Question

Refer to Example \(11.52\).

a. Find the margin of error for the estimate of \(p\).

b. Obtain a sample size that will ensure a margin of error of at most \(0.02\) for a \(95%\) confidence interval without making a guess for the observed value of \(\hat{p}\).

c. Find a \(95%\) confidence interval for \(p\) if for sample of size determined in part (b), \(36.0%\) of those sampled drink alcoholic beverages at least once a week.

d. Determine the margin of error for the estimate in part (c) and compare it to the margin of error specified in part (b).

e. Repeat part (b)-(d) if you can reasonably presume that the percentage of adults sampled who drink alcoholic beverages at least once a week will be at most \(40%\)

f. Compare your results you obtained in part (b)-(d) with those obtained in part (e).

Short answer

Part a. Margin of error for the estimate of population proportion \(p\) is \(0.0241\).

Part b. The sample size required to restrict margin of error at most \(0.02\) is \(n=2401\)

Part c. \(95%\) confidence interval is \(0.3408\) to \(0.3792\).

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

Part e. \(n=2401, 95% CI=0.3804\) to \(0.4196, E=0.0196\)

Step-by-step solution

Part a. Step 1. Given information

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 \(95%\).

Part a. Step 2. Calculation

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 \(0.0241\).

Part b. Step 1. Calculation

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 \(n=2401\).

Part c. Step 1. Calculation

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 \(0.3792\)

Part d. Step 1. Calculation

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 \(0.02\).

Part e. Step 1. Calculation

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

Part f. Step 1. Calculation

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 \(0.4196, E=0.0196\)