Question

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism, which 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:

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

At the 0.05 Significance level, can he conclude that the number of absences has declined?

a. State the decision rule for 0.02 significance level: H₀ : μ_d ≤ 0; H₁ : μ_d > 0. (3 Decimal places)

Reject H₀ if t > _________.

b. Compute the value of the test statistic. (3 decimal places)

c. ESTIMATE the P-Value (4 decimal places)

Answer 1

Solution :

The null and alternative hypotheses would are as follows :

a) Our test is right-tailed. So we must obtain right-tailed critical t value.

Significance level = 0.05

Degrees of freedom = (n - 1) = (8 - 1) = 7

The right tailed critical t value at 0.05 significance level and 7 degrees of freedom is given by,

Critical value = t(0.05, 7) = 1.895

Decision rule at 0.05 significance level :

Reject H​​​​​​0 if t > 1.895.

b) To test the hypothesis the most appropriate test is paired t-test. The test statistic is given as follows :

Where, is sample mean of differences, n is sample size and s is sample standard deviation of the differences.

From the above table we have,

The value of the test statistic is 1.386.

c) Our test is right-tailed test. So, right-tailed p-value for the test statistic is given as follows :

p-value = P(T > t)

p-value = P(T > 1.386)

p-value = 0.1041

The p-value is 0.1041.

Significance level = 0.05

(0.1041 > 0.05)

Since, p-value is greater than the significance level of 0.05, therefore we shall be fail to reject the null hypothesis (H​​​​​​0) at 0.05 significance level.

Conclusion : At 0.05 significance level, there is not sufficient evidence to conclude that the number of absences has declined.