# 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.

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.)

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Wribreeminsl

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) $$\displaystyle{P}={0.5}$$
$$\displaystyle{E}={0.05}$$
$$\displaystyle{Z}_{{{c}}}={2.576}$$
$$\displaystyle{n}=?$$
$$\displaystyle{n}={P}{\left({1}-{p}\right)}\times{\left({\frac{{{7}{c}}}{{{E}}}}\right)}^{{{2}}}$$
$$\displaystyle={0.5}{\left({0.5}\right)}\times{\left({\frac{{{2.576}}}{{{0.05}}}}\right)}^{{{2}}}$$
$$\displaystyle{n}={664}$$