91Ó°ÊÓ

In a survey, \(985\) out of \(1516\) randomly selected Americans said they drank beer, wine, or hard liquor at least occasionally.

Confidence level is of \(95%\).

In a survey, \(985\) out of \(1516\) randomly selected Americans said they drank beer, wine, or hard liquor at least occasionally.

Therefore, number of success is \(x=985\) and sample population size is \(n=1516\).

So, sample population proportion would be \(\hat{p}=\frac{x}{n}=\frac{985}{1516}=0.6497\)

For a \(95%\) confidence level, \(\alpha =0.05\).

So, \(Z_{\alpha/2}=1.96\)

The margin of error of the estimate of population proportion,\(p\), is given as -

\(E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

\(E=1.96\sqrt{\frac{0.6467(1-0.6467)}{1516}}\)

\(\Rightarrow E=1.96 \times 0.0123\)

\(\Rightarrow E= 0.0241\)

Hence, margin of error for the estimate of population proportion \(p\) is \(0.0241\).

A \((1-\alpha)\) level confidence interval for a population that has margin of error of at most \(E\), sample size is obtained by -

\(n=0.25\cdot \left ( \frac{Z_{\alpha/2}}{E} \right )^{2}\)

Here, for a \(95%\) confidence level, \(\alpha=0.05\). So, \(z_{\alpha/2}=1.96\)

\(E=0.02\)

Therefore, required sample size \(n\) is calculated as -

\(n=0.25\cdot \left ( \frac{1.96}{0.02} \right )^{2}\)

\(\Rightarrow n=2401\)

Hence, the sample size required to restrict margin of error at most \(0.02\) is \(n=2401\).

For a \(95%\) confidence level, \(\alpha =0.05\). So, \(Z_{\alpha/2}=1.96\)

So, \(95%\) confidence interval will be given as -

\(CI=\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

\(\Rightarrow CI=0.36\pm 1.96\sqrt{\frac{0.36(1-0.36)}{2401}}\)

\(\Rightarrow CI=0.36\pm 0.0192\)

\(\Rightarrow CI=0.3408\) to \(0.3792\)

Hence, \(95%\) confidence interval is \(0.3408\) to \(0.3792\)

From part (c), \(95%\) confidence interval is \(0.3408\) to \(0.3792\).

Margin of error is calculated by dividing the width of confidence interval by \(2\).

So, \(E=\frac{0.3408-0.3792}{2}=0.0192\)

This margin of error \(0.0192\) is less than margin of error specified in part (b) which is \(0.02\).

A \((1-\alpha)\) level confidence interval for a population that has margin of error of at most \(E\), sample size is obtained by -

\(n=0.25\cdot \left ( \frac{z_{\alpha/2}}{E} \right )^{2}\)

Here, for a \(95%\) confidence level, \(\alpha =0.05\). So, \(Z_{\alpha/2}=1.96\)

\(E=0.02\)

Therefore, required sample size \(n\) is calculated as -

\(n=0.25\left ( \frac{1.96}{0.02} \right )^{2}\)

\(\Rightarrow n=2401\)

Hence, the sample size required to restrict margin of error at most \(0.02\) is \(n=2401\)

For a \(95%\) confidence level, \(\alpha =0.05\). So, \(Z_{\alpha/2}=1.96\)

So, \(95%\) confidence interval will be given as -

\(CI=\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

\(\Rightarrow CI=0.40\pm 1.96\sqrt{\frac{0.40(1-0.40)}{2401}}\)

\(\Rightarrow CI=0.40\pm 0.0196\)

\(\Rightarrow CI=0.3804\) to \(0.4196\)

Hence, \(95%\) confidence interval is \(0.3804\) to \(0.4196\)

Margin of error is calculated by dividing the width of confidence interval by \(2\).

S0, \(E=\frac{0.4192-0.3804}{2}=0.0196\)

Results from part (b)-(d): \(n=2401, 95% CI=0.3408\) to \(0.3792, E=0.0192\)

Results from part (e): \(n=2401, 95% CI=0.3804\) to \(0.4196, E=0.0196\)