A certain reported that in a survey of 2006 American adults, 24\% said the

tabita57i 2021-09-22 Answered

A certain reported that in a survey of 2006 American adults, 24% said they believed in astrology.
a) Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. (Round your answers to three decimal places.)
(_______, _______)
b) What sample size would be required for the width of a 99% CI to be at most 0.05 irresoective of the value of p^? (Round your answer up to the nearest integer.)

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Expert Answer

escumantsu
Answered 2021-09-23 Author has 98 answers

a) Given data,
n=2006
P=24%=0.24
Z-value at 99% confidence is Zc=2.576
Margin of error (E)=Zc×p(1p)n
=2.576×0.24(10.24)2006
=2.576×9.0927218344e5
E=0.052
The 99% is p^+E
=0.24±0.052
=(0.188,0.292)
Step 2
b) P=0.5
E=0.05
Zc=2.576
n=?
n=P(1p)×(7cE)2
=0.5(0.5)×(2.5760.05)2
n=664

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