In: Statistics and Probability
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 σ ≈ 3.7 thousand dollars.
3.5 |
6.6 |
3.7 |
8.4 |
8.5 |
5.8 |
8.9 |
6.4 |
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 μ, 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 μ, 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.
Solution :-
Given data,
σ ≈ 3.7 thousand dollars
3.5 |
6.6 |
3.7 |
8.4 |
8.5 |
5.8 |
8.9 |
6.4 |
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 |
From the data we can get the following as,
Now, we can calculate the given questions as,
(a)
The average annual profits per employee (in units of one thousand dollars per employee) for companies in retail sales is 5.0257 thosand dollars per employee.
(b)
80% confidence interval for μ, the average annual profit per employee for retail sales is C.I = (4.22, 5.823)
C.I = 5.0257 + 1.285 × 0.6254
C.I = 5.0257 + 0.80366
C.I = (4.222, 5.8294)
lower limit = 4.22
upper limit = 5.83
(e)
95% confidence interval for μ, the average annual profit per employee for retail sales is C.I = (3.80, 6.25).
C.I = 5.0257 + 1.96 × 0.6254
C.I = 5.0257 + 1.22578
C.I = (3.80, 6.25)
lower limit = 3.80
upper limit = 6.25
Hence solved.