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

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

df = n1 + n2 - 2 = 6              
              
tcrit = 1.943180281

Hence, we reject Ho if t > 1.94. [ANSWER]

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Calculating the means of each group,              
              
X1 =    13.25          
X2 =    12          
              
Calculating the standard deviations of each group,              
              
s1 =    3.593976442          
s2 =    3.16227766          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    4          
n2 = sample size of group 2 =    4          
Thus, df = n1 + n2 - 2 =    6          
Also, sD =    2.393567769          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    0.522   [ANSWER, TEST STATISTIC]      
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where uD = hypothesized difference =    0          
              
Also, using p values,              
              
p =    0.310109983   [ANSWER, P VALUE]

This p value is greater than 0.1.

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Comparing this to the significance level,    WE DO NOT REJECT THE NULL HYPOTHESIS.