Part (a)

Consider the \(\displaystyle{n}={1000}\) and \(\displaystyle{x}={843}\). Then, the point estimate is,

\(\displaystyle\hat{{{P}}}={1}-{0.843}\)

\(\displaystyle={0.157}\)

Then, at \(\displaystyle{95}\%\) confidence level.

\(\displaystyle\alpha={1}-{0.95}\)

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

\(\displaystyle{\frac{{\alpha}}{{{2}}}}={0.025}\)

\(\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={Z}_{{{0.025}}}\)

\(\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={1.960}\)

Then, the margin of the error is,

\(\displaystyle{E}={Z}_{{{\frac{{\alpha}}{{{2}}}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

\(=1.96\sqrt{\frac{0.843(0.157)}{1000}}\)

\(\displaystyle={0.023}\)

Thus, the \(\displaystyle{95}\%\) confidence interval for the population p is,

\(\displaystyle{0.843}-{0.023}\leq{p}\leq{0.843}+{0.023}\)

\(\displaystyle{0.820}\leq{p}\leq{0.866}\)

Part (b)

Consider the margin error is,

\(\displaystyle{E}={0.03}\)

Then, the sample size is,

\(\displaystyle{n}={\left({\frac{{{Z}_{{{\frac{{\alpha}}{{{2}}}}}}}}{{{E}}}}\right)}^{{{2}}}\hat{{{P}}}{\left({1}-\hat{{{P}}}\right)}\)

\(\displaystyle={\left({\frac{{{1.96}}}{{{0.03}}}}\right)}^{{{2}}}{0.843}\times{0.157}\)

\(\displaystyle={564.93}\)

\(\displaystyle={565}\)

Part(c) For p less than 0.03, the sample size is obtained as,

\(\displaystyle{n}={\left({\frac{{{Z}_{{{\frac{{\alpha}}{{{2}}}}}}}}{{{E}}}}\right)}^{{{2}}}\hat{{{P}}}{\left({1}-\hat{{{P}}}\right)}\)

\(\displaystyle={\left({\frac{{{1.96}}}{{{0.03}}}}\right)}^{{{2}}}{0.5}\times{0.5}\)

\(\displaystyle={1067.11}\)

\(\displaystyle={1068}\)