The following sample information shows the number of defecti

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

At the .05 significance level, can we conclude there are more defects produced on the day shift?

Solution

Let

day shift = group 2
afternoon shift = group 1

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.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    0.668557923          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    0.334278962          
              
Calculating the mean of the differences (third column):              
              
XD =    1.25          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    3.739 [ANSWER, TEST STATISTIC]