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

  Click here for the Excel Data File

At the 0.025 significance level, can he conclude that the number of absences has declined? Estimate the p-value.

1. State the decision rule for 0.025 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.

• Reject H₀

• Fail to reject H₀

hypothesis. Reject H0 Fail to reject H0

Answer 1

Solution-A:

if p<0.025,reject null hypothesis.

if p>0.025,do not reject null hypothesis

Solution-b:

Compute the test statistic.

Employee Before After diff dbar dev devsq
1 1 5 3 2 2.375 -0.375 0.140625
2 2 5 6 -1 2.375 -3.375 11.390625
3 3 6 2 4 2.375 1.625 2.640625
4 4 7 7 0 2.375 -2.375 5.640625
5 5 4 3 1 2.375 -1.375 1.890625
6 6 5 2 3 2.375 0.625 0.390625
7 7 7 1 6 2.375 3.625 13.140625
8 8 6 2 4 2.375 1.625 2.640625

Totals are

Before After diff dbar dev devsq
45.000 26.000 19.000 19.000 0.000 37.875

difference in means=dbar=2.375

difference in standard deviation=s=sqrt(37.875 /8-1)=

test statistic

t=dbar/s/sqrt(n)

=2.375/(2.326094/sqrt(8))

= 2.887895

test statistic=2.888

t=2.888

Solution-c:

df=n-1=8-1=7

p value ine xcel

==T.DIST.2T( 2.887895,7)

=0.02338642

p=0.0234

Between 0.01 And 0.025

Solution-d:

p<0.025

Reject Ho

REJECT HO

Same t test output in excel

T Test: Paired Two Sample for Means
	Before	After
Mean	5.625	3.25
Variance	1.125	4.5
Observations	8	8
Pearson Correlation	0.047619048
Hypothesized Mean Difference	0
df	7
t Stat	2.887894388
P(T<=t) one-tail	0.01169322
t Critical one-tail	1.894578584
P(T<=t) two-tail	0.02338644
t Critical two-tail	2.364624236