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 6 months before the exercise program began and in the 6 months following the exercise program. Following are the results.
Employee Before After
1 7 7
2 7 1
3 2 6
4 5 4
5 7 3
6 6 3
7 4 1
8 1 4
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 : μd ≤ 0; H1 : μd > 0. b. Compute the value of the test statistic.
c. Estimate the p-value? d. What is your decision regarding H0?

Answer 1

The hypothesis is given as

H0:  μd< 0

H1: μd > 0

How to solve for the hypothesis.

The null hypothesis is given as

H0:  μd< 0

The alternative hypothesis is given as

H1: μd > 0

SWe have to find the value of s

we would use this formula s = sqrt [ (Σ(di - d)^2 / (n - 1) ]

This gives us  3.454

Next we have to determine the standard error

s / √(n)

3.454/2.8284

= 1.22

Degree of freedom = 8 - 1

= 7

t = (x1 - x2) - D / S.E

=  1.025

Find t critical at 0.10

= 1.895

P-value = 0.1697

d. Given that p value is greater thatn 0.1 we fail to reject the null hypothesis.

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