared. If provided with a linear model, we might like to describe how closely the data clusteraround the linear fit.

Family Income

$0 $50k $100k $150k $200k $250k

$10k

$20k

$30k

Gift

Aid

Fro

m U

nive

rsity

Figure 8.15: Gift aid and family income for a random sample of 50 freshmanstudents from Elmhurst College, shown with the least squares regression line.

12www.cc.com/video-clips/l4nkoq

8.2. LEAST SQUARES REGRESSION 323

The R2 of a linear model describes the amount of variation in the response that is explainedby the least squares line. For example, consider the Elmhurst data, shown in Figure 8.15. Thevariance of the response variable, aid received, is about s2

aid ≈ 29.8 million. However, if we applyour least squares line, then this model reduces our uncertainty in predicting aid using a student’sfamily income. The variability in the residuals describes how much variation remains after using themodel: s2

RES≈ 22.4 million. In short, there was a reduction of

s2aid − s2

RES

s2aid

=29,800,000− 22,400,000

29,800,000=

7,500,000

29,800,000= 0.25

or about 25% in the data’s variation by using information about family income for predicting aidusing a linear model. This corresponds exactly to the R-squared value:

R = −0.499 R2 = 0.25

GUIDED PRACTICE 8.15

If a linear model has a very strong negative relationship with a correlation of -0.97, how much ofthe variation in the response is explained by the explanatory variable?13

8.2.8 Categorical predictors with two levels

Categorical variables are also useful in predicting outcomes. Here we consider a categoricalpredictor with two levels (recall that a level is the same as a category). We’ll consider Ebay auctionsfor a video game, Mario Kart for the Nintendo Wii, where both the total price of the auction and thecondition of the game were recorded. Here we want to predict total price based on game condition,which takes values used and new. A plot of the auction data is shown in Figure 8.16.

0(used)

1(new)

Tota

l Pric

price = 42.87 + 10.90 cond_new

Figure 8.16: Total auction prices for the video game Mario Kart, divided into used(x = 0) and new (x = 1) condition games. The least squares regression line is alsoshown.

To incorporate the game condition variable into a regression equation, we must convert thecategories into a numerical form. We will do so using an indicator variable called cond new,which takes value 1 when the game is new and 0 when the game is used. Using this indicatorvariable, the linear model may be written as

p̂rice = β0 + β1 × cond new

13About R2 = (−0.97)2 = 0.94 or 94% of the variation is explained by the linear model.

324 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

Estimate Std. Error t value Pr(>|t|)(Intercept) 42.87 0.81 52.67 <0.0001cond new 10.90 1.26 8.66 <0.0001

Figure 8.17: Least squares regression summary for the final auction price againstthe condition of the game.

The parameter estimates are given in Figure 8.17, and the model equation can be summarized as

p̂rice = 42.87 + 10.90× cond new

For categorical predictors with just two levels, the linearity assumption will always be satisfied.However, we must evaluate whether the residuals in each group are approximately normal and haveapproximately equal variance. As can be seen in Figure 8.16, both of these conditions are reasonablysatisfied by the auction data.

EXAMPLE 8.16

Interpret the two parameters estimated in the model for the price of Mario Kart in eBay auctions.

The intercept is the estimated price when cond new takes value 0, i.e. when the game is in usedcondition. That is, the average selling price of a used version of the game is $42.87.

The slope indicates that, on average, new games sell for about $10.90 more than used games.

INTERPRETING MODEL ESTIMATES FOR CATEGORICAL PREDICTORS

The estimated intercept is the value of the response variable for the first category (i.e. thecategory corresponding to an indicator value of 0). The estimated slope is the average changein the response variable between the two categories.

We’ll elaborate further on this topic in Chapter 9, where we examine the influence of manypredictor variables simultaneously using multiple regression.

8.2. LEAST SQUARES REGRESSION 325

Exercises

8.17 Units of regression. Consider a regression predicting weight (kg) from height (cm) for a sample ofadult males. What are the units of the correlation coefficient, the intercept, and the slope?

8.18 Which is higher? Determine if I or II is higher or if they are equal. Explain your reasoning. For aregression line, the uncertainty associated with the slope estimate, b1, is higher when

I. there is a lot of scatter around the regression line or

II. there is very little scatter around the regression line

8.19 Over-under, Part I. Suppose we fit a regression line to predict the shelf life of an apple based on itsweight. For a particular apple, we predict the shelf life to be 4.6 days. The apple’s residual is -0.6 days. Didwe over or under estimate the shelf-life of the apple? Explain your reasoning.

8.20 Over-under, Part II. Suppose we fit a regression line to predict the number of incidents of skin cancerper 1,000 people from the number of sunny days in a year. For a particular year, we predict the incidence ofskin cancer to be 1.5 per 1,000 people, and the residual for this year is 0.5. Did we over or under estimatethe incidence of skin cancer? Explain your reasoning.

8.21 Tourism spending. The Association of Turkish Travel Agencies reports the number of foreign touristsvisiting Turkey and tourist spending by year.14 Three plots are provided: scatterplot showing the relationshipbetween these two variables along with the least squares fit, residuals plot, and histogram of residuals.

0 5000 15000 25000

050

0010

000

1500

Number of tourists (thousands)

Spe

ndin

g (m

illio

n $)

Number of tourists (thousands)

Res

idua

0 5000 15000 25000

−10

000

1000

Residuals−1500 −750 0 750 1500

(a) Describe the relationship between number of tourists and spending.

(b) What are the explanatory and response variables?

(d) Do the data meet the conditions required for fitting a least squares line? In addition to the scatterplot,use the residual plot and histogram to answer this question.

14Association of Turkish Travel Agencies, Foreign Visitors Figure & Tourist Spendings By Years.

326 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.22 Nutrition at Starbucks, Part I. The scatterplot below shows the relationship between the number ofcalories and amount of carbohydrates (in grams) Starbucks food menu items contain.15 Since Starbucks onlylists the number of calories on the display items, we are interested in predicting the amount of carbs a menuitem has based on its calorie content.

Calories

Car

bs (

gram

100 200 300 400 500

Calories

Res

idua

ls100 200 300 400 500

−20

Residuals−40 −20 0 20 40

(a) Describe the relationship between number of calories and amount of carbohydrates (in grams) thatStarbucks food menu items contain.

(b) In this scenario, what are the explanatory and response variables?

(d) Do these data meet the conditions required for fitting a least squares line?

8.23 The Coast Starlight, Part II. Exercise 8.11 introduces data on the Coast Starlight Amtrak train thatruns from Seattle to Los Angeles. The mean travel time from one stop to the next on the Coast Starlightis 129 mins, with a standard deviation of 113 minutes. The mean distance traveled from one stop to thenext is 108 miles with a standard deviation of 99 miles. The correlation between travel time and distance is0.636.

(a) Write the equation of the regression line for predicting travel time.

(b) Interpret the slope and the intercept in this context.

(c) Calculate R2 of the regression line for predicting travel time from distance traveled for the CoastStarlight, and interpret R2 in the context of the application.

(d) The distance between Santa Barbara and Los Angeles is 103 miles. Use the model to estimate the timeit takes for the Starlight to travel between these two cities.

(e) It actually takes the Coast Starlight about 168 mins to travel from Santa Barbara to Los Angeles.Calculate the residual and explain the meaning of this residual value.

(f) Suppose Amtrak is considering adding a stop to the Coast Starlight 500 miles away from Los Angeles.Would it be appropriate to use this linear model to predict the travel time from Los Angeles to thispoint?

8.24 Body measurements, Part III. Exercise 8.13 introduces data on shoulder girth and height of a groupof individuals. The mean shoulder girth is 107.20 cm with a standard deviation of 10.37 cm. The meanheight is 171.14 cm with a standard deviation of 9.41 cm. The correlation between height and shoulder girthis 0.67.

(a) Write the equation of the regression line for predicting height.

(b) Interpret the slope and the intercept in this context.

(c) Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in thecontext of the application.

(d) A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of thisstudent using the model.

(e) The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.

(f) A one year old has a shoulder girth of 56 cm. Would it be appropriate to use this linear model to predictthe height of this child?

15Source: Starbucks.com, collected on March 10, 2011,www.starbucks.com/menu/nutrition.

8.2. LEAST SQUARES REGRESSION 327

8.25 Murders and poverty, Part I. The following regression output is for predicting annual murders permillion from percentage living in poverty in a random sample of 20 metropolitan areas.

Estimate Std. Error t value Pr(>|t|)(Intercept) -29.901 7.789 -3.839 0.001

poverty% 2.559 0.390 6.562 0.000

s = 5.512 R2 = 70.52% R2adj = 68.89%

(a) Write out the linear model.

(b) Interpret the intercept.

(d) Interpret R2.

(e) Calculate the correlation coefficient.

●

●●

●

Percent in Poverty

Ann

ual M

urde

rs p

er M

illio

14% 18% 22% 26%

8.26 Cats, Part I. The following regression output is for predicting the heart weight (in g) of cats fromtheir body weight (in kg). The coefficients are estimated using a dataset of 144 domestic cats.

Estimate Std. Error t value Pr(>|t|)(Intercept) -0.357 0.692 -0.515 0.607

body wt 4.034 0.250 16.119 0.000

s = 1.452 R2 = 64.66% R2adj = 64.41%

(a) Write out the linear model.

(b) Interpret the intercept.

(d) Interpret R2.

(e) Calculate the correlation coefficient.

Body weight (kg)

Hea

rt w

eigh

t (g)

2.0 2.5 3.0 3.5 4.05

328 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.3 Types of outliers in linear regression

In this section, we identify criteria for determining which outliers are important and influential.Outliers in regression are observations that fall far from the cloud of points. These points areespecially important because they can have a strong influence on the least squares line.

EXAMPLE 8.17

There are six plots shown in Figure 8.18 along with the least squares line and residual plots. For eachscatterplot and residual plot pair, identify the outliers and note how they influence the least squaresline. Recall that an outlier is any point that doesn’t appear to belong with the vast majority of theother points.

(1) There is one outlier far from the other points, though it only appears to slightly influencethe line.

(2) There is one outlier on the right, though it is quite close to the least squares line, whichsuggests it wasn’t very influential.

(3) There is one point far away from the cloud, and this outlier appears to pull the least squaresline up on the right; examine how the line around the primary cloud doesn’t appear to fitvery well.

(4) There is a primary cloud and then a small secondary cloud of four outliers. The secondarycloud appears to be influencing the line somewhat strongly, making the least square line fitpoorly almost everywhere. There might be an interesting explanation for the dual clouds,which is something that could be investigated.

(5) There is no obvious trend in the main cloud of points and the outlier on the right appears tolargely control the slope of the least squares line.

(6) There is one outlier far from the cloud. However, it falls quite close to the least squares lineand does not appear to be very influential.

Examine the residual plots in Figure 8.18. You will probably find that there is some trend inthe main clouds of (3) and (4). In these cases, the outliers influenced the slope of the least squareslines. In (5), data with no clear trend were assigned a line with a large trend simply due to oneoutlier (!).

LEVERAGE

Points that fall horizontally away from the center of the cloud tend to pull harder on the line,so we call them points with high leverage.

Points that fall horizontally far from the line are points of high leverage; these points canstrongly influence the slope of the least squares line. If one of these high leverage points does appearto actually invoke its influence on the slope of the line – as in cases (3), (4), and (5) of Example 8.17– then we call it an influential point. Usually we can say a point is influential if, had we fitted theline without it, the influential point would have been unusually far from the least squares line.

It is tempting to remove outliers. Don’t do this without a very good reason. Models that ignoreexceptional (and interesting) cases often perform poorly. For instance, if a financial firm ignored thelargest market swings – the “outliers” – they would soon go bankrupt by making poorly thought-outinvestments.

8.3. TYPES OF OUTLIERS IN LINEAR REGRESSION 329

(1) (2) (3)

(4) (5) (6)

Figure 8.18: Six plots, each with a least squares line and residual plot. All datasets have at least one outlier.

330 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

Exercises

8.27 Outliers, Part I. Identify the outliers in the scatterplots shown below, and determine what type ofoutliers they are. Explain your reasoning.

(a) (b) (c)

8.28 Outliers, Part II. Identify the outliers in the scatterplots shown below and determine what type ofoutliers they are. Explain your reasoning.

(a) (b) (c)

8.29 Urban homeowners, Part I. The scatterplot below shows the percent of families who own their homevs. the percent of the population living in urban areas.16 There are 52 observations, each corresponding toa state in the US. Puerto Rico and District of Columbia are also included.

(a) Describe the relationship between the percent offamilies who own their home and the percent ofthe population living in urban areas.

(b) The outlier at the bottom right corner is Districtof Columbia, where 100% of the population is con-sidered urban. What type of an outlier is this ob-servation?

Percent Urban Population40% 60% 80% 100%

45%

55%

65%

75%

Per

cent

n T

heir

Hom

8.30 Crawling babies, Part II. Exercise 8.12 introduces data on the average monthly temperature duringthe month babies first try to crawl (about 6 months after birth) and the average first crawling age for babiesborn in a given month. A scatterplot of these two variables reveals a potential outlying month when theaverage temperature is about 53◦F and average crawling age is about 28.5 weeks. Does this point have highleverage? Is it an influential point?

16United States Census Bureau, 2010 Census Urban and Rural Classification and Urban Area Criteria and HousingCharacteristics: 2010.

8.4. INFERENCE FOR LINEAR REGRESSION 331

8.4 Inference for linear regression

In this section, we discuss uncertainty in the estimates of the slope and y-intercept for aregression line. Just as we identified standard errors for point estimates in previous chapters, wefirst discuss standard errors for these new estimates.

8.4.1 Midterm elections and unemployment

Elections for members of the United States House of Representatives occur every two years,coinciding every four years with the U.S. Presidential election. The set of House elections occurringduring the middle of a Presidential term are called midterm elections. In America’s two-partysystem, one political theory suggests the higher the unemployment rate, the worse the President’sparty will do in the midterm elections.

To assess the validity of this claim, we can compile historical data and look for a connection.We consider every midterm election from 1898 to 2018, with the exception of those elections duringthe Great Depression. Figure 8.19 shows these data and the least-squares regression line:

% change in House seats for President’s party

= −7.36− 0.89× (unemployment rate)

We consider the percent change in the number of seats of the President’s party (e.g. percent changein the number of seats for Republicans in 2018) against the unemployment rate.

Examining the data, there are no clear deviations from linearity, the constant variance con-dition, or substantial outliers. While the data are collected sequentially, a separate analysis wasused to check for any apparent correlation between successive observations; no such correlation wasfound.

Per

cent

Cha

nge

in S

eats

Pre

side

nt's

Par

ty in

Hou

se o

f Rep

Percent Unemployment

●

4% 8% 12%

−30%

−20%

−10%

10%

McKinley1898

Reagan1982

Clinton1994

Bush2002

Obama2010

Trump2018

● DemocratRepublican

Figure 8.19: The percent change in House seats for the President’s party in eachelection from 1898 to 2010 plotted against the unemployment rate. The two pointsfor the Great Depression have been removed, and a least squares regression linehas been fit to the data.

332 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

GUIDED PRACTICE 8.18

The data for the Great Depression (1934 and 1938) were removed because the unemployment ratewas 21% and 18%, respectively. Do you agree that they should be removed for this investigation?Why or why not?17

There is a negative slope in the line shown in Figure 8.19. However, this slope (and the y-intercept) are only estimates of the parameter values. We might wonder, is this convincing evidencethat the “true” linear model has a negative slope? That is, do the data provide strong evidence thatthe political theory is accurate, where the unemployment rate is a useful predictor of the midtermelection? We can frame this investigation into a statistical hypothesis test:

H0: β1 = 0. The true linear model has slope zero.

HA: β1 6= 0. The true linear model has a slope different than zero. The unemployment is predictiveof whether the President’s party wins or loses seats in the House of Representatives.

We would reject H0 in favor of HA if the data provide strong evidence that the true slope parameteris different than zero. To assess the hypotheses, we identify a standard error for the estimate,compute an appropriate test statistic, and identify the p-value.

8.4.2 Understanding regression output from software

Just like other point estimates we have seen before, we can compute a standard error and teststatistic for b1. We will generally label the test statistic using a T , since it follows the t-distribution.

We will rely on statistical software to compute the standard error and leave the explanationof how this standard error is determined to a second or third statistics course. Figure 8.20 showssoftware output for the least squares regression line in Figure 8.19. The row labeled unemp includesthe point estimate and other hypothesis test information for the slope, which is the coefficient of theunemployment variable.

Estimate Std. Error t value Pr(>|t|)(Intercept) -7.3644 5.1553 -1.43 0.1646

unemp -0.8897 0.8350 -1.07 0.2961df = 27

Figure 8.20: Output from statistical software for the regression line modeling themidterm election losses for the President’s party as a response to unemployment.

EXAMPLE 8.19

What do the first and second columns of Figure 8.20 represent?

The entries in the first column represent the least squares estimates, b0 and b1, and the values in thesecond column correspond to the standard errors of each estimate. Using the estimates, we couldwrite the equation for the least square regression line as

ŷ = −7.3644− 0.8897x

where ŷ in this case represents the predicted change in the number of seats for the president’s party,and x represents the unemployment rate.

17We will provide two considerations. Each of these points would have very high leverage on any least-squaresregression line, and years with such high unemployment may not help us understand what would happen in otheryears where the unemployment is only modestly high. On the other hand, these are exceptional cases, and we wouldbe discarding important information if we exclude them from a final analysis.

8.4. INFERENCE FOR LINEAR REGRESSION 333

We previously used a t-test statistic for hypothesis testing in the context of numerical data.Regression is very similar. In the hypotheses we consider, the null value for the slope is 0, so we cancompute the test statistic using the T (or Z) score formula:

T =estimate− null value

SE=−0.8897− 0

0.8350= −1.07

This corresponds to the third column of Figure 8.20.

EXAMPLE 8.20

Use the table in Figure 8.20 to determine the p-value for the hypothesis test.

The last column of the table gives the p-value for the two-sided hypothesis test for the coefficient ofthe unemployment rate: 0.2961. That is, the data do not provide convincing evidence that a higherunemployment rate has any correspondence with smaller or larger losses for the President’s party inthe House of Representatives in midterm elections.

INFERENCE FOR REGRESSION

We usually rely on statistical software to identify point estimates, standard errors, test statistics,and p-values in practice. However, be aware that software will not generally check whether themethod is appropriate, meaning we must still verify conditions are met.

EXAMPLE 8.21

Examine Figure 8.15 on page 322, which relates the Elmhurst College aid and student family income.How sure are you that the slope is statistically significantly different from zero? That is, do youthink a formal hypothesis test would reject the claim that the true slope of the line should be zero?

While the relationship between the variables is not perfect, there is an evident decreasing trend inthe data. This suggests the hypothesis test will reject the null claim that the slope is zero.

GUIDED PRACTICE 8.22

Figure 8.21 shows statistical software output from fitting the least squares regression line shown inFigure 8.15. Use this output to formally evaluate the following hypotheses.18

H0: The true coefficient for family income is zero.

HA: The true coefficient for family income is not zero.

Estimate Std. Error t value Pr(>|t|)(Intercept) 24319.3 1291.5 18.83 <0.0001

family income -0.0431 0.0108 -3.98 0.0002df = 48

Figure 8.21: Summary of least squares fit for the Elmhurst College data, where weare predicting the gift aid by the university based on the family income of students.

18We look in the second row corresponding to the family income variable. We see the point estimate of the slopeof the line is -0.0431, the standard error of this estimate is 0.0108, and the t-test statistic is T = −3.98. The p-valuecorresponds exactly to the two-sided test we are interested in: 0.0002. The p-value is so small that we reject the nullhypothesis and conclude that family income and financial aid at Elmhurst College for freshman entering in the year2011 are negatively correlated and the true slope parameter is indeed less than 0, just as we believed in Example 8.21.

334 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.4.3 Confidence interval for a coefficient

Similar to how we can conduct a hypothesis test for a model coefficient using regression output,we can also construct a confidence interval for that coefficient.

EXAMPLE 8.23

Compute the 95% confidence interval for the family income coefficient using the regression outputfrom Table 8.21.

The point estimate is -0.0431 and the standard error is SE = 0.0108. When constructing a confidenceinterval for a model coefficient, we generally use a t-distribution. The degrees of freedom for thedistribution are noted in the regression output, df = 48, allowing us to identify t?48 = 2.01 for use inthe confidence interval.

We can now construct the confidence interval in the usual way:

point estimate± t?48 × SE → −0.0431± 2.01× 0.0108 → (−0.0648,−0.0214)

We are 95% confident that with each dollar increase in family income, the university’s gift aid ispredicted to decrease on average by $0.0214 to $0.0648.

CONFIDENCE INTERVALS FOR COEFFICIENTS

Confidence intervals for model coefficients can be computed using the t-distribution:

bi ± t?df × SEbi

where t?df is the appropriate t-value corresponding to the confidence level with the model’sdegrees of freedom.

On the topic of intervals in this book, we’ve focused exclusively on confidence intervals formodel parameters. However, there are other types of intervals that may be of interest, includingprediction intervals for a response value and also confidence intervals for a mean response value inthe context of regression. These two interval types are introduced in an online extra that you maydownload at

www.openintro.org/d?file=stat extra linear regression supp

8.4. INFERENCE FOR LINEAR REGRESSION 335

Exercises

In the following exercises, visually check the conditions for fitting a least squares regression line.However, you do not need to report these conditions in your solutions.

8.31 Body measurements, Part IV. The scatterplot and least squares summary below show the relation-ship between weight measured in kilograms and height measured in centimeters of 507 physically activeindividuals.

Height (cm)

Wei

ght (

kg)

150 175 200

110

Estimate Std. Error t value Pr(>|t|)(Intercept) -105.0113 7.5394 -13.93 0.0000

height 1.0176 0.0440 23.13 0.0000

(a) Describe the relationship between height and weight.

(b) Write the equation of the regression line. Interpret the slope and intercept in context.

(c) Do the data provide strong evidence that an increase in height is associated with an increase in weight?State the null and alternative hypotheses, report the p-value, and state your conclusion.

(d) The correlation coefficient for height and weight is 0.72. Calculate R2 and interpret it in context.

8.32 Beer and blood alcohol content. Many people believe that gender, weight, drinking habits, and manyother factors are much more important in predicting blood alcohol content (BAC) than simply consideringthe number of drinks a person consumed. Here we examine data from sixteen student volunteers at OhioState University who each drank a randomly assigned number of cans of beer. These students were evenlydivided between men and women, and they differed in weight and drinking habits. Thirty minutes later, apolice officer measured their blood alcohol content (BAC) in grams of alcohol per deciliter of blood.19 Thescatterplot and regression table summarize the findings.

●

●●

●

2 4 6 8

0.05

0.10

0.15

Cans of beer

C (

gram

s / d

ecili

ter)

Estimate Std. Error t value Pr(>|t|)(Intercept) -0.0127 0.0126 -1.00 0.3320

beers 0.0180 0.0024 7.48 0.0000

(a) Describe the relationship between the number of cans of beer and BAC.

(b) Write the equation of the regression line. Interpret the slope and intercept in context.

(c) Do the data provide strong evidence that drinking more cans of beer is associated with an increase inblood alcohol? State the null and alternative hypotheses, report the p-value, and state your conclusion.

(d) The correlation coefficient for number of cans of beer and BAC is 0.89. Calculate R2 and interpret it incontext.

(e) Suppose we visit a bar, ask people how many drinks they have had, and also take their BAC. Do youthink the relationship between number of drinks and BAC would be as strong as the relationship foundin the Ohio State study?

19J. Malkevitch and L.M. Lesser. For All Practical Purposes: Mathematical Literacy in Today’s World. WHFreeman & Co, 2008.

336 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.33 Husbands and wives, Part II. The scatterplot below summarizes husbands’ and wives’ heights in arandom sample of 170 married couples in Britain, where both partners’ ages are below 65 years. Summaryoutput of the least squares fit for predicting wife’s height from husband’s height is also provided in the table.

Husband's height (in inches)

Wife

's h

eigh

t (in

inch

es)

60 65 70 7555

Estimate Std. Error t value Pr(>|t|)(Intercept) 43.5755 4.6842 9.30 0.0000

height husband 0.2863 0.0686 4.17 0.0000

(a) Is there strong evidence that taller men marry taller women? State the hypotheses and include anyinformation used to conduct the test.

(b) Write the equation of the regression line for predicting wife’s height from husband’s height.

(d) Given that R2 = 0.09, what is the correlation of heights in this data set?

(e) You meet a married man from Britain who is 5’9” (69 inches). What would you predict his wife’s heightto be? How reliable is this prediction?

(f) You meet another married man from Britain who is 6’7” (79 inches). Would it be wise to use the samelinear model to predict his wife’s height? Why or why not?

8.34 Urban homeowners, Part II. Exercise 8.29 gives a scatterplot displaying the relationship between thepercent of families that own their home and the percent of the population living in urban areas. Below is asimilar scatterplot, excluding District of Columbia, as well as the residuals plot. There were 51 cases.

(a) For these data, R2 = 0.28. What is thecorrelation? How can you tell if it ispositive or negative?

(b) Examine the residual plot. What do youobserve? Is a simple least squares fitappropriate for these data?

% Urban population

% W

ho o

hom

40 60 80

5560

6570

75−

100

8.4. INFERENCE FOR LINEAR REGRESSION 337

8.35 Murders and poverty, Part II. Exercise 8.25 presents regression output from a model for predictingannual murders per million from percentage living in poverty based on a random sample of 20 metropolitanareas. The model output is also provided below.

Estimate Std. Error t value Pr(>|t|)(Intercept) -29.901 7.789 -3.839 0.001

poverty% 2.559 0.390 6.562 0.000

s = 5.512 R2 = 70.52% R2adj = 68.89%

(a) What are the hypotheses for evaluating whether poverty percentage is a significant predictor of murderrate?

(b) State the conclusion of the hypothesis test from part (a) in context of the data.

(d) Do your results from the hypothesis test and the confidence interval agree? Explain.

8.36 Babies. Is the gestational age (time between conception and birth) of a low birth-weight baby usefulin predicting head circumference at birth? Twenty-five low birth-weight babies were studied at a Harvardteaching hospital; the investigators calculated the regression of head circumference (measured in centimeters)against gestational age (measured in weeks). The estimated regression line is

̂head circumference = 3.91 + 0.78× gestational age

(a) What is the predicted head circumference for a baby whose gestational age is 28 weeks?

(b) The standard error for the coefficient of gestational age is 0. 35, which is associated with df = 23. Doesthe model provide strong evidence that gestational age is significantly associated with head circumfer-ence?

338 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

Chapter exercises

8.37 True / False. Determine if the following statements are true or false. If false, explain why.

(a) A correlation coefficient of -0.90 indicates a stronger linear relationship than a correlation of 0.5.

(b) Correlation is a measure of the association between any two variables.

8.38 Trees. The scatterplots below show the relationship between height, diameter, and volume of timberin 31 felled black cherry trees. The diameter of the tree is measured 4.5 feet above the ground.20

●●●

●● ●

●●

●●●●

●●

●

●●●

●●

●

● ●●

●●

●

Height (feet)

Vol

ume

(cub

ic fe

et)

60 70 80 90

●●●

●●

●●●●

●●

●

●●●

●●

●

●●●

●●

●

Diameter (inches)

Vol

ume

(cub

ic fe

et)

8 12 16 20

(a) Describe the relationship between volume and height of these trees.

(b) Describe the relationship between volume and diameter of these trees.

(c) Suppose you have height and diameter measurements for another black cherry tree. Which of thesevariables would be preferable to use to predict the volume of timber in this tree using a simple linearregression model? Explain your reasoning.

8.39 Husbands and wives, Part III. Exercise 8.33 presents a scatterplot displaying the relationship betweenhusbands’ and wives’ ages in a random sample of 170 married couples in Britain, where both partners’ agesare below 65 years. Given below is summary output of the least squares fit for predicting wife’s age fromhusband’s age.

Husband's age (in years)

Wife

's a

ge (

in y

ears

)

20 40 60

Estimate Std. Error t value Pr(>|t|)(Intercept) 1.5740 1.1501 1.37 0.1730

age husband 0.9112 0.0259 35.25 0.0000df = 168

(a) We might wonder, is the age difference between husbands and wives consistent across ages? If this werethe case, then the slope parameter would be β1 = 1. Use the information above to evaluate if there isstrong evidence that the difference in husband and wife ages differs for different ages.

(b) Write the equation of the regression line for predicting wife’s age from husband’s age.

(d) Given that R2 = 0.88, what is the correlation of ages in this data set?

(e) You meet a married man from Britain who is 55 years old. What would you predict his wife’s age tobe? How reliable is this prediction?

(f) You meet another married man from Britain who is 85 years old. Would it be wise to use the samelinear model to predict his wife’s age? Explain.

20Source: R Dataset, stat.ethz.ch/R-manual/R-patched/library/datasets/html/trees.html.

8.4. INFERENCE FOR LINEAR REGRESSION 339

8.40 Cats, Part II. Exercise 8.26 presents regression output from a model for predicting the heart weight(in g) of cats from their body weight (in kg). The coefficients are estimated using a dataset of 144 domesticcat. The model output is also provided below.

Estimate Std. Error t value Pr(>|t|)(Intercept) -0.357 0.692 -0.515 0.607

body wt 4.034 0.250 16.119 0.000

s = 1.452 R2 = 64.66% R2adj = 64.41%

(a) We see that the point estimate for the slope is positive. What are the hypotheses for evaluating whetherbody weight is positively associated with heart weight in cats?

(b) State the conclusion of the hypothesis test from part (a) in context of the data.

(d) Do your results from the hypothesis test and the confidence interval agree? Explain.

8.41 Nutrition at Starbucks, Part II. Exercise 8.22 introduced a data set on nutrition information onStarbucks food menu items. Based on the scatterplot and the residual plot provided, describe the relationshipbetween the protein content and calories of these menu items, and determine if a simple linear model isappropriate to predict amount of protein from the number of calories.

Calories

Pro

tein

(gr

ams)

100 200 300 400 500

−20

8.42 Helmets and lunches. The scatterplot shows the relationship between socioeconomic status measuredas the percentage of children in a neighborhood receiving reduced-fee lunches at school (lunch) and thepercentage of bike riders in the neighborhood wearing helmets (helmet). The average percentage of childrenreceiving reduced-fee lunches is 30.8% with a standard deviation of 26.7% and the average percentage ofbike riders wearing helmets is 38.8% with a standard deviation of 16.9%.

(a) If the R2 for the least-squares regression line forthese data is 72%, what is the correlationbetween lunch and helmet?

(b) Calculate the slope and intercept for theleast-squares regression line for these data.

(d) Interpret the slope of the least-squares regressionline in the context of the application.

(e) What would the value of the residual be for aneighborhood where 40% of the children receivereduced-fee lunches and 40% of the bike riderswear helmets? Interpret the meaning of thisresidual in the context of the application.

●

●●

●

● ●

●

Rate of Receiving a Reduced−Fee Lunch

0% 20% 40% 60% 80%0%

20%

40%

60%

Rat

e of

Wea

ring

a H

elm

8.43 Match the correlation, Part III. Match each correlation to the corresponding scatterplot.

(a) r = −0.72

(b) r = 0.07

(d) r = 0.99

(1) (2) (3) (4)

340 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION

8.44 Rate my professor. Many college courses conclude by giving students the opportunity to evaluatethe course and the instructor anonymously. However, the use of these student evaluations as an indicator ofcourse quality and teaching effectiveness is often criticized because these measures may reflect the influenceof non-teaching related characteristics, such as the physical appearance of the instructor. Researchers atUniversity of Texas, Austin collected data on teaching evaluation score (higher score means better) andstandardized beauty score (a score of 0 means average, negative score means below average, and a positivescore means above average) for a sample of 463 professors.21 The scatterplot below shows the relationshipbetween these variables, and regression output is provided for predicting teaching evaluation score frombeauty score.

Estimate Std. Error t value Pr(>|t|)(Intercept) 4.010 0.0255 157.21 0.0000

beauty Cell 1 0.0322 4.13 0.0000

(a) Given that the average standardized beautyscore is -0.0883 and average teachingevaluation score is 3.9983, calculate theslope. Alternatively, the slope may becomputed using just the informationprovided in the model summary table.

(b) Do these data provide convincing evidencethat the slope of the relationship betweenteaching evaluation and beauty is positive?Explain your reasoning.

(c) List the conditions required for linearregression and check if each one is satisfiedfor this model based on the followingdiagnostic plots.

Beauty

Teac

hing

eva

luat

ion

−1 0 1 22

Beauty

Res

idua

−1 0 1 2

−1

Residuals

−2 −1 0 1 2

050

100

150

Order of data collection

Res

idua

0 100 200 300 400

−1

21Daniel S Hamermesh and Amy Parker. “Beauty in the classroom: Instructors’ pulchritude and putative peda-gogical productivity”. In: Economics of Education Review 24.4 (2005), pp. 369–376.

341

Chapter 9Multiple and logistic

regression

9.1 Introduction to multiple regression

9.2 Model selection

9.3 Checking model conditions using graphs

9.4 Multiple regression case study: Mario Kart

9.5 Introduction to logistic regression

342

The principles of simple linear regression lay the foundation for more

sophisticated regression models used in a wide range of challenging

settings. In Chapter 9, we explore multiple regression, which introduces

the possibility of more than one predictor in a linear model, and logistic

regression, a technique for predicting categorical outcomes with two

levels.

For videos, slides, and other resources, please visit

www.openintro.org/os

9.1. INTRODUCTION TO MULTIPLE REGRESSION 343

9.1 Introduction to multiple regression

Multiple regression extends simple two-variable regression to the case that still has one responsebut many predictors (denoted x1, x2, x3, …). The method is motivated by scenarios where manyvariables may be simultaneously connected to an output.

We will consider data about loans from the peer-to-peer lender, Lending Club, which is a dataset we first encountered in Chapters 1 and 2. The loan data includes terms of the loan as well asinformation about the borrower. The outcome variable we would like to better understand is theinterest rate assigned to the loan. For instance, all other characteristics held constant, does it matterhow much debt someone already has? Does it matter if their income has been verified? Multipleregression will help us answer these and other questions.

The data set loans includes results from 10,000 loans, and we’ll be looking at a subset of theavailable variables, some of which will be new from those we saw in earlier chapters. The first sixobservations in the data set are shown in Figure 9.1, and descriptions for each variable are shown inFigure 9.2. Notice that the past bankruptcy variable (bankruptcy) is an indicator variable, whereit takes the value 1 if the borrower had a past bankruptcy in their record and 0 if not. Usingan indicator variable in place of a category name allows for these variables to be directly used inregression. Two of the other variables are categorical (income ver and issued), each of which cantake one of a few different non-numerical values; we’ll discuss how these are handled in the modelin Section 9.1.1.

interest rate income ver debt to income credit util bankruptcy term issued credit checks1 14.07 verified 18.01 0.55 0 60 Mar2018 62 12.61 not 5.04 0.15 1 36 Feb2018 13 17.09 source only 21.15 0.66 0 36 Feb2018 44 6.72 not 10.16 0.20 0 36 Jan2018 05 14.07 verified 57.96 0.75 0 36 Mar2018 76 6.72 not 6.46 0.09 0 36 Jan2018 6…

……

Figure 9.1: First six rows from the loans data set.

variable description

interest rate Interest rate for the loan.income ver Categorical variable describing whether the borrower’s income source and amount

have been verified, with levels verified, source only, and not.debt to income Debt-to-income ratio, which is the percentage of total debt of the borrower divided

by their total income.credit util Of all the credit available to the borrower, what fraction are they utilizing. For

example, the credit utilization on a credit card would be the card’s balance dividedby the card’s credit limit.

bankruptcy An indicator variable for whether the borrower has a past bankruptcy in herrecord. This variable takes a value of 1 if the answer is “yes” and 0 if the answeris “no”.

term The length of the loan, in months.issued The month and year the loan was issued, which for these loans is always during

the first quarter of 2018.credit checks Number of credit checks in the last 12 months. For example, when filing an appli-

cation for a credit card, it is common for the company receiving the applicationto run a credit check.

Figure 9.2: Variables and their descriptions for the loans data set.

344 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.1.1 Indicator and categorical variables as predictors

Let’s start by fitting a linear regression model for interest rate with a single predictor indicatingwhether or not a person has a bankruptcy in their record:

r̂ate = 12.33 + 0.74× bankruptcy

Results of this model are shown in Figure 9.3.

Estimate Std. Error t value Pr(>|t|)(Intercept) 12.3380 0.0533 231.49 <0.0001bankruptcy 0.7368 0.1529 4.82 <0.0001

df = 9998

Figure 9.3: Summary of a linear model for predicting interest rate based on whetherthe borrower has a bankruptcy in their record.

EXAMPLE 9.1

Interpret the coefficient for the past bankruptcy variable in the model. Is this coefficient significantlydifferent from 0?

The bankruptcy variable takes one of two values: 1 when the borrower has a bankruptcy in theirhistory and 0 otherwise. A slope of 0.74 means that the model predicts a 0.74% higher interestrate for those borrowers with a bankruptcy in their record. (See Section 8.2.8 for a review of theinterpretation for two-level categorical predictor variables.) Examining the regression output inFigure 9.3, we can see that the p-value for bankruptcy is very close to zero, indicating there isstrong evidence the coefficient is different from zero when using this simple one-predictor model.

Suppose we had fit a model using a 3-level categorical variable, such as income ver. The outputfrom software is shown in Figure 9.4. This regression output provides multiple rows for the income

ver variable. Each row represents the relative difference for each level of income ver. However, weare missing one of the levels: not (for not verified). The missing level is called the reference level,and it represents the default level that other levels are measured against.

Estimate Std. Error t value Pr(>|t|)(Intercept) 11.0995 0.0809 137.18 <0.0001income ver: source only 1.4160 0.1107 12.79 <0.0001income ver: verified 3.2543 0.1297 25.09 <0.0001

df = 9998

Figure 9.4: Summary of a linear model for predicting interest rate based on whetherthe borrower’s income source and amount has been verified. This predictor hasthree levels, which results in 2 rows in the regression output.

EXAMPLE 9.2

How would we write an equation for this regression model?

The equation for the regression model may be written as a model with two predictors:

r̂ate = 11.10 + 1.42× income versource only + 3.25× income ververified

We use the notation variablelevel to represent indicator variables for when the categorical variabletakes a particular value. For example, income versource only would take a value of 1 if income ver

was source only for a loan, and it would take a value of 0 otherwise. Likewise, income ververifiedwould take a value of 1 if income ver took a value of verified and 0 if it took any other value.

9.1. INTRODUCTION TO MULTIPLE REGRESSION 345

The notation used in Example 9.2 may feel a bit confusing. Let’s figure out how to use theequation for each level of the income ver variable.

EXAMPLE 9.3

Using the model from Example 9.2, compute the average interest rate for borrowers whose incomesource and amount are both unverified.

When income ver takes a value of not, then both indicator functions in the equation from Exam-ple 9.2 are set to zero:

r̂ate = 11.10 + 1.42× 0 + 3.25× 0

= 11.10

The average interest rate for these borrowers is 11.1%. Because the not level does not have its owncoefficient and it is the reference value, the indicators for the other levels for this variable all dropout.

EXAMPLE 9.4

Using the model from Example 9.2, compute the average interest rate for borrowers whose incomesource is verified but the amount is not.

When income ver takes a value of source only, then the corresponding variable takes a value of 1while the other (income ververified) is 0:

r̂ate = 11.10 + 1.42× 1 + 3.25× 0

= 12.52

The average interest rate for these borrowers is 12.52%.

GUIDED PRACTICE 9.5

Compute the average interest rate for borrowers whose income source and amount are both verified.1

PREDICTORS WITH SEVERAL CATEGORIES

When fitting a regression model with a categorical variable that has k levels where k > 2,software will provide a coefficient for k − 1 of those levels. For the last level that does notreceive a coefficient, this is the reference level, and the coefficients listed for the other levelsare all considered relative to this reference level.

1When income ver takes a value of verified, then the corresponding variable takes a value of 1 while the other(income versource only) is 0:

r̂ate = 11.10 + 1.42× 0 + 3.25× 1

= 14.35

The average interest rate for these borrowers is 14.35%.

346 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

GUIDED PRACTICE 9.6

Interpret the coefficients in the income ver model.2

The higher interest rate for borrowers who have verified their income source or amount issurprising. Intuitively, we’d think that a loan would look less risky if the borrower’s income hasbeen verified. However, note that the situation may be more complex, and there may be confoundingvariables that we didn’t account for. For example, perhaps lender require borrowers with poor creditto verify their income. That is, verifying income in our data set might be a signal of some concernsabout the borrower rather than a reassurance that the borrower will pay back the loan. For thisreason, the borrower could be deemed higher risk, resulting in a higher interest rate. (What otherconfounding variables might explain this counter-intuitive relationship suggested by the model?)

GUIDED PRACTICE 9.7

How much larger of an interest rate would we expect for a borrower who has verified their incomesource and amount vs a borrower whose income source has only been verified?3

9.1.2 Including and assessing many variables in a model

The world is complex, and it can be helpful to consider many factors at once in statisticalmodeling. For example, we might like to use the full context of borrower to predict the interest ratethey receive rather than using a single variable. This is the strategy used in multiple regression.While we remain cautious about making any causal interpretations using multiple regression onobservational data, such models are a common first step in gaining insights or providing someevidence of a causal connection.

We want to construct a model that accounts for not only for any past bankruptcy or whetherthe borrower had their income source or amount verified, but simultaneously accounts for all thevariables in the data set: income ver, debt to income, credit util, bankruptcy, term, issued,and credit checks.

r̂ate = β0 + β1 × income versource only + β2 × income ververified + β3 × debt to income

+ β4 × credit util + β5 × bankruptcy + β6 × term

+ β7 × issuedJan2018 + β8 × issuedMar2018 + β9 × credit checks

This equation represents a holistic approach for modeling all of the variables simultaneously. Noticethat there are two coefficients for income ver and also two coefficients for issued, since both are3-level categorical variables.

We estimate the parameters β0, β1, β2, …, β9 in the same way as we did in the case of a singlepredictor. We select b0, b1, b2, …, b9 that minimize the sum of the squared residuals:

SSE = e21 + e2

2 + · · ·+ e210000 =

10000∑i=1

e2i =

10000∑i=1

(yi − ŷi)2(9.8)

where yi and ŷi represent the observed interest rates and their estimated values according to themodel, respectively. 10,000 residuals are calculated, one for each observation. We typically use acomputer to minimize the sum of squares and compute point estimates, as shown in the sampleoutput in Figure 9.5. Using this output, we identify the point estimates bi of each βi, just as we didin the one-predictor case.

2Each of the coefficients gives the incremental interest rate for the corresponding level relative to the not level,which is the reference level. For example, for a borrower whose income source and amount have been verified, themodel predicts that they will have a 3.25% higher interest rate than a borrower who has not had their income sourceor amount verified.

3Relative to the not category, the verified category has an interest rate of 3.25% higher, while the source only

category is only 1.42% higher. Thus, verified borrowers will tend to get an interest rate about 3.25%−1.42% = 1.83%higher than source only borrowers.

9.1. INTRODUCTION TO MULTIPLE REGRESSION 347

Estimate Std. Error t value Pr(>|t|)(Intercept) 1.9251 0.2102 9.16 <0.0001

income ver: source only 0.9750 0.0991 9.83 <0.0001income ver: verified 2.5374 0.1172 21.65 <0.0001

debt to income 0.0211 0.0029 7.18 <0.0001credit util 4.8959 0.1619 30.24 <0.0001

bankruptcy 0.3864 0.1324 2.92 0.0035term 0.1537 0.0039 38.96 <0.0001

issued: Jan2018 0.0276 0.1081 0.26 0.7981issued: Mar2018 -0.0397 0.1065 -0.37 0.7093

credit checks 0.2282 0.0182 12.51 <0.0001df = 9990

Figure 9.5: Output for the regression model, where interest rate is the outcomeand the variables listed are the predictors.

MULTIPLE REGRESSION MODEL

A multiple regression model is a linear model with many predictors. In general, we write themodel as

ŷ = β0 + β1×1 + β2×2 + · · ·+ βkxk

when there are k predictors. We always estimate the βi parameters using statistical software.

EXAMPLE 9.9

Write out the regression model using the point estimates from Figure 9.5. How many predictors arethere in this model?

The fitted model for the interest rate is given by:

r̂ate = 1.925 + 0.975× income versource only + 2.537× income ververified + 0.021× debt to income

+ 4.896× credit util + 0.386× bankruptcy + 0.154× term

+ 0.028× issuedJan2018 − 0.040× issuedMar2018 + 0.228× credit checks

If we count up the number of predictor coefficients, we get the effective number of predictors in themodel: k = 9. Notice that the issued categorical predictor counts as two, once for the two levelsshown in the model. In general, a categorical predictor with p different levels will be represented byp− 1 terms in a multiple regression model.

GUIDED PRACTICE 9.10

What does β4, the coefficient of variable credit util, represent? What is the point estimate of β4?4

4β4 represents the change in interest rate we would expect if someone’s credit utilization was 0 and went to 1, allother factors held even. The point estimate is b4 = 4.90%.

348 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

EXAMPLE 9.11

Compute the residual of the first observation in Figure 9.1 on page 343 using the equation identifiedin Guided Practice 9.9.

To compute the residual, we first need the predicted value, which we compute by plugging val-ues into the equation from Example 9.9. For example, income versource only takes a value of 0,income ververified takes a value of 1 (since the borrower’s income source and amount were verified),debt to income was 18.01, and so on. This leads to a prediction of r̂ate1 = 18.09. The observedinterest rate was 14.07%, which leads to a residual of e1 = 14.07− 18.09 = −4.02.

EXAMPLE 9.12

We estimated a coefficient for bankruptcy in Section 9.1.1 of b4 = 0.74 with a standard error ofSEb1 = 0.15 when using simple linear regression. Why is there a difference between that estimateand the estimated coefficient of 0.39 in the multiple regression setting?

If we examined the data carefully, we would see that some predictors are correlated. For instance,when we estimated the connection of the outcome interest rate and predictor bankruptcy usingsimple linear regression, we were unable to control for other variables like whether the borrower hadher income verified, the borrower’s debt-to-income ratio, and other variables. That original modelwas constructed in a vacuum and did not consider the full context. When we include all of thevariables, underlying and unintentional bias that was missed by these other variables is reduced oreliminated. Of course, bias can still exist from other confounding variables.

Example 9.12 describes a common issue in multiple regression: correlation among predictorvariables. We say the two predictor variables are collinear (pronounced as co-linear) when theyare correlated, and this collinearity complicates model estimation. While it is impossible to preventcollinearity from arising in observational data, experiments are usually designed to prevent predictorsfrom being collinear.

GUIDED PRACTICE 9.13

The estimated value of the intercept is 1.925, and one might be tempted to make some interpretationof this coefficient, such as, it is the model’s predicted price when each of the variables take valuezero: income source is not verified, the borrower has no debt (debt-to-income and credit utilizationare zero), and so on. Is this reasonable? Is there any value gained by making this interpretation?5

5Many of the variables do take a value 0 for at least one data point, and for those variables, it is reasonable.However, one variable never takes a value of zero: term, which describes the length of the loan, in months. If term

is set to zero, then the loan must be paid back immediately; the borrower must give the money back as soon as shereceives it, which means it is not a real loan. Ultimately, the interpretation of the intercept in this setting is notinsightful.

9.1. INTRODUCTION TO MULTIPLE REGRESSION 349

9.1.3 Adjusted R2R2R2 as a better tool for multiple regression

We first used R2 in Section 8.2 to determine the amount of variability in the response that wasexplained by the model:

R2 = 1− variability in residuals

variability in the outcome= 1− V ar(ei)

V ar(yi)

where ei represents the residuals of the model and yi the outcomes. This equation remains validin the multiple regression framework, but a small enhancement can make it even more informativewhen comparing models.

GUIDED PRACTICE 9.14

The variance of the residuals for the model given in Guided Practice 9.9 is 18.53, and the varianceof the total price in all the auctions is 25.01. Calculate R2 for this model.6

This strategy for estimating R2 is acceptable when there is just a single variable. However,it becomes less helpful when there are many variables. The regular R2 is a biased estimate of theamount of variability explained by the model when applied to a new sample of data. To get a betterestimate, we use the adjusted R2.

ADJUSTEDR2R2R2 AS A TOOL FOR MODEL ASSESSMENT

The adjusted R2R2R2 is computed as

R2adj = 1− s2

residuals/(n− k − 1)

s2outcome/(n− 1)

= 1− s2residuals

s2outcome

× n− 1

n− k − 1

where n is the number of cases used to fit the model and k is the number of predictor variablesin the model. Remember that a categorical predictor with p levels will contribute p− 1 to thenumber of variables in the model.

Because k is never negative, the adjusted R2 will be smaller – often times just a little smaller– than the unadjusted R2. The reasoning behind the adjusted R2 lies in the degrees of freedomassociated with each variance, which is equal to n− k − 1 for the multiple regression context. If wewere to make predictions for new data using our current model, we would find that the unadjustedR2 would tend to be slightly overly optimistic, while the adjusted R2 formula helps correct this bias.

GUIDED PRACTICE 9.15

There were n = 10000 auctions in the loans data set and k = 9 predictor variables in the model.Use n, k, and the variances from Guided Practice 9.14 to calculate R2

adj for the interest rate model.7

GUIDED PRACTICE 9.16

Suppose you added another predictor to the model, but the variance of the errors V ar(ei) didn’t godown. What would happen to the R2? What would happen to the adjusted R2? 8

Adjusted R2 could have been used in Chapter 8. However, when there is only k = 1 predictors,adjusted R2 is very close to regular R2, so this nuance isn’t typically important when the model hasonly one predictor.

6R2 = 1− 18.5325.01

= 0.2591.7R2

adj = 1 − 18.5325.01

× 10000−11000−9−1

= 0.2584. While the difference is very small, it will be important when we fine

tune the model in the next section.8The unadjusted R2 would stay the same and the adjusted R2 would go down.

350 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

Exercises

9.1 Baby weights, Part I. The Child Health and Development Studies investigate a range of topics. Onestudy considered all pregnancies between 1960 and 1967 among women in the Kaiser Foundation HealthPlan in the San Francisco East Bay area. Here, we study the relationship between smoking and weight ofthe baby. The variable smoke is coded 1 if the mother is a smoker, and 0 if not. The summary table belowshows the results of a linear regression model for predicting the average birth weight of babies, measured inounces, based on the smoking status of the mother.9

Estimate Std. Error t value Pr(>|t|)(Intercept) 123.05 0.65 189.60 0.0000

smoke -8.94 1.03 -8.65 0.0000

The variability within the smokers and non-smokers are about equal and the distributions are symmetric.With these conditions satisfied, it is reasonable to apply the model. (Note that we don’t need to checklinearity since the predictor has only two levels.)

(a) Write the equation of the regression model.

(b) Interpret the slope in this context, and calculate the predicted birth weight of babies born to smokerand non-smoker mothers.

9.2 Baby weights, Part II. Exercise 9.1 introduces a data set on birth weight of babies. Another variablewe consider is parity, which is 1 if the child is the first born, and 0 otherwise. The summary table belowshows the results of a linear regression model for predicting the average birth weight of babies, measured inounces, from parity.

Estimate Std. Error t value Pr(>|t|)(Intercept) 120.07 0.60 199.94 0.0000

parity -1.93 1.19 -1.62 0.1052

(a) Write the equation of the regression model.

(b) Interpret the slope in this context, and calculate the predicted birth weight of first borns and others.

9Child Health and Development Studies, Baby weights data set.

9.1. INTRODUCTION TO MULTIPLE REGRESSION 351

9.3 Baby weights, Part III. We considered the variables smoke and parity, one at a time, in modelingbirth weights of babies in Exercises 9.1 and 9.2. A more realistic approach to modeling infant weights isto consider all possibly related variables at once. Other variables of interest include length of pregnancy indays (gestation), mother’s age in years (age), mother’s height in inches (height), and mother’s pregnancyweight in pounds (weight). Below are three observations from this data set.

bwt gestation parity age height weight smoke

1 120 284 0 27 62 100 02 113 282 0 33 64 135 0…

……

…1236 117 297 0 38 65 129 0

The summary table below shows the results of a regression model for predicting the average birth weight ofbabies based on all of the variables included in the data set.

Estimate Std. Error t value Pr(>|t|)(Intercept) -80.41 14.35 -5.60 0.0000

gestation 0.44 0.03 15.26 0.0000parity -3.33 1.13 -2.95 0.0033

age -0.01 0.09 -0.10 0.9170height 1.15 0.21 5.63 0.0000weight 0.05 0.03 1.99 0.0471smoke -8.40 0.95 -8.81 0.0000

(a) Write the equation of the regression model that includes all of the variables.

(b) Interpret the slopes of gestation and age in this context.

(c) The coefficient for parity is different than in the linear model shown in Exercise 9.2. Why might therebe a difference?

(d) Calculate the residual for the first observation in the data set.

(e) The variance of the residuals is 249.28, and the variance of the birth weights of all babies in the dataset is 332.57. Calculate the R2 and the adjusted R2. Note that there are 1,236 observations in the dataset.

352 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.4 Absenteeism, Part I. Researchers interested in the relationship between absenteeism from school andcertain demographic characteristics of children collected data from 146 randomly sampled students in ruralNew South Wales, Australia, in a particular school year. Below are three observations from this data set.

eth sex lrn days

1 0 1 1 22 0 1 1 11…

……

146 1 0 0 37

The summary table below shows the results of a linear regression model for predicting the average numberof days absent based on ethnic background (eth: 0 – aboriginal, 1 – not aboriginal), sex (sex: 0 – female, 1- male), and learner status (lrn: 0 – average learner, 1 – slow learner).10

Estimate Std. Error t value Pr(>|t|)(Intercept) 18.93 2.57 7.37 0.0000

eth -9.11 2.60 -3.51 0.0000sex 3.10 2.64 1.18 0.2411lrn 2.15 2.65 0.81 0.4177

(a) Write the equation of the regression model.

(b) Interpret each one of the slopes in this context.

(c) Calculate the residual for the first observation in the data set: a student who is aboriginal, male, a slowlearner, and missed 2 days of school.

(d) The variance of the residuals is 240.57, and the variance of the number of absent days for all studentsin the data set is 264.17. Calculate the R2 and the adjusted R2. Note that there are 146 observationsin the data set.

9.5 GPA. A survey of 55 Duke University students asked about their GPA, number of hours they study atnight, number of nights they go out, and their gender. Summary output of the regression model is shownbelow. Note that male is coded as 1.

Estimate Std. Error t value Pr(>|t|)(Intercept) 3.45 0.35 9.85 0.00studyweek 0.00 0.00 0.27 0.79sleepnight 0.01 0.05 0.11 0.91

outnight 0.05 0.05 1.01 0.32gender -0.08 0.12 -0.68 0.50

(a) Calculate a 95% confidence interval for the coefficient of gender in the model, and interpret it in thecontext of the data.

(b) Would you expect a 95% confidence interval for the slope of the remaining variables to include 0? Explain

9.6 Cherry trees. Timber yield is approximately equal to the volume of a tree, however, this value is difficultto measure without first cutting the tree down. Instead, other variables, such as height and diameter, maybe used to predict a tree’s volume and yield. Researchers wanting to understand the relationship betweenthese variables for black cherry trees collected data from 31 such trees in the Allegheny National Forest,Pennsylvania. Height is measured in feet, diameter in inches (at 54 inches above ground), and volume incubic feet.11

Estimate Std. Error t value Pr(>|t|)(Intercept) -57.99 8.64 -6.71 0.00

height 0.34 0.13 2.61 0.01diameter 4.71 0.26 17.82 0.00

(a) Calculate a 95% confidence interval for the coefficient of height, and interpret it in the context of thedata.

(b) One tree in this sample is 79 feet tall, has a diameter of 11.3 inches, and is 24.2 cubic feet in volume.Determine if the model overestimates or underestimates the volume of this tree, and by how much.

10W. N. Venables and B. D. Ripley. Modern Applied Statistics with S. Fourth Edition. Data can also be found inthe R MASS package. New York: Springer, 2002.

11D.J. Hand. A handbook of small data sets. Chapman & Hall/CRC, 1994.

9.2. MODEL SELECTION 353

9.2 Model selection

The best model is not always the most complicated. Sometimes including variables that arenot evidently important can actually reduce the accuracy of predictions. In this section, we discussmodel selection strategies, which will help us eliminate variables from the model that are found tobe less important. It’s common (and hip, at least in the statistical world) to refer to models thathave undergone such variable pruning as parsimonious.

In practice, the model that includes all available explanatory variables is often referred to asthe full model. The full model may not be the best model, and if it isn’t, we want to identify asmaller model that is preferable.

9.2.1 Identifying variables in the model that may not be helpful

Adjusted R2 describes the strength of a model fit, and it is a useful tool for evaluating whichpredictors are adding value to the model, where adding value means they are (likely) improving theaccuracy in predicting future outcomes.

Let’s consider two models, which are shown in Tables 9.6 and 9.7. The first table summarizesthe full model since it includes all predictors, while the second does not include the issued variable.

Estimate Std. Error t value Pr(>|t|)(Intercept) 1.9251 0.2102 9.16 <0.0001

income ver: source only 0.9750 0.0991 9.83 <0.0001income ver: verified 2.5374 0.1172 21.65 <0.0001

debt to income 0.0211 0.0029 7.18 <0.0001credit util 4.8959 0.1619 30.24 <0.0001

bankruptcy 0.3864 0.1324 2.92 0.0035term 0.1537 0.0039 38.96 <0.0001

issued: Jan2018 0.0276 0.1081 0.26 0.7981issued: Mar2018 -0.0397 0.1065 -0.37 0.7093

credit checks 0.2282 0.0182 12.51 <0.0001R2adj = 0.25843 df = 9990

Figure 9.6: The fit for the full regression model, including the adjusted R2.

Estimate Std. Error t value Pr(>|t|)(Intercept) 1.9213 0.1982 9.69 <0.0001

income ver: source only 0.9740 0.0991 9.83 <0.0001income ver: verified 2.5355 0.1172 21.64 <0.0001

debt to income 0.0211 0.0029 7.19 <0.0001credit util 4.8958 0.1619 30.25 <0.0001

bankruptcy 0.3869 0.1324 2.92 0.0035term 0.1537 0.0039 38.97 <0.0001

credit checks 0.2283 0.0182 12.51 <0.0001

R2adj = 0.25854 df = 9992

Figure 9.7: The fit for the regression model after dropping the issued variable.

EXAMPLE 9.17

Which of the two models is better?

We compare the adjusted R2 of each model to determine which to choose. Since the first model hasan R2

adj smaller than the R2adj of the second model, we prefer the second model to the first.

354 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

Will the model without issued be better than the model with issued? We cannot know forsure, but based on the adjusted R2, this is our best assessment.

9.2.2 Two model selection strategies

Two common strategies for adding or removing variables in a multiple regression model arecalled backward elimination and forward selection. These techniques are often referred to as step-wise model selection strategies, because they add or delete one variable at a time as they “step”through the candidate predictors.

Backward elimination starts with the model that includes all potential predictor variables.Variables are eliminated one-at-a-time from the model until we cannot improve the adjusted R2.The strategy within each elimination step is to eliminate the variable that leads to the largestimprovement in adjusted R2.

EXAMPLE 9.18

Results corresponding to the full model for the loans data are shown in Figure 9.6. How should weproceed under the backward elimination strategy?

Our baseline adjusted R2 from the full model is R2adj = 0.25843, and we need to determine whether

dropping a predictor will improve the adjusted R2. To check, we fit models that each drop a differentpredictor, and we record the adjusted R2:

Exclude … income ver debt to income credit util bankruptcy

R2adj = 0.22380 R2

adj = 0.25468 R2adj = 0.19063 R2

adj = 0.25787

term issued credit checks

R2adj = 0.14581 R2

adj = 0.25854 R2adj = 0.24689

The model without issued has the highest adjusted R2 of 0.25854, higher than the adjusted R2 forthe full model. Because eliminating issued leads to a model with a higher adjusted R2, we dropissued from the model.

Since we eliminated a predictor from the model in the first step, we see whether we should eliminateany additional predictors. Our baseline adjusted R2 is now R2

adj = 0.25854. We now fit new models,which consider eliminating each of the remaining predictors in addition to issued:

Exclude issued and … income ver debt to income credit util

R2adj = 0.22395 R2

adj = 0.25479 R2adj = 0.19074

bankruptcy term credit checks

R2adj = 0.25798 R2

adj = 0.14592 R2adj = 0.24701

None of these models lead to an improvement in adjusted R2, so we do not eliminate any of theremaining predictors. That is, after backward elimination, we are left with the model that keeps allpredictors except issued, which we can summarize using the coefficients from Figure 9.7:

r̂ate = 1.921 + 0.974× income versource only + 2.535× income ververified

+ 0.021× debt to income + 4.896× credit util + 0.387× bankruptcy

+ 0.154× term + 0.228× credit check

The forward selection strategy is the reverse of the backward elimination technique. Insteadof eliminating variables one-at-a-time, we add variables one-at-a-time until we cannot find anyvariables that improve the model (as measured by adjusted R2).

9.2. MODEL SELECTION 355

EXAMPLE 9.19

Construct a model for the loans data set using the forward selection strategy.

We start with the model that includes no variables. Then we fit each of the possible models withjust one variable. That is, we fit the model including just income ver, then the model including justdebt to income, then a model with just credit util, and so on. Then we examine the adjustedR2 for each of these models:

Add … income ver debt to income credit util bankruptcy

R2adj = 0.05926 R2

adj = 0.01946 R2adj = 0.06452 R2

adj = 0.00222

term issued credit checks

R2adj = 0.12855 R2

adj = −¡0.00018 R2adj = 0.01711

In this first step, we compare the adjusted R2 against a baseline model that has no predictors.The no-predictors model always has R2

adj = 0. The model with one predictor that has the largest

adjusted R2 is the model with the term predictor, and because this adjusted R2 is larger than theadjusted R2 from the model with no predictors (R2

adj = 0), we will add this variable to our model.

We repeat the process again, this time considering 2-predictor models where one of the predictorsis term and with a new baseline of R2

adj = 0.12855:

Add term and … income ver debt to income credit util

R2adj = 0.16851 R2

adj = 0.14368 R2adj = 0.20046

bankruptcy issued credit checks

R2adj = 0.13070 R2

adj = 0.12840 R2adj = 0.14294

The best second predictor, credit util, has a higher adjusted R2 (0.20046) than the baseline(0.12855), so we also add credit util to the model.

Since we have again added a variable to the model, we continue and see whether it would be beneficialto add a third variable:

Add term, credit util, and … income ver debt to income

R2adj = 0.24183 R2

adj = 0.20810

bankruptcy issued credit checks

R2adj = 0.20169 R2

adj = 0.20031 R2adj = 0.21629

The model adding income ver improved adjusted R2 (0.24183 to 0.20046), so we add income ver

to the model.

We continue on in this way, next adding debt to income, then credit checks, and bankruptcy.At this point, we come again to the issued variable: adding this variable leads to R2

adj = 0.25843,

while keeping all the other variables but excluding issued leads to a higher R2adj = 0.25854. This

means we do not add issued. In this example, we have arrived at the same model that we identifiedfrom backward elimination.

MODEL SELECTION STRATEGIES

Backward elimination begins with the model having the largest number of predictors and elim-inates variables one-by-one until we are satisfied that all remaining variables are important tothe model. Forward selection starts with no variables included in the model, then it adds invariables according to their importance until no other important variables are found.

Backward elimination and forward selection sometimes arrive at different final models. If tryingboth techniques and this happens, it’s common to choose the model with the larger R2

adj .

356 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.2.3 The p-value approach, an alternative to adjusted R2R2R2

The p-value may be used as an alternative to R2adj for model selection:

Backward elimination with the p-value approach. In backward elimination, we would iden-tify the predictor corresponding to the largest p-value. If the p-value is above the significancelevel, usually α = 0.05, then we would drop that variable, refit the model, and repeat the pro-cess. If the largest p-value is less than α = 0.05, then we would not eliminate any predictorsand the current model would be our best-fitting model.

Forward selection with the p-value approach. In forward selection with p-values, we reversethe process. We begin with a model that has no predictors, then we fit a model for each possiblepredictor, identifying the model where the corresponding predictor’s p-value is smallest. If thatp-value is smaller than α = 0.05, we add it to the model and repeat the process, consideringwhether to add more variables one-at-a-time. When none of the remaining predictors can beadded to the model and have a p-value less than 0.05, then we stop adding variables and thecurrent model would be our best-fitting model.

GUIDED PRACTICE 9.20

Examine Figure 9.7 on page 353, which considers the model including all variables except the variablefor the month the loan was issued. If we were using the p-value approach with backward eliminationand we were considering this model, which of these variables would be up for elimination? Wouldwe drop that variable, or would we keep it in the model?12

While the adjusted R2 and p-value approaches are similar, they sometimes lead to differentmodels, with the R2

adj approach tending to include more predictors in the final model.

ADJUSTEDR2R2R2 VS P-VALUE APPROACH

When the sole goal is to improve prediction accuracy, use R2adj . This is commonly the case in

machine learning applications.

When we care about understanding which variables are statistically significant predictors of theresponse, or if there is interest in producing a simpler model at the potential cost of a littleprediction accuracy, then the p-value approach is preferred.

Regardless of whether you use R2adj or the p-value approach, or if you use the backward elimi-

nation of forward selection strategy, our job is not done after variable selection. We must still verifythe model conditions are reasonable.

12The bankruptcy predictor is up for elimination since it has the largest p-value. However, since that p-value issmaller than 0.05, we would still keep it in the model.

9.2. MODEL SELECTION 357

Exercises

9.7 Baby weights, Part IV. Exercise 9.3 considers a model that predicts a newborn’s weight using severalpredictors (gestation length, parity, age of mother, height of mother, weight of mother, smoking status ofmother). The table below shows the adjusted R-squared for the full model as well as adjusted R-squaredvalues for all models we evaluate in the first step of the backwards elimination process.

Model Adjusted R2

1 Full model 0.25412 No gestation 0.10313 No parity 0.24924 No age 0.25475 No height 0.23116 No weight 0.25367 No smoking status 0.2072

Which, if any, variable should be removed from the model first?

9.8 Absenteeism, Part II. Exercise 9.4 considers a model that predicts the number of days absent usingthree predictors: ethnic background (eth), gender (sex), and learner status (lrn). The table below showsthe adjusted R-squared for the model as well as adjusted R-squared values for all models we evaluate in thefirst step of the backwards elimination process.

Model Adjusted R2

1 Full model 0.07012 No ethnicity -0.00333 No sex 0.06764 No learner status 0.0723

Which, if any, variable should be removed from the model first?

9.9 Baby weights, Part V. Exercise 9.3 provides regression output for the full model (including all explana-tory variables available in the data set) for predicting birth weight of babies. In this exercise we considera forward-selection algorithm and add variables to the model one-at-a-time. The table below shows thep-value and adjusted R2 of each model where we include only the corresponding predictor. Based on thistable, which variable should be added to the model first?

variable gestation parity age height weight smoke

p-value 2.2× 10−16 0.1052 0.2375 2.97× 10−12 8.2× 10−8 2.2× 10−16

R2adj 0.1657 0.0013 0.0003 0.0386 0.0229 0.0569

9.10 Absenteeism, Part III. Exercise 9.4 provides regression output for the full model, including all ex-planatory variables available in the data set, for predicting the number of days absent from school. In thisexercise we consider a forward-selection algorithm and add variables to the model one-at-a-time. The tablebelow shows the p-value and adjusted R2 of each model where we include only the corresponding predictor.Based on this table, which variable should be added to the model first?

variable ethnicity sex learner status

p-value 0.0007 0.3142 0.5870R2adj 0.0714 0.0001 0

9.11 Movie lovers, Part I. Suppose a social scientist is interested in studying what makes audiences loveor hate a movie. She collects a random sample of movies (genre, length, cast, director, budget, etc.) aswell as a measure of the success of the movie (score on a film review aggregator website). If as part of herresearch she is interested in finding out which variables are significant predictors of movie success, what typeof model selection method should she use?

9.12 Movie lovers, Part II. Suppose an online media streaming company is interested in building a movierecommendation system. The website maintains data on the movies in their database (genre, length, cast,director, budget, etc.) and additionally collects data from their subscribers ( demographic information,previously watched movies, how they rated previously watched movies, etc.). The recommendation sys-tem will be deemed successful if subscribers actually watch, and rate highly, the movies recommended tothem. Should the company use the adjusted R2 or the p-value approach in selecting variables for theirrecommendation system?

358 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.3 Checking model conditions using graphs

Multiple regression methods using the model

ŷ = β0 + β1×1 + β2×2 + · · ·+ βkxk

generally depend on the following four conditions:

1. the residuals of the model are nearly normal (less important for larger data sets),

2. the variability of the residuals is nearly constant,

3. the residuals are independent, and

4. each variable is linearly related to the outcome.

9.3.1 Diagnostic plots

Diagnostic plots can be used to check each of these conditions. We will consider the modelfrom the Lending Club loans data, and check whether there are any notable concerns:

r̂ate = 1.921 + 0.974× income versource only + 2.535× income ververified

+ 0.021× debt to income + 4.896× credit util + 0.387× bankruptcy

+ 0.154× term + 0.228× credit check

Check for outliers. In theory, the distribution of the residuals should be nearly normal; in prac-tice, normality can be relaxed for most applications. Instead, we examine a histogram of theresiduals to check if there are any outliers: Figure 9.8 is a histogram of these outliers. Sincethis is a very large data set, only particularly extreme observations would be a concern in thisparticular case. There are no extreme observations that might cause a concern.

If we intended to construct what are called prediction intervals for future observations, wewould be more strict and require the residuals to be nearly normal. Prediction intervals arefurther discussed in an online extra on the OpenIntro website:

www.openintro.org/d?id=stat extra linear regression supp

Residuals

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9.3. CHECKING MODEL CONDITIONS USING GRAPHS 359

Absolute values of residuals against fitted values. A plot of the absolute value of the resid-uals against their corresponding fitted values (ŷi) is shown in Figure 9.9. This plot is helpfulto check the condition that the variance of the residuals is approximately constant, and asmoothed line has been added to represent the approximate trend in this plot. There is moreevident variability for fitted values that are larger, which we’ll discuss further.

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Figure 9.9: Comparing the absolute value of the residuals against the fitted values(ŷi) is helpful in identifying deviations from the constant variance assumption.

Residuals in order of their data collection. This type of plot can be helpful when observationswere collected in a sequence. Such a plot is helpful in identifying any connection between casesthat are close to one another. The loans in this data set were issued over a 3 month period,and the month the loan was issued was not found to be important, suggesting this is not aconcern for this data set. In cases where a data set does show some pattern for this check,time series methods may be useful.

Residuals against each predictor variable. We consider a plot of the residuals against each ofthe predictors in Figure 9.10. For those instances where there are only 2-3 groups, box plotsare shown. For the numerical outcomes, a smoothed line has been fit to the data to makeit easier to review. Ultimately, we are looking for any notable change in variability betweengroups or pattern in the data.

Here are the things of importance from these plots:

• There is some minor differences in variability between the verified income groups.

• There is a very clear pattern for the debt-to-income variable. What also stands out isthat this variable is very strongly right skewed: there are few observations with very highdebt-to-income ratios.

• The downward curve on the right side of the credit utilization and credit check plotssuggests some minor misfitting for those larger values.

Having reviewed the diagnostic plots, there are two options. The first option is to, if we’re notconcerned about the issues observed, use this as the final model; if going this route, it is importantto still note any abnormalities observed in the diagnostics. The second option is to try to improvethe model, which is what we’ll try to do with this particular model fit.

360 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

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Figure 9.10: Diagnostic plots for residuals against each of the predictors. Forthe box plots, we’re looking for notable differences in variability. For numericalpredictors, we also check for trends or other structure in the data.

9.3. CHECKING MODEL CONDITIONS USING GRAPHS 361

9.3.2 Options for improving the model fit

There are several options for improvement a model, including transforming variables, seekingout additional variables to fill model gaps, or using more advanced methods that would accountfor challenges around inconsistent variability or nonlinear relationships between predictors and theoutcome.

The main concern for the initial model is that there is a notable nonlinear relationship betweenthe debt-to-income variable observed in Figure 9.10. To resolve this issue, we’re going to consider acouple strategies for adjusting the relationship between the predictor variable and the outcome.

Let’s start by taking a look at a histogram of debt to income in Figure 9.11. The variable isextremely skewed, and upper values will have a lot of leverage on the fit. Below are several options:

• log transformation (log x),

• square root transformation (√x),

• inverse transformation (1/x),

• truncation (cap the max value possible)

If we inspected the data more closely, we’d observe some instances where the variable takes a valueof 0, and since log(0) and 1/x are undefined when x = 0, we’ll exclude these transformations fromfurther consideration.13 A square root transformation is valid for all values the variable takes, andtruncating some of the larger observations is also a valid approach. We’ll consider both of theseapproaches.

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Figure 9.11: Histogram of debt to income, where extreme skew is evident.

To try transforming the variable, we make two new variables representing the transformedversions:

Square root. We create a new variable, sqrt debt to income, where all the values are simply thesquare roots of the values in debt to income, and then refit the model as before. The resultis shown in the left panel of Figure 9.12. The square root pulled in the higher values a bit, butthe fit still doesn’t look great since the smoothed line is still wavy.

Truncate at 50. We create a new variable, debt to income 50, where any values in debt to

income that are greater than 50 are shrunk to exactly 50. Refitting the model once more, thediagnostic plot for this new variable is shown in the right panel of Figure 9.12. Here the fitlooks much more reasonable, so this appears to be a reasonable approach.

The downside of using transformations is that it reduces the ease of interpreting the results. For-tunately, since the truncation transformation only affects a relatively small number of cases, theinterpretation isn’t dramatically impacted.

13There are ways to make them work, but we’ll leave those options to a later course.

362 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

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Figure 9.12: Histogram of debt to income, where extreme skew is evident.

As a next step, we’d evaluate the new model using the truncated version of debt to income,we would complete all the same procedures as before. The other two issues noted while inspectingdiagnostics in Section 9.3.1 are still present in the updated model. If we choose to report this model,we would want to also discuss these shortcomings to be transparent in our work. Depending onwhat the model will be used, we could either try to bring those under control, or we could stop sincethose issues aren’t severe. Had the non-constant variance been a little more dramatic, it would be ahigher priority. Ultimately we decided that the model was reasonable, and we report its final formhere:

r̂ate = 1.562 + 1.002× income versource only + 2.436× income ververified

+ 0.048× debt to income 50 + 4.694× credit util + 0.394× bankruptcy

+ 0.153× term + 0.223× credit check

A sharp eye would notice that the coefficient for debt to income 50 is more than twice as large aswhat the coefficient had been for the debt to income variable in the earlier model. This suggeststhose larger values not only were points with high leverage, but they were influential points thatwere dramatically impacting the coefficient.

“ALL MODELS ARE WRONG, BUT SOME ARE USEFUL” -GEORGE E.P. BOX

The truth is that no model is perfect. However, even imperfect models can be useful. Reportinga flawed model can be reasonable so long as we are clear and report the model’s shortcomings.

Don’t report results when conditions are grossly violated. While there is a little leeway in modelconditions, don’t go too far. If model conditions are very clearly violated, consider a new model,even if it means learning more statistical methods or hiring someone who can help. To help youget started, we’ve developed a couple additional sections that you may find on OpenIntro’s website.These sections provide a light introduction to what are called interaction terms and to fittingnonlinear curves to data, respectively:

www.openintro.org/d?file=stat extra interaction effects

www.openintro.org/d?file=stat extra nonlinear relationships

9.3. CHECKING MODEL CONDITIONS USING GRAPHS 363

Exercises

9.13 Baby weights, Part VI. Exercise 9.3 presents a regression model for predicting the average birth weightof babies based on length of gestation, parity, height, weight, and smoking status of the mother. Determineif the model assumptions are met using the plots below. If not, describe how to proceed with the analysis.

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364 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.14 Movie returns, Part I. A FiveThirtyEight.com article reports that “Horror movies get nowhere nearas much draw at the box office as the big-time summer blockbusters or action/adventure movies … butthere’s a huge incentive for studios to continue pushing them out. The return-on-investment potential forhorror movies is absurd.” To investigate how the return-on-investment compares between genres and howthis relationship has changed over time, an introductory statistics student fit a model predicting the ratio ofgross revenue of movies from genre and release year for 1,070 movies released between 2000 and 2018. Usingthe plots given below, determine if this regression model is appropriate for these data.14

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14FiveThirtyEight, Scary Movies Are The Best Investment In Hollywood.

9.4. MULTIPLE REGRESSION CASE STUDY: MARIO KART 365

9.4 Multiple regression case study: Mario Kart

We’ll consider Ebay auctions of a video game called Mario Kart for the Nintendo Wii. The outcomevariable of interest is the total price of an auction, which is the highest bid plus the shippingcost. We will try to determine how total price is related to each characteristic in an auction whilesimultaneously controlling for other variables. For instance, all other characteristics held constant,are longer auctions associated with higher or lower prices? And, on average, how much more dobuyers tend to pay for additional Wii wheels (plastic steering wheels that attach to the Wii controller)in auctions? Multiple regression will help us answer these and other questions.

9.4.1 Data set and the full model

The mariokart data set includes results from 141 auctions. Four observations from this dataset are shown in Figure 9.13, and descriptions for each variable are shown in Figure 9.14. Noticethat the condition and stock photo variables are indicator variables, similar to bankruptcy in theloan data set.

price cond new stock photo duration wheels1 51.55 1 1 3 12 37.04 0 1 7 1…

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Figure 9.13: Four observations from the mariokart data set.

variable description

price Final auction price plus shipping costs, in US dollars.cond new Indicator variable for if the game is new (1) or used (0).stock photo Indicator variable for if the auction’s main photo is a stock photo.duration The length of the auction, in days, taking values from 1 to 10.wheels The number of Wii wheels included with the auction. A Wii wheel

is an optional steering wheel accessory that holds the Wii controller.

Figure 9.14: Variables and their descriptions for the mariokart data set.

366 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

GUIDED PRACTICE 9.21

We fit a linear regression model with the game’s condition as a predictor of auction price. Resultsof this model are summarized below:

Estimate Std. Error t value Pr(>|t|)(Intercept) 42.8711 0.8140 52.67 <0.0001cond new 10.8996 1.2583 8.66 <0.0001

df = 139

Write down the equation for the model, note whether the slope is statistically different from zero,and interpret the coefficient.15

Sometimes there are underlying structures or relationships between predictor variables. Forinstance, new games sold on Ebay tend to come with more Wii wheels, which may have led tohigher prices for those auctions. We would like to fit a model that includes all potentially importantvariables simultaneously. This would help us evaluate the relationship between a predictor variableand the outcome while controlling for the potential influence of other variables.

We want to construct a model that accounts for not only the game condition, as in GuidedPractice 9.21, but simultaneously accounts for three other variables:

̂price = β0 + β1 × cond new + β2 × stock photo

+ β3 × duration + β4 × wheels

Figure 9.15 summarizes the full model. Using this output, we identify the point estimates of eachcoefficient.

Estimate Std. Error t value Pr(>|t|)(Intercept) 36.2110 1.5140 23.92 <0.0001cond new 5.1306 1.0511 4.88 <0.0001

stock photo 1.0803 1.0568 1.02 0.3085duration -0.0268 0.1904 -0.14 0.8882

wheels 7.2852 0.5547 13.13 <0.0001df = 136

Figure 9.15: Output for the regression model where price is the outcome andcond new, stock photo, duration, and wheels are the predictors.

GUIDED PRACTICE 9.22

Write out the model’s equation using the point estimates from Figure 9.15. How many predictorsare there in this model?16

GUIDED PRACTICE 9.23

What does β4, the coefficient of variable x4 (Wii wheels), represent? What is the point estimate ofβ4?17

15The equation for the line may be written as

p̂rice = 42.87 + 10.90× cond new

Examining the regression output in Guided Practice 9.21, we can see that the p-value for cond new is very close tozero, indicating there is strong evidence that the coefficient is different from zero when using this simple one-variablemodel.

The cond new is a two-level categorical variable that takes value 1 when the game is new and value 0 when thegame is used. This means the 10.90 model coefficient predicts an extra $10.90 for those games that are new versusthose that are used.

16p̂rice = 36.21+5.13×cond new+1.08×stock photo−0.03×duration+7.29×wheels, with the k = 4 predictors.17It is the average difference in auction price for each additional Wii wheel included when holding the other variables

constant. The point estimate is b4 = 7.29.

9.4. MULTIPLE REGRESSION CASE STUDY: MARIO KART 367

GUIDED PRACTICE 9.24

Compute the residual of the first observation in Figure 9.13 using the equation identified in GuidedPractice 9.22.18

EXAMPLE 9.25

We estimated a coefficient for cond new in Section 9.21 of b1 = 10.90 with a standard error ofSEb1 = 1.26 when using simple linear regression. Why might there be a difference between thatestimate and the one in the multiple regression setting?

If we examined the data carefully, we would see that there is collinearity among some predictors.For instance, when we estimated the connection of the outcome price and predictor cond new usingsimple linear regression, we were unable to control for other variables like the number of Wii wheelsincluded in the auction. That model was biased by the confounding variable wheels. When we useboth variables, this particular underlying and unintentional bias is reduced or eliminated (thoughbias from other confounding variables may still remain).

9.4.2 Model selection

Let’s revisit the model for the Mario Kart auction and complete model selection using backwardsselection. Recall that the full model took the following form:

p̂rice = 36.21 + 5.13× cond new + 1.08× stock photo− 0.03× duration + 7.29× wheels

EXAMPLE 9.26

Results corresponding to the full model for the mariokart data were shown in Figure 9.15 on thefacing page. For this model, we consider what would happen if dropping each of the variables in themodel:

Exclude … cond new stock photo duration wheels

R2adj = 0.6626 R2

adj = 0.7107 R2adj = 0.7128 R2

adj = 0.3487

For the full model, R2adj = 0.7108. How should we proceed under the backward elimination strategy?

The third model without duration has the highest R2adj of 0.7128, so we compare it to R2

adj for the

full model. Because eliminating duration leads to a model with a higher R2adj , we drop duration

from the model.

GUIDED PRACTICE 9.27

In Example 9.26, we eliminated the duration variable, which resulted in a model with R2adj = 0.7128.

Let’s look at if we would eliminate another variable from the model using backwards elimination:

Exclude duration and … cond new stock photo wheels

R2adj = 0.6587 R2

adj = 0.7124 R2adj = 0.3414

Should we eliminate any additional variable, and if so, which variable should we eliminate?19

18ei = yi − ŷi = 51.55− 49.62 = 1.93, where 49.62 was computed using the variables values from the observationand the equation identified in Guided Practice 9.22.

19Removing any of the three remaining variables would lead to a decrease in R2adj , so we should not remove any

additional variables from the model after we removed duration.

368 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

GUIDED PRACTICE 9.28

After eliminating the auction’s duration from the model, we are left with the following reducedmodel:

p̂rice = 36.05 + 5.18× cond new + 1.12× stock photo + 7.30× wheels

How much would you predict for the total price for the Mario Kart game if it was used, used a stockphoto, and included two wheels and put up for auction during the time period that the Mario Kartdata were collected?20

GUIDED PRACTICE 9.29

Would you be surprised if the seller from Guided Practice 9.28 didn’t get the exact price predicted?21

9.4.3 Checking model conditions using graphs

Let’s take a closer look at the diagnostics for the Mario Kart model to check if the model we haveidentified is reasonable.

Check for outliers. A histogram of the residuals is shown in Figure 9.16. With a data set wellover a hundred, we’re primarily looking for major outliers. While one minor outlier appearson the upper end, it is not a concern for this large of a data set.

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Absolute values of residuals against fitted values. A plot of the absolute value of the resid-uals against their corresponding fitted values (ŷi) is shown in Figure 9.17. We don’t see anyobvious deviations from constant variance in this example.

Residuals in order of their data collection. A plot of the residuals in the order their corre-sponding auctions were observed is shown in Figure 9.18. Here we see no structure thatindicates a problem.

Residuals against each predictor variable. We consider a plot of the residuals against thecond new variable, the residuals against the stock photo variable, and the residuals againstthe wheels variable. These plots are shown in Figure 9.19. For the two-level condition vari-able, we are guaranteed not to see any remaining trend, and instead we are checking that thevariability doesn’t fluctuate across groups, which it does not. However, looking at the stock

20We would plug in 0 for cond new 1 for stock photo, and 2 for wheels into the equation, which would return$51.77, which is the total price we would expect for the auction.

21No. The model provides the average auction price we would expect, and the price for one auction to the nextwill continue to vary a bit (but less than what our prediction would be without the model).

9.4. MULTIPLE REGRESSION CASE STUDY: MARIO KART 369

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photo variable, we find that there is some difference in the variability of the residuals in the twogroups. Additionally, when we consider the residuals against the wheels variable, we see somepossible structure. There appears to be curvature in the residuals, indicating the relationshipis probably not linear.

As with the loans analysis, we would summarize diagnostics when reporting the model results.In the case of this auction data, we would report that there appears to be non-constant variance inthe stock photo variable and that there may be a nonlinear relationship between the total price andthe number of wheels included for an auction. This information would be important to buyers andsellers who may review the analysis, and omitting this information could be a setback to the verypeople who the model might assist.

Note: there are no exercises for this section.

370 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

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9.5. INTRODUCTION TO LOGISTIC REGRESSION 371

9.5 Introduction to logistic regression

In this section we introduce logistic regression as a tool for building models when there is acategorical response variable with two levels, e.g. yes and no. Logistic regression is a type ofgeneralized linear model (GLM) for response variables where regular multiple regression doesnot work very well. In particular, the response variable in these settings often takes a form whereresiduals look completely different from the normal distribution.

GLMs can be thought of as a two-stage modeling approach. We first model the responsevariable using a probability distribution, such as the binomial or Poisson distribution. Second, wemodel the parameter of the distribution using a collection of predictors and a special form of multipleregression. Ultimately, the application of a GLM will feel very similar to multiple regression, evenif some of the details are different.

9.5.1 Resume data

We will consider experiment data from a study that sought to understand the effect of race andsex on job application callback rates; details of the study and a link to the data set may be foundin Appendix B.9. To evaluate which factors were important, job postings were identified in Bostonand Chicago for the study, and researchers created many fake resumes to send off to these jobs tosee which would elicit a callback. The researchers enumerated important characteristics, such asyears of experience and education details, and they used these characteristics to randomly generatethe resumes. Finally, they randomly assigned a name to each resume, where the name would implythe applicant’s sex and race.

The first names that were used and randomly assigned in this experiment were selected so thatthey would predominantly be recognized as belonging to Black or White individuals; other raceswere not considered in this study. While no name would definitively be inferred as pertaining toa Black individual or to a White individual, the researchers conducted a survey to check for racialassociation of the names; names that did not pass this survey check were excluded from usage inthe experiment. You can find the full set of names that did pass the survey test and were ultimatelyused in the study in Figure 9.20. For example, Lakisha was a name that their survey indicatedwould be interpreted as a Black woman, while Greg was a name that would generally be interpretedto be associated with a White male.

first name race sex first name race sex first name race sexAisha black female Hakim black male Laurie white femaleAllison white female Jamal black male Leroy black maleAnne white female Jay white male Matthew white maleBrad white male Jermaine black male Meredith white femaleBrendan white male Jill white female Neil white maleBrett white male Kareem black male Rasheed black maleCarrie white female Keisha black female Sarah white femaleDarnell black male Kenya black female Tamika black femaleEbony black female Kristen white female Tanisha black femaleEmily white female Lakisha black female Todd white maleGeoffrey white male Latonya black female Tremayne black maleGreg white male Latoya black female Tyrone black male

Figure 9.20: List of all 36 unique names along with the commonly inferred raceand sex associated with these names.

The response variable of interest is whether or not there was a callback from the employer forthe applicant, and there were 8 attributes that were randomly assigned that we’ll consider, withspecial interest in the race and sex variables. Race and sex are protected classes in the UnitedStates, meaning they are not legally permitted factors for hiring or employment decisions. The fullset of attributes considered is provided in Figure 9.21.

372 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

variable description

callback Specifies whether the employer called the applicant following submission of theapplication for the job.

job city City where the job was located: Boston or Chicago.college degree An indicator for whether the resume listed a college degree.years experience Number of years of experience listed on the resume.honors Indicator for the resume listing some sort of honors, e.g. employee of the month.military Indicator for if the resume listed any military experience.email address Indicator for if the resume listed an email address for the applicant.race Race of the applicant, implied by their first name listed on the resume.sex Sex of the applicant (limited to only male and female in this study), implied

by the first name listed on the resume.

Figure 9.21: Descriptions for the callback variable along with 8 other variables inthe resume data set. Many of the variables are indicator variables, meaning theytake the value 1 if the specified characteristic is present and 0 otherwise.

All of the attributes listed on each resume were randomly assigned. This means that noattributes that might be favorable or detrimental to employment would favor one demographic overanother on these resumes. Importantly, due to the experimental nature of this study, we can infercausation between these variables and the callback rate, if the variable is statistically significant.Our analysis will allow us to compare the practical importance of each of the variables relative toeach other.

9.5.2 Modeling the probability of an event

Logistic regression is a generalized linear model where the outcome is a two-level categoricalvariable. The outcome, Yi, takes the value 1 (in our application, this represents a callback for theresume) with probability pi and the value 0 with probability 1− pi. Because each observation has aslightly different context, e.g. different education level or a different number of years of experience,the probability pi will differ for each observation. Ultimately, it is this probability that we modelin relation to the predictor variables: we will examine which resume characteristics correspond tohigher or lower callback rates.

NOTATION FOR A LOGISTIC REGRESSION MODEL

The outcome variable for a GLM is denoted by Yi, where the index i is used to representobservation i. In the resume application, Yi will be used to represent whether resume i receiveda callback (Yi = 1) or not (Yi = 0).

The predictor variables are represented as follows: x1,i is the value of variable 1 for observationi, x2,i is the value of variable 2 for observation i, and so on.

The logistic regression model relates the probability a resume would receive a callback (pi) tothe predictors x1,i, x2,i, …, xk,i through a framework much like that of multiple regression:

transformation(pi) = β0 + β1×1,i + β2×2,i + · · ·+ βkxk,i (9.30)

We want to choose a transformation in the equation that makes practical and mathematical sense.For example, we want a transformation that makes the range of possibilities on the left hand sideof the equation equal to the range of possibilities for the right hand side; if there was no transfor-mation for this equation, the left hand side could only take values between 0 and 1, but the righthand side could take values outside of this range. A common transformation for pi is the logittransformation, which may be written as

logit(pi) = loge

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)The logit transformation is shown in Figure 9.22. Below, we rewrite the equation relating Yi to its

9.5. INTRODUCTION TO LOGISTIC REGRESSION 373

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(pi

1− pi

)= β0 + β1×1,i + β2×2,i + · · ·+ βkxk,i

In our resume example, there are 8 predictor variables, so k = 8. While the precise choice of alogit function isn’t intuitive, it is based on theory that underpins generalized linear models, whichis beyond the scope of this book. Fortunately, once we fit a model using software, it will start tofeel like we’re back in the multiple regression context, even if the interpretation of the coefficients ismore complex.

EXAMPLE 9.31

We start by fitting a model with a single predictor: honors. This variable indicates whether theapplicant had any type of honors listed on their resume, such as employee of the month. Thefollowing logistic regression model was fit using statistical software:

log

(pi

1− pi

)= −2.4998 + 0.8668× honors

(a) If a resume is randomly selected from the study and it does not have any honors listed, what isthe probability resulted in a callback?(b) What would the probability be if the resume did list some honors?

(a) If a randomly chosen resume from those sent out is considered, and it does not list honors,then honors takes value 0 and the right side of the model equation equals -2.4998. Solving for pi:e−2.4998

1+e−2.4998 = 0.076. Just as we labeled a fitted value of yi with a “hat” in single-variable and multipleregression, we do the same for this probability: p̂i = 0.076.

(b) If the resume had listed some honors, then the right side of the model equation is −2.4998 +0.8668× 1 = −1.6330, which corresponds to a probability p̂i = 0.163.

Notice that we could examine -2.4998 and -1.6330 in Figure 9.22 to estimate the probability beforeformally calculating the value.

374 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

To convert from values on the logistic regression scale (e.g. -2.4998 and -1.6330 in Exam-ple 9.31), use the following formula, which is the result of solving for pi in the regression model:

pi =eβ0+β1×1,i+···+βkxk,i

1 + eβ0+β1×1,i+···+βkxk,i

As with most applied data problems, we substitute the point estimates for the parameters (the βi)so that we can make use of this formula. In Example 9.31, the probabilities were calculated as

e−2.4998

1 + e−2.4998= 0.076

e−2.4998+0.8668

1 + e−2.4998+0.8668= 0.163

While knowing whether a resume listed honors provides some signal when predicting whether or notthe employer would call, we would like to account for many different variables at once to understandhow each of the different resume characteristics affected the chance of a callback.

9.5.3 Building the logistic model with many variables

We used statistical software to fit the logistic regression model with all 8 predictors describedin Figure 9.21. Like multiple regression, the result may be presented in a summary table, which isshown in Figure 9.23. The structure of this table is almost identical to that of multiple regression;the only notable difference is that the p-values are calculated using the normal distribution ratherthan the t-distribution.

Estimate Std. Error z value Pr(>|z|)(Intercept) -2.6632 0.1820 -14.64 <0.0001job city: Chicago -0.4403 0.1142 -3.85 0.0001college degree -0.0666 0.1211 -0.55 0.5821years experience 0.0200 0.0102 1.96 0.0503honors 0.7694 0.1858 4.14 <0.0001military -0.3422 0.2157 -1.59 0.1127email address 0.2183 0.1133 1.93 0.0541race: white 0.4424 0.1080 4.10 <0.0001sex: male -0.1818 0.1376 -1.32 0.1863

Figure 9.23: Summary table for the full logistic regression model for the resumecallback example.

Just like multiple regression, we could trim some variables from the model. Here we’ll use astatistic called Akaike information criterion (AIC), which is an analog to how we used adjustedR-squared in multiple regression, and we look for models with a lower AIC through a backwardelimination strategy. After using this criteria, the college degree variable is eliminated, givingthe smaller model summarized in Figure 9.24, which is what we’ll rely on for the remainder of thissection.

Estimate Std. Error z value Pr(>|z|)(Intercept) -2.7162 0.1551 -17.51 <0.0001job city: Chicago -0.4364 0.1141 -3.83 0.0001years experience 0.0206 0.0102 2.02 0.0430honors 0.7634 0.1852 4.12 <0.0001military -0.3443 0.2157 -1.60 0.1105email address 0.2221 0.1130 1.97 0.0494race: white 0.4429 0.1080 4.10 <0.0001sex: male -0.1959 0.1352 -1.45 0.1473

Figure 9.24: Summary table for the logistic regression model for the resume call-back example, where variable selection has been performed using AIC.

9.5. INTRODUCTION TO LOGISTIC REGRESSION 375

EXAMPLE 9.32

The race variable had taken only two levels: black and white. Based on the model results, wasrace a meaningful factor for if a prospective employer would call back?

We see that the p-value for this coefficient is very small (very nearly zero), which implies that raceplayed a statistically significant role in whether a candidate received a callback. Additionally, wesee that the coefficient shown corresponds to the level of white, and it is positive. This positivecoefficient reflects a positive gain in callback rate for resumes where the candidate’s first nameimplied they were White. The data provide very strong evidence of racism by prospective employersthat favors resumes where the first name is typically interpreted to be White.

The coefficient of racewhite in the full model in Figure 9.23, is nearly identical to the modelshown in Figure 9.24. The predictors in this experiment were thoughtfully laid out so that thecoefficient estimates would typically not be much influenced by which other predictors were in themodel, which aligned with the motivation of the study to tease out which effects were important togetting a callback. In most observational data, it’s common for point estimates to change a little,and sometimes a lot, depending on which other variables are included in the model.

EXAMPLE 9.33

Use the model summarized in Figure 9.24 to estimate the probability of receiving a callback fora job in Chicago where the candidate lists 14 years experience, no honors, no military experience,includes an email address, and has a first name that implies they are a White male.

We can start by writing out the equation using the coefficients from the model, then we can add inthe corresponding values of each variable for this individual:

log

1− p

)= −2.7162− 0.4364× job cityChicago + 0.0206× years experience + 0.7634× honors

− 0.3443× military + 0.2221× email + 0.4429× racewhite − 0.1959× sexmale

= −2.7162− 0.4364× 1 + 0.0206× 14 + 0.7634× 0

− 0.3443× 0 + 0.2221× 1 + 0.4429× 1− 0.1959× 1

= −2.3955

We can now back-solve for p: the chance such an individual will receive a callback is about 8.35%.

EXAMPLE 9.34

Compute the probability of a callback for an individual with a name commonly inferred to be froma Black male but who otherwise has the same characteristics as the one described in Example 9.33.

We can complete the same steps for an individual with the same characteristics who is Black, wherethe only difference in the calculation is that the indicator variable racewhite will take a value of 0.Doing so yields a probability of 0.0553. Let’s compare the results with those of Example 9.33.

In practical terms, an individual perceived as White based on their first name would need to applyto 1

0.0835 ≈ 12 jobs on average to receive a callback, while an individual perceived as Black based ontheir first name would need to apply to 1

0.0553 ≈ 18 jobs on average to receive a callback. That is,applicants who are perceived as Black need to apply to 50% more employers to receive a callbackthan someone who is perceived as White based on their first name for jobs like those in the study.

What we’ve quantified in this section is alarming and disturbing. However, one aspect thatmakes this racism so difficult to address is that the experiment, as well-designed as it is, cannotsend us much signal about which employers are discriminating. It is only possible to say thatdiscrimination is happening, even if we cannot say which particular callbacks – or non-callbacks– represent discrimination. Finding strong evidence of racism for individual cases is a persistentchallenge in enforcing anti-discrimination laws.

376 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.5.4 Diagnostics for the callback rate model

LOGISTIC REGRESSION CONDITIONS

There are two key conditions for fitting a logistic regression model:

1. Each outcome Yi is independent of the other outcomes.

2. Each predictor xi is linearly related to logit(pi) if all other predictors are held constant.

The first logistic regression model condition – independence of the outcomes – is reasonable forthe experiment since characteristics of resumes were randomly assigned to the resumes that weresent out.

The second condition of the logistic regression model is not easily checked without a fairlysizable amount of data. Luckily, we have 4870 resume submissions in the data set! Let’s firstvisualize these data by plotting the true classification of the resumes against the model’s fittedprobabilities, as shown in Figure 9.25.

Predicted Probability

0.0 0.2 0.4 0.6 0.8 1.0

0 (No Callback)

1 (Callback)

Figure 9.25: The predicted probability that each of the 4870 resumes results in acallback. Noise (small, random vertical shifts) have been added to each point sopoints with nearly identical values aren’t plotted exactly on top of one another.

9.5. INTRODUCTION TO LOGISTIC REGRESSION 377

Predicted Probability

Trut

0.0 0.2 0.4 0.6 0.8 1.0

0 (No Callback)

0.2

0.4

0.6

0.8

1 (Callback)

● ●●

●●

●

Observations are bucketed,then we compute the observed probability in each bucket (y)against the average predicted probability (x)for each of the buckets with 95% confidence intervals.

Figure 9.26: The dashed line is within the confidence bound of the 95% confidenceintervals of each of the buckets, suggesting the logistic fit is reasonable.

We’d like to assess the quality of the model. For example, we might ask: if we look at resumesthat we modeled as having a 10% chance of getting a callback, do we find about 10% of them actuallyreceive a callback? We can check this for groups of the data by constructing a plot as follows:

1. Bucket the data into groups based on their predicted probabilities.

2. Compute the average predicted probability for each group.

3. Compute the observed probability for each group, along with a 95% confidence interval.

4. Plot the observed probabilities (with 95% confidence intervals) against the average predictedprobabilities for each group.

The points plotted should fall close to the line y = x, since the predicted probabilities should besimilar to the observed probabilities. We can use the confidence intervals to roughly gauge whetheranything might be amiss. Such a plot is shown in Figure 9.26.

Additional diagnostics may be created that are similar to those featured in Section 9.3. Forinstance, we could compute residuals as the observed outcome minus the expected outcome (ei =Yi − p̂i), and then we could create plots of these residuals against each predictor. We might alsocreate a plot like that in Figure 9.26 to better understand the deviations.

378 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.5.5 Exploring discrimination between groups of different sizes

Any form of discrimination is concerning, and this is why we decided it was so important todiscuss this topic using data. The resume study also only examined discrimination in a single aspect:whether a prospective employer would call a candidate who submitted their resume. There was a50% higher barrier for resumes simply when the candidate had a first name that was perceived tobe from a Black individual. It’s unlikely that discrimination would stop there.

EXAMPLE 9.35

Let’s consider a sex-imbalanced company that consists of 20% women and 80% men,22 and we’llsuppose that the company is very large, consisting of perhaps 20,000 employees. Suppose whensomeone goes up for promotion at this company, 5 of their colleagues are randomly chosen toprovide feedback on their work.

Now let’s imagine that 10% of the people in the company are prejudiced against the other sex.That is, 10% of men are prejudiced against women, and similarly, 10% of women are prejudicedagainst men.

Who is discriminated against more at the company, men or women?

Let’s suppose we took 100 men who have gone up for promotion in the past few years. For thesemen, 5 × 100 = 500 random colleagues will be tapped for their feedback, of which about 20% willbe women (100 women). Of these 100 women, 10 are expected to be biased against the man theyare reviewing. Then, of the 500 colleagues reviewing them, men will experience discrimination byabout 2% of their colleagues when they go up for promotion.

Let’s do a similar calculation for 100 women who have gone up for promotion in the last few years.They will also have 500 random colleagues providing feedback, of which about 400 (80%) will bemen. Of these 400 men, about 40 (10%) hold a bias against women. Of the 500 colleagues providingfeedback on the promotion packet for these women, 8% of the colleagues hold a bias against thewomen.

Example 9.35 highlights something profound: even in a hypothetical setting where each de-mographic has the same degree of prejudice against the other demographic, the smaller group ex-periences the negative effects more frequently. Additionally, if we would complete a handful ofexamples like the one above with different numbers, we’d learn that the greater the imbalance inthe population groups, the more the smaller group is disproportionately impacted.23

Of course, there are other considerable real-world omissions from the hypothetical example.For example, studies have found instances where people from an oppressed group also discriminateagainst others within their own oppressed group. As another example, there are also instances wherea majority group can be oppressed, with apartheid in South Africa being one such historic example.Ultimately, discrimination is complex, and there are many factors at play beyond the mathematicsproperty we observed in Example 9.35.

We close this book on this serious topic, and we hope it inspires you to think about the powerof reasoning with data. Whether it is with a formal statistical model or by using critical thinkingskills to structure a problem, we hope the ideas you have learned will help you do more and dobetter in life.

22A more thoughtful example would include non-binary individuals.23If a proportion p of a company are women and the rest of the company consists of men, then under the hypothetical

situation the ratio of rates of discrimination against women vs men would be given by 1−pp

; this ratio is always greater

than 1 when p < 0.5.

9.5. INTRODUCTION TO LOGISTIC REGRESSION 379

Exercises

9.15 Possum classification, Part I. The common brushtail possum of the Australia region is a bit cuterthan its distant cousin, the American opossum (see Figure 8.4 on page 307). We consider 104 brushtailpossums from two regions in Australia, where the possums may be considered a random sample fromthe population. The first region is Victoria, which is in the eastern half of Australia and traverses thesouthern coast. The second region consists of New South Wales and Queensland, which make up easternand northeastern Australia. We use logistic regression to differentiate between possums in these two regions.The outcome variable, called population, takes value 1 when a possum is from Victoria and 0 when it isfrom New South Wales or Queensland. We consider five predictors: sex male (an indicator for a possumbeing male), head length, skull width, total length, and tail length. Each variable is summarized ina histogram. The full logistic regression model and a reduced model after variable selection are summarizedin the table.

Fre

quen

sex_male

0(Female)

1(Male)

head_length (in mm)

Fre

quen

85 90 95 100

skull_width (in mm)

Fre

quen

50 55 60 65

total_length (in cm)

Fre

quen

75 80 85 90 95

tail_length (in cm)

Fre

quen

32 34 36 38 40 42

Fre

quen

0(Not Victoria)

1(Victoria)

population

Full Model Reduced Model

Estimate SE Z Pr(>|Z|) Estimate SE Z Pr(>|Z|)(Intercept) 39.2349 11.5368 3.40 0.0007 33.5095 9.9053 3.38 0.0007

sex male -1.2376 0.6662 -1.86 0.0632 -1.4207 0.6457 -2.20 0.0278head length -0.1601 0.1386 -1.16 0.2480skull width -0.2012 0.1327 -1.52 0.1294 -0.2787 0.1226 -2.27 0.0231total length 0.6488 0.1531 4.24 0.0000 0.5687 0.1322 4.30 0.0000

tail length -1.8708 0.3741 -5.00 0.0000 -1.8057 0.3599 -5.02 0.0000

(a) Examine each of the predictors. Are there any outliers that are likely to have a very large influence onthe logistic regression model?

(b) The summary table for the full model indicates that at least one variable should be eliminated whenusing the p-value approach for variable selection: head length. The second component of the tablesummarizes the reduced model following variable selection. Explain why the remaining estimates changebetween the two models.

380 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

9.16 Challenger disaster, Part I. On January 28, 1986, a routine launch was anticipated for the Challengerspace shuttle. Seventy-three seconds into the flight, disaster happened: the shuttle broke apart, killing allseven crew members on board. An investigation into the cause of the disaster focused on a critical sealcalled an O-ring, and it is believed that damage to these O-rings during a shuttle launch may be related tothe ambient temperature during the launch. The table below summarizes observational data on O-rings for23 shuttle missions, where the mission order is based on the temperature at the time of the launch. Tempgives the temperature in Fahrenheit, Damaged represents the number of damaged O- rings, and Undamagedrepresents the number of O-rings that were not damaged.

Shuttle Mission 1 2 3 4 5 6 7 8 9 10 11 12

Temperature 53 57 58 63 66 67 67 67 68 69 70 70Damaged 5 1 1 1 0 0 0 0 0 0 1 0Undamaged 1 5 5 5 6 6 6 6 6 6 5 6

Shuttle Mission 13 14 15 16 17 18 19 20 21 22 23

Temperature 70 70 72 73 75 75 76 76 78 79 81Damaged 1 0 0 0 0 1 0 0 0 0 0Undamaged 5 6 6 6 6 5 6 6 6 6 6

(a) Each column of the table above represents a different shuttle mission. Examine these data and describewhat you observe with respect to the relationship between temperatures and damaged O-rings.

(b) Failures have been coded as 1 for a damaged O-ring and 0 for an undamaged O-ring, and a logisticregression model was fit to these data. A summary of this model is given below. Describe the keycomponents of this summary table in words.

Estimate Std. Error z value Pr(>|z|)(Intercept) 11.6630 3.2963 3.54 0.0004

Temperature -0.2162 0.0532 -4.07 0.0000

(d) Based on the model, do you think concerns regarding O-rings are justified? Explain.

9.17 Possum classification, Part II. A logistic regression model was proposed for classifying commonbrushtail possums into their two regions in Exercise 9.15. The outcome variable took value 1 if the possumwas from Victoria and 0 otherwise.

Estimate SE Z Pr(>|Z|)(Intercept) 33.5095 9.9053 3.38 0.0007

sex male -1.4207 0.6457 -2.20 0.0278skull width -0.2787 0.1226 -2.27 0.0231total length 0.5687 0.1322 4.30 0.0000

tail length -1.8057 0.3599 -5.02 0.0000

(a) Write out the form of the model. Also identify which of the variables are positively associated whencontrolling for other variables.

(b) Suppose we see a brushtail possum at a zoo in the US, and a sign says the possum had been captured inthe wild in Australia, but it doesn’t say which part of Australia. However, the sign does indicate thatthe possum is male, its skull is about 63 mm wide, its tail is 37 cm long, and its total length is 83 cm.What is the reduced model’s computed probability that this possum is from Victoria? How confidentare you in the model’s accuracy of this probability calculation?

9.5. INTRODUCTION TO LOGISTIC REGRESSION 381

9.18 Challenger disaster, Part II. Exercise 9.16 introduced us to O-rings that were identified as a plausibleexplanation for the breakup of the Challenger space shuttle 73 seconds into takeoff in 1986. The investigationfound that the ambient temperature at the time of the shuttle launch was closely related to the damage ofO-rings, which are a critical component of the shuttle. See this earlier exercise if you would like to browsethe original data.

50 55 60 65 70 75 80

0.0

0.2

0.4

0.6

0.8

1.0P

roba

bilit

y of

dam

age

Temperature (Fahrenheit)

(a) The data provided in the previous exercise are shown in the plot. The logistic model fit to these datamay be written as

log

(p̂

1− p̂

)= 11.6630− 0.2162× Temperature

where p̂ is the model-estimated probability that an O-ring will become damaged. Use the model tocalculate the probability that an O-ring will become damaged at each of the following ambient tem-peratures: 51, 53, and 55 degrees Fahrenheit. The model-estimated probabilities for several additionalambient temperatures are provided below, where subscripts indicate the temperature:

p̂57 = 0.341 p̂59 = 0.251 p̂61 = 0.179 p̂63 = 0.124

p̂65 = 0.084 p̂67 = 0.056 p̂69 = 0.037 p̂71 = 0.024

(b) Add the model-estimated probabilities from part (a) on the plot, then connect these dots using a smoothcurve to represent the model-estimated probabilities.

(c) Describe any concerns you may have regarding applying logistic regression in this application, and noteany assumptions that are required to accept the model’s validity.

382 CHAPTER 9. MULTIPLE AND LOGISTIC REGRESSION

Chapter exercises

9.19 Multiple regression fact checking. Determine which of the following statements are true and false.For each statement that is false, explain why it is false.

(a) If predictors are collinear, then removing one variable will have no influence on the point estimate ofanother variable’s coefficient.

(b) Suppose a numerical variable x has a coefficient of b1 = 2.5 in the multiple regression model. Supposealso that the first observation has x1 = 7.2, the second observation has a value of x1 = 8.2, and thesetwo observations have the same values for all other predictors. Then the predicted value of the secondobservation will be 2.5 higher than the prediction of the first observation based on the multiple regressionmodel.

(c) If a regression model’s first variable has a coefficient of b1 = 5.7, then if we are able to influence thedata so that an observation will have its x1 be 1 larger than it would otherwise, the value y1 for thisobservation would increase by 5.7.

(d) Suppose we fit a multiple regression model based on a data set of 472 observations. We also noticethat the distribution of the residuals includes some skew but does not include any particularly extremeoutliers. Because the residuals are not nearly normal, we should not use this model and require moreadvanced methods to model these data.

9.20 Logistic regression fact checking. Determine which of the following statements are true and false.For each statement that is false, explain why it is false.

(a) Suppose we consider the first two observations based on a logistic regression model, where the firstvariable in observation 1 takes a value of x1 = 6 and observation 2 has x1 = 4. Suppose we realized wemade an error for these two observations, and the first observation was actually x1 = 7 (instead of 6)and the second observation actually had x1 = 5 (instead of 4). Then the predicted probability from thelogistic regression model would increase the same amount for each observation after we correct thesevariables.

(b) When using a logistic regression model, it is impossible for the model to predict a probability that isnegative or a probability that is greater than 1.

(c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic re-gression model need not be independent.

(d) When fitting logistic regression, we typically complete model selection using adjusted R2.

9.21 Spam filtering, Part I. Spam filters are built on principles similar to those used in logistic regression.We fit a probability that each message is spam or not spam. We have several email variables for thisproblem: to multiple, cc, attach, dollar, winner, inherit, password, format, re subj, exclaim subj,and sent email. We won’t describe what each variable means here for the sake of brevity, but each is eithera numerical or indicator variable.

(a) For variable selection, we fit the full model, which includes all variables, and then we also fit each modelwhere we’ve dropped exactly one of the variables. In each of these reduced models, the AIC value forthe model is reported below. Based on these results, which variable, if any, should we drop as part ofmodel selection? Explain.

Variable Dropped AIC

None Dropped 1863.50to multiple 2023.50cc 1863.18attach 1871.89dollar 1879.70winner 1885.03inherit 1865.55password 1879.31format 2008.85re subj 1904.60exclaim subj 1862.76sent email 1958.18

See the next page for part (b).

9.5. INTRODUCTION TO LOGISTIC REGRESSION 383

(b) Consider the following model selection stage. Here again we’ve computed the AIC for each leave-one-variable-out model. Based on the results, which variable, if any, should we drop as part of modelselection? Explain.

Variable Dropped AIC

None Dropped 1862.41to multiple 2019.55attach 1871.17dollar 1877.73winner 1884.95inherit 1864.52password 1878.19format 2007.45re subj 1902.94sent email 1957.56

9.22 Movie returns, Part II. The student from Exercise 9.14 analyzed return-on-investment (ROI) formovies based on release year and genre of movies. The plots below show the predicted ROI vs. actual ROIfor each of the genres separately. Do these figures support the comment in the FiveThirtyEight.com articlethat states, “The return-on-investment potential for horror movies is absurd.” Note that the x-axis rangevaries for each plot.

Drama Horror

Action Adventure Comedy

0 20 40 0 25 50 75 100

0 5 10 15 0 5 10 15 0 10 20 30 40

2.5

5.0

7.5

10.0

2.5

5.0

7.5

10.0

Actual ROI

Pre

dict

ed R

9.23 Spam filtering, Part II. In Exercise 9.21, we encountered a data set where we applied logistic regressionto aid in spam classification for individual emails. In this exercise, we’ve taken a small set of these variablesand fit a formal model with the following output:

Estimate Std. Error z value Pr(>|z|)(Intercept) -0.8124 0.0870 -9.34 0.0000to multiple -2.6351 0.3036 -8.68 0.0000

winner 1.6272 0.3185 5.11 0.0000format -1.5881 0.1196 -13.28 0.0000re subj -3.0467 0.3625 -8.40 0.0000

(a) Write down the model using the coefficients from the model fit.

(b) Suppose we have an observation where to multiple = 0, winner = 1, format = 0, and re subj = 0.What is the predicted probability that this message is spam?

(c) Put yourself in the shoes of a data scientist working on a spam filter. For a given message, how highmust the probability a message is spam be before you think it would be reasonable to put it in a spambox(which the user is unlikely to check)? What tradeoffs might you consider? Any ideas about how youmight make your spam-filtering system even better from the perspective of someone using your emailservice?

384

Appendix A

Exercise solutions

1 Introduction to data

1.1 (a) Treatment: 10/43 = 0.23→ 23%.

(b) Control: 2/46 = 0.04 → 4%. (c) A higher per-

centage of patients in the treatment group were pain

free 24 hours after receiving acupuncture. (d) It is

possible that the observed difference between the two

group percentages is due to chance.

1.3 (a) “Is there an association between air pol-

lution exposure and preterm births?” (b) 143,196

births in Southern California between 1989 and 1993.

ide, ozone, and particulate matter less than 10µg/m3

(PM10) collected at air-quality-monitoring stations

as well as length of gestation. Continuous numerical

variables.

1.5 (a) “Does explicitly telling children not to cheat

affect their likelihood to cheat?”. (b) 160 children

between the ages of 5 and 15. (c) Four variables: (1)

age (numerical, continuous), (2) sex (categorical), (3)

whether they were an only child or not (categorical),

(4) whether they cheated or not (categorical).

1.7 Explanatory: acupuncture or not. Response: if

the patient was pain free or not.

1.9 (a) 50×3 = 150. (b) Four continuous numerical

variables: sepal length, sepal width, petal length, and

petal width. (c) One categorical variable, species,

with three levels: setosa, versicolor, and virginica.

1.11 (a) Airport ownership status (public/private),

airport usage status (public/private), latitude, and

longitude. (b) Airport ownership status: categori-

cal, not ordinal. Airport usage status: categorical,

not ordinal. Latitude: numerical, continuous. Lon-

gitude: numerical, continuous.

1.13 (a) Population: all births, sample: 143,196

births between 1989 and 1993 in Southern Califor-

nia. (b) If births in this time span at the geography

can be considered to be representative of all births,

then the results are generalizable to the population

of Southern California. However, since the study is

observational the findings cannot be used to establish

causal relationships.

1.15 (a) Population: all asthma patients aged 18-69

who rely on medication for asthma treatment. Sam-

ple: 600 such patients. (b) If the patients in this

sample, who are likely not randomly sampled, can be

considered to be representative of all asthma patients

aged 18-69 who rely on medication for asthma treat-

ment, then the results are generalizable to the pop-

ulation defined above. Additionally, since the study

is experimental, the findings can be used to establish

causal relationships.

1.17 (a) Observation. (b) Variable. (c) Sample

statistic (mean). (d) Population parameter (mean).

1.19 (a) Observational. (b) Use stratified sampling

to randomly sample a fixed number of students, say

10, from each section for a total sample size of 40

students.

1.21 (a) Positive, non-linear, somewhat strong.

Countries in which a higher percentage of the popu-

lation have access to the internet also tend to have

higher average life expectancies, however rise in life

expectancy trails off before around 80 years old.

(b) Observational. (c) Wealth: countries with indi-

viduals who can widely afford the internet can prob-

ably also afford basic medical care. (Note: Answers

may vary.)

1.23 (a) Simple random sampling is okay. In fact,

it’s rare for simple random sampling to not be a

reasonable sampling method! (b) The student opin-

ions may vary by field of study, so the stratifying by

this variable makes sense and would be reasonable.

have more similar opinions, and we want clusters to

be diverse with respect to the outcome of interest,

so this would not be a good approach. (Additional

thought: the clusters in this case may also have very

different numbers of people, which can also create

unexpected sample sizes.)

385

1.25 (a) The cases are 200 randomly sampled men

and women. (b) The response variable is attitude to-

wards a fictional microwave oven. (c) The explana-

tory variable is dispositional attitude. (d) Yes, the

cases are sampled randomly. (e) This is an observa-

tional study since there is no random assignment to

treatments. (f) No, we cannot establish a causal link

between the explanatory and response variables since

the study is observational. (g) Yes, the results of the

study can be generalized to the population at large

since the sample is random.

1.27 (a) Simple random sample. Non-response bias,

if only those people who have strong opinions about

the survey responds his sample may not be repre-

sentative of the population. (b) Convenience sample.

Under coverage bias, his sample may not be represen-

tative of the population since it consists only of his

friends. It is also possible that the study will have

non-response bias if some choose to not bring back

the survey. (c) Convenience sample. This will have

a similar issues to handing out surveys to friends.

(d) Multi-stage sampling. If the classes are similar

to each other with respect to student composition

this approach should not introduce bias, other than

potential non-response bias.

1.29 (a) Exam performance. (b) Light level: fluo-

rescent overhead lighting, yellow overhead lighting,

no overhead lighting (only desk lamps). (c) Sex:

man, woman.

1.31 (a) Exam performance. (b) Light level (over-

head lighting, yellow overhead lighting, no overhead

lighting) and noise level (no noise, construction noise,

and human chatter noise). (c) Since the researchers

want to ensure equal gender representation, sex will

be a blocking variable.

1.33 Need randomization and blinding. One possi-

ble outline: (1) Prepare two cups for each participant,

one containing regular Coke and the other contain-

ing Diet Coke. Make sure the cups are identical and

contain equal amounts of soda. Label the cups A

(regular) and B (diet). (Be sure to randomize A and

B for each trial!) (2) Give each participant the two

cups, one cup at a time, in random order, and ask

the participant to record a value that indicates how

much she liked the beverage. Be sure that neither

the participant nor the person handing out the cups

knows the identity of the beverage to make this a

double- blind experiment. (Answers may vary.)

1.35 (a) Observational study. (b) Dog: Lucy. Cat:

Luna. (c) Oliver and Lily. (d) Positive, as the popu-

larity of a name for dogs increases, so does the pop-

ularity of that name for cats.

1.37 (a) Experiment. (b) Treatment: 25 grams of

chia seeds twice a day, control: placebo. (c) Yes,

gender. (d) Yes, single blind since the patients were

blinded to the treatment they received. (e) Since this

is an experiment, we can make a causal statement.

However, since the sample is not random, the causal

statement cannot be generalized to the population at

large.

1.39 (a) Non-responders may have a different re-

sponse to this question, e.g. parents who returned

the surveys likely don’t have difficulty spending time

with their children. (b) It is unlikely that the women

who were reached at the same address 3 years later

are a random sample. These missing responders are

probably renters (as opposed to homeowners) which

means that they might be in a lower socio- economic

status than the respondents. (c) There is no control

group in this study, this is an observational study,

and there may be confounding variables, e.g. these

people may go running because they are generally

healthier and/or do other exercises.

1.41 (a) Randomized controlled experiment.

(b) Explanatory: treatment group (categorical, with

3 levels). Response variable: Psychological well-

being. (c) No, because the participants were volun-

teers. (d) Yes, because it was an experiment. (e) The

statement should say “evidence” instead of “proof”.

1.43 (a) County, state, driver’s race, whether the

car was searched or not, and whether the driver was

arrested or not. (b) All categorical, non-ordinal.

Explanatory: race of the driver.

2 Summarizing data

2.1 (a) Positive association: mammals with longer

gestation periods tend to live longer as well. (b) As-

sociation would still be positive. (c) No, they are not

independent. See part (a).

2.3 The graph below shows a ramp up period. There

may also be a period of exponential growth at the

start before the size of the petri dish becomes a fac-

tor in slowing growth.

time

num

ber

of b

acte

ria c

ells

2.5 (a) Population mean, µ2007 = 52; sample mean,

x̄2008 = 58. (b) Population mean, µ2001 = 3.37; sam-

ple mean, x̄2012 = 3.59.

2.7 Any 10 employees whose average number of days

off is between the minimum and the mean number of

days off for the entire workforce at this plant.

386 APPENDIX A. EXERCISE SOLUTIONS

2.9 (a) Dist 2 has a higher mean since 20 > 13, and a

higher standard deviation since 20 is further from the

rest of the data than 13. (b) Dist 1 has a higher mean

since −20 > −40, and Dist 2 has a higher standard

deviation since -40 is farther away from the rest of

the data than -20. (c) Dist 2 has a higher mean since

all values in this distribution are higher than those

in Dist 1, but both distribution have the same stan-

dard deviation since they are equally variable around

their respective means. (d) Both distributions have

the same mean since they’re both centered at 300,

but Dist 2 has a higher standard deviation since the

observations are farther from the mean than in Dist 1.

2.11 (a) About 30. (b) Since the distribution is

right skewed the mean is higher than the median.

IQR: about 20. (d) Values that are considered to be

unusually low or high lie more than 1.5×IQR away

from the quartiles. Upper fence: Q3 + 1.5 × IQR =

37.5 + 1.5× 20 = 67.5; Lower fence: Q1 – 1.5 × IQR

= 17.5− 1.5× 20 = −12.5; The lowest AQI recorded

is not lower than 5 and the highest AQI recorded is

not higher than 65, which are both within the fences.

Therefore none of the days in this sample would be

considered to have an unusually low or high AQI.

2.13 The histogram shows that the distribution is

bimodal, which is not apparent in the box plot. The

box plot makes it easy to identify more precise values

of observations outside of the whiskers.

2.15 (a) The distribution of number of pets per

household is likely right skewed as there is a natural

boundary at 0 and only a few people have many pets.

Therefore the center would be best described by the

median, and variability would be best described by

the IQR. (b) The distribution of number of distance

to work is likely right skewed as there is a natural

boundary at 0 and only a few people live a very long

distance from work. Therefore the center would be

best described by the median, and variability would

be best described by the IQR. (c) The distribution

of heights of males is likely symmetric. Therefore

the center would be best described by the mean, and

variability would be best described by the standard

deviation.

2.17 (a) The median is a much better measure of

the typical amount earned by these 42 people. The

mean is much higher than the income of 40 of the

42 people. This is because the mean is an arithmetic

average and gets affected by the two extreme obser-

vations. The median does not get effected as much

since it is robust to outliers. (b) The IQR is a much

better measure of variability in the amounts earned

by nearly all of the 42 people. The standard devi-

ation gets affected greatly by the two high salaries,

but the IQR is robust to these extreme observations.

2.19 (a) The distribution is unimodal and symmet-

ric with a mean of about 25 minutes and a standard

deviation of about 5 minutes. There does not ap-

pear to be any counties with unusually high or low

mean travel times. Since the distribution is already

unimodal and symmetric, a log transformation is not

necessary. (b) Answers will vary. There are pockets

of longer travel time around DC, Southeastern NY,

Chicago, Minneapolis, Los Angeles, and many other

big cities. There is also a large section of shorter

average commute times that overlap with farmland

in the Midwest. Many farmers’ homes are adjacent

to their farmland, so their commute would be brief,

which may explain why the average commute time

for these counties is relatively low.

2.21 (a) We see the order of the categories and the

relative frequencies in the bar plot. (b) There are no

features that are apparent in the pie chart but not

in the bar plot. (c) We usually prefer to use a bar

plot as we can also see the relative frequencies of the

categories in this graph.

2.23 The vertical locations at which the ideologi-

cal groups break into the Yes, No, and Not Sure

categories differ, which indicates that likelihood of

supporting the DREAM act varies by political ide-

ology. This suggests that the two variables may be

dependent.

387

2.25 (a) (i) False. Instead of comparing counts, we

should compare percentages of people in each group

who suffered cardiovascular problems. (ii) True.

(iii) False. Association does not imply causation. We

cannot infer a causal relationship based on an obser-

vational study. The difference from part (ii) is subtle.

(iv) True.

(b) Proportion of all patients who had cardiovascular

problems: 7,979227,571

≈ 0.035

rosiglitazone group, if having cardiovascular prob-

lems and treatment were independent, can be cal-

culated as the number of patients in that group mul-

tiplied by the overall cardiovascular problem rate in

the study: 67, 593 ∗ 7,979227,571

≈ 2370.

(d) (i) H0: The treatment and cardiovascular prob-

lems are independent. They have no relation-

ship, and the difference in incidence rates between

the rosiglitazone and pioglitazone groups is due to

chance. HA: The treatment and cardiovascular prob-

lems are not independent. The difference in the in-

cidence rates between the rosiglitazone and pioglita-

zone groups is not due to chance and rosiglitazone

is associated with an increased risk of serious car-

diovascular problems. (ii) A higher number of pa-

tients with cardiovascular problems than expected

under the assumption of independence would provide

support for the alternative hypothesis as this would

suggest that rosiglitazone increases the risk of such

problems. (iii) In the actual study, we observed 2,593

cardiovascular events in the rosiglitazone group. In

the 1,000 simulations under the independence model,

we observed somewhat less than 2,593 in every single

simulation, which suggests that the actual results did

not come from the independence model. That is, the

variables do not appear to be independent, and we

reject the independence model in favor of the alterna-

tive. The study’s results provide convincing evidence

that rosiglitazone is associated with an increased risk

of cardiovascular problems.

2.27 (a) Decrease: the new score is smaller than

the mean of the 24 previous scores. (b) Calculate a

weighted mean. Use a weight of 24 for the old mean

and 1 for the new mean: (24×74+1×64)/(24+1) =

73.6. (c) The new score is more than 1 standard de-

viation away from the previous mean, so increase.

2.29 No, we would expect this distribution to be

right skewed. There are two reasons for this:

(1) there is a natural boundary at 0 (it is not possible

to watch less than 0 hours of TV), (2) the standard

deviation of the distribution is very large compared

to the mean.

2.31 The distribution of ages of best actress win-

ners are right skewed with a median around 30 years.

The distribution of ages of best actor winners is also

right skewed, though less so, with a median around

40 years. The difference between the peaks of these

distributions suggest that best actress winners are

typically younger than best actor winners. The ages

of best actress winners are more variable than the

ages of best actor winners. There are potential out-

liers on the higher end of both of the distributions.

2.33

●

60 70 80 90Scores

3 Probability

3.1 (a) False. These are independent trials.

(b) False. There are red face cards. (c) True. A

card cannot be both a face card and an ace.

3.3 (a) 10 tosses. Fewer tosses mean more variabil-

ity in the sample fraction of heads, meaning there’s a

better chance of getting at least 60% heads. (b) 100

tosses. More flips means the observed proportion

of heads would often be closer to the average, 0.50,

and therefore also above 0.40. (c) 100 tosses. With

more flips, the observed proportion of heads would

often be closer to the average, 0.50. (d) 10 tosses.

Fewer flips would increase variability in the fraction

of tosses that are heads.

3.5 (a) 0.510 = 0.00098. (b) 0.510 = 0.00098.

3.7 (a) No, there are voters who are both indepen-

dent and swing voters.

(b)

not. Since 35% of voters are Independents and 11%

are both Independent and swing voters, the other

24% must not be swing voters. (d) 0.47. (e) 0.53.

(f) P(Independent) × P(swing) = 0.35×0.23 = 0.08,

which does not equal P(Independent and swing) =

0.11, so the events are dependent.

388 APPENDIX A. EXERCISE SOLUTIONS

3.9 (a) If the class is not graded on a curve, they

are independent. If graded on a curve, then neither

independent nor disjoint – unless the instructor will

only give one A, which is a situation we will ignore

in parts (b) and (c). (b) They are probably not in-

dependent: if you study together, your study habits

would be related, which suggests your course perfor-

mances are also related. (c) No. See the answer to

part (a) when the course is not graded on a curve.

More generally: if two things are unrelated (indepen-

dent), then one occurring does not preclude the other

from occurring.

3.11 (a) 0.16 + 0.09 = 0.25. (b) 0.17 + 0.09 = 0.26.

band and wife are independent: 0.25× 0.26 = 0.065.

You might also notice we actually made a second as-

sumption: that the decision to get married is unre-

lated to education level. (d) The husband/wife in-

dependence assumption is probably not reasonable,

because people often marry another person with a

comparable level of education. We will leave it to

you to think about whether the second assumption

noted in part (c) is reasonable.

3.13 (a) No, but we could if A and B are indepen-

dent. (b-i) 0.21. (b-ii) 0.79. (b-iii) 0.3. (c) No, be-

cause 0.1 6= 0.21, where 0.21 was the value computed

under independence from part (a). (d) 0.143.

3.15 (a) No, 0.18 of respondents fall into this combi-

nation. (b) 0.60+0.20−0.18 = 0.62. (c) 0.18/0.20 =

0.9. (d) 0.11/0.33 ≈ 0.33. (e) No, otherwise

the answers to (c) and (d) would be the same.

(f) 0.06/0.34 ≈ 0.18.

3.17 (a) No. There are 6 females who like Five Guys

Burgers. (b) 162/248 = 0.65. (c) 181/252 = 0.72.

(d) Under the assumption of a dating choices be-

ing independent of hamburger preference, which on

the surface seems reasonable: 0.65 × 0.72 = 0.468.

(e) (252 + 6− 1)/500 = 0.514.

3.19 (a)Can construct

box plots?Passed?

yes, 0.8

Yes, 0.860.8*0.86 = 0.688

No, 0.140.8*0.14 = 0.112

no, 0.2

Yes, 0.650.2*0.65 = 0.13

No, 0.350.2*0.35 = 0.07

(b) 0.84

3.21 0.0714. Even when a patient tests positive forlupus, there is only a 7.14% chance that he actuallyhas lupus. House may be right.

Lupus? Result

yes, 0.02

positive, 0.980.02*0.98 = 0.0196

negative, 0.020.02*0.02 = 0.0004

no, 0.98

positive, 0.260.98*0.26 = 0.2548

negative, 0.740.98*0.74 = 0.7252

3.23 (a) 0.3. (b) 0.3. (c) 0.3. (d) 0.3 × 0.3 = 0.09.

(e) Yes, the population that is being sampled from is

identical in each draw.

3.25 (a) 2/9 ≈ 0.22. (b) 3/9 ≈ 0.33. (c) 310× 2

9≈

0.067. (d) No, e.g. in this exercise, removing one chip

meaningfully changes the probability of what might

be drawn next.

3.27 P (1leggings, 2jeans, 3jeans) = 524× 7

23× 6

22=

0.0173. However, the person with leggings could have

come 2nd or 3rd, and these each have this same prob-

ability, so 3× 0.0173 = 0.0519.

3.29 (a) 13. (b) No, these 27 students are not a ran-

dom sample from the university’s student population.

For example, it might be argued that the proportion

of smokers among students who go to the gym at 9

am on a Saturday morning would be lower than the

proportion of smokers in the university as a whole.

3.31 (a) E(X) = 3.59. SD(X) = 9.64. (b) E(X) =

-1.41. SD(X) = 9.64. (c) No, the expected net profit

is negative, so on average you expect to lose money.

3.33 5% increase in value.

3.35 E = -0.0526. SD = 0.9986.

3.37 Approximate answers are OK.

(a) (29 + 32)/144 = 0.42. (b) 21/144 = 0.15.

3.39 (a) Invalid. Sum is greater than 1. (b) Valid.

Probabilities are between 0 and 1, and they sum to

1. In this class, every student gets a C. (c) Invalid.

Sum is less than 1. (d) Invalid. There is a negative

probability. (e) Valid. Probabilities are between 0

and 1, and they sum to 1. (f) Invalid. There is a

negative probability.

3.41 0.8247.HIV? Result

yes, 0.259

positive, 0.9970.259*0.997 = 0.2582

negative, 0.0030.259*0.003 = 0.0008

no, 0.741

positive, 0.0740.741*0.074 = 0.0548

negative, 0.9260.741*0.926 = 0.6862

3.43 (a) E = $3.90. SD = $0.34.

(b) E = $27.30. SD = $0.89.

3.45 V ar(X1+X2

)= V ar

(X12

+ X22

)= V ar(X1)

22 + V ar(X2)

= σ2

4+ σ2

= σ2/2

3.47 V ar(X1+X2+···+Xn

)= V ar

(X1n

+ X2n

+ · · ·+ Xnn

)= V ar(X1)

n2 + V ar(X2)

n2 + · · ·+ V ar(Xn)

= σ2

n2 + σ2

n2 + · · ·+ σ2

n2 (there are n of these terms)

= nσ2

= σ2/n

389

4 Distributions of random variables

4.1 (a) 8.85%. (b) 6.94%. (c) 58.86%. (d) 4.56%.

(a)−1.35 0

(b)0 1.48

(d)−2 0 2

4.3 (a) Verbal: N(µ = 151, σ = 7), Quant: N(µ =

153, σ = 7.67). (b) ZV R = 1.29, ZQR = 0.52.

Z = 1.29

Z = 0.52

mean on the Verbal Reasoning section and 0.52 stan-

dard deviations above the mean on the Quantita-

tive Reasoning section. (d) She did better on the

Verbal Reasoning section since her Z-score on that

section was higher. (e) PercV R = 0.9007 ≈ 90%,

PercQR = 0.6990 ≈ 70%. (f) 100%−90% = 10% did

better than her on VR, and 100%− 70% = 30% did

better than her on QR. (g) We cannot compare the

raw scores since they are on different scales. Com-

paring her percentile scores is more appropriate when

comparing her performance to others. (h) Answer to

part (b) would not change as Z-scores can be calcu-

lated for distributions that are not normal. However,

we could not answer parts (d)-(f) since we cannot use

the normal probability table to calculate probabilities

and percentiles without a normal model.

4.5 (a) Z = 0.84, which corresponds to approxi-

mately 159 on QR. (b) Z = −0.52, which corresponds

to approximately 147 on VR.

4.7 (a) Z = 1.2, P (Z > 1.2) = 0.1151.

(b) Z = −1.28→ 70.6◦F or colder.

4.9 (a) N(25, 2.78). (b) Z = 1.08, P (Z > 1.08) =

0.1401. (c) The answers are very close because only

the units were changed. (The only reason why they

differ at all because 28◦C is 82.4◦F, not precisely

83◦F.) (d) Since IQR = Q3 − Q1, we first need to

find Q3 and Q1 and take the difference between the

two. Remember that Q3 is the 75th and Q1 is the

25th percentile of a distribution. Q1 = 23.13, Q3 =

26.86, IQR = 26. 86 – 23.13 = 3.73.

4.11 (a) No. The cards are not independent. For

example, if the first card is an ace of clubs, that im-

plies the second card cannot be an ace of clubs. Ad-

ditionally, there are many possible categories, which

would need to be simplified. (b) No. There are six

events under consideration. The Bernoulli distribu-

tion allows for only two events or categories. Note

that rolling a die could be a Bernoulli trial if we sim-

ply to two events, e.g. rolling a 6 and not rolling a 6,

though specifying such details would be necessary.

4.13 (a) 0.8752×0.125 = 0.096. (b) µ = 8, σ = 7.48.

4.15 If p is the probability of a success, then the

mean of a Bernoulli random variable X is given by

µ = E[X] = P (X = 0)× 0 + P (X = 1)× 1

= (1− p)× 0 + p× 1 = 0 + p = p

4.17 (a) Binomial conditions are met: (1) Indepen-

dent trials: In a random sample, whether or not one

18-20 year old has consumed alcohol does not de-

pend on whether or not another one has. (2) Fixed

number of trials: n = 10. (3) Only two outcomes

at each trial: Consumed or did not consume alcohol.

(4) Probability of a success is the same for each trial:

p = 0.697. (b) 0.203. (c) 0.203. (d) 0.167. (e) 0.997.

4.19 (a) µ34.85, σ = 3.25 (b) Z = 45−34.853.25

= 3.12.

45 is more than 3 standard deviations away from the

mean, we can assume that it is an unusual observa-

tion. Therefore yes, we would be surprised. (c) Using

the normal approximation, 0.0009. With 0.5 correc-

tion, 0.0015.

4.21 (a) 1− 0.753 = 0.5781. (b) 0.1406. (c) 0.4219.

(d) 1− 0.253 = 0.9844.

4.23 (a) Geometric distribution: 0.109. (b) Bino-

mial: 0.219. (c) Binomial: 0.137. (d) 1 − 0.8756 =

0.551. (e) Geometric: 0.084. (f) Using a binomial

distribution with n = 6 and p = 0.75, we see that

µ = 4.5, σ = 1.06, and Z = 2.36. Since this is not

within 2 SD, it may be considered unusual.

4.25 (a)Anna

1/5 ×Ben

1/4 ×Carl

1/3 ×Damian

1/2 ×Eddy

1/1 =

1/5! = 1/120. (b) Since the probabilities must add

to 1, there must be 5! = 120 possible orderings.

4.27 (a) 0.0804. (b) 0.0322. (c) 0.0193.

4.29 (a) Negative binomial with n = 4 and p = 0.55,

where a success is defined here as a female student.

The negative binomial setting is appropriate since

the last trial is fixed but the order of the first 3 tri-

als is unknown. (b) 0.1838. (c)(

)= 3. (d) In the

binomial model there are no restrictions on the out-

come of the last trial. In the negative binomial model

the last trial is fixed. Therefore we are interested in

the number of ways of orderings of the other k − 1

successes in the first n− 1 trials.

390 APPENDIX A. EXERCISE SOLUTIONS

4.31 (a) Poisson with λ = 75. (b) µ = λ = 75,

σ =√λ = 8.66. (c) Z = −1.73. Since 60 is within

2 standard deviations of the mean, it would not gen-

erally be considered unusual. Note that we often use

this rule of thumb even when the normal model does

not apply. (d) Using Poisson with λ = 75: 0.0402.

4.33 (a) λk×e−λk!

= 6.55×e−6.5

5!= 0.1454

(b) The probability will come to 0.0015 + 0.0098 +

0.0318 = 0.0431 (0.0430 if no rounding error).

meaning people are coming in small clusters. That

is, if one person arrives, there’s a chance that they

brought one or more other people in their vehicle.

This means individuals (the people) are not indepen-

dent, even if the car arrivals are independent, and this

breaks a core assumption for the Poisson distribution.

That is, the number of people visiting between 2pm

and 3pm would not follow a Poisson distribution.

4.35 0 wins (-$3): 0.1458. 1 win (-$1): 0.3936. 2

wins (+$1): 0.3543. 3 wins (+$3): 0.1063.

4.37 Want to find the probability that there will

be 1,786 or more enrollees. Using the normal ap-

proximation, with µ = np = 2, 500 × 0.7 = 1750

and σ =√np(1− p) =

√2, 500× 0.7× 0.3 ≈ 23,

Z = 1.61, and P (Z > 1.61) = 0.0537. With a 0.5

correction: 0.0559.

4.39 (a) Z = 0.67. (b) µ = $1650, x = $1800.

→ σ = $223.88.

4.41 (a) (1−0.471)2×0.471 = 0.1318. (b) 0.4713 =

0.1045. (c) µ = 1/0.471 = 2.12, σ =√

2.38 = 1.54.

(d) µ = 1/0.30 = 3.33, σ = 2.79. (e) When p is

smaller, the event is rarer, meaning the expected

number of trials before a success and the standard

deviation of the waiting time are higher.

4.43 Z = 1.56, P (Z > 1.56) = 0.0594, i.e. 6%.

4.45 (a) Z = 0.73, P (Z > 0.73) = 0.2327. (b) If you

are bidding on only one auction and set a low maxi-

mum bid price, someone will probably outbid you. If

you set a high maximum bid price, you may win the

auction but pay more than is necessary. If bidding

on more than one auction, and you set your maxi-

mum bid price very low, you probably won’t win any

of the auctions. However, if the maximum bid price

is even modestly high, you are likely to win multiple

auctions. (c) An answer roughly equal to the 10th

percentile would be reasonable. Regrettably, no per-

centile cutoff point guarantees beyond any possible

event that you win at least one auction. However, you

may pick a higher percentile if you want to be more

sure of winning an auction. (d) Answers will vary a

little but should correspond to the answer in part (c).

We use the 10th percentile: Z = −1.28→ $69.80.

4.47 (a) Z = 3.5, upper tail is 0.0002. (More precise

value: 0.000233, but we’ll use 0.0002 for the calcula-

tions here.)

(b) 0.0002× 2000 = 0.4. We would expect about 0.4

10 year olds who are 76 inches or taller to show up.

(c)(

20000

)(0.0002)0(1− 0.0002)2000 = 0.67029.

(d) 0.40×e−0.4

0!= 1×e−0.4

1= 0.67032.

5 Foundations for inference

5.1 (a) Mean. Each student reports a numerical

value: a number of hours. (b) Mean. Each student

reports a number, which is a percentage, and we can

average over these percentages. (c) Proportion. Each

student reports Yes or No, so this is a categorical

variable and we use a proportion. (d) Mean. Each

student reports a number, which is a percentage like

in part (b). (e) Proportion. Each student reports

whether or not s/he expects to get a job, so this is a

categorical variable and we use a proportion.

5.3 (a) The sample is from all computer chips manu-

factured at the factory during the week of production.

We might be tempted to generalize the population to

represent all weeks, but we should exercise caution

here since the rate of defects may change over time.

(b) The fraction of computer chips manufactured at

the factory during the week of production that had

defects. (c) Estimate the parameter using the data:

p̂ = 27212

= 0.127. (d) Standard error (or SE).

(e) Compute the SE using p̂ = 0.127 in place of p:

SE ≈√

p̂(1−p̂)n

=√

0.127(1−0.127)212

= 0.023. (f) The

standard error is the standard deviation of p̂. A value

of 0.10 would be about one standard error away from

the observed value, which would not represent a very

uncommon deviation. (Usually beyond about 2 stan-

dard errors is a good rule of thumb.) The engineer

should not be surprised. (g) Recomputed standard

error using p = 0.1: SE =√

0.1(1−0.1)212

= 0.021. This

value isn’t very different, which is typical when the

standard error is computed using relatively similar

proportions (and even sometimes when those propor-

tions are quite different!).

5.5 (a) Sampling distribution. (b) If the popula-

tion proportion is in the 5-30% range, the success-

failure condition would be satisfied and the sampling

distribution would be symmetric. (c) We use the

formula for the standard error: SE =√

p(1−p)n

=√0.08(1−0.08)

800= 0.0096. (d) Standard error. (e) The

distribution will tend to be more variable when we

have fewer observations per sample.

391

5.7 Recall that the general formula is

point estimate ± z? × SE. First, identify the

three different values. The point estimate is

45%, z? = 1.96 for a 95% confidence level, and

SE = 1.2%. Then, plug the values into the formula:

45% ± 1.96 × 1.2% → (42.6%, 47.4%) We are

95% confident that the proportion of US adults who

live with one or more chronic conditions is between

42.6% and 47.4%.

5.9 (a) False. Confidence intervals provide a range

of plausible values, and sometimes the truth is

missed. A 95% confidence interval “misses” about

5% of the time. (b) True. Notice that the descrip-

tion focuses on the true population value. (c) True.

If we examine the 95% confidence interval computed

in Exercise 5.9, we can see that 50% is not included

in this interval. This means that in a hypothesis test,

we would reject the null hypothesis that the propor-

tion is 0.5. (d) False. The standard error describes

the uncertainty in the overall estimate from natural

fluctuations due to randomness, not the uncertainty

corresponding to individuals’ responses.

5.11 (a) False. Inference is made on the population

parameter, not the point estimate. The point esti-

mate is always in the confidence interval. (b) True.

sample mean. (d) False. To be more confident that

we capture the parameter, we need a wider interval.

Think about needing a bigger net to be more sure of

catching a fish in a murky lake. (e) True. Optional

explanation: This is true since the normal model was

used to model the sample mean. The margin of er-

ror is half the width of the interval, and the sample

mean is the midpoint of the interval. (f) False. In

the calculation of the standard error, we divide the

standard deviation by the square root of the sample

size. To cut the SE (or margin of error) in half, we

would need to sample 22 = 4 times the number of

people in the initial sample.

5.13 (a) The visitors are from a simple randomsample, so independence is satisfied. The success-failure condition is also satisfied, with both 64 and752 − 64 = 688 above 10. Therefore, we can usea normal distribution to model p̂ and construct aconfidence interval. (b) The sample proportion isp̂ = 64

752= 0.085. The standard error is

SE =

√p(1− p)

n≈√p̂(1− p̂)

√0.085(1− 0.085)

752= 0.010

The confidence interval is 0.085 ± 1.65 × 0.010 →(0.0685, 0.1015). We are 90% confident that 6.85%

to 10.15% of first-time site visitors will register using

the new design.

5.15 (a) H0 : p = 0.5 (Neither a majority nor mi-

nority of students’ grades improved) HA : p 6= 0.5

(Either a majority or a minority of students’ grades

improved)

(b) H0 : µ = 15 (The average amount of company

time each employee spends not working is 15 min-

utes for March Madness.) HA : µ 6= 15 (The aver-

age amount of company time each employee spends

not working is different than 15 minutes for March

Madness.)

5.17 (1) The hypotheses should be about the pop-

ulation proportion (p), not the sample proportion.

(2) The null hypothesis should have an equal sign.

(3) The alternative hypothesis should have a not-

equals sign, and (4) it should reference the null

value, p0 = 0.6, not the observed sample propor-

tion. The correct way to set up these hypotheses

is: H0 : p = 0.6 and HA : p 6= 0.6.

5.19 (a) This claim is reasonable, since the entire in-

terval lies above 50%. (b) The value of 70% lies out-

side of the interval, so we have convincing evidence

that the researcher’s conjecture is wrong. (c) A 90%

confidence interval will be narrower than a 95% confi-

dence interval. Even without calculating the interval,

we can tell that 70% would not fall in the interval,

and we would reject the researcher’s conjecture based

on a 90% confidence level as well.

5.21 (i) Set up hypotheses. H0: p = 0.5, HA:

p 6= 0.5. We will use a significance level of α = 0.05.

(ii) Check conditions: simple random sample gets

us independence, and the success-failure conditions

is satisfied since 0.5 × 1000 = 500 for each group

is at least 10. (iii) Next, we calculate: SE =√0.5(1− 0.5)/1000 = 0.016. Z = 0.42−0.5

0.016= −5,

which has a one-tail area of about 0.0000003, so

the p-value is twice this one-tail area at 0.0000006.

(iv) Make a conclusion: Because the p-value is less

than α = 0.05, we reject the null hypothesis and

conclude that the fraction of US adults who believe

raising the minimum wage will help the economy is

not 50%. Because the observed value is less than

50% and we have rejected the null hypothesis, we can

conclude that this belief is held by fewer than 50%

of US adults. (For reference, the survey also explores

support for changing the minimum wage, which is a

different question than if it will help the economy.)

5.23 If the p-value is 0.05, this means the test statis-

tic would be either Z = −1.96 or Z = 1.96. We’ll

show the calculations for Z = 1.96. Standard er-

ror: SE =√

0.3(1− 0.3)/90 = 0.048. Finally, set

up the test statistic formula and solve for p̂: 1.96 =p̂−0.30.048

→ p̂ = 0.394 Alternatively, if Z = −1.96 was

used: p̂ = 0.206.

392 APPENDIX A. EXERCISE SOLUTIONS

5.25 (a) H0: Anti-depressants do not affect the

symptoms of Fibromyalgia. HA: Anti-depressants do

affect the symptoms of Fibromyalgia (either helping

or harming). (b) Concluding that anti-depressants

either help or worsen Fibromyalgia symptoms when

they actually do neither. (c) Concluding that anti-

depressants do not affect Fibromyalgia symptoms

when they actually do.

5.27 (a) We are 95% confident that Americans

spend an average of 1.38 to 1.92 hours per day relax-

ing or pursuing activities they enjoy. (b) Their con-

fidence level must be higher as the width of the con-

fidence interval increases as the confidence level in-

creases. (c) The new margin of error will be smaller,

since as the sample size increases, the standard error

decreases, which will decrease the margin of error.

5.29 (a) H0: The restaurant meets food safety and

sanitation regulations. HA: The restaurant does not

meet food safety and sanitation regulations. (b) The

food safety inspector concludes that the restaurant

does not meet food safety and sanitation regulations

and shuts down the restaurant when the restaurant is

actually safe. (c) The food safety inspector concludes

that the restaurant meets food safety and sanitation

regulations and the restaurant stays open when the

restaurant is actually not safe. (d) A Type 1 Error

may be more problematic for the restaurant owner

since his restaurant gets shut down even though it

meets the food safety and sanitation regulations.

(e) A Type 2 Error may be more problematic for

diners since the restaurant deemed safe by the in-

spector is actually not. (f) Strong evidence. Diners

would rather a restaurant that meet the regulations

get shut down than a restaurant that doesn’t meet

the regulations not get shut down.

5.31 (a) H0 : punemp = punderemp: The propor-

tions of unemployed and underemployed people who

are having relationship problems are equal. HA :

punemp 6= punderemp: The proportions of unem-

ployed and underemployed people who are having re-

lationship problems are different. (b) If in fact the

two population proportions are equal, the probabil-

ity of observing at least a 2% difference between the

sample proportions is approximately 0.35. Since this

is a high probability we fail to reject the null hypoth-

esis. The data do not provide convincing evidence

that the proportion of of unemployed and underem-

ployed people who are having relationship problems

are different.

5.33 Because 130 is inside the confidence interval,

we do not have convincing evidence that the true av-

erage is any different than what the nutrition label

suggests.

5.35 True. If the sample size gets ever larger, then

the standard error will become ever smaller. Even-

tually, when the sample size is large enough and the

standard error is tiny, we can find statistically sig-

nificant yet very small differences between the null

value and point estimate (assuming they are not ex-

actly equal).

5.37 (a) In effect, we’re checking whether men arepaid more than women (or vice-versa), and we’d ex-pect these outcomes with either chance under the nullhypothesis:

H0 : p = 0.5 HA : p 6= 0.5

We’ll use p to represent the fraction of cases wheremen are paid more than women.(b) Below is the completion of the hypothesis test.

• There isn’t a good way to check independencehere since the jobs are not a simple randomsample. However, independence doesn’t seemunreasonable, since the individuals in each jobare different from each other. The success-failure condition is met since we check it usingthe null proportion: p0n = (1− p0)n = 10.5 isgreater than 10.

• We can compute the sample proportion, SE,and test statistic:

p̂ = 19/21 = 0.905

SE =

√0.5× (1− 0.5)

21= 0.109

Z =0.905− 0.5

0.109= 3.72

The test statistic Z corresponds to an uppertail area of about 0.0001, so the p-value is 2times this value: 0.0002.

• Because the p-value is smaller than 0.05, wereject the notion that all these gender pay dis-parities are due to chance. Because we observethat men are paid more in a higher proportionof cases and we have rejected H0, we can con-clude that men are being paid higher amountsin ways not explainable by chance alone.

If you’re curious for more info around this topic, in-

cluding a discussion about adjusting for additional

factors that affect pay, please see the following video

by Healthcare Triage: youtu.be/aVhgKSULNQA.

393

6 Inference for categorical data

6.1 (a) False. Doesn’t satisfy success-failure condi-

tion. (b) True. The success-failure condition is not

satisfied. In most samples we would expect p̂ to be

close to 0.08, the true population proportion. While

p̂ can be much above 0.08, it is bound below by 0,

suggesting it would take on a right skewed shape.

Plotting the sampling distribution would confirm this

suspicion. (c) False. SEp̂ = 0.0243, and p̂ = 0.12 is

only 0.12−0.080.0243

= 1.65 SEs away from the mean, which

would not be considered unusual. (d) True. p̂ = 0.12

is 2.32 standard errors away from the mean, which

is often considered unusual. (e) False. Decreases the

SE by a factor of 1/√

6.3 (a) True. See the reasoning of 6.1(b). (b) True.

We take the square root of the sample size in the SE

formula. (c) True. The independence and success-

failure conditions are satisfied. (d) True. The inde-

pendence and success-failure conditions are satisfied.

6.5 (a) False. A confidence interval is constructed to

estimate the population proportion, not the sample

proportion. (b) True. 95% CI: 82% ± 2%. (c) True.

By the definition of the confidence level. (d) True.

Quadrupling the sample size decreases the SE and

ME by a factor of 1/√

4. (e) True. The 95% CI is

entirely above 50%.

6.7 With a random sample, independence is satis-

fied. The success-failure condition is also satisfied.

ME = z?√

p̂(1−p̂)n

= 1.96√

0.56×0.44600

= 0.0397 ≈ 4%

6.9 (a) No. The sample only represents students

who took the SAT, and this was also an online sur-

vey. (b) (0.5289, 0.5711). We are 90% confident

that 53% to 57% of high school seniors who took the

SAT are fairly certain that they will participate in

a study abroad program in college. (c) 90% of such

random samples would produce a 90% confidence in-

terval that includes the true proportion. (d) Yes.

The interval lies entirely above 50%.

6.11 (a) We want to check for a majority (or minor-ity), so we use the following hypotheses:

H0 : p = 0.5 HA : p 6= 0.5

We have a sample proportion of p̂ = 0.55 and a sam-ple size of n = 617 independents.Since this is a random sample, independence is sat-isfied. The success-failure condition is also satisfied:617× 0.5 and 617× (1− 0.5) are both at least 10 (weuse the null proportion p0 = 0.5 for this check in aone-proportion hypothesis test).Therefore, we can model p̂ using a normal distribu-tion with a standard error of

SE =

√p(1− p)

n= 0.02

(We use the null proportion p0 = 0.5 to compute thestandard error for a one-proportion hypothesis test.)

Next, we compute the test statistic:

Z =0.55− 0.5

0.02= 2.5

This yields a one-tail area of 0.0062, and a p-value of

2× 0.0062 = 0.0124.

Because the p-value is smaller than 0.05, we reject

the null hypothesis. We have strong evidence that

the support is different from 0.5, and since the data

provide a point estimate above 0.5, we have strong

evidence to support this claim by the TV pundit.

(b) No. Generally we expect a hypothesis test and

a confidence interval to align, so we would expect

the confidence interval to show a range of plausible

values entirely above 0.5. However, if the confidence

level is misaligned (e.g. a 99% confidence level and

a α = 0.05 significance level), then this is no longer

generally true.

6.13 (a) H0 : p = 0.5. HA : p 6= 0.5. Independence

(random sample) is satisfied, as is the success-failure

conditions (using p0 = 0.5, we expect 40 successes

and 40 failures). Z = 2.91 → the one tail area is

0.0018, so the p-value is 0.0036. Since the p-value

< 0.05, we reject the null hypothesis. Since we re-

jected H0 and the point estimate suggests people are

better than random guessing, we can conclude the

rate of correctly identifying a soda for these people

is significantly better than just by random guessing.

(b) If in fact people cannot tell the difference between

diet and regular soda and they were randomly guess-

ing, the probability of getting a random sample of

80 people where 53 or more identify a soda correctly

(or 53 or more identify a soda incorrectly) would be

0.0036.

6.15 Since a sample proportion (p̂ = 0.55) is avail-able, we use this for the sample size calculations.The margin of error for a 90% confidence interval

is 1.65 × SE = 1.65 ×√

p(1−p)n

. We want this to be

less than 0.01, where we use p̂ in place of p:

1.65×√

0.55(1− 0.55)

n≤ 0.01

1.652 0.55(1− 0.55)

0.012≤ n

From this, we get that n must be at least 6739.

6.17 This is not a randomized experiment, and it

is unclear whether people would be affected by the

behavior of their peers. That is, independence may

not hold. Additionally, there are only 5 interven-

tions under the provocative scenario, so the success-

failure condition does not hold. Even if we consider

a hypothesis test where we pool the proportions, the

success-failure condition will not be satisfied. Since

one condition is questionable and the other is not sat-

isfied, the difference in sample proportions will not

follow a nearly normal distribution.

394 APPENDIX A. EXERCISE SOLUTIONS

6.19 (a) False. The entire confidence interval is

above 0. (b) True. (c) True. (d) True. (e) False. It

is simply the negated and reordered values: (-0.06,-

0.02).

6.21 (a) Standard error:

SE =

√0.79(1− 0.79)

347+

0.55(1− 0.55)

617= 0.03

Using z? = 1.96, we get:

0.79− 0.55± 1.96× 0.03→ (0.181, 0.299)

We are 95% confident that the proportion of

Democrats who support the plan is 18.1% to 29.9%

higher than the proportion of Independents who sup-

port the plan. (b) True.

6.23 (a) College grads: 23.7%. Non-college grads:

33.7%. (b) Let pCG and pNCG represent the pro-

portion of college graduates and non-college gradu-

ates who responded “do not know”. H0 : pCG =

pNCG. HA : pCG 6= pNCG. Independence is sat-

isfied (random sample), and the success-failure con-

dition, which we would check using the pooled pro-

portion (p̂pool = 235/827 = 0.284), is also satisfied.

Z = −3.18 → p-value = 0.0014. Since the p-value is

very small, we reject H0. The data provide strong ev-

idence that the proportion of college graduates who

do not have an opinion on this issue is different than

that of non-college graduates. The data also indicate

that fewer college grads say they “do not know” than

non-college grads (i.e. the data indicate the direction

after we reject H0).

6.25 (a) College grads: 35.2%. Non-college grads:

33.9%. (b) Let pCG and pNCG represent the pro-

portion of college graduates and non-college grads

who support offshore drilling. H0 : pCG = pNCG.

HA : pCG 6= pNCG. Independence is satisfied

(random sample), and the success-failure condition,

which we would check using the pooled proportion

(p̂pool = 286/827 = 0.346), is also satisfied. Z = 0.39

→ p-value = 0.6966. Since the p-value > α (0.05),

we fail to reject H0. The data do not provide strong

evidence of a difference between the proportions of

college graduates and non-college graduates who sup-

port off-shore drilling in California.

6.27 Subscript C means control group. Subscript Tmeans truck drivers. H0 : pC = pT . HA : pC 6= pT .

Independence is satisfied (random samples), as is the

success-failure condition, which we would check us-

ing the pooled proportion (p̂pool = 70/495 = 0.141).

Z = −1.65 → p-value = 0.0989. Since the p-value

is high (default to alpha = 0.05), we fail to reject

H0. The data do not provide strong evidence that

the rates of sleep deprivation are different for non-

transportation workers and truck drivers.

6.29 (a) Summary of the study:

Virol. failureYes No Total

TreatmentNevaripine 26 94 120Lopinavir 10 110 120Total 36 204 240

(b) H0 : pN = pL. There is no difference in virologic

failure rates between the Nevaripine and Lopinavir

groups. HA : pN 6= pL. There is some difference

in virologic failure rates between the Nevaripine and

Lopinavir groups. (c) Random assignment was used,

so the observations in each group are independent. If

the patients in the study are representative of those

in the general population (something impossible to

check with the given information), then we can also

confidently generalize the findings to the population.

The success-failure condition, which we would check

using the pooled proportion (p̂pool = 36/240 = 0.15),

is satisfied. Z = 2.89 → p-value = 0.0039. Since

the p-value is low, we reject H0. There is strong evi-

dence of a difference in virologic failure rates between

the Nevaripine and Lopinavir groups. Treatment and

virologic failure do not appear to be independent.

6.31 (a) False. The chi-square distribution has

one parameter called degrees of freedom. (b) True.

creases, the shape of the chi-square distribution be-

comes more symmetric.

6.33 (a) H0: The distribution of the format of the

book used by the students follows the professor’s pre-

dictions. HA: The distribution of the format of the

book used by the students does not follow the profes-

sor’s predictions. (b) Ehard copy = 126×0.60 = 75.6.

Eprint = 126 × 0.25 = 31.5. Eonline = 126 × 0.15 =

18.9. (c) Independence: The sample is not ran-

dom. However, if the professor has reason to be-

lieve that the proportions are stable from one term to

the next and students are not affecting each other’s

study habits, independence is probably reasonable.

Sample size: All expected counts are at least 5.

(d) χ2 = 2.32, df = 2, p-value = 0.313. (e) Since

the p-value is large, we fail to reject H0. The data do

not provide strong evidence indicating the professor’s

predictions were statistically inaccurate.

6.35 (a) Two-way table:

QuitTreatment Yes No TotalPatch + support group 40 110 150Only patch 30 120 150Total 70 230 300

(b-i) Erow1,col1 = (row 1 total)×(col 1 total)table total

= 35. This

is lower than the observed value.

(b-ii) Erow2,col2 = (row 2 total)×(col 2 total)table total

= 115.

This is lower than the observed value.

395

6.37 H0: The opinion of college grads and non-gradsis not different on the topic of drilling for oil and nat-ural gas off the coast of California. HA: Opinionsregarding the drilling for oil and natural gas off thecoast of California has an association with earning acollege degree.

Erow 1,col 1 = 151.5 Erow 1,col 2 = 134.5

Erow 2,col 1 = 162.1 Erow 2,col 2 = 143.9

Erow 3,col 1 = 124.5 Erow 3,col 2 = 110.5

Independence: The samples are both random, un-

related, and from less than 10% of the population,

so independence between observations is reasonable.

Sample size: All expected counts are at least 5.

χ2 = 11.47, df = 2 → p-value = 0.003. Since the

p-value < α, we reject H0. There is strong evidence

that there is an association between support for off-

shore drilling and having a college degree.

6.39 No. The samples at the beginning and at the

end of the semester are not independent since the

survey is conducted on the same students.

6.41 (a) H0: The age of Los Angeles residents is

independent of shipping carrier preference variable.

HA: The age of Los Angeles residents is associ-

ated with the shipping carrier preference variable.

(b) The conditions are not satisfied since some ex-

pected counts are below 5.

6.43 (a) Independence is satisfied (random sample),

as is the success-failure condition (40 smokers, 160

non-smokers). The 95% CI: (0.145, 0.255). We are

95% confident that 14.5% to 25.5% of all students at

this university smoke. (b) We want z?SE to be no

larger than 0.02 for a 95% confidence level. We use

z? = 1.96 and plug in the point estimate p̂ = 0.2

within the SE formula: 1.96√

0.2(1− 0.2)/n ≤ 0.02.

The sample size n should be at least 1,537.

6.45 (a) Proportion of graduates from this univer-

sity who found a job within one year of graduating.

p̂ = 348/400 = 0.87. (b) This is a random sample,

so the observations are independent. Success-failure

condition is satisfied: 348 successes, 52 failures, both

well above 10. (c) (0.8371, 0.9029). We are 95%

confident that approximately 84% to 90% of gradu-

ates from this university found a job within one year

of completing their undergraduate degree. (d) 95%

of such random samples would produce a 95% con-

fidence interval that includes the true proportion of

students at this university who found a job within one

year of graduating from college. (e) (0.8267, 0.9133).

Similar interpretation as before. (f) 99% CI is wider,

as we are more confident that the true proportion

is within the interval and so need to cover a wider

range.

6.47 Use a chi-squared goodness of fit test. H0:

Each option is equally likely. HA: Some options are

preferred over others. Total sample size: 99. Ex-

pected counts: (1/3) * 99 = 33 for each option. These

are all above 5, so conditions are satisfied. df =

3− 1 = 2 and χ2 = (43−33)2

33+ (21−33)2

33+ (35−33)2

33=

7.52 → p-value = 0.023. Since the p-value is less

than 5%, we reject H0. The data provide convincing

evidence that some options are preferred over others.

6.49 (a) H0 : p = 0.38. HA : p 6= 0.38. Inde-

pendence (random sample) and the success-failure

condition are satisfied. Z = −20.5 → p-value ≈ 0.

Since the p-value is very small, we reject H0. The

data provide strong evidence that the proportion of

Americans who only use their cell phones to access

the internet is different than the Chinese proportion

of 38%, and the data indicate that the proportion is

lower in the US. (b) If in fact 38% of Americans used

their cell phones as a primary access point to the in-

ternet, the probability of obtaining a random sample

of 2,254 Americans where 17% or less or 59% or more

use their only their cell phones to access the inter-

net would be approximately 0. (c) (0.1545, 0.1855).

We are 95% confident that approximately 15.5% to

18.6% of all Americans primarily use their cell phones

to browse the internet.

7 Inference for numerical data

7.1 (a) df = 6− 1 = 5, t?5 = 2.02 (column with two

tails of 0.10, row with df = 5). (b) df = 21− 1 = 20,

t?20 = 2.53 (column with two tails of 0.02, row with

df = 20). (c) df = 28, t?28 = 2.05. (d) df = 11,

t?11 = 3.11.

7.3 (a) 0.085, do not reject H0. (b) 0.003, reject H0.

7.5 The mean is the midpoint: x̄ = 20. Identify the

margin of error: ME = 1.015, then use t?35 = 2.03

and SE = s/√n in the formula for margin of error

to identify s = 3.

7.7 (a) H0: µ = 8 (New Yorkers sleep 8 hrs per

night on average.) HA: µ 6= 8 (New Yorkers sleep

less or more than 8 hrs per night on average.) (b) In-

dependence: The sample is random. The min/max

suggest there are no concerning outliers. T = −1.75.

df = 25 − 1 = 24. (c) p-value = 0.093. If in fact

the true population mean of the amount New Yorkers

sleep per night was 8 hours, the probability of getting

a random sample of 25 New Yorkers where the aver-

age amount of sleep is 7.73 hours per night or less

(or 8.27 hours or more) is 0.093. (d) Since p-value

> 0.05, do not reject H0. The data do not provide

strong evidence that New Yorkers sleep more or less

than 8 hours per night on average. (e) No, since the

p-value is smaller than 1− 0.90 = 0.10.

396 APPENDIX A. EXERCISE SOLUTIONS

7.9 T is either -2.09 or 2.09. Then x̄ is one of thefollowing:

−2.09 =x̄− 60

8√20

→ x̄ = 56.26

2.09 =x̄− 60

8√20

→ x̄ = 63.74

7.11 (a) We will conduct a 1-sample t-test. H0:

µ = 5. HA: µ 6= 5. We’ll use α = 0.05. This

is a random sample, so the observations are inde-

pendent. To proceed, we assume the distribution

of years of piano lessons is approximately normal.

SE = 2.2/√

20 = 0.4919. The test statistic is

T = (4.6 − 5)/SE = −0.81. df = 20 − 1 = 19. The

one-tail area is about 0.21, so the p-value is about

0.42, which is bigger than α = 0.05 and we do not

reject H0. That is, we do not have sufficiently strong

evidence to reject the notion that the average is 5

years.

(b) Using SE = 0.4919 and t?df=19 = 2.093, the con-

fidence interval is (3.57, 5.63). We are 95% confident

that the average number of years a child takes piano

lessons in this city is 3.57 to 5.63 years.

pothesis and the null value of 5 was in the t-interval.

7.13 If the sample is large, then the margin of error

will be about 1.96× 100/√n. We want this value to

be less than 10, which leads to n ≥ 384.16, meaning

we need a sample size of at least 385 (round up for

sample size calculations!).

7.15 Paired, data are recorded in the same cities at

two different time points. The temperature in a city

at one point is not independent of the temperature

in the same city at another time point.

7.17 (a) Since it’s the same students at the begin-

ning and the end of the semester, there is a pairing

between the datasets, for a given student their be-

ginning and end of semester grades are dependent.

(b) Since the subjects were sampled randomly, each

observation in the men’s group does not have a spe-

cial correspondence with exactly one observation in

the other (women’s) group. (c) Since it’s the same

subjects at the beginning and the end of the study,

there is a pairing between the datasets, for a subject

student their beginning and end of semester artery

thickness are dependent. (d) Since it’s the same sub-

jects at the beginning and the end of the study, there

is a pairing between the datasets, for a subject stu-

dent their beginning and end of semester weights are

dependent.

7.19 (a) For each observation in one data set, there

is exactly one specially corresponding observation in

the other data set for the same geographic location.

The data are paired. (b) H0 : µdiff = 0 (There is no

difference in average number of days exceeding 90°F

in 1948 and 2018 for NOAA stations.) HA : µdiff 6= 0

(There is a difference.) (c) Locations were randomly

sampled, so independence is reasonable. The sample

size is at least 30, so we’re just looking for partic-

ularly extreme outliers: none are present (the ob-

servation off left in the histogram would be con-

sidered a clear outlier, but not a particularly ex-

treme one). Therefore, the conditions are satisfied.

(d) SE = 17.2/√

197 = 1.23. T = 2.9−01.23

= 2.36 with

degrees of freedom df = 197− 1 = 196. This leads to

a one-tail area of 0.0096 and a p-value of about 0.019.

(e) Since the p-value is less than 0.05, we reject H0.

The data provide strong evidence that NOAA sta-

tions observed more 90°F days in 2018 than in 1948.

(f) Type 1 Error, since we may have incorrectly re-

jected H0. This error would mean that NOAA sta-

tions did not actually observe a decrease, but the

sample we took just so happened to make it appear

that this was the case. (g) No, since we rejected H0,

which had a null value of 0.

7.21 (a) SE = 1.23 and t? = 1.65. 2.9 ± 1.65 ×1.23→ (0.87, 4.93).

(b) We are 90% confident that there was an increase

of 0.87 to 4.93 in the average number of days that hit

90°F in 2018 relative to 1948 for NOAA stations.

7.23 (a) These data are paired. For example, the

Friday the 13th in say, September 1991, would prob-

ably be more similar to the Friday the 6th in Septem-

ber 1991 than to Friday the 6th in another month or

year.

(b) Let µdiff = µsixth − µthirteenth. H0 : µdiff = 0.

HA : µdiff 6= 0.

dom. However, if we think these dates are roughly

equivalent to a simple random sample of all such Fri-

day 6th/13th date pairs, then independence is rea-

sonable. To proceed, we must make this strong as-

sumption, though we should note this assumption in

any reported results. Normality: With fewer than 10

observations, we would need to see clear outliers to

be concerned. There is a borderline outlier on the

right of the histogram of the differences, so we would

want to report this in formal analysis results.

(d) T = 4.93 for df = 10− 1 = 9 → p-value = 0.001.

(e) Since p-value < 0.05, reject H0. The data provide

strong evidence that the average number of cars at

the intersection is higher on Friday the 6th than on

Friday the 13th. (We should exercise caution about

generalizing the interpretation to all intersections or

roads.)

(f) If the average number of cars passing the inter-

section actually was the same on Friday the 6th and

13th, then the probability that we would observe a

test statistic so far from zero is less than 0.01.

(g) We might have made a Type 1 Error, i.e. incor-

rectly rejected the null hypothesis.

397

7.25 (a) H0 : µdiff = 0. HA : µdiff 6= 0.

T = −2.71. df = 5. p-value = 0.042. Since p-

value < 0.05, reject H0. The data provide strong

evidence that the average number of traffic accident

related emergency room admissions are different be-

tween Friday the 6th and Friday the 13th. Further-

more, the data indicate that the direction of that

difference is that accidents are lower on Friday the

6th relative to Friday the 13th.

(b) (-6.49, -0.17).

so we cannot so easily infer a causal intervention im-

plied by this statement. It is true that there is a

difference. However, for example, this does not mean

that a responsible adult going out on Friday the 13th

has a higher chance of harm than on any other night.

7.27 (a) Chicken fed linseed weighed an average of

218.75 grams while those fed horsebean weighed an

average of 160.20 grams. Both distributions are rela-

tively symmetric with no apparent outliers. There is

more variability in the weights of chicken fed linseed.

(b) H0 : µls = µhb. HA : µls 6= µhb.

We leave the conditions to you to consider.

T = 3.02, df = min(11, 9) = 9 → p-value = 0.014.

Since p-value < 0.05, reject H0. The data provide

strong evidence that there is a significant difference

between the average weights of chickens that were fed

linseed and horsebean.

(d) Yes, since p-value > 0.01, we would not have re-

jected H0.

7.29 H0 : µC = µS . HA : µC 6= µS . T = 3.27,

df = 11 → p-value = 0.007. Since p-value < 0.05,

reject H0. The data provide strong evidence that the

average weight of chickens that were fed casein is dif-

ferent than the average weight of chickens that were

fed soybean (with weights from casein being higher).

Since this is a randomized experiment, the observed

difference can be attributed to the diet.

7.31 Let µdiff = µpre − µpost. H0 : µdiff = 0:

Treatment has no effect. HA : µdiff 6= 0: Treat-

ment has an effect on P.D.T. scores, either positive

or negative. Conditions: The subjects are randomly

assigned to treatments, so independence within and

between groups is satisfied. All three sample sizes

are smaller than 30, so we look for clear outliers.

There is a borderline outlier in the first treatment

group. Since it is borderline, we will proceed, but

we should report this caveat with any results. For

all three groups: df = 13. T1 = 1.89 → p-value =

0.081, T2 = 1.35→ p-value = 0.200), T3 = −1.40→(p-value = 0.185). We do not reject the null hypoth-

esis for any of these groups. As earlier noted, there

is some uncertainty about if the method applied is

reasonable for the first group.

7.33 Difference we care about: 40. Single tail of

90%: 1.28 × SE. Rejection region bounds: ±1.96 ×SE (if 5% significance level). Setting 3.24×SE = 40,

subbing in SE =√

942

n+ 942

n, and solving for the

sample size n gives 116 plots of land for each fertilizer.

7.35 Alternative.

7.37 H0: µ1 = µ2 = · · · = µ6. HA: The average

weight varies across some (or all) groups. Indepen-

dence: Chicks are randomly assigned to feed types

(presumably kept separate from one another), there-

fore independence of observations is reasonable. Ap-

prox. normal: the distributions of weights within

each feed type appear to be fairly symmetric. Con-

stant variance: Based on the side-by-side box plots,

the constant variance assumption appears to be rea-

sonable. There are differences in the actual com-

puted standard deviations, but these might be due

to chance as these are quite small samples. F5,65 =

15.36 and the p-value is approximately 0. With such

a small p-value, we reject H0. The data provide con-

vincing evidence that the average weight of chicks

varies across some (or all) feed supplement groups.

7.39 (a) H0: The population mean of MET for eachgroup is equal to the others. HA: At least one pairof means is different. (b) Independence: We don’thave any information on how the data were collected,so we cannot assess independence. To proceed, wemust assume the subjects in each group are indepen-dent. In practice, we would inquire for more details.Normality: The data are bound below by zero andthe standard deviations are larger than the means,indicating very strong skew. However, since the sam-ple sizes are extremely large, even extreme skew isacceptable. Constant variance: This condition issufficiently met, as the standard deviations are rea-sonably consistent across groups. (c) See below, withthe last column omitted:

Df Sum Sq Mean Sq F value

coffee 4 10508 2627 5.2Residuals 50734 25564819 504Total 50738 25575327

(d) Since p-value is very small, reject H0. The data

provide convincing evidence that the average MET

differs between at least one pair of groups.

7.41 (a) H0: Average GPA is the same for all ma-

jors. HA: At least one pair of means are different.

(b) Since p-value > 0.05, fail to reject H0. The data

do not provide convincing evidence of a difference be-

tween the average GPAs across three groups of ma-

jors. (c) The total degrees of freedom is 195+2 = 197,

so the sample size is 197 + 1 = 198.

7.43 (a) False. As the number of groups increases,

so does the number of comparisons and hence the

modified significance level decreases. (b) True.

independent regardless of sample size.

398 APPENDIX A. EXERCISE SOLUTIONS

7.45 (a) H0: Average score difference is the same

for all treatments. HA: At least one pair of means

are different. (b) We should check conditions. If we

look back to the earlier exercise, we will see that the

patients were randomized, so independence is satis-

fied. There are some minor concerns about skew, es-

pecially with the third group, though this may be ac-

ceptable. The standard deviations across the groups

are reasonably similar. Since the p-value is less than

0.05, reject H0. The data provide convincing evi-

dence of a difference between the average reduction

in score among treatments. (c) We determined that

at least two means are different in part (b), so we

now conduct K = 3 × 2/2 = 3 pairwise t-tests that

each use α = 0.05/3 = 0.0167 for a significance level.

Use the following hypotheses for each pairwise test.

H0: The two means are equal. HA: The two means

are different. The sample sizes are equal and we use

the pooled SD, so we can compute SE = 3.7 with

the pooled df = 39. The p-value for Trmt 1 vs. Trmt

3 is the only one under 0.05: p-value = 0.035 (or

0.024 if using spooled in place of s1 and s3, though

this won’t affect the final conclusion). The p-value is

larger than 0.05/3 = 1.67, so we do not have strong

evidence to conclude that it is this particular pair of

groups that are different. That is, we cannot identify

if which particular pair of groups are actually differ-

ent, even though we’ve rejected the notion that they

are all the same!

7.47 H0 : µT = µC . HA : µT 6= µC . T = 2.24,

df = 21 → p-value = 0.036. Since p-value < 0.05,

reject H0. The data provide strong evidence that

the average food consumption by the patients in the

treatment and control groups are different. Further-

more, the data indicate patients in the distracted

eating (treatment) group consume more food than

patients in the control group.

7.49 False. While it is true that paired analysis re-

quires equal sample sizes, only having the equal sam-

ple sizes isn’t, on its own, sufficient for doing a paired

test. Paired tests require that there be a special cor-

respondence between each pair of observations in the

two groups.

7.51 (a) We are building a distribution of sample

statistics, in this case the sample mean. Such a dis-

tribution is called a sampling distribution. (b) Be-

cause we are dealing with the distribution of sample

means, we need to check to see if the Central Limit

Theorem applies. Our sample size is greater than 30,

and we are told that random sampling is employed.

With these conditions met, we expect that the dis-

tribution of the sample mean will be nearly normal

and therefore symmetric. (c) Because we are dealing

with a sampling distribution, we measure its variabil-

ity with the standard error. SE = 18.2/√

45 = 2.713.

(d) The sample means will be more variable with the

smaller sample size.

7.53 (a) We should set 1.0% equal to 2.84 standard

errors: 2.84×SEdesired = 1.0% (see Example 7.37 on

page 282 for details). This means the standard error

should be about SE = 0.35% to achieve the desired

statistical power.

(b) The margin of error was 0.5×(2.6%−(−0.2%)) =

1.4%, so the standard error in the experiment must

have been 1.96 × SEoriginal = 1.4% → SEoriginal =

0.71%.

of the sample size, so we should increase the sample

size by a factor of 2.032 = 4.12.

(d) The team should run an experiment 4.12 times

larger, so they should have a random sample of 4.12%

of their users in each of the experiment arms in the

new experiment.

7.55 Independence: it is a random sample, so we

can assume that the students in this sample are in-

dependent of each other with respect to number of

exclusive relationships they have been in. Notice

that there are no students who have had no exclu-

sive relationships in the sample, which suggests some

student responses are likely missing (perhaps only

positive values were reported). The sample size is at

least 30, and there are no particularly extreme out-

liers, so the normality condition is reasonable. 90%

CI: (2.97, 3.43). We are 90% confident that under-

graduate students have been in 2.97 to 3.43 exclusive

relationships, on average.

7.57 The hypotheses should be about the popula-tion mean (µ), not the sample mean. The null hy-pothesis should have an equal sign and the alterna-tive hypothesis should be about the null hypothesizedvalue, not the observed sample mean. Correction:

H0 : µ = 10 hours

HA : µ 6= 10 hours

A two-sided test allows us to consider the possibility

that the data show us something that we would find

surprising.

399

8 Introduction to linear regression

8.1 (a) The residual plot will show randomly dis-

tributed residuals around 0. The variance is also ap-

proximately constant. (b) The residuals will show

a fan shape, with higher variability for smaller x.

There will also be many points on the right above

the line. There is trouble with the model being fit

here.

8.3 (a) Strong relationship, but a straight line would

not fit the data. (b) Strong relationship, and a linear

fit would be reasonable. (c) Weak relationship, and

trying a linear fit would be reasonable. (d) Moder-

ate relationship, but a straight line would not fit the

data. (e) Strong relationship, and a linear fit would

be reasonable. (f) Weak relationship, and trying a

linear fit would be reasonable.

8.5 (a) Exam 2 since there is less of a scatter in the

plot of final exam grade versus exam 2. Notice that

the relationship between Exam 1 and the Final Exam

appears to be slightly nonlinear. (b) Exam 2 and the

final are relatively close to each other chronologically,

or Exam 2 may be cumulative so has greater similari-

ties in material to the final exam. Answers may vary.

8.7 (a) r = −0.7 → (4). (b) r = 0.45 → (3).

8.9 (a) The relationship is positive, weak, and pos-

sibly linear. However, there do appear to be some

anomalous observations along the left where sev-

eral students have the same height that is notably

far from the cloud of the other points. Addition-

ally, there are many students who appear not to

have driven a car, and they are represented by a

set of points along the bottom of the scatterplot.

(b) There is no obvious explanation why simply be-

ing tall should lead a person to drive faster. How-

ever, one confounding factor is gender. Males tend

to be taller than females on average, and personal ex-

periences (anecdotal) may suggest they drive faster.

If we were to follow-up on this suspicion, we would

find that sociological studies confirm this suspicion.

The gender variable is indeed an important confound-

ing variable.

8.11 (a) There is a somewhat weak, positive, pos-

sibly linear relationship between the distance trav-

eled and travel time. There is clustering near the

lower left corner that we should take special note of.

(b) Changing the units will not change the form, di-

rection or strength of the relationship between the

two variables. If longer distances measured in miles

are associated with longer travel time measured in

minutes, longer distances measured in kilometers will

be associated with longer travel time measured in

hours. (c) Changing units doesn’t affect correlation:

r = 0.636.

8.13 (a) There is a moderate, positive, and lin-

ear relationship between shoulder girth and height.

(b) Changing the units, even if just for one of the vari-

ables, will not change the form, direction or strength

of the relationship between the two variables.

8.15 In each part, we can write the husband ages as

a linear function of the wife ages.

(a) ageH = ageW + 3.

(b) ageH = ageW − 2.

Since the slopes are positive and these are perfect

linear relationships, the correlation will be exactly 1

in all three parts. An alternative way to gain insight

into this solution is to create a mock data set, e.g.

5 women aged 26, 27, 28, 29, and 30, then find the

husband ages for each wife in each part and create a

scatterplot.

8.17 Correlation: no units. Intercept: kg. Slope:

kg/cm.

8.19 Over-estimate. Since the residual is calculated

as observed − predicted, a negative residual means

that the predicted value is higher than the observed

value.

8.21 (a) There is a positive, very strong, linear as-

sociation between the number of tourists and spend-

ing. (b) Explanatory: number of tourists (in thou-

sands). Response: spending (in millions of US dol-

lars). (c) We can predict spending for a given number

of tourists using a regression line. This may be use-

ful information for determining how much the coun-

try may want to spend in advertising abroad, or to

forecast expected revenues from tourism. (d) Even

though the relationship appears linear in the scatter-

plot, the residual plot actually shows a nonlinear re-

lationship. This is not a contradiction: residual plots

can show divergences from linearity that can be diffi-

cult to see in a scatterplot. A simple linear model is

inadequate for modeling these data. It is also impor-

tant to consider that these data are observed sequen-

tially, which means there may be a hidden structure

not evident in the current plots but that is important

to consider.

400 APPENDIX A. EXERCISE SOLUTIONS

8.23 (a) First calculate the slope: b1 = R×sy/sx =

0.636 × 113/99 = 0.726. Next, make use of the

fact that the regression line passes through the point

(x̄, ȳ): ȳ = b0 +b1× x̄. Plug in x̄, ȳ, and b1, and solve

for b0: 51. Solution: ̂travel time = 51 + 0.726 ×distance. (b) b1: For each additional mile in dis-

tance, the model predicts an additional 0.726 minutes

in travel time. b0: When the distance traveled is 0

miles, the travel time is expected to be 51 minutes.

It does not make sense to have a travel distance of

0 miles in this context. Here, the y-intercept serves

only to adjust the height of the line and is mean-

ingless by itself. (c) R2 = 0.6362 = 0.40. About

40% of the variability in travel time is accounted for

by the model, i.e. explained by the distance trav-

eled. (d) ̂travel time = 51 + 0.726 × distance =

51+0.726×103 ≈ 126 minutes. (Note: we should be

cautious in our predictions with this model since we

have not yet evaluated whether it is a well-fit model.)

(e) ei = yi − ŷi = 168 − 126 = 42 minutes. A pos-

itive residual means that the model underestimates

the travel time. (f) No, this calculation would require

extrapolation.

8.25 (a) ̂murder = −29.901 + 2.559 × poverty%.

(b) Expected murder rate in metropolitan areas with

no poverty is -29. 901 per million. This is obvi-

ously not a meaningful value, it just serves to ad-

just the height of the regression line. (c) For each

additional percentage increase in poverty, we expect

murders per million to be higher on average by 2.559.

(d) Poverty level explains 70.52% of the variability in

murder rates in metropolitan areas. (e)√

0.7052 =

0.8398.

8.27 (a) There is an outlier in the bottom right.

Since it is far from the center of the data, it is a

point with high leverage. It is also an influential

point since, without that observation, the regression

line would have a very different slope.

(b) There is an outlier in the bottom right. Since it is

far from the center of the data, it is a point with high

leverage. However, it does not appear to be affecting

the line much, so it is not an influential point.

the x-axis direction), so this point does not have high

leverage. This means the point won’t have much ef-

fect on the slope of the line and so is not an influential

point.

8.29 (a) There is a negative, moderate-to-strong,

somewhat linear relationship between percent of fam-

ilies who own their home and the percent of the pop-

ulation living in urban areas in 2010. There is one

outlier: a state where 100% of the population is ur-

ban. The variability in the percent of homeownership

also increases as we move from left to right in the plot.

(b) The outlier is located in the bottom right corner,

horizontally far from the center of the other points,

so it is a point with high leverage. It is an influen-

tial point since excluding this point from the analysis

would greatly affect the slope of the regression line.

8.31 (a) The relationship is positive, moderate-to-

strong, and linear. There are a few outliers but no

points that appear to be influential.

(b) ̂weight = −105.0113 + 1.0176× height.Slope: For each additional centimeter in height, the

model predicts the average weight to be 1.0176 addi-

tional kilograms (about 2.2 pounds).

Intercept: People who are 0 centimeters tall are ex-

pected to weigh – 105.0113 kilograms. This is obvi-

ously not possible. Here, the y- intercept serves only

to adjust the height of the line and is meaningless by

itself.

(β1 = 0).

HA: The true slope coefficient of height is different

than zero (β1 6= 0).

The p-value for the two-sided alternative hypothe-

sis (β1 6= 0) is incredibly small, so we reject H0.

The data provide convincing evidence that height and

weight are positively correlated. The true slope pa-

rameter is indeed greater than 0.

(d) R2 = 0.722 = 0.52. Approximately 52% of the

variability in weight can be explained by the height

of individuals.

8.33 (a) H0: β1 = 0. HA: β1 6= 0. The p-value,

as reported in the table, is incredibly small and is

smaller than 0.05, so we reject H0. The data pro-

vide convincing evidence that wives’ and husbands’

heights are positively correlated.

(b) ̂heightW = 43.5755 + 0.2863× heightH .

height, the average wife’s height is expected to be

an additional 0.2863 inches on average. Intercept:

Men who are 0 inches tall are expected to have wives

who are, on average, 43.5755 inches tall. The inter-

cept here is meaningless, and it serves only to adjust

the height of the line.

(d) The slope is positive, so r must also be positive.

r =√

0.09 = 0.30.

(e) 63.2612. Since R2 is low, the prediction based on

this regression model is not very reliable.

(f) No, we should avoid extrapolating.

8.35 (a) H0 : β1 = 0;HA : β1 6= 0 (b) The p-

value for this test is approximately 0, therefore we

reject H0. The data provide convincing evidence

that poverty percentage is a significant predictor of

murder rate. (c) n = 20, df = 18, T ∗18 = 2.10;

2.559±2.10×0.390 = (1.74, 3.378); For each percent-

age point poverty is higher, murder rate is expected

to be higher on average by 1.74 to 3.378 per million.

(d) Yes, we rejected H0 and the confidence interval

does not include 0.

8.37 (a) True. (b) False, correlation is a measure

of the linear association between any two numerical

variables.

401

8.39 (a) The point estimate and standard error are

b1 = 0.9112 and SE = 0.0259. We can compute

a T-score: T = (0.9112 − 1)/0.0259 = −3.43. Us-

ing df = 168, the p-value is about 0.001, which

is less than α = 0.05. That is, the data provide

strong evidence that the average difference between

husbands’ and wives’ ages has actually changed over

time. (b) âgeW = 1.5740 + 0.9112×ageH . (c) Slope:

For each additional year in husband’s age, the model

predicts an additional 0.9112 years in wife’s age. This

means that wives’ ages tend to be lower for later

ages, suggesting the average gap of husband and

wife age is larger for older people. Intercept: Men

who are 0 years old are expected to have wives who

are on average 1.5740 years old. The intercept here

is meaningless and serves only to adjust the height

of the line. (d) R =√

0.88 = 0.94. The regres-

sion of wives’ ages on husbands’ ages has a positive

slope, so the correlation coefficient will be positive.

(e) âgeW = 1.5740 + 0.9112 × 55 = 51.69. Since R2

is pretty high, the prediction based on this regres-

sion model is reliable. (f) No, we shouldn’t use the

same model to predict an 85 year old man’s wife’s

age. This would require extrapolation. The scatter-

plot from an earlier exercise shows that husbands in

this data set are approximately 20 to 65 years old.

The regression model may not be reasonable outside

of this range.

8.41 There is an upwards trend. However, the vari-

ability is higher for higher calorie counts, and it looks

like there might be two clusters of observations above

and below the line on the right, so we should be cau-

tious about fitting a linear model to these data.

8.43 (a) r = −0.72 → (2) (b) r = 0.07 → (4)

9 Multiple and logistic regression

9.1 (a) ̂baby weight = 123.05−8.94×smoke (b) The

estimated body weight of babies born to smoking

mothers is 8.94 ounces lower than babies born to non-

smoking mothers. Smoker: 123.05−8.94×1 = 114.11

ounces. Non-smoker: 123.05 − 8.94 × 0 = 123.05

ounces. (c) H0: β1 = 0. HA: β1 6= 0. T = −8.65,

and the p-value is approximately 0. Since the p-value

is very small, we reject H0. The data provide strong

evidence that the true slope parameter is different

than 0 and that there is an association between birth

weight and smoking. Furthermore, having rejected

H0, we can conclude that smoking is associated with

lower birth weights.

9.3 (a) ̂baby weight = −80.41 + 0.44 × gestation −3.33 × parity − 0.01 × age + 1.15 × height + 0.05 ×weight − 8.40 × smoke. (b) βgestation: The model

predicts a 0.44 ounce increase in the birth weight of

the baby for each additional day of pregnancy, all

else held constant. βage: The model predicts a 0.01

ounce decrease in the birth weight of the baby for

each additional year in mother’s age, all else held

constant. (c) Parity might be correlated with one of

the other variables in the model, which complicates

model estimation. (d) ̂baby weight = 120.58. e =

120 − 120.58 = −0.58. The model over-predicts this

baby’s birth weight. (e) R2 = 0.2504. R2adj = 0.2468.

9.5 (a) (-0.32, 0.16). We are 95% confident that

male students on average have GPAs 0.32 points

lower to 0.16 points higher than females when con-

trolling for the other variables in the model. (b) Yes,

since the p-value is larger than 0.05 in all cases (not

including the intercept).

9.7 Remove age.

9.9 Based on the p-value alone, either gestation or

smoke should be added to the model first. However,

since the adjusted R2 for the model with gestation is

higher, it would be preferable to add gestation in the

first step of the forward- selection algorithm. (Other

explanations are possible. For instance, it would be

reasonable to only use the adjusted R2.)

9.11 She should use p-value selection since she is

interested in finding out about significant predictors,

not just optimizing predictions.

9.13 Nearly normal residuals: With so many obser-

vations in the data set, we look for particularly ex-

treme outliers in the histogram and do not see any.

variability of residuals: The scatterplot of the resid-

uals versus the fitted values does not show any over-

all structure. However, values that have very low or

very high fitted values appear to also have somewhat

larger outliers. In addition, the residuals do appear

to have constant variability between the two parity

and smoking status groups, though these items are

relatively minor.

Independent residuals: The scatterplot of residuals

versus the order of data collection shows a random

scatter, suggesting that there is no apparent struc-

tures related to the order the data were collected.

Linear relationships between the response variable

and numerical explanatory variables: The residuals

vs. height and weight of mother are randomly dis-

tributed around 0. The residuals vs. length of ges-

tation plot also does not show any clear or strong

remaining structures, with the possible exception of

very short or long gestations. The rest of the residu-

als do appear to be randomly distributed around 0.

All concerns raised here are relatively mild. There

are some outliers, but there is so much data that the

influence of such observations will be minor.

402 APPENDIX A. EXERCISE SOLUTIONS

9.15 (a) There are a few potential outliers, e.g. on

the left in the total length variable, but nothing

that will be of serious concern in a data set this

large. (b) When coefficient estimates are sensitive

to which variables are included in the model, this

typically indicates that some variables are collinear.

For example, a possum’s gender may be related to

its head length, which would explain why the coef-

ficient (and p-value) for sex male changed when we

removed the head length variable. Likewise, a pos-

sum’s skull width is likely to be related to its head

length, probably even much more closely related than

the head length was to gender.

9.17 (a) The logistic model relating p̂i to the pre-

dictors may be written as log(

p̂i1−p̂i

)= 33.5095 −

1.4207×sex malei−0.2787×skull widthi+0.5687×total lengthi − 1.8057 × tail lengthi. Only total

length has a positive association with a possum be-

ing from Victoria. (b) p̂ = 0.0062. While the proba-

bility is very near zero, we have not run diagnostics

on the model. We might also be a little skeptical that

the model will remain accurate for a possum found

in a US zoo. For example, perhaps the zoo selected a

possum with specific characteristics but only looked

in one region. On the other hand, it is encouraging

that the possum was caught in the wild. (Answers

regarding the reliability of the model probability will

vary.)

9.19 (a) False. When predictors are collinear, it

means they are correlated, and the inclusion of one

variable can have a substantial influence on the point

estimate (and standard error) of another. (b) True.

from an experiment and x1 was one of the variables

set by the researchers. (Multiple regression can be

useful for forming hypotheses about causal relation-

ships, but it offers zero guarantees.) (d) False. We

should check normality like we would for inference

for a single mean: we look for particularly extreme

outliers if n ≥ 30 or for clear outliers if n < 30.

9.21 (a) exclaim subj should be removed, since it’s

removal reduces AIC the most (and the resulting

model has lower AIC than the None Dropped model).

(b) Removing any variable will increase AIC, so we

should not remove any variables from this set.

9.23 (a) The equation is:

log

(pi

1− pi

)= −0.8124

− 2.6351× to multiple

+ 1.6272× winner

− 1.5881× format

− 3.0467× re subj

(b) First find log(

p1−p

), then solve for p:

log

1− p

)= −0.8124− 2.6351× 0 + 1.6272× 1

− 1.5881× 0− 3.0467× 0

= 0.8148p

1− p = e0.8148 → p = 0.693

be very disruptive to the person using the email ser-

vice if they are missing emails that aren’t spam. Even

only a 90% chance that a message is spam is probably

enough to warrant keeping it in the inbox. Maybe a

probability of 99% would be a reasonable cutoff. As

for other ideas to make it even better, it may be worth

building a second model that tries to classify the im-

portance of an email message. If we have both the

spam model and the importance model, we now have

a better way to think about cost-benefit tradeoffs.

For instance, perhaps we would be willing to have

a lower probability-of-spam threshold for messages

we were confident were not important, and perhaps

we want an even higher probability threshold (e.g.

99.99%) for emails we are pretty sure are important.

403

Appendix B

Data sets within the text

Each data set within the text is described in this appendix, and there is a corresponding page foreach of these data sets at openintro.org/data. This page also includes additional data sets thatcan be used for honing your skills. Each data set has its own page with the following information:

• List of the data set’s variables.

• CSV download.

• R object file download.

B.1 Introduction to data

1.1 stent30, stent365 → The stent data is split across two data sets, one for days 0-30 resultsand one for days 0-365 results.Chimowitz MI, Lynn MJ, Derdeyn CP, et al. 2011. Stenting versus Aggressive MedicalTherapy for Intracranial Arterial Stenosis. New England Journal of Medicine 365:993-1003.www.nejm.org/doi/full/10.1056/NEJMoa1105335.NY Times article: www.nytimes.com/2011/09/08/health/research/08stent.html.

1.2 loan50, loans full schema→ This data comes from Lending Club (lendingclub.com), whichprovides a large set of data on the people who received loans through their platform. The dataused in the textbook comes from a sample of the loans made in Q1 (Jan, Feb, March) 2018.

1.2 county, county complete → These data come from several government sources. For thosevariables included in the county data set, only the most recent data is reported, as of whatwas available in late 2018. Data prior to 2011 is all from census.gov, where the specific QuickFacts page providing the data is no longer available. The more recent data comes from USDA(ers.usda.gov), Bureau of Labor Statistics (bls.gov/lau), SAIPE (census.gov/did/www/saipe),and American Community Survey (census.gov/programs-surveys/acs).

1.3 Nurses’ Health Study → For more information on this data set, seewww.channing.harvard.edu/nhs

1.4 The study we had in mind when discussing the simple randomization (no blocking) study wasAnturane Reinfarction Trial Research Group. 1980. Sulfinpyrazone in the prevention of suddendeath after myocardial infarction. New England Journal of Medicine 302(5):250-256.

B.2 Summarizing data

2.1 loan50, county → These data sets are described in Data Appendix B.1.

2.2 loan50, county → These data sets are described in Data Appendix B.1.

2.3 malaria → Lyke et al. 2017. PfSPZ vaccine induces strain-transcending T cells and durableprotection against heterologous controlled human malaria infection. PNAS 114(10):2711-2716.www.pnas.org/content/114/10/2711

404 APPENDIX B. DATA SETS WITHIN THE TEXT

B.3 Probability

3.1 loan50, county → These data sets are described in Data Appendix B.1.

3.1 playing cards → Data set describing the 52 cards in a standard deck.

3.2 family college → Simulated data based on real population summaries atnces.ed.gov/pubs2001/2001126.pdf.

3.2 smallpox → Fenner F. 1988. Smallpox and Its Eradication (History of International PublicHealth, No. 6). Geneva: World Health Organization. ISBN 92-4-156110-6.

3.2 Mammogram screening, probabilities→ The probabilities reported were obtained using studiesreported at www.breastcancer.org and www.ncbi.nlm.nih.gov/pmc/articles/PMC1173421.

3.2 Jose campus visits, probabilities → This example was made up.

3.3 No data sets were described in this section.

3.4 Course material purchases and probabilities → This example was made up.

3.4 Auctions for TV and toaster → This example was made up.

3.4 stocks 18 → Monthly returns for Caterpillar, Exxon Mobil Corp, and Google for November2015 to October 2018.

3.5 fcid → This sample can be considered a simple random sample from the US population. Itrelies on the USDA Food Commodity Intake Database.

B.4 Distributions of random variables

4.1 SAT and ACT score distributions → The SAT score data comes from the 2018 distribution,which is provided atreports.collegeboard.org/pdf/2018-total-group-sat-suite-assessments-annual-report.pdf

The ACT score data is available atact.org/content/dam/act/unsecured/documents/cccr2018/P 99 999999 N S N00 ACT-GCPR National.pdf

We also acknowledge that the actual ACT score distribution is not nearly normal. However,since the topic is very accessible, we decided to keep the context and examples.

4.1 Male heights → The distribution is based on the USDA Food Commodity Intake Database.

4.1 possum → The distribution parameters are based on a sample of possums from Australiaand New Guinea. The original source of this data is as follows. Lindenmayer DB, et al.1995. Morphological variation among columns of the mountain brushtail possum, Trichosuruscaninus Ogilby (Phalangeridae: Marsupiala). Australian Journal of Zoology 43: 449-458.

4.2 Exceeding insurance deductible → These statistics were made up but are possible values onemight observe for low-deductible plans.

4.3 Exceeding insurance deductible → These statistics were made up but are possible values onemight observe for low-deductible plans.

4.3 Smoking friends→ Unfortunately, we don’t currently have additional information on the sourcefor the 30% statistic, so don’t consider this one as fact since we cannot verify it was from areputable source.

4.3 US smoking rate → The 15% smoking rate in the US figure is close to the value from theCenters for Disease Control and Prevention website, which reports a value of 14% as of the2017 estimate:cdc.gov/tobacco/data statistics/fact sheets/adult data/cig smoking/index.htm

4.4 Football kicker → This example was made up.

4.4 Heart attack admissions → This example was made up, though the heart attack admissionsare realistic for some hospitals.

4.5 ami occurrences→ This is a simulated data set but resembles actual AMI data for New YorkCity based on typical AMI incidence rates.

405

B.5 Foundations for inference

5.1 pew energy 2018→ The actual data has more observations than were referenced in this chap-ter. That is, we used a subsample since it helped smooth some of the examples to have a bitmore variability. The pew energy 2018 data set represents the full data set for each of the dif-ferent energy source questions, which covers solar, wind, offshore drilling, hydrolic fracturing,and nuclear energy. The statistics used to construct the data are from the following page:

www.pewinternet.org/2018/05/14/majorities-see-government-efforts-to-protect-the-environment-as-

insufficient/

5.2 pew energy 2018 → See the details for this data set above in the Section 5.1 data section.

5.2 ebola survey → In New York City on October 23rd, 2014, a doctor who had recently beentreating Ebola patients in Guinea went to the hospital with a slight fever and was subsequentlydiagnosed with Ebola. Soon thereafter, an NBC 4 New York/The Wall Street Journal/MaristPoll found that 82% of New Yorkers favored a “mandatory 21-day quarantine for anyone whohas come in contact with an Ebola patient”. This poll included responses of 1,042 New Yorkadults between Oct 26th and 28th, 2014. Poll ID NY141026 on maristpoll.marist.edu.

5.3 pew energy 2018 → See the details for this data set above in the Section 5.1 data section.

5.3 Rosling questions → We noted much smaller samples than the Roslings’ describe in theirbook, Factfulness, The samples we describe are similar but not the same as the actual rates.The approximate rates for the correct answers for the two questions for (sometimes different)populations discussed in the book, as reported in Factfulness, are

– 80% of the world’s 1 year olds have been vaccinated against some disease: 13% get thiscorrect (17% in the US). gapm.io/q9

– Number of children in the world in 2100: 9% correct. gapm.io/q5

Here are a few more questions and a rough percent of people who get them correct:

– In all low-income countries across the world today, how many girls finish primary school:20%, 40%, or 60%? Answer: 60%. About 7% of people get this question correct.gapm.io/q1

– What is the life expectancy of the world today: 50 years, 60 years, or 70 years? Answer:70 years. In the US, about 43% of people get this question correct. gapm.io/q4

– In 1996, tigers, giant pandas, and black rhinos were all listed as endangered. How manyof these three species are more critically endangered today: two of them, one of them,none of them? Answer: none of them. About 7% of people get this question correct.gapm.io/q11

– How many people in the world have some access to electricity? 20%, 50%, 80%. Answer:80%. About 22% of people get this correct. gapm.io/q12

For more information, check out the book, Factfulness.

5.3 pew energy 2018 → See the details for this data set above in the Section 5.1 data section.

5.3 nuclear survey → A simple random sample of 1,028 US adults in March 2013 found that56% of US adults support nuclear arms reduction.www.gallup.com/poll/161198/favor-russian-nuclear-arms-reductions.aspx

5.3 Car manufacturing → This example was made up.

5.3 stent30, stent365 → These data sets are described in Data Appendix B.1.

406 APPENDIX B. DATA SETS WITHIN THE TEXT

B.6 Inference for categorical data

6.1 Payday loans → The statistics come from the following source:pewtrusts.org/-/media/assets/2017/04/payday-loan-customers-want-more-protections-methodology.pdf

6.1 Tire factory → This example was made up.

6.2 cpr → Böttiger et al. Efficacy and safety of thrombolytic therapy after initially unsuccessfulcardiopulmonary resuscitation: a prospective clinical trial. The Lancet, 2001.

6.2 fish oil 18 → Manson JE, et al. 2018. Marine n-3 Fatty Acids and Prevention of Cardio-vascular Disease and Cancer. NEJMoa1811403.

6.2 mammogram → Miller AB. 2014. Twenty five year follow-up for breast cancer incidence andmortality of the Canadian National Breast Screening Study: randomised screening trial. BMJ2014;348:g366.

6.2 drone blades → The quality control data set for quadcopter drone blades is a made-up dataset for an example. We provide the simulated data in the drone blades data set.

6.3 jury→ The jury data set for examining discrimination is a made-up data set an example. Weprovide the simulated data in the jury data set.

6.3 sp500 1950 2018 → Data is sourced from finance.yahoo.com.

6.4 ask → Minson JA, Ruedy NE, Schweitzer ME. There is such a thing as a stupid question:Question disclosure in strategic communication.opim.wharton.upenn.edu/DPlab/papers/workingPapers/

Minson working Ask%20(the%20Right%20Way)%20and%20You%20Shall%20Receive.pdf

6.4 diabetes2 → Zeitler P, et al. 2012. A Clinical Trial to Maintain Glycemic Control in Youthwith Type 2 Diabetes. N Engl J Med.

B.7 Inference for numerical data

7.1 Risso’s dolphins → Endo T and Haraguchi K. 2009. High mercury levels in hair samples fromresidents of Taiji, a Japanese whaling town. Marine Pollution Bulletin 60(5):743-747.

Taiji was featured in the movie The Cove, and it is a significant source of dolphin and whalemeat in Japan. Thousands of dolphins pass through the Taiji area annually, and we assumesthese 19 dolphins reasonably represent a simple random sample from those dolphins.

7.1 Croaker white fish → fda.gov/food/foodborneillnesscontaminants/metals/ucm115644.htm

7.1 run17 → www.cherryblossom.org

7.2 textbooks, ucla textbooks f18→ Data were collected by OpenIntro staff in 2010 and againin 2018. For the 2018 sample, we sampled 201 UCLA courses. Of those, 68 required booksthat could be found on Amazon. The websites where information was retrieved:sa.ucla.edu/ro/public/soc, ucla.verbacompare.com, and amazon.com.

7.3 stem cells→Menard C, et al. 2005. Transplantation of cardiac-committed mouse embryonicstem cells to infarcted sheep myocardium: a preclinical study. The Lancet: 366:9490, p1005-1012.

7.3 ncbirths → Birth records released by North Carolina in 2004. Unfortunately, we don’t cur-rently have additional information on the source for this data set.

7.3 Exam versions → This example was made up.

7.4 Blood pressure statistics → The blood pressure standard deviation for patients with bloodpressure ranging from 140 to 180 mmHg is guessed and may be a little (but likely not dramat-ically) imprecise from what we’d observe in actual data.

7.5 toy anova → Data used for Figure 7.19, where this data was made up.

7.5 mlb players 18 → Data were retrieved from mlb.mlb.com/stats. Only players with at least100 at bats were considered during the analysis.

7.5 classdata → This example was made up.

407

B.8 Introduction to linear regression

8.1 simulated scatter → Fake data used for the first three plots. The perfect linear plot usesgroup 4 data, where group variable in the data set (Figure 8.1). The group of 3 imperfectlinear plots use groups 1-3 (Figure 8.2). The sinusoidal curve uses group 5 data (Figure 8.3).The group of 3 scatterplots with residual plots use groups 6-8 (Figure 8.8). The correlationplots uses groups 9-19 data (Figures 8.9 and 8.10).

8.1 possum → This data set is described in Data Appendix B.4.

8.2 elmhurst→ These data were sampled from a table of data for all freshman from the 2011 classat Elmhurst College that accompanied an article titled What Students Really Pay to Go toCollege published online by The Chronicle of Higher Education: chronicle.com/article/What-Students-Really-Pay-to-Go/131435.

8.2 simulated scatter→ The plots for things that can go wrong uses groups 20-23 (Figure 8.12).

8.2 mariokart → Auction data from Ebay (ebay.com) for the game Mario Kart for the NintendoWii. This data set was collected in early October, 2009.

8.3 simulated scatter → The plots for types of outliers uses groups 24-29 (Figure 8.18).

8.4 midterms house → Data was retrieved from Wikipedia.

B.9 Multiple and logistic regression

9.1 loans full schema → This data set is described in Data Appendix B.1.

9.2 loans full schema → This data set is described in Data Appendix B.1.

9.3 loans full schema → This data set is described in Data Appendix B.1.

9.4 mariokart → This data set is described in Data Appendix B.8.

9.5 resume → Bertrand M, Mullainathan S. 2004. Are Emily and Greg More Employable thanLakisha and Jamal? A Field Experiment on Labor Market Discrimination. The AmericanEconomic Review 94:4 (991-1013). www.nber.org/papers/w9873

We did omit discussion of some structure in the data for the analysis presented: the exper-iment design included blocking, where typically four resumes were sent to each job: one foreach inferred race/sex combination (as inferred based on the first name). We did not worryabout this blocking aspect, since accounting for the blocking would reduce the standard er-ror without notably changing the point estimates for the race and sex variables versus theanalysis performed in the section. That is, the most interesting conclusions in the study areunaffected even when completing a more sophisticated analysis.

408

Appendix C

Distribution tables

C.1 Normal Probability Table

A normal probability table may be used to find percentiles of a normal distribution using aZ-score, or vice-versa. Such a table lists Z-scores and the corresponding percentiles. An abbreviatedprobability table is provided in Figure C.1 that we’ll use for the examples in this appendix. A fulltable may be found on page 410.

Second decimal place of ZZ 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

……

…

Figure C.1: A section of the normal probability table. The percentile for a normalrandom variable with Z = 1.00 has been highlighted, and the percentile closest to0.8000 has also been highlighted.

When using a normal probability table to find a percentile for Z (rounded to two decimals),identify the proper row in the normal probability table up through the first decimal, and thendetermine the column representing the second decimal value. The intersection of this row andcolumn is the percentile of the observation. For instance, the percentile of Z = 0.45 is shown in row0.4 and column 0.05 in Figure C.1: 0.6736, or the 67.36th percentile.

Negative Z Positive Z

Figure C.2: The area to the left of Z represents the percentile of the observation.

409

EXAMPLE C.1

SAT scores follow a normal distribution, N(1100, 200). Ann earned a score of 1300 on her SAT witha corresponding Z-score of Z = 1. She would like to know what percentile she falls in among allSAT test-takers.

Ann’s percentile is the percentage of people who earned a lower SAT score than her. We shadethe area representing those individuals in the following graph:

500 700 900 1100 1300 1500 1700

The total area under the normal curve is always equal to 1, and the proportion of people who scoredbelow Ann on the SAT is equal to the area shaded in the graph. We find this area by looking inrow 1.0 and column 0.00 in the normal probability table: 0.8413. In other words, Ann is in the 84th

percentile of SAT takers.

EXAMPLE C.2

How do we find an upper tail area?

The normal probability table always gives the area to the left. This means that if we want the areato the right, we first find the lower tail and then subtract it from 1. For instance, 84.13% of SATtakers scored below Ann, which means 15.87% of test takers scored higher than Ann.

We can also find the Z-score associated with a percentile. For example, to identify Z for the80th percentile, we look for the value closest to 0.8000 in the middle portion of the table: 0.7995.We determine the Z-score for the 80th percentile by combining the row and column Z values: 0.84.

EXAMPLE C.3

Find the SAT score for the 80th percentile.

We look for the are to the value in the table closest to 0.8000. The closest value is 0.7995, whichcorresponds to Z = 0.84, where 0.8 comes from the row value and 0.04 comes from the column value.Next, we set up the equation for the Z-score and the unknown value x as follows, and then we solvefor x:

Z = 0.84 =x− 1100

200→ x = 1268

The College Board scales scores to increments of 10, so the 80th percentile is 1270. (Reporting 1268would have been perfectly okay for our purposes.)

For additional details about working with the normal distribution and the normal probabilitytable, see Section 4.1, which starts on page 133.

410 APPENDIX C. DISTRIBUTION TABLES

Negative Z

Second decimal place of Z0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 Z

0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 −3.40.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 −3.30.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 −3.20.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 −3.10.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 −3.0

0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 −2.90.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.0026 −2.80.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 −2.70.0036 0.0037 0.0038 0.0039 0.0040 0.0041 0.0043 0.0044 0.0045 0.0047 −2.60.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.0062 −2.5

0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.0082 −2.40.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.0107 −2.30.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.0139 −2.20.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179 −2.10.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.0228 −2.0

0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.0287 −1.90.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.0359 −1.80.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 0.0436 0.0446 −1.70.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.0548 −1.60.0559 0.0571 0.0582 0.0594 0.0606 0.0618 0.0630 0.0643 0.0655 0.0668 −1.5

0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.0808 −1.40.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 −1.30.0985 0.1003 0.1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.1151 −1.20.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.1357 −1.10.1379 0.1401 0.1423 0.1446 0.1469 0.1492 0.1515 0.1539 0.1562 0.1587 −1.0

0.1611 0.1635 0.1660 0.1685 0.1711 0.1736 0.1762 0.1788 0.1814 0.1841 −0.90.1867 0.1894 0.1922 0.1949 0.1977 0.2005 0.2033 0.2061 0.2090 0.2119 −0.80.2148 0.2177 0.2206 0.2236 0.2266 0.2296 0.2327 0.2358 0.2389 0.2420 −0.70.2451 0.2483 0.2514 0.2546 0.2578 0.2611 0.2643 0.2676 0.2709 0.2743 −0.60.2776 0.2810 0.2843 0.2877 0.2912 0.2946 0.2981 0.3015 0.3050 0.3085 −0.5

0.3121 0.3156 0.3192 0.3228 0.3264 0.3300 0.3336 0.3372 0.3409 0.3446 −0.40.3483 0.3520 0.3557 0.3594 0.3632 0.3669 0.3707 0.3745 0.3783 0.3821 −0.30.3859 0.3897 0.3936 0.3974 0.4013 0.4052 0.4090 0.4129 0.4168 0.4207 −0.20.4247 0.4286 0.4325 0.4364 0.4404 0.4443 0.4483 0.4522 0.4562 0.4602 −0.10.4641 0.4681 0.4721 0.4761 0.4801 0.4840 0.4880 0.4920 0.4960 0.5000 −0.0∗For Z ≤ −3.50, the probability is less than or equal to 0.0002.

411

Positive Z