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 last six months following the exercise program. Below are the results.

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

At the 0.01 significance level, can he conclude that the number of absences has declined? Hint: For the calculations, assume the "Before" data as the first sample.

Reject H0 if t >  . (Round your answer to 3 decimal places.)

The test statistic is   . (Round your answer to 3 decimal places.)

The p-value is  (Click to select)  between 0.01 and 0.025  between 0.05 and 0.1  between 0.025 and 0.05  .

Decision:  (Click to select)  Do not reject  Reject  H0.

Answer 1

Solution:

Here, we have to use paired t test.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: the number of absences has not declined.

Alternative hypothesis: Ha: the number of absences has declined.

H0: µd = 0 versus Ha: µd > 0

This is a right tailed test.

We take difference as before minus after.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = 2

Sd = 2.2039

n = 8

df = n – 1 = 7

α = 0.01

Critical value = 2.9980

Reject H0 if t > 2.998

Test statistic is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (2 - 0)/[ 2.2039/sqrt(8)]

t = 2.5668

The test statistic is 2.567.

The p-value by using t-table is given as below:

P-value = 0.0186

The p-value is between 0.01 and 0.025.

P-value > α = 0.01

So, we do not reject the null hypothesis

Decision: Do not reject H0.

There is not sufficient evidence to conclude that the number of absences has declined.