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: Determine the calculated value oft. Was the training effective in improving customer relationships when tested at a 0.05 level of significance?
A) -2.542; Reject the null hypothesis and conclude that the training was effective
B) 2.542; Reject the null hypothesis and conclude that the training was ineffective
C) 2.447; Fail to reject the null hypothesis and conclude that mean survey scores are the same
D) 2.542; Fail to reject the null hypothesis and conclude that the mean survey scores are not equal.
E) 2.447; Reject the null hypothesis and conclude that the training was effective

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:  Determine the calculated value oft. Was the training effective in improving customer relationships when tested at a 0.05 level of significance?
A) -2.542; Reject the null hypothesis and conclude that the training was effective 
B) 2.542; Reject the null hypothesis and conclude that the training was ineffective 
C) 2.447; Fail to reject the null hypothesis and conclude that mean survey scores are the same 
D) 2.542; Fail to reject the null hypothesis and conclude that the mean survey scores are not equal. 
E) 2.447; Reject the null hypothesis and conclude that the training was effective

Quizplus · Accepted Answer

To calculate the t-value, we need to use the formula:

t = (x̄diff - μdiff) / (sDiff / √n)

where x̄diff is the mean of the differences in survey scores, μdiff is the hypothesized population mean difference (which is 0 since we are testing whether the training was effective), sDiff is the standard deviation of the differences, and n is the sample size.

Using the given data, we find that:

x̄diff = (8 - 2)/7 = 0.857
sDiff = 3.559
n = 7

Plugging these values into the formula, we get:

t = (0.857 - 0) / (3.559 / √7) = -2.542

Looking up the t-value in the t-distribution table with 6 degrees of freedom (because df = n-1 = 7-1 = 6), at the 0.05 level of significance, we get a critical t-value of ±2.447.

Since the calculated t-value (-2.542) is less than the critical t-value (-2.447), 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