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 last six months following the exercise program. Below are the results.

Employee	Before	After
1	6	3
2	6	2
3	7	1
4	7	2
5	4	2
6	6	6
7	6	3
8	6	7

At the 0.10 significance level, can he conclude that the number of absences has declined? Hint: For the calculations, assume the "Before" data as the first sample.

Reject H0 if t >  . (Round your answer to 3 decimal places.)

The test statistic is   . (Round your answer to 3 decimal places.)

The p-value is  (Click to select)  between 0.005 and 0.01  between 0.0005 and 0.005  between 0.01 and 0.05  .

Decision:  (Click to select)  Reject  Do not reject  H0.

Answer 1

The given problem can be tested using paired sample t-test

Employee	Before(x)	After(y)	Difference d=x-y
1	6	3	3
2	6	2	4
3	7	1	6
4	7	2	5
5	4	2	2
6	6	6	0
7	6	3	3
8	6	7	-1

Here the null hypothesis H0  is there is no significant difference between averages of  before values and after values .

Now the test statistic for paired t test is

where mean value of differences

s = standard deviation of differences

n = sample size

For the given data   2.75 s = 2.222 n =8

t = 3.274

At 0.10 significance level the table value of t is 1.415

Thus we reject if t > 1.415

The test statistic t = 3.274

p-value is 0.0068 (i.e; between 0.005 and 0.01)

Here the t satistic value is greater than critical value. So we reject the null hypothesis H0.