SOLUTION: CAN SOMEONE PLEASE HELP ME IN STATISTICS? THANK YOU SO MUCH!!! Jane Doe is Vice President for Human Resources for a large manufacturing company. In recent years she has noticed

Question 188225: CAN SOMEONE PLEASE HELP ME IN STATISTICS? THANK YOU SO MUCH!!!
Jane Doe is Vice President for Human Resources for a large manufacturing company. In recent years she has noticed an increase in absenteeism that she thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, she began a fitness program in which employees exercise during their lunch hour. To evaluate the program, she 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. Analyze the results using the proper two-sample test. At the .01 significance level, can he conclude that the number of absences has declined significantly? Explain.

Employee

1
2
3
4
5
6
7
8

Before
6
7
6
7
6
5
4
5

After
4
2
4
4
3
7
2
4

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
To evaluate the program, she 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. Analyze the results using the proper two-sample test. At the .01 significance level, can he conclude that the number of absences has declined significantly? Explain.
Employee
1
2
3
4
5
6
7
8
Before
6
7
6
7
6
5
4
5
After
4
2
4
4
3
7
2
4
----------------
Please post the data in row/column form.
----------------
Since the data is in pairs (before/after) the data is dependent.
You need to find and analyze the mean of the paired differences.
--------------------
Ho: u(difference) = 0
Ha: u(before)-u(after) > 0
--------------
The mean of the paired differences = 2
The standard deviation of the paired differences = 2
---
Critical Value: t = 2.998
Decision Rule: Reject Ho if test statistic is greater than 2.998
test statistic: t = [dbar - 0]s/sqrt(n)
t = [2-0]/[2/sqrt(8)] = 2.8284
------------------
Conclusion:
Sine the test statistic does not fall in the reject interval, Fail
to reject Ho.
There has been no statistically significant difference in the days
of absense.
======================
Cheers,
Stan H.
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