Question 214372
Please show work on how you solved confidence interval problem.  A sample of 20 pages was taken withreplacement 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.
  
   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 percent confidence innterval for the true mean.

sample mean = 325.35 standard deviation = 180.34

standard error: t[s/sqrt(n)] = 2.086*[180.34/sqrt(21) = 82.09

Therefore 95% CI: 325.35-82.09 < u < 325.35+82.09

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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?
Ans: n = [z*s/E]^2 = [2.576*180.34/10]^2 = 2157.84
Rounding up you get 2158
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d.  If this is not a reasonble suggest one that is.
Ans: n can be reduced by reducing the level of confidence and/or from
increasing the level of error allowed.

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Cheers,
Stan H.