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

At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.

a. State the decision rule for 0.05 significance level: H₀ : μ_d≤ 0; H₁ : μ_d > 0. (Round the final answer to 3 decimal places.)

Reject H₀ if t >  .

b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)

Value of the test statistic            

c. Estimate the p-value? (Round the final answer to 4 decimal places.)

d. What is your decision regarding H₀?

At the 0.05 significance level,  (Click to select)  do not reject  reject  H₀.

Answer 1

Part a)

Reject null hypothesis if t > t(α)
Critical value t(α) = t(0.05) = 1.895
t > 1.895

part b)

Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(89.875 / 8-1) = 3.5832
d̅ = Σ di/n = 3 / 8 = 0.375
t = d̅ / ( S(d) / √(n) )
t = 0.375 / ( 3.5832 / √(8) )
t = 0.296

Part c)

P - value = P ( t > 0.296 ) = 0.3879

Part d)

Test Criteria :-
Reject null hypothesis if t > t(α)
Critical value t(α) = t(0.05) = 1.8946
t < t(α) = 0.296 < 1.8946
Result :- Fail to reject null hypothesis

Do not reject H0.