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.
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 | ||||||||
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Since 0.0163 is less than 0.05, we reject the null hypothesis | ||||||||
Accept the training is effective. | |||||||||
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