Question 188256This question is from textbook Applied Statistics in Business and Economics
: 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 a large block of multicolored illustrations, maps, and text). The data (in square 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 percent 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 percent confidence?
(d) If this is not a reasonable requirement, suggest one that is.
This question is from textbook Applied Statistics in Business and Economics
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 a large block of multicolored illustrations, maps, and text). The data (in square 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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The sample mean is 346.5
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(a) Construct a 95 percent confidence interval for the true mean.
E = 1.96 = 74.6716
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95% C.I.: 346.5 - 74.6716 < u < 346.5 + 74.6716
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(b) Why might normality be an issue here?
I'll leave that to you.
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(c) What sample size would be needed to obtain an error of ±10 square
millimeters with 99 percent confidence?
E = z
n = [z*s/E]^2
n = [1.96*170.38/10]^2 = 1115
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(d) If this is not a reasonable requirement, suggest one that is.
Increase the error-limit requirement and/or reduce the confidence limit.
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Cheers,
Stan H.
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