A certain reported that in a survey of 2006 American adults, 24\% said they believed in astrology.Calculate a confidence interval at the 99\% confidence level for the proportion of all adult Americans who believe in astrology.

UkusakazaL 2021-08-06 Answered

A certain reported that in a survey of 2006 American adults, \(\displaystyle{24}\%\) said they believed in astrology.
a) Calculate a confidence interval at the \(\displaystyle{99}\%\) 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 \(\displaystyle{99}\%\) CI to be at most 0.05 irresoective of the value of \(\displaystyle\hat{{{p}}}\)? (Round your answer up to the nearest integer.)

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Expert Answer

Wribreeminsl
Answered 2021-08-08 Author has 20222 answers

a) Given data,
\(\displaystyle{n}={2006}\)
\(\displaystyle{P}={24}\%={0.24}\)
Z-value at \(\displaystyle{99}\%\) confidence is \(\displaystyle{Z}_{{{c}}}={2.576}\)
Margin of error \(\displaystyle{\left({E}\right)}={Z}_{{{c}}}\times\sqrt{{{\frac{{{p}{\left({1}-{p}\right)}}}{{{n}}}}}}\)
\(\displaystyle={2.576}\times\sqrt{{{\frac{{{0.24}{\left({1}-{0.24}\right)}}}{{{2006}}}}}}\)
\(\displaystyle={2.576}\times\sqrt{{{9.0927218344}{e}^{{-{5}}}}}\)
\(\displaystyle{E}={0.052}\)
The \(\displaystyle{99}\%\) is \(\displaystyle\hat{{{p}}}+{E}\)
\(\displaystyle={0.24}\pm{0.052}\)
\(\displaystyle={\left({0.188},{0.292}\right)}\)
Step 2
b) \(\displaystyle{P}={0.5}\)
\(\displaystyle{E}={0.05}\)
\(\displaystyle{Z}_{{{c}}}={2.576}\)
\(\displaystyle{n}=?\)
\(\displaystyle{n}={P}{\left({1}-{p}\right)}\times{\left({\frac{{{7}{c}}}{{{E}}}}\right)}^{{{2}}}\)
\(\displaystyle={0.5}{\left({0.5}\right)}\times{\left({\frac{{{2.576}}}{{{0.05}}}}\right)}^{{{2}}}\)
\(\displaystyle{n}={664}\)

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