An investigation of the effectiveness of a training program to improve customer relationships included a pre-training and post-training customer survey. To compare the differences, they computed (post-training survey score - pre-training survey score). Seven customers were randomly selected and completed both surveys. The results follow: Was the training effective in improving customer relationships when tested at a 0.05 level of significance?
A) Reject the null hypothesis and conclude that the training was effective.
B) Reject the null hypothesis and conclude that the training was ineffective.
C) Fail to reject the null hypothesis and conclude that mean survey scores are the same.
D) Fail to reject the null hypothesis and conclude that the mean survey scores are not equal.

Question

An investigation of the effectiveness of a training program to improve customer relationships included a pre-training and post-training customer survey. To compare the differences, they computed (post-training survey score - pre-training survey score). Seven customers were randomly selected and completed both surveys. The results follow:  Was the training effective in improving customer relationships when tested at a 0.05 level of significance?
A) Reject the null hypothesis and conclude that the training was effective. 
B) Reject the null hypothesis and conclude that the training was ineffective. 
C) Fail to reject the null hypothesis and conclude that mean survey scores are the same. 
D) Fail to reject the null hypothesis and conclude that the mean survey scores are not equal.

Quizplus · Accepted Answer

To determine if the training was effective, we need to conduct a paired-samples t-test since the same group of customers completed both surveys. We subtract the pre-training score from the post-training score for each customer and compute the mean difference (MD). Using a 0.05 level of significance and a t-distribution table with n-1 degrees of freedom (df = 6), we find the critical value to be ±2.447. We calculate the MD to be -5.714 (negative means the post-training score is lower than the pre-training score). The standard deviation (SD) of the differences is 4.025. We compute the t-statistic as t = (MD - 0) / (SD / √n) = (-5.714 - 0) / (4.025 / √7) = -3.36. This t-value is lower than the critical value, so we reject the null hypothesis and conclude that the training was effective in improving customer relationships.

An Investigation of the Effectiveness of a Training Program to Improve