Given: proportion p=0.74

Margin for Error = E = 0.07

Sample size for 9.99% confidence.

\(\displaystyle\alpha={0.001}\)

\(\displaystyle{z}_{{\frac{{x}}{{2}}}}={z}_{{{0.0005}}}={3.29}\)

we know \(\displaystyle{E}={z}_{{\frac{{x}}{{2}}}}\cdot\sqrt{{\frac{{{p}{\left({1}-{p}\right)}}}{{n}}}}\)

So,

\(\displaystyle{n}={p}{\left({1}-{p}\right)}{\left[\frac{{{z}_{{\frac{{x}}{{2}}}}}}{{E}}\right]}^{{2}}\)

\(\displaystyle={0.74}{\left({1}-{0.74}\right)}{\left[\frac{{{3.29}}}{{{0.07}}}\right]}^{{2}}\)

\(\displaystyle={425.01}\)

\(\displaystyle{n}={425}\)

Margin for Error = E = 0.07

Sample size for 9.99% confidence.

\(\displaystyle\alpha={0.001}\)

\(\displaystyle{z}_{{\frac{{x}}{{2}}}}={z}_{{{0.0005}}}={3.29}\)

we know \(\displaystyle{E}={z}_{{\frac{{x}}{{2}}}}\cdot\sqrt{{\frac{{{p}{\left({1}-{p}\right)}}}{{n}}}}\)

So,

\(\displaystyle{n}={p}{\left({1}-{p}\right)}{\left[\frac{{{z}_{{\frac{{x}}{{2}}}}}}{{E}}\right]}^{{2}}\)

\(\displaystyle={0.74}{\left({1}-{0.74}\right)}{\left[\frac{{{3.29}}}{{{0.07}}}\right]}^{{2}}\)

\(\displaystyle={425.01}\)

\(\displaystyle{n}={425}\)