Question

The results of a sample of 372 subscribers toWiredmagazine shows the time spent using the Internet...

  1. The results of a sample of 372 subscribers toWiredmagazine 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?
    2. Develop a 95% confidence interval for the population mean
    3. If the editors of Wiredwanted to have the error be no more than 1 hour, using the same 95% confidence interval, what size of sample would be required.

Excel Data

27
33
26
16
7
25
10
32
35
35
8
35
24
13
18
34
19
21
10
6
38
22
10
17
21
10
32
26
19
23
16
27
36
9
19
32
31
36
16
12
13
17
25
30
37
18
19
5
24
19
28
36
17
37
16
33
21
4
5
9
34
29
32
23
5
7
16
33
5
36
2
19
6
32
36
25
15
6
33
29
3
14
21
15
8
31
26
17
31
35
38
21
6
26
24
30
13
22
18
2
2
29
5
8
11
10
35
25
20
15
28
6
6
11
14
36
13
28
5
24
26
10
29
6
31
21
15
2
22
27
25
16
19
18
33
6
36
22
9
30
24
20
16
8
37
33
33
5
25
11
32
27
6
35
4
28
34
11
37
9
9
35
35
17
15
16
16
24
24
26
14
20
16
21
19
36
27
3
34
35
21
22
29
16
32
26
24
37
21
38
5
6
17
16
10
31
10
19
30
8
30
30
12
26
2
21
7
4
21
6
37
35
9
14
20
3
36
24
3
20
28
23
2
10
5
5
33
20
16
28
35
11
30
28
2
16
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12
4
4
16
16
2
22
5
7
16
37
31
21
17
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2
2
4
2
38
17
30
9
5
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25
32
22
35
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3
12
30
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31
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31
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13
6
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2
2
37
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Math   Statistics-And-Probability   
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Homework Answers

Answer #1

∑x = 7347

n = 372

Mean , x̅ = Ʃx/n = 7347/372 = 19.75

µ = 19.75, σ = 10.95, n = 372

P(X̅ < 19) =

= P( (X̅-μ)/(σ/√n) < (19-19.75)/(10.95/√372) )

= P(z < -1.321)

Using excel function:

= NORM.S.DIST(-1.321, 1)

= 0.0932

----

Population standard deviation, σ = 10.95

Margin of error, E = 1

Confidence interval, CL = 0.95

Significance level, α = 1-CL = 0.05

Critical value, z = NORM.S.INV(0.05/2) = 1.9600

Sample size, n = (z * σ / E)² = (1.96 * 10.95 / 1)² = 460.6 = 461

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