Question

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he 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 six months following the exercise program. Below are the results.

Employee	Before	After
1	5	3
2	5	6
3	6	2
4	7	7
5	4	3
6	5	2
7	7	1
8	6	2

At the 0.025 significance level, can he conclude that the number of absences has declined? Estimate the p-value.

1. State the decision rule for 0.025 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t > _________?

2. Compute the test statistic. (Round your answer to 3 decimal places.)

3. The p-value is

• Between 0.01 And 0.025

• Between 0.001 And 0.005

• Between 0.005 And 0.01 (WHICH ONE?)

4. State your decision about the null hypothesis.

• Reject H₀

• Fail to reject H_{0 (whıch one?)}

Answer 1

Ho: No of absences is the same. ie. = 0

Ha: No of absences has declined. ie. > 0

(a) significance level = 0.025, df = 7

t critical (right tailed test) = 2.36462425

Reject Ho if p value is less than 0.025

Reject Ho if t stat is greater than 2.365

(b) test statistics

employee	Before	After	Diff (Bef-Aft)	Dev (diff - mean)	Sq deviation
1	5	3	2	-0.4	0.1
2	5	6	-1	-3.4	11.4
3	6	2	4	1.6	2.6
4	7	7	0	-2.4	5.6
5	4	3	1	-1.4	1.9
6	5	2	3	0.6	0.4
7	7	1	6	3.6	13.1
8	6	2	4	1.6	2.6
Total	45	26	19.0	0.0	37.9

sample diff mean = = 19/ 8 = 2.375

sample diff sd =

t stat = 2.888

(c) p value = 0.0117

p value is between 0.01 and 0.025

(d) As t stat (2.888) is greater than 2.365 we reject the Null hypothesis.