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 results1,6,3,2,6,2,3,7,1,4,7,2,5,4,2,6,6,6,7,6,3,8,8,7State the decision rule for 0.100 significance level.Compute the test Statistics

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Asked Nov 24, 2019
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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

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

State the decision rule for 0.100 significance level.

Compute the test Statistics

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Expert Answer

Step 1

Degrees of freedom:

For student’s t distribution, the degrees of freedom is n –1.

The number of observations is 8.

Thus, degrees of freedom is 7.

Critical value:        

The critical value for α = 0.100 and 7 degrees of freedom using t-distribution is 1.415.

Step 2

Decision rule for right-tailed test at α = 0.05:

We reject H0 if t > 1.415.

The test statistic is t which is calculated as,

d
do
t =
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d do t =

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Step 3

where d-bar is the mean of the differences, µd0 is the hypothesized mean difference, sd-bar  is the standard deviation of the differences.

The dif...

Σd
24
= 3
Σ(4-a)
п-1
п
(3-3)(4-3)(6-3)+(5-3+(2-3)+(0-3)+(3-3)+(1-3)')
_
=0.7071
7
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Σd 24 = 3 Σ(4-a) п-1 п (3-3)(4-3)(6-3)+(5-3+(2-3)+(0-3)+(3-3)+(1-3)') _ =0.7071 7

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