Question 458331: 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. (Data are from a project by MBA student Daniel R. Dalach.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A sample of 20 pages was taken without replacement fromthe 1,591-page phone directory Ameritech Pages Plus Yellow pages
On each page, the mean area devoted to display ads was measured (a display as 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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a) construct a 95% confidence interval for the true mean
x-bar = 346.5
E = t*s/sqrt(n)
E = 2.093=79.738
95% CI: 346.5-79.738 < u < 346.5+79.738
95% CI: 266.762 < u < 426.24
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b Why might normallity be an issue here?
All the pages were taken from the same phone book.
The CI is a statement about the whole population of phone books.
The sample might very well not be representative of the whole population.
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c. What sample size would be needed to obtain an error of _+ 10 square millimeters with 99% confidence
Since E = t*s/(sqrt(n))
sqrt(n) = t*s/E
n = [t*s/E]^2 = [2.093*170.378/10]^2 = 1271.64
Rounded up you get n=1272
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d. If this is not a reasonable requirement, suggest one that is.
You could lessen the sample size by decreasing the confidence
or increasing the error limit.
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Cheers,
Stan H.
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