Question

7. A health psychologist knew that corporate executives in general have an average score of 80 with a standard deviation of 12 on a stress inventory and that the scores are normally distributed. In order to learn whether corporate executives who exercise regularly have lower stress scores, the psychologist measured the stress of 55 exercising executives and found them to have a mean score of 72. Was there a significant reduction (at the .01 level)? a. What type of test would this be and why? (1pt) b. Show all steps for hypothesis testing and include a graph with your response.

Answer 1

Solution:

Given: the scores of corporate executives are normally distributed with  an average score of 80 with a standard deviation of 12 on a stress inventory.

Thus

Sample size = n = 55

Sample mean =

We have to test if there a significant reduction   stress scores.

Level of significance = 0.01

Part a)

Since Population is normally distributed with known standard deviation , we use one sample z test for mean.

Part b)

Step 1) State H0 and H1:

Step 2) Z test statistic:

Step 3) Find z critical value:

Since this is left tailed test,

look in z table for area = 0.0100 or its closest area and find z value.

Area 0.0099 is closest to 0.0100 and it corresponds to -2.3 and 0.03

thus z critical value = -2.33

Step 4) Decision Rule:

Reject H0, if z test statistic value < z critical value, otherwise we fail to reject H0.

Since z test statistic value = z = -4.94 < z critical value = -2.33 , we reject H0.

Step 5) Conclusion:

Since we have rejected null hypothesis H0, there sufficient evidence to conclude that: there a significant reduction   stress scores of corporate executives who exercise regularly .