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	6	2
2	6	6
3	7	3
4	7	7
5	4	5
6	7	3
7	5	1
8	5	2

  Click here for the Excel Data File

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.)

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

Answer 1

the necessary calculation table :-

Before(xi)	After(yi)
6	2	4	3.0625
6	6	0	5.0625
7	3	4	3.0625
7	7	0	5.0625
4	5	-1	10.5625
7	3	4	3.0625
5	1	4	3.0625
5	2	3	0.5625
		sum = 18	sum=33.5

sample size (n) = 8

hypothesis:-

[ claim ]

where, .

a).degrees of freedom = (n-1) = (8-1) = 7

t critical value for alpha=0.025 , df = 7 , right tailed test be:-

  [ using t distribution table ]

b). the test statistic be:-

c).decision:-

so, we reject the null hypothesis. We conclude that there is enough evidence to claim that the number of absences has declined.

*** as in question it is mentioned to calculate the p value i am calculating this below***

the p value is :-

[in any blank cell of excel type =T.DIST.RT(2.909,7) ]

*** if you have any doubt regarding the problem ,please write it in the comment box...if satisfied,please UPVOTE.