Question

A sample of 126 corporate managers and another sample of 168 college professors produced mean job-
related stress scores of 7.35 for the managers and 6.86 for the professors. Suppose that the standard
deviations of the stress scores are 1.12 for the managers and 1.82 for the professors. The null hypothesis
is that the mean stress scores are the same for corporate managers and college professors, while the
alternative hypothesis is that the mean stress score for managers is different from the mean stress score
for professors.

1. Derive the corresponding 90% confidence interval for the difference between the mean stress
scores for all corporate managers and college professors, rounded to three decimal places.
Part A: The lower limit is
Part B: The upper limit is

2. The significance level for the test is 1%. What are the critical values of z?
A) -2.58 and 2.58 B) -1.96 and 1.96 C) -2.33 and 2.33 D) -3.09 and 3.09

3. What is the value of the test statistic, rounded to three decimal places?

4. What is the p-value for this test, rounded to four decimal places?

5. Do you reject or fail to reject the null hypothesis at the 1% significance level? State your answer
as "reject" or "fail to reject", but don't include the quotation marks.

Answer 1

The statistical software output for this problem is:

Hence,

1. 90% confidence interval:

Lower limit = 0.207

Upper limit = 0.773

2. Critical values = -2.58 and 2.58

Option A is correct.

3. Test statistic = 2.845

4. p-value = 0.004

5. Reject