Lester Hollar is vice president for human resources for a large manufacturing co

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.

At the 0.05 significance level, can he conclude that the number of absences has declined? Hint: For the calculations, assume the "Before" data as the first sample.

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.

Explanation / Answer

Let ud = u2 - u1.              

u1 = after group
u2 = before group

Let ud = u2 - u1.              
Formulating the null and alternative hypotheses,              
              
Ho:   ud   <=   0  
Ha:   ud   >   0  
At level of significance =    0.05          
As we can see, this is a    right   tailed test.      

As df = n - 1 =    7          
              
Then the critical value of t is              
              
tcrit = 2.364624252  

Hence, we reject Ho if t > 2.365 [ANSWER, CRITICAL VALUE]

              
Calculating the standard deviation of the differences (third column):              
              
s =    2.108783938          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    0.745567711          
              
Calculating the mean of the differences (third column):              
              
XD =    2.375          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    3.185492027   [ANSWER, TEST STATISTIC]

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