Step 1

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.

Which of the following is true about this comparison?

Answer:

The confidence interval is smaller for Survey B.

Explanation:

From the Standard Normal Table, the z* value for \(\displaystyle{95}\%\) level is 1.96.

Step 2

The \(\displaystyle{95}\%\) confidence interval for survey A is below:

Sample statistic \(\displaystyle\pm{M}{E}=\hat{{{p}}}\pm{z}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

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

\(\displaystyle={0.88}\pm{\left({1.96}\times{0.0103}\right)}\)

\(\displaystyle={0.88}\pm{0.0201}\)

\(\displaystyle={\left({0.8599},{0.9001}\right)}\)

Thus, the \(\displaystyle{95}\%\) confidence interval for survey A is (0.8599, 0.9001)

The \(\displaystyle{95}\%\) confidence interval for survey B is below:

Sample statistic \(\displaystyle\pm{M}{E}=\hat{{{p}}}\pm{z}\cdot\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)

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

\(\displaystyle={0.82}\pm{\left({1.96}\times{0.0096}\right)}\)

\(\displaystyle={0.82}\pm{0.0188}\)

\(\displaystyle={\left({0.8012},{0.8388}\right)}\)

Thus, the \(\displaystyle{95}\%\) confidence interval for survey B is (0.8012, 0.8388)

Answer:

The confidence interval is smaller for Survey B.