2) in a random sample of 230 men (18 or older), 157 are married. construct a 92% confidence interval to…

Explanation:

Step1: Calculate the sample proportion ($\hat{p}$)

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 157$ (number of married men) and $n=230$ (sample size). So, $\hat{p}=\frac{157}{230}\approx0.6826$.

Step2: Find the critical value ($z$)

For a $92%$ confidence interval, the significance level $\alpha=1 - 0.92=0.08$. Then $\frac{\alpha}{2}=0.04$. Using the standard normal distribution table or a calculator, the $z$-score $z_{\frac{\alpha}{2}}=z_{0.04}\approx1.75$.

Step3: Calculate the margin of error ($E$)

The formula for the margin of error for a proportion is $E = z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.6826$, $n = 230$, and $z_{\frac{\alpha}{2}}=1.75$ into the formula: [ \begin{align*} E&=1.75\sqrt{\frac{0.6826\times(1 - 0.6826)}{230}}\ &=1.75\sqrt{\frac{0.6826\times0.3174}{230}}\ &=1.75\sqrt{\frac{0.2167}{230}}\ &=1.75\sqrt{0.000942}\ &=1.75\times0.0307\ &=0.0537 \end{align*} ]

Step4: Calculate the confidence interval

The confidence interval is $\hat{p}-E<p<\hat{p} + E$. Substitute $\hat{p}=0.6826$ and $E = 0.0537$: $0.6826- 0.0537=0.6289\approx0.629$ and $0.6826 + 0.0537=0.7363\approx0.736$.

Answer:

C) $(0.629,0.736)$