Question

1. in a random sample of 230 men (18 or older), 157 are married. construct a 92% confidence interval to estimate the true proportion of married men (18 or older).

a) (0.02, 0.98)
b) (0.629, 0.726)
c) (0.629, 0.736)
d) (0.652, 0.713)
e) (0.639, 0.726)
f) none of the above.

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:

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Step4: Calculate the confidence interval

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

Answer:

C) $(0.629,0.736)$