Question

____________________________________________ The results of a sample of 372 subscribers to Wired magazine shows the time spent...

____________________________________________

  1. The results of a sample of 372 subscribers to Wired magazine shows the time spent using the Internet during the week. Previous surveys have revealed that the population standard deviation is 10.95 hours. The sample data can be found in the Excel test data file.

  1. What is the probability that another sample of 372 subscribers spends less than 19.00 hours per week using the Internet?

____________________________________________

  1. Develop a 95% confidence interval for the population mean

____________________________________________

  1. If the editors of Wired wanted to have the error be no more than 1 hour, using the same 95% confidence interval, what size of sample would be required.

____________________________________________

35
14
7
24
2
22
14
15
36
13
37
25
33
14
11
8
7
15
19
6
3
27
29
35
6
32
10
5
2
12
21
9
38
32
24
10
15
25
10
7
32
12
18
26
35
38
37
11
25
24
32
35
11
31
4
25
15
13
37
5
21
36
34
22
13
12
12
38
23
13
15
11
10
19
9
22
29
17
26
36
35
3
14
32
20
13
23
3
10
15
32
8
21
31
31
8
5
17
18
28
19
8
33
7
34
23
27
4
5
23
37
34
25
21
37
27
4
6
21
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24
12
21
3
10
20
14
11
18
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4
4
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29
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5
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35
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31
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30
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2

Homework Answers

Answer #1

Given,

Mean 18.54301
Sd 10.8051
n 372

σ = 10.95

a)

P(X bar < 19)

Z = (X bar - Mean)/(σ/SQRT(n)) = (19-18.543)/(10.95/SQRT(372)) = 0.81

P(Z < 0.8050) = 0.7910

b)

95% CI

Given
X bar 18.543
S 10.95
n 372
α= 0.05
CI
Zc 1.959964 +/-   NORM.S.INV(α/2)
Upper 19.65573 X bar + Zc*(S/SQRT(n))
Lower 17.43027 X bar - Zc*(S/SQRT(n))

CI = (17.4303, 19.6557)

c)

σ = 10.95

ME = 1

C = 0.95

Zc = 1.96

n >= (Zc*σ/ME)^2 = (1.96*10.95/1)^2 = 460.62 = 461

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