Attempting to improve the quality of services provided to customers, the owner of a chain of high-fashion department stores randomly selected a number of clerks for special training in customer relations. Of this group, only 10% were the subject of complaints to the store manager during the 3 months following the training. On the other hand, 15% of a sample of untrained clerks were mentioned in customer complaints to the manager during this same period. The data are in file Clerks, with data for each group coded as 1 = not mentioned in a complaint and 2 = mentioned in a complaint. Using the 0.05 level of significance, does the training appear to be effective in reducing the incidence of customer dissatisfaction with sales personnel? Identify and interpret the p-value for the test.   2-Sample Z-Test, Comparing Proportions from Independent Samples         When the Hypothesized Difference is 0                       Calculated Values:               pbar = 0.125           Summary of Sample Data:   Std. Error = 0.023           Proportion for Sample 1 (p1) 0.1000 z = -2.138            Size of Sample 1 (n1) 400     p-value if the test is:           Proportion for Sample 2 (p2) 0.1500 Left-Tail              Size of Sample 2 (n2) 400 0.0163                               Confidence level desired 0.95               Lower confidence limit -0.096               Upper confidence limit -0.004               The standard error for the CI calc is 0.02332                                     Since 0.0163 is less than 0.05, we reject the null hypothesis       Accept the training is effective.

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Attempting to improve the quality of services provided to customers, the owner of a chain of high-fashion department stores randomly selected a number of clerks for special training in customer relations. Of this group, only 10% were the subject of complaints to the store manager during the 3 months following the training. On the other hand, 15% of a sample of untrained clerks were mentioned in customer complaints to the manager during this same period. The data are in file Clerks, with data for each group coded as 1 = not mentioned in a complaint and 2 = mentioned in a complaint. Using the 0.05 level of significance, does the training appear to be effective in reducing the incidence of customer dissatisfaction with sales personnel? Identify and interpret the p-value for the test.

 

2-Sample Z-Test, Comparing Proportions from Independent Samples        
When the Hypothesized Difference is 0              
        Calculated Values:          
    pbar = 0.125          
Summary of Sample Data:   Std. Error = 0.023          
Proportion for Sample 1 (p1) 0.1000 z = -2.138          
 Size of Sample 1 (n1) 400     p-value if the test is:          
Proportion for Sample 2 (p2) 0.1500 Left-Tail            
 Size of Sample 2 (n2) 400 0.0163            
                 
Confidence level desired 0.95              
Lower confidence limit -0.096              
Upper confidence limit -0.004              
The standard error for the CI calc is 0.02332              
                 
 
  Since 0.0163 is less than 0.05, we reject the null hypothesis  
    Accept the training is effective.        
 
               
                 
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