Step 1

Given information

Error = 3% = 0.02

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.9600}{\left({F}{r}{o}{m}{E}{x}{c}{e}{l}={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.025}\right)}\right)}\)

Population proportion = 0.30

Margin of error formula is given by

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

a)

\(\displaystyle{n}={p}{\left({1}-{p}\right)}\times{\left(\frac{{{Z}_{{\frac{\alpha}{{2}}}}}}{{E}}\right)}^{{2}}={0.30}\times{0.70}\times{\left(\frac{{1.9600}}{{0.02}}\right)}^{{2}}={2016.84}\approx{2017}\)

Sample size = 2017

Step 2

b)

x=470

n=2017

\(\displaystyle\hat{{{p}}}=\frac{{x}}{{n}}=\frac{{470}}{{2017}}={0.2330}\)

c)

95% Confidence interval given by

\(\displaystyle\hat{{{p}}}\pm{Z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}\)

\(\displaystyle{0.2330}\pm{1.9600}\sqrt{{\frac{{{0.2330}{\left({1}-{0.2330}\right)}}}{{2017}}}}={\left({0.2146},{0.2514}\right)}\)

\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.9600}{\left({F}{r}{o}{m}{E}{x}{c}{e}{l}={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.025}\right)}\right)}\)

Given information

Error = 3% = 0.02

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.9600}{\left({F}{r}{o}{m}{E}{x}{c}{e}{l}={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.025}\right)}\right)}\)

Population proportion = 0.30

Margin of error formula is given by

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

a)

\(\displaystyle{n}={p}{\left({1}-{p}\right)}\times{\left(\frac{{{Z}_{{\frac{\alpha}{{2}}}}}}{{E}}\right)}^{{2}}={0.30}\times{0.70}\times{\left(\frac{{1.9600}}{{0.02}}\right)}^{{2}}={2016.84}\approx{2017}\)

Sample size = 2017

Step 2

b)

x=470

n=2017

\(\displaystyle\hat{{{p}}}=\frac{{x}}{{n}}=\frac{{470}}{{2017}}={0.2330}\)

c)

95% Confidence interval given by

\(\displaystyle\hat{{{p}}}\pm{Z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}\)

\(\displaystyle{0.2330}\pm{1.9600}\sqrt{{\frac{{{0.2330}{\left({1}-{0.2330}\right)}}}{{2017}}}}={\left({0.2146},{0.2514}\right)}\)

\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.9600}{\left({F}{r}{o}{m}{E}{x}{c}{e}{l}={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.025}\right)}\right)}\)