Question

Jobs and productivity! How do retail stores rate? One way to answer this question is to...

Jobs and productivity! How do retail stores rate? One way to answer this question is to examine annual profits per employee. The following data give annual profits per employee (in units of one thousand dollars per employee) for companies in retail sales. Assume ? ? 3.6 thousand dollars.

3.7

6.9

3.6

8.0

7.5

5.4

8.8

5.5

2.6

2.9

8.1

?1.9

11.9

8.2

6.4

4.7

5.5

4.8

3.0

4.3

?6.0

1.5

2.9

4.8

?1.7

9.4

5.5

5.8

4.7

6.2

15.0

4.1

3.7

5.1

4.2

(a) Use a calculator or appropriate computer software to find x for the preceding data. (Round your answer to two decimal places.)
thousand dollars per employee

(b) Let us say that the preceding data are representative of the entire sector of retail sales companies. Find an 80% confidence interval for ?, the average annual profit per employee for retail sales. (Round your answers to two decimal places.)

lower limit     thousand dollars
upper limit     thousand dollars


(c) Let us say that you are the manager of a retail store with a large number of employees. Suppose the annual profits are less than 3 thousand dollars per employee. Do you think this might be low compared with other retail stores? Explain by referring to the confidence interval you computed in part (b).

Yes. This confidence interval suggests that the profits per employee are less than those of other retail stores.No. This confidence interval suggests that the profits per employee do not differ from those of other retail stores.     


(d) Suppose the annual profits are more than 6.5 thousand dollars per employee. As store manager, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b).

Yes. This confidence interval suggests that the profits per employee are greater than those of other retail stores.No. This confidence interval suggests that the profits per employee do not differ from those of other retail stores.     


(e) Find an 95% confidence interval for ?, the average annual profit per employee for retail sales. (Round your answers to two decimal places.)

lower limit     thousand dollars
upper limit     thousand dollars


Let us say that you are the manager of a retail store with a large number of employees. Suppose the annual profits are less than 3 thousand dollars per employee. Do you think this might be low compared with other retail stores? Explain by referring to the confidence interval you computed in part (b).

Yes. This confidence interval suggests that the profits per employee are less than those of other retail stores.No. This confidence interval suggests that the profits per employee do not differ from those of other retail stores.     


Suppose the annual profits are more than 6.5 thousand dollars per employee. As store manager, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b).

Yes. This confidence interval suggests that the profits per employee are greater than those of other retail stores.No. This confidence interval suggests that the profits per employee do not differ from those of other retail stores.   

Homework Answers

Answer #1

here sample size=n=35, sample mean=5 and standard deviation=sd=3.6 ( given)

(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

(b) 80% confidence interval for population mean=sample mean±z(0.2/2)*sd/sqrt(n)

=5±1.28*3.6/sqrt(35)=5±0.78=(4.22, 5.78)

lower limit=4.22 thousand dollar

upper limit=5.78 thousand dollar

(c) Yes. This confidence interval suggests that the profits per employee are less than those of other retail stores.

since the 3 thousand is below the lower limit of confidence interval=4.22 thousand

(d) Yes. This confidence interval suggests that the profits per employee are greater than those of other retail stores.

since the 6.5 thousand is above the upper limit of confidence interval=5.78 thousand

(e) 95% confidence interval for population mean=sample mean±z(0.5/2)*sd/sqrt(n)

=5±1.96*3.6/sqrt(35)=5±1.19=(3.81, 6.19)

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