Paired t test: Absenteeism of Employees-MCQs

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