SOLUTION: A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yello Pages. On each page, the mean area devoted to display ads was mea
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Question 183325: A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yello Pages. On each page, the mean area devoted to display ads was measured (a display ad is large block of multicolored illustrations, maps, and text). The data (in sguare millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
(a)Construct a 95% confidence interval for the true mean. (b) why might normality be an issue here? (c) What sample size would be needed to obtain an error of +,- 10 square millimeters with 99% confidence? (d) If this is not a reasonable requirement, suggest one that is.
Construct
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured (a display ad is large block of multicolored illustrations, maps, and text). The data (in sguare millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
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(a)Construct a 95% confidence interval for the true mean.
sample mean = 346.5
s = 170.38 ; n= 20
E = 1.96*170.38/sqrt(20) = 74.67
95% C.I.: 346.5-74.67 < u < 346.5+74.67
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(b) why might normality be an issue here?
The sample was not necessarily representative of the set of all Yellow Pages.
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(c) What sample size would be needed to obtain an error of +,- 10 square millimeters with 99% confidence?
Since E = z*s/sqrt(n)
n = [z*s/E]^2
Your problem:
n = [2.576*170.38/10]^2 = 1926
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(d) If this is not a reasonable requirement, suggest one that is.
Lessen the level of confidence that is required or increase the
margin of error (E)
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Cheers,
Stan H.
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