Given data:

Survey A asked 1000 people how they liked the new movie Avengers: Endgame and \(\displaystyle{88}\%\) said they did enjoy it.

Survey B also concluded that \(\displaystyle{82}\%\) of people liked the move but they asked 1600 total moviegoers.

Confidence interval for Survey A

CI(Proportion) \(\displaystyle={\left(\hat{{{p}}}-{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}},\hat{{{p}}}+{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\right)}\)

\(\displaystyle={\left({0.88}-{1.96}\times\sqrt{{{\frac{{{0.88}{\left({1}-{0.88}\right)}}}{{{1000}}}}}},{0.88}+{1.96}\times\sqrt{{{\frac{{{0.88}{\left({1}-{0.88}\right)}}}{{{1000}}}}}}\right)}\)

\(\displaystyle={\left({0.86},{0.9}\right)}\)

Confidence interval for Survey B

CI(Proportion) \(\displaystyle={\left(\hat{{{p}}}-{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}},\hat{{{p}}}+{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\right)}\)

\(\displaystyle={\left({0.82}-{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}},{0.82}+{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}}\right)}\)

\(\displaystyle={\left({0.801},{0.839}\right)}\)

Answer:

2. The confidence interval is smaller for Survey B.