SOLUTION: 3. A telephone poll of 800 American adults asked "where would you rather go in your spare time?" One response, by 275 adults, was "a movie". Compute and explain a 90% confidence i

Question 1100467: 3. A telephone poll of 800 American adults asked "where would you rather go in your spare time?" One response, by 275 adults, was "a movie". Compute and explain a 90% confidence interval estimate of the proportion of all American adults who would respond "a movie".
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
90% CI half interval is 1.645*sqrt(0.34375*0.65625)/800; 0.34375 is 275/800.
=1.645*0.0168
0.0276
the mean is 0.34375
The CI is the mean +/- the half interval
(0.316, 0.371)We are 90% confident that the true proportion, which we don't know and never will lies in this interval. All values in this interval are plausible.

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