a) Given \(\displaystyle{x}={17}\) and \(\displaystyle{n}={75}\)

\(\displaystyle\hat{{{p}}}={\frac{{{x}}}{{{n}}}}={\frac{{{17}}}{{{75}}}}={0.2267}\)

The critical value corresponding with \(\displaystyle{95}\%\) confidence level is 1.96

\(\displaystyle{95}\%{C}{I}=\hat{{{p}}}\pm\ {z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

\(\displaystyle={0.2267}\pm{1.96}\sqrt{{{\frac{{{0.2267}{\left({1}-{0.2267}\right)}}}{{{75}}}}}}\)

\(\displaystyle={\left({0.1319},\ {0.3215}\right)}\)

b) The critical value corresponding with \(\displaystyle{98}\%\) confidence level is 2.33

\(\displaystyle{95}\%{C}{I}=\hat{{{p}}}\pm\ {z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

\(\displaystyle={0.2267}\pm{2.33}\sqrt{{{\frac{{{0.2267}{\left({1}-{0.2267}\right)}}}{{{75}}}}}}\)

\(\displaystyle={\left({0.1141},\ {0.3393}\right)}\)

c) The critical value corresponding with \(\displaystyle{95}\%\) confidence level is 1.96

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

sample size \(\displaystyle={\left({\frac{{{z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}}}{{{M}{E}}}}\right)}^{{{2}}}{p}{\left({1}-{p}\right)}\)

\(\displaystyle={\left({\frac{{{1.96}}}{{{0.05}}}}\right)}^{{{2}}}{0.5}{\left({1}-{0.5}\right)}\) Let \(\displaystyle{p}={0.5}\)

\(\displaystyle={348.16}\)

\(\displaystyle\approx{384}\)