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. For a 0.05 significance level, what is the decision regarding the hypothesis that the training was effective in improving customer relationships?
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.  For a 0.05 significance level, what is the decision regarding the hypothesis that the training was effective in improving customer relationships?
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

We need to perform a two-sample t-test to determine if there is a significant difference between the pre-training and post-training survey scores. The null hypothesis is that there is no difference between the mean survey scores, and the alternative hypothesis is that there is a significant difference.
We can calculate the mean difference between the pre-training and post-training survey scores as (17-4)/7 = 1.86.
With a sample size of 7 and a significance level of 0.05, the degrees of freedom are 6. Using a t-distribution table or calculator, we find that the critical value for a two-tailed test is 2.447.
The t-statistic for the mean difference is (1.86-0)/(3.38/sqrt(7)) = 2.22.
Since the t-statistic is greater than the critical value, 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