Using TI-84 plus calculator A company psychologist wanted to test if company exe

Question

Using TI-84 plus calculator

A company psychologist wanted to test if company executives have job related stress scores higher than those of university professors. He took a sample of 45 executives and 60 college professors and tested them for job related stress. The sample of 45 executives gave a mean stress score of 7.7 while the sample of 60 professors gave a mean stress score of 5.6. Assume that the standard deviations of the two Populations are 0.9 and 1.2 respectively. a) Construct 95% confidence interval for the difference between the mean stress scores of all executives and all professors. b.) Test at the 5% significance level whether the mean stress score of all executives is higher than that of all professors.

Explanation / Answer

1(A)

(a) 95 % confidence interal for the difference of mean test scores.

Standerd Error of the difference x1-x2 = sqrt [ 21 / n1 + 22 / n2 ]= sqrt [ 0.92 /45 + 1.22 /60] = 0.2049

95% CI = (xexecutive- bar - Xprofessors-bar) + Z0.05 * x1-x2

= (7.7 - 5.6) +- 1.96 * 0.2049 = 2.1 + - 1.96 * 0.2049 = (1.698, 2.502)

(b) Here alpha = 0.05

Test Statistic:

Z = ((xexecutive- bar - Xprofessors-bar)/  x1-x2 = 2.1 /0.2049 = 10.25

Zcritical = 1.645

so Z > Zcritical so we can say that mean stress score of all executives are higher than college professors.

Executives (E) College professors (P) Sample Size 45 60 Mean Stress Score xbar 7.7 5.6 Standard Deviation 0.9 1.2

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