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) \(\displaystyle{P}={0.5}\)

\(\displaystyle{E}={0.05}\)

\(\displaystyle{Z}_{{{c}}}={2.576}\)

\(\displaystyle{n}=?\)

\(\displaystyle{n}={P}{\left({1}-{p}\right)}\times{\left({\frac{{{7}{c}}}{{{E}}}}\right)}^{{{2}}}\)

\(\displaystyle={0.5}{\left({0.5}\right)}\times{\left({\frac{{{2.576}}}{{{0.05}}}}\right)}^{{{2}}}\)

\(\displaystyle{n}={664}\)