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T test for employee absenteeism

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The mean (m) number of employee absences in company X is 6 days annually, according to previous studies conducted by the Human Resources Department.
Five years have passed without determining whether this mean continues to be the same. Therefore, the company's industrial psychologist selects a sample of 9 employees and determine the number of absences they have accumulated over a period of 12 months. He obtained a mean of 9 days, with a standard deviation of 1.658, using the following data:
12
10
10
9
9
9
8
8
6
In this problem, you will reach a conclusion about whether or not we can determine that the mean for employee absences has increased significantly. Before figuring your response, you must decide what hypothesis test you will use (for example, single sample t-test, dependent samples t-test, independent samples t-test, ANOVA, etc.) Then using the 5% level of significance, determine the following:

a. the null and research hypotheses

b. the comparison distribution used (for example, a t-distribution of 20 degrees of freedom, a t-distribution of 32 degrees of freedom, etc.) and whether the test is unilateral or bilateral

c. the cutoff score on the comparison distribution (for example, a t-critical or "cutoff t" of 1.5 etc.)

d. your sample's test score (for example, a t-score of 2.4)

e. your conclusion on whether to accept or reject the null hypothesis (you must show how the comparison of your cutoff score with your sample's test score leads to your conclusion).

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Solution Summary

The solution provides step by step method for the calculation of T test for employee absenteeism . Formula for the calculation and Interpretations of the results are also included.

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Statistics: Is a significant decrease in the absences after the fitness program?

Robyn is the Vice President for Human Resources for a large manufacturing company. In recent years she has noticed an increase in absenteeism that she thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, she began a fitness program in which employees exercise during their lunch hour. To evaluate the program, she selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. At the .05 significance level, can she conclude there is a significant decrease in the absences after the program?

Employee Before After
A 6 3
B 6 2
C 7 1
D 7 3
E 4 2
F 3 4
G 5 3
H 4 5

1. What test is most appropriate for this problem?
A. Z-test of means
B. T-test of paired samples
C. T-test of independent samples
D. Chi-square - goodness of fit
E. Chi-square - test of independence
F. ANOVA - single factor
Before After
Mean 5.25 2.88
Variance 2.21 1.55
Observations 8 8
Pearson Correlation -0.60
Hypothesized Mean Difference 0
Df 7
t Stat 2.75
P(T<=t) one-tail 0.01
t Critical one-tail 1.89
P(T<=t) two-tail 0.03
t Critical two-tail 2.36

2. What is the null hypothesis?

A. &#956;Before = &#956;After
B. &#956;Before &#8805; &#956;After
C. &#956;Before &#8804; &#956;After
D. &#956;Before &#8800; &#956;After
E. None of the above

3. What is the test value?

A. 2.75
B. 1.89
C. 2.36
D. 2.25
E. 2.88

4. What is the decision?

A. The test value is less than the critical value, therefore, accept the null of no difference.
B. The test value is less than the critical value, therefore, reject the null of no difference.
C. The p-value is greater than the critical value, therefore, reject the null of no difference.
D. The p-value is greater than the critical value, therefore, reject the null of no difference.
E. Something else

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