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

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 0.1 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.)

p-value: ____________-

d. What is your decision regarding H₀?

At the 0.05 significance level,  (reject or do not reject?)  H₀.

Answer 1

We use t test for testing mean of dependent samples, that is we use t test for testing mean of difference between values of two samples is 0 or not.

The hypothesis to be tested is :

H0 : mean of difference <=0

V/s

H1 : mean of difference > 0

A.

We reject H0,

If t > tn-1,significance level

If t > t7, 0.1

If t > 1.895

B.

Test statistic is given by,

t = dbar /(Sd/(n)

dbar = mean of di's

Where di = Xi - Yi

Xi = number of absenties before exercise program of i th participant

Yi = number of absenties after exercise program of same i th participant

dbar = 0.625

Sd = standard deviation of di's

Sd = {(di - dbar) /n-1}

Sd = 4.4058

Therefore test statistic is,

t = 0.401

C.

For greater than type alternative hypothesis,

(since critical region for H0 is, t > test statistic)

P-value = P( t < t test statistic)

P - value = 1 - P(t > 0.401)

P-value = 0.3502

D.

P-value = 0.3502 > 0.05,

Hence we accept H0 at 5% significance level.

Conclusion :

There may be no effect of exercise program on absenties.