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 last six months. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.
Employee Before After
1 6 5
2 6 2
3 7 1
4 7 3
5 4 3
6 3 6
7 5 3
8 6 7

Question

12-08-2020
Mathematics

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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 last six months. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.
Employee Before After
1 6 5
2 6 2
3 7 1
4 7 3
5 4 3
6 3 6
7 5 3
8 6 7

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akiran007 akiran007 · Answer 1 · 2020-08-12T12:22:04+02:00

Answer:

t >± 1.895

t= 0.1705

Step-by-step explanation:

The null and alternative hypotheses are

H0: μd=0 Ha: μd>0

Significance level is set at ∝= 0.05

The critical region for t df=7 t >± 1.895

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Employee After Before d = after - before d²

1 6 5 1 1

2 6 2 4 16

3 7 1 6 36

4 7 3 4 16

5 4 3 1 1

6 3 6 -3 9

7 5 3 2 4

8 6 7 -1 1

∑ 14 84

d`= ∑d/n= 14/8= 1.75

sd²= 1/8( 84- 14²/8) = 1/8 ( 84 - 24.5) = 59.5

sd= 7.7136

t= 3/ 7.7136/ √8

t= 0.1705

Since the calculated value of t= 0.1705 < ± 1.895 therefore reject the null hypothesis at 5 % significance level . On the basis of this we cannot conclude that the number of absences has declined.