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

  Click here for the Excel Data File

At the 0.010 significance level, can he conclude that the number of absences has declined? Estimate the p-value.

1. State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.)

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

4. State your decision about the null hypothesis.

• Fail to reject H₀

• Reject H₀

Answer 1

Null and alternative hypotheses:

The null and alternative hypotheses are as follows:

a. Decision rule:

If p-value is less than the significance level of 0.010, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

b. Test statistic:

Consider the preliminary calculations:

c. P-value:

Using Excel formula “=T.DIST.RT(2.961,7)”, the p-value is obtained as 0.011.

The p-value is between 0.01 and 0.025.

d. Conclusion:

Since p-value is greater than 0.010, the null hypothesis is not rejected at 1% level of significance (Fail to reject H0). Therefore, the test does not support the conclusion that the number of absences has declined.