Step 1

Solution:

a.

From the given information, confidence level is 0.95, p=0.30 and E=0.02.

\(\displaystyle{1}-\alpha={0.95}\)

\(\displaystyle\alpha={1}-{0.95}\)

\(\displaystyle\alpha={0.05}\)

Step 2

Then,

\(\displaystyle{n}={\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}{p}{\left({1}-{p}\right)}\)

\(\displaystyle={\left(\frac{{Z}_{{\frac{{0.05}}{{2}}}}}{{0.02}}\right)}^{{2}}{0.30}{\left({1}-{0.30}\right)}\)

\(\displaystyle={\left(\frac{{Z}_{{0.025}}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}\)

\(\displaystyle={\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}{\left[{U}{\sin{{g}}}{e}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n}={N}{O}{R}{M}.{I}{N}{V}{\left({0.025},{0},{1}\right)}\right]}\)

=2016.84

\(\displaystyle\approx{2017}\)

Thus, the required sample size is 2017.

Solution:

a.

From the given information, confidence level is 0.95, p=0.30 and E=0.02.

\(\displaystyle{1}-\alpha={0.95}\)

\(\displaystyle\alpha={1}-{0.95}\)

\(\displaystyle\alpha={0.05}\)

Step 2

Then,

\(\displaystyle{n}={\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}{p}{\left({1}-{p}\right)}\)

\(\displaystyle={\left(\frac{{Z}_{{\frac{{0.05}}{{2}}}}}{{0.02}}\right)}^{{2}}{0.30}{\left({1}-{0.30}\right)}\)

\(\displaystyle={\left(\frac{{Z}_{{0.025}}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}\)

\(\displaystyle={\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}{\left({0.21}\right)}{\left[{U}{\sin{{g}}}{e}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n}={N}{O}{R}{M}.{I}{N}{V}{\left({0.025},{0},{1}\right)}\right]}\)

=2016.84

\(\displaystyle\approx{2017}\)

Thus, the required sample size is 2017.