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

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

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