Question # Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88\% said they did enjoy it. Survey B also concluded that 82\% of people liked the move but they asked 1600 total moviegoers. Which of the following is true about this comparison?

Confidence intervals
ANSWERED 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.
1. The margin of error is the same for both.
2. The confidence interval is smaller for Survey B.
3. Increasing the number of people asked does not change the $$\displaystyle{95}\%$$ confidence interval.
4. Survey A is more accurate since the percentage is higher.
5. Survey A has more approvals than Survey B.
6. Survey A is better than Survey B since it has a higher percentage. 2021-08-02

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) $$\displaystyle={\left(\hat{{{p}}}-{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}},\hat{{{p}}}+{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\right)}$$
$$\displaystyle={\left({0.82}-{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}},{0.82}+{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}}\right)}$$
$$\displaystyle={\left({0.801},{0.839}\right)}$$