Instructions
Use the accompanying datasets BUS-7106 Data Set Week 2 Assignment Excel file and BUS-7106 Data Set Week 2 Assignment SPSS file to address the following assignment.
In a firm with 500 employees, the CEO is interested in knowing what level of commitment the company’s employees have in the firm. The CEO suspects that approximately 250 full-time employees are more committed (e.g., emotional attachment to the firm) than the nearly 250 contracted personnel. The CEO asked the human resources director to survey all employees using an established measurement scale that taps into organizational commitment. The dataset represents the average scores of each worker on this scale. Test the null hypothesis that there is not a statistically significant difference between the two groups of workers using the independent samples t-test in both Excel and SPSS.
In your write-up of this study, please briefly describe the purpose of the study, describe the use of the t-test to explore this hypothesis, and report the findings in detail. Last, please explain the results in narrative form. In other words, tell the story of the study beyond simply calculating the t-value.
Length: 3-5 pages
References: At least 3 resources (remember that analysis software counts as source materials)
Week 2
| Job Status | Commitment |
| 0 | 5 |
| 0 | 3 |
| 0 | 2.5 |
| 0 | 4 |
| 0 | 4.5 |
| 0 | 3 |
| 0 | 2.5 |
| 0 | 4.25 |
| 0 | 4 |
| 0 | 4.25 |
| 0 | 3 |
| 0 | 5 |
| 0 | 4.5 |
| 0 | 4 |
| 0 | 2.5 |
| 0 | 3 |
| 0 | 4 |
| 0 | 3 |
| 0 | 4.25 |
| 0 | 2.5 |
| 0 | 4.25 |
| 0 | 3 |
| 0 | 5 |
| 0 | 4 |
| 0 | 4.25 |
| 0 | 3 |
| 0 | 4.5 |
| 0 | 2.5 |
| 0 | 3 |
| 0 | 4 |
| 1 | 5 |
| 1 | 3.5 |
| 1 | 2.5 |
| 1 | 2.5 |
| 1 | 3 |
| 1 | 4 |
| 1 | 3.25 |
| 1 | 4.5 |
| 1 | 4 |
| 1 | 5 |
| 1 | 3.5 |
| 1 | 3.25 |
| 1 | 3 |
| 1 | 4 |
| 1 | 2.5 |
| 1 | 4.5 |
| 1 | 5 |
| 1 | 3.5 |
| 1 | 5 |
| 1 | 3.25 |
| 1 | 4 |
| 1 | 3 |
| 1 | 3.5 |
| 1 | 5 |
| 1 | 3.25 |
| 1 | 4 |
| 1 | 3.5 |
| 1 | 5 |
| 1 | 3.75 |
| 1 | 3 |
Apply the T-Test
School of Business, Northcentral University.
January, 2022
Introduction
This paper describes what an independent T test is, how to use it properly, how it was first derived, what each part of the formula represents, the T test’s strengths, weaknesses, limitations, and when to question its results. As per the assignment, two theoretical datasets within Excel and SPSS are analyzed using an independent samples T test examining the differences in level of commitment or emotional attachment to a firm between 250 contracted workers and 250 full-time employees. I describe the purpose of the study, the limitations, and why the T-test was used in the first place, with a narrative describing the overarching themes of the assignment. The next sections highlight how a T-test can be applied to COVID data in SPSS and MATLAB to illustrate this week’s material’s practical application to research. This might not be possible given the assumption within the t-test, but this is tested within this paper.
Independent T-Test Detailed Overview
When a researcher wants to compare the means between two unrelated groups with the same continuous, dependent variable an independent samples t-test (independent t-test), can be performed (). Two examples are, seeing if second-year performance in a graduate program in business varied based upon gender, where the dependent variable is second-year performance and the independent variable is gender with two groups, male, and female (). The second example is there a difference in performance on the GMAT based upon educational level with two groups current undergraduate students those who graduated (). Test performance is the dependent variable and educational level is the independent variable that can be manipulated ().
The independent t-test has several assumptions that must be followed in order to be considered reliable and valid which are:
T-Test: Analyzing, Applying, and Interpreting
1.) The dependent variable must take the form of interval or ratio within a continuous scale, like graded test performance, IQ score, or weight.
2.) The IV needs to consist of two categorical, independent groups, known as a binary classification, like gender, employer status, or drinker, yes, or no.
3.) The participants of one group must be 100% independent of the other group with zero overlap.
4.) There cannot be any significant outliers, but in SPSS the software easily shows the outliers and illustrates their effects.
5.) The dependent variable need not be exactly normally distributed, but it needs to be approximately normally distributed.
6.) There must be a homogeneity of variances (). This will be discussed in more detail later in the paper. Suffice to say, for now, within SPSS one can test for the homogeneity of variances via Levene’s test for equality of variances ().
The independent sample t-test is also known as the two sample t-test, or student’s t-test, taking the form of an inferential statistical test where: The null hypothesis takes the form, H0: u1 = u2, where the population means from two unrelated groups equate, and the alternative hypothesis takes the form, HA: u1 ≠ u2 where the two population means from each unrelated group are not equal (). In most cases it is HA being looked for to reject the null hypothesis, and a significance level of alpha is set, and is usually 0.05, but it can be higher or lower based upon a given data structure, sample size, and researcher’s assumptions ().
T-test history and Development
In 1908 an English chemist working for Guinness, William Sealy Gosset developed the t- distribution, a probability distribution of curves, and the t-test under the pseudonym of student ().
While Gosset created the statistical curves and tests to compare the means of two independent batches of alcohol for Guinness, he published his findings in the journal Biometrika under the pseudonym of student as a requirement of his employer at the time (). To put it more precisely the t-distribution are a family of curves that approximate the normal curve but they are a little shorter and fatter due to them being used on smaller sample sizes that do not quite generate a normal bell shaped curve, and when the sample size is >20 the curve is almost exactly the same as a normal curve (). At >=30 sample size the distinguishing features are non-existent or not relevant (). The t distribution is leptokurtic with a kurtosis greater than three, where a normal distribution has a kurtosis of three (). However, there are in some cases, separate calculations for moment of kurtosis versus moment of skewness (), but this is outside the scope of this paper.
High kurtosis indicates heavier tails, meaning more outliers, however, an effective t-test based upon the t-distribution, while having fatter tails than a normal distribution’s sample still tends to be low enough on outliers to be effective if the data is appropriate for a t-test (). Kurtosis is also
referred to as skewness, or level of skewness, where the general formula takes the form: µ4/ σ4 = Kurtosis, or Kurt, for short (). µ4 represents the fourth central moment and σ4 represents the standard deviation (). If kurtosis exceeds the limits allowed for a given data set, and outside the
upper bounds allowed for the t-distribution, and thus, the t-test, then the results would be invalid due toa significant number of outliers (). The full treatment of calculating moment coefficient of skewness or kurtosis, is beyond the scope of this paper, however, a general treatment is included as it relates to the t-test. For skewness checks the general calculation is
(
2
3
2
)skewness:
g1 = m3 / m 3/2 m = ∑(x−x)3 / n and m = ∑(x−x)̅ 2 / n,
With xbar as the mean, and n the sample size, and m3 the third moment within the data set, with four moments in total which are, the mean, the variance, and standard deviation, skewness, and
kurtosis (). The standard deviation, the square root of the variance is more commonly used because its derivation represents an easy to understand probability along a normal or approximately normal curve with an easily depicted spread (). The formula for the population standard deviation is:
Where sigma is the population standard deviation, N is the population size, xi is the is each value from the population and mu is the population mean (). For the sample standard deviation, the formula is:
A full mathematical treatment of S.D. and the t-test is outside the scope of this paper, but keep in mind under different assumptions of a population or sample, and equal between group variance, or asymmetrical variance, the t-test would be calculated differently and the tails would become either flatter or fatter due to alterations in kurtosis (). In general, unequal variance is not recommended to apply a t-test to, but there are methods to analyze the asymmetrical variance and normalize the results, but there is statistical skepticism to this approach ().
Clarification and Discussion
The reason more advanced topics were touched upon, some in detail and others at a surface level is because what is taught in undergraduate statistics courses and most business graduate statistics courses leave out important information for advanced research, journal submission, and proper quantitative based dissertations. 60%-72% of all research studies suffer
being unable to be replicated or suffer from regression to the mean (). Ivey League and other tier 1 Universities are not immune to these issues (). As a scholar practitioner, PhD student, published researcher, and aspiring University professor, I want to produce the highest quality research with my collaborators as is possible and as an independent researcher as well. I may never teach at Stanford or Harvard, but I want to produce research that gets published in the high impact journals and influences the science of statistics and financial econometrics.
Excel Data T-Test Analysis
Below is the Excel analysis of the assigned data using the independent sample t-test.
First an F test for two groups is performed to see if the variance is equal between each group with the results below:
|
F-Test Two-Sample for Variances |
||
|
Variable 1 |
Variable 2 |
|
|
Mean |
0.5 |
3.71666 7 |
|
Variance |
0.25423 7 |
0.66624 3 |
|
Observations |
60 |
60 |
|
df |
59 |
59 |
|
F |
0.38159 8 |
|
|
P(F<=f) one- tail |
0.00014 9 |
|
|
F Critical one-tail |
0.64936 9 |
The variance is not exactly equal, so a two-sample assumed unequal variances is applied. Recall earlier it was stated in this paper that an unequal variance approach is more controversial than an equal variance independent sample t-test. This is not a mainstream view but a controversy between statisticians and mathematicians, and the simple explanation is the assumptions within pooling usually work fine and unequal variance can sometimes unnecessarily skew results, but unequal variance measures can avoid making untrue assumptions about data with wide variance differences. Having said that, for the purposes of this course an F test was used to test for unequal variance and then the ‘appropriate’ t-test was applied for this assignment.
Here is the result of the t-test assuming unequal variance:
|
t-Test: Two-Sample Assuming Unequal Variances |
||
|
Variable 1 |
Variable 2 |
|
|
Mean |
0.5 |
3.71666 7 |
|
Variance |
0.25423 7 |
0.66624 3 |
|
Observations |
60 |
60 |
|
Hypothesized Mean Difference |
0 |
|
|
df |
98 |
|
|
t Stat |
-25.9701 |
|
|
P(T<=t) one-tail |
4.98E-46 |
|
|
t Critical one-tail |
1.66055 1 |
|
|
P(T<=t) two-tail |
9.97E-46 |
|
|
t Critical two-tail |
1.98446 7 |
Here is the t-test assuming equal variance:
|
t-Test: Two-Sample Assuming Equal Variances |
||
|
Variable 1 |
Variable 2 |
|
|
Mean |
0.5 |
3.71666 7 |
|
Variance |
0.25423 7 |
0.66624 3 |
|
Observations |
60 |
60 |
|
Pooled Variance |
0.46024 |
|
|
Hypothesized Mean Difference |
0 |
|
|
df |
118 |
|
|
t Stat |
-25.9701 |
|
|
P(T<=t) one-tail |
6.31E-51 |
|
|
t Critical one-tail |
1.65787 |
|
|
P(T<=t) two-tail |
1.26E-50 |
|
|
t Critical two-tail |
1.98027 2 |
The output is almost identical with the main difference of there being a pooled variance which when applying stochastic calculus makes more sense than not having a pooled variance, and while outside the scope of this paper, I take the minority view that the equal variance assumption can be more detailed and robust than unequal variance assumption; to be safe I would always apply both methods to the data. Just to anticipate future assignments, an ANOVA is just a modified F test to compare three or more between group means which is more accurate than multiple t-tests since it controls for a Type I error more robustly ().
SPSS DATA Analysis
Below is the SPSS analysis of the assigned data using the independent sample t-test.
Group Statistics
|
Full or part time N |
Mean |
Std. Deviation |
Std. Error Mean |
||
|
Level of commitment to the firm |
Part-time employee |
30 |
3.7583 |
.81071 |
.14802 |
|
2.00 |
0a |
. |
. |
. |
a. t cannot be computed because at least one of the groups is empty. At the first attempt here is the incomplete analysis:
For a paired samples T test here are the results:
Paired Samples Statistics
|
Mean |
N |
Std. Deviation |
Std. Error Mean |
||
|
Pair 1 |
Full or part time |
.5000 |
60 |
.50422 |
.06509 |
|
Level of commitment to the firm |
3.7167 |
60 |
.81624 |
.10538 |
Paired Samples Statistics
(
Pair 1
Full or
part
time &
Level
of
commitment
to
the
firm
60
.051
) N Correlation
Paired Samples Test
(
95%
Confidence
Interval
of
th
Difference
Lower
Upper
)Paired Differences
Mean
Std. Deviation
Std. Error Mean
(
Pair 1
Full or
part
time
–
Level of
commitment
to
the
firm
-3.21667
.93707
.12098
-3.45874
-2.97459
)For an ANOVA we see:
ANOVA
Level of commitment to the firm
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
|
Between Groups |
.104 |
1 |
.104 |
.154 |
.696 |
|
Within Groups |
39.204 |
58 |
.676 |
||
|
Total |
39.308 |
59 |
Finally, applying Levene’s test in SPSS, we see:
Test of Homogeneity of Variances
|
Levene Statistic |
df1 |
df2 |
Sig. |
||
|
Level of commitment to the firm |
Based on Mean |
.582 |
1 |
58 |
.449 |
|
Based on Median |
.133 |
1 |
58 |
.716 |
|
|
Based on Median and with adjusted df |
.133 |
1 |
57.991 |
.716 |
|
|
Based on trimmed mean |
.601 |
1 |
58 |
.442 |
Variances are close to equal, and none of the P values get to or below 0.05. Thus we do not reject the null hypothesis, and the variances are not significantly different, which resembles the conclusion within Excel using assuming equal variance with a pooled variance and the results from an one-way ANOVA (). Job status is not a significant contributor to level of job commitment overall, but there are subtle seemingly random differences. If one were inclined to apply principles components analysis and partition functions there could possibly be relevant differences, but on a small scale within latent variables.
SPSS Analysis of COVID Data T-Test
MATLAB Data of COVID Data T-Test
Apply the T
–
Test
School of Business,
Northcentral University
.
BUS
–
7106 v2: Statistics II
January
, 202
2
Apply the T-Test
School of Business, Northcentral University.
BUS-7106 v2: Statistics II
January, 2022
1
Running Head: T-TEST
Apply the T-test
Nikki Owens
BUS-7106: Statistics II
Northcentral University
Dr. Fred Rispoli
May 9, 2021
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2
T-TEST
Apply the T-test
Introduction
“T-test is a statistical test that is used to compare the means of two groups. It is often used
in hypothesis testing to determine whether a process or treatment actually influences the
population or interest, or whether two groups are different from one another” (Bevans, 2020).
This paper will analyze a t-test for a CEO of a firm with 500 employees. The CEO is
performing a study to determine how committed the employees are to the firm. A t-test is
performed to test the null hypothesis which is there is no statistically significant difference
between the two groups of workers in the dataset. The t-test was chosen to analyze the results of
the hypothesis because the two variables have an unknown variance which can be known by the
t-test. The analysis information that was found during the test is discussed below.
T-test Excel
T-Test
t-Test: Two-Sample Assuming Equal Variances
Job Status
Commitme
nt
Mean 0.5 3.71666667
Variance
0.2542372
9 0.66624294
Observations 60 60
Pooled Variance
0.4602401
1
Hypothesized Mean
Difference 0
df 118
t Stat -25.97015
P(T<=t) one-tail 6.312E-51
t Critical one-tail
1.6578695
2
P(T<=t) two-tail
1.2624E-
50
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3
T-TEST
t Critical two-tail
1.9802722
5
Based on the results in the analysis above the most important number that is really
needed is the P-value. The p-value is what determines if the null hypothesis is accepted or
rejected. The p-value was found to be 1.2624E-50 and the significant alpha value used was 0.05.
Since the P-value of 1.2624E-50 is less than 0.05, the variables of job status and commitment do
have a statistically significant difference and therefore the null hypothesis would be rejected
since the CEO wanted to test the variables for no significant difference amoung them.
T-test SPSS
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4
T-TEST
Based on the results in the analysis above, the P-value was found to be 0.696 and the
significant alpha value used was 0.05. Since the P-value of 0.696 is greater than 0.05, the
variables of job status and commitment do not have a statistically significant difference and
therefore the null hypothesis would be accepted since the results was true based on what the
CEO was testing in the study.
Results
When it comes to calculating the t-value, I would simply use the formula t=mean-
theoretical mean value divided by standard deviation divided by the square root of the variable
dataset size (n). This is how I would calculate the t-value manually, however, I used Microsoft
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5
T-TEST
excel and SPSS software to calculate the t-value for the study which was a lot faster and less
time consuming. When I tested the null hypothesis t-test in Microsoft excel it was determined
that the variables had a difference, so the null hypothesis was rejected. When I tested the null
hypothesis t-test in SPSS software it was determined that the variables were the same and did not
have a difference, so the null hypothesis was accepted. Both software’s had different results but
used the same information that was collected from a survey, but the SPSS was more accurate for
the study and was able to help the CEO determine how committed the employees were based on
their job status.
Conclusion
In conclusion, based on the SPSS software, if I was the CEO, I would accept the null
hypothesis and come up with a plan to keep my employees committed to the firm by determining
where they are most experienced and place them in that position because the two variables used
from the survey had no significant differences between each other due to they had similar mean
values and the p-value was greater than the significant value of 0.05.
References
Bevans, R. (2020). An introduction to t-test. Retrieved from https://www.scribbr.com/statistics/t-
test/
Microsoft Excel Software
SPSS Software
Weiers, R. M. (2011). Introduction to business statistics (7th ed.). Mason, OH: South-
Western, Cengage Learning
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6
T-TEST
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BUS-7106-2
NORTHCENTRAL UNIVERSITY
ASSIGNMENT COVER SHEET
Student Date: 11/10/2019
Course ID # BUS-7106 V1 Professor:
Assignment Title: Apply the T-Test Assignment Number: 2
1
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BUS-7106-2
Introduction
In this study, I analysis the null hypothesis and apply the T-Test for the dataset in a company
between two groups of employees. By figuring out the mean, the average of all data samples will
be calculated and it looks like the average in field Two is higher than the average in field one.
However, it will be the only part of the subject which the mean will show us since we might have
various distributions based on those distributions and the samples’ variance, there might be an
obvious difference among them and the t- value would be the solution. In order to find the
differences between the two samples, company should calculate the ratio of samples by finding the
difference between the absolute values and the means. Since the standard deviation showing us
how the data spreading from the mean, I am not only calculating the standard deviation, but also, I
will square SD in order to show the variance.
If the variance gets increased, the value will be decreased. I have increased the sample’
numbers basically, would increase the data and will give us the higher number of t-value (Hughes,
L. W., & Palmer, D.K. 2007).
I test the null hypothesis to show statistically that there is not significantly difference among
the data which is considered as critical value since, the t-value result would be less than that.
However, I won’t reject the null hypothesis, and if the t-value is more than the critical value, the
null hypothesis would be rejected. By considering the alternate Hypothesis, there might be a kind
of situation among the two datasets.
As we know we have
sd 1
2
n 1
√(¿
+ sd 2
2
n 2
)
t −value=
mean 2−mean 1
¿
. Based on that, I calculate the results.
2
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BUS-7106-2
Hypothesis:
Null hypothesis: H0:
Alternative hypothesis: H1:
Using SPSS software hypothesis testing is:
3
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BUS-7106-2
here significance value is 0.449>0.05 Then I accept the H0 so there is no statistically significant
different among both groups.
Using MS excel:
F-Test Two-Sample for Variances
Variable 1 Variable 2
Mean 3.675 3.758333
Variance 0.694612 0.657256
Observations 30 30
df 29 29
F 1.056837
P(F<=f) one-tail 0.441345
F Critical one-tail 1.860811
here also 0.441345>0.05 so we accept H0 so there is no difference
4
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BUS-7106-2
conclusion:
Here we accept H0 so both groups have not any differences and they have equal means.
About independent t-test:
1) The two-sample t-test is utilized for comparing the two means of two independent populations
denoted by
2) samples of n1 and n2 observations are randomly selected from the two populations
3) The greater the difference between sample means the greater the evidence against the hypothesis
of equality of populations means are untrue.
5
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BUS-7106-2
References
Hughes, L. W., & Palmer, D.K. (2007). An investigation of the effects of psychological contract
and organization-based self-esteem on organizational commitment in a sample of permanent
workers.
https://eds-a-ebscohost-com.proxy1.ncu.edu/eds/detail/detail?vid=0&sid=eab533f1-4943-42e6-
9ff2-999ee4065f0b%40sdc-v-sessmgr01&bdata=JnNpdGU9ZWRzLWxpdmU%3d#AN=2009-
08811-005&db=psyh
Field, A. (Academic). (2012). T-Test. SAGE Research Methods.
Steckler, A., & McLeroy, K. R. (2008). The importance of external validity.
American Journal of Public Health
Gepp, A., Kumar, K. (2012). Business failure prediction using statistical techniques: A review.
In Kumar, K., & Chaturvedi, A. (Eds.). Some Recent Developments in Statistical Theory and
Applications.
Sommer, B. (n.d.). Reliability and validity. UC Davis. Retrieved from
ht t p: / /psc.dss.u c d a vis.edu/so m me r b/so m me r d e mo / in t ro/v a l i di t y .htm
The Association for Educational Communications and Technology (2001). What is descriptive
research. Retrieved from ht t p: / /ww w . aec t.o r g /edt e c h/ed1/41/4 1 – 01.ht m l
Maheshwari, T., Reganti, A. N., Gupta, S., Jamatia, A., Kumar, U., Gamback, B., & Das, A.
(2017). A societal sentiment analysis: Predicting the values and ethics of individuals by
analyzing social media content. Proceedings of the 15th Conference of the European
Chapter of the Association for Computational Linguistics
6
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1
`
Apply the T-Test
School of Business, Northcentral University
BUS-7106 v2: Statistics II
August 29, 2021
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2
Apply the T-Test
Introduction
The company CEO requested research to measure the level of commitment of the 500
employees. The company employs 250 full-time employees and 250 contracted employees. The
CEO believes that the full-time employees are more committed than the contracted ones. The HR
director surveyed samples of thirty from each group.
The null hypothesis is that there is no statistical difference between each group’s
commitment. The alternate hypothesis is that there is a significant difference between the
commitment of each group.
T-Test
The t-Test is a statistical test that compares the mean of two different groups (Weiers,
Ronald M., 2008). it’s used to test hypotheses. In this test, the first group is the full-time
employees, and the second group is the contracted employees. The test will decide if there is a
significant difference between the two groups that will reject the null hypothesis that there is no
significant difference. The results will show a non-significant difference between the two to
reject the alternate hypothesis. The formula for the test is:
X is the mean of each group, S is the combined standard error of the means, and n is the
population of each sample. For the company t-test, SPSS and Excel will be used to perform the
test.
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3
SPSS Results
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4
Excel Results
Results Interpretation
The two systems showed similar test results; for the first group (full-time employees), the
mean was 3.67, Standard deviation of .83. While group two (contracted employees), the mean
was 3.75, with a standard deviation of .81.
T= -.39 and p= .35, which showed a significant difference, will reject the null hypothesis
that there is no difference in commitment between full-time and contracted employees.
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5
Conclusion
When the CEO asked the HR director to perform the survey between employees to
determine the level of commitment to the company, the best way to determine the results was by
using the t-test. Which is a statistical test that compares the means of two different groups. The
groups should be independent, normally distributed, and have a similar number of variances.
The director chose thirty employees from each group to conduct the survey, which is
within the range to perform a good t-test (between 20-30). The t result was .39, and p was .35,
which is well above the significant level of .05. The results allowed the HR manager to reject the
null hypothesis and accept the alternate hypothesis that full-time employees are more committed
to the company, as the CEO predicted.
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6
References
Frank, Julian, & Klar, Bernhard. (2016). Methods to test for equality of two normal distributions.
Statistical Methods and Applications, 25(4), 581–599. https://doi.org/10.1007/s10260-
016-0353-z
Weiers, Ronald M. (2008). Introduction to business statistics (6th ed.). South-Weestern cengage
learning. https://proxy1.ncu.edu/login?url=https://search-ebscohost-
com.proxy1.ncu.edu/login.aspx?
direct=true&db=edsbvb&AN=edsbvb.BV022756900&site=eds-live
Xu, Manfei, Fralick, Drew, Zheng, Julia Z., Wang, Bokai, Tu, Xin M., & Feng, Changyong.
(2017). The differences and similarities between two-sample t-test and paired t-test.
Shanghai Archives of Psychiatry, 29(3), 184. https://doi.org/10.11919/j.issn.1002-
0829.217070
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