SOLUTION: I need help in statistics. Can someone help me with this? Thank you!
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus
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Question 186937: I need help in statistics. Can someone help me with this? Thank you!
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.
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
(a) Construct a 95 percent confidence interval for the true mean.
Sample mean = 346.5; Sample standard deviation = 170.38
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E = 2.086 = 79.47
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95% CI: 346.5-79.47 < u < 346.5+79.47
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(b) Why might normality be an issue here?
The sample is limited to a single source and may not be representative
of the population of yellow pages.
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(c) What sample size would be needed to obtain an error of }10 square millimeters with 99 percent confidence?
n = [z*s/E]^2
n = [2.58.96*170.38/E]^2 = ?
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Comment: Your posting has garbled the value of "E".
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(d) If this is not a reasonable requirement, suggest one that is.
??
Don't know the answer to c so I don't know the answer to d.
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Cheers,
Stan H.
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