Question

Dr Susan Benner is an industrial Psychologist. She is currently studying stress among executives of internet companies. She has developed a questionnaire that she believes measures stress. A score above 80 indicates stress at a dangerous level. A random sample of 15 executives revealed the following stress level scores.

94	78	83	90	78	99	97	90
97	90	93	94	100	75	84

a. Find the mean stress level for this sample. What is the point estimate of the population mean?

The sample mean of all the 15 executives equated to (89.4667) 89.47. The point of estimate of the population mean is 80 - 89.4667 = 9.4667.

b. Construct a 95% confidence interval for the population mean.

95% confidence interval is between 84.99 and 93.94, found by 89.4667±1.76 (8.08/(sqrt)15)

c. According to Dr Benner's test, is it reasonable to conclude that the mean stress level of internet executives is 80? Explain.

It is perfectly reasonable to conclude that a score above 80% is indicative of a dangerous stress level. It also should be noted that the lower stress level scores are still above the 80 thresholds.

(Instructor's Comments)

The point of estimate of the population mean is 80 - 89.4667 = 9.4667. Are you sure? Is 89.467 the point estimate?

Check the t-score value again. Explain how you obtained the value of 1.76? Correct it and then revise the limits. Check your conclusion as well.

Please attach your excel spreadsheet to support the results.

(STUDENT NOTES)

I have completed this assignment but didn't complete it correctly. If someone can assist me in understanding what i did wrong USING EXCEL (NOT MINITAB) that  would be appreciated!!

Answer 1

Here you use the point estimate of population mean as the difference between the sample mean and population mean i e 80 - 89.4667 = 9.4667. Which is completely wrong because from the Infrential statistics, we know that the sample mean is the only point estimate of population mean so the point estimate of population mean will be 89.4667.

And mistake you done is with t score, we have total 15 observation so degree of freedom is 14. And t at 0.05 with 14 degree of freedom is 2.145. Which is the value from two tailed but you have taken value from one tailed table for constructing the confidence limits which is wrong.

So 95% confidence interval is , found by 89.4667±2.145×(8.08/(sqrt)15 =84.99, 93.94

According to Dr Benner's test, it is not reasonable to conclude that the mean stress level of internet executives is 80 because the value 80 is not contained in confidence interval