A randomly selected group of 130 U.S. residents aged 15 and over were surveyed and 42 reported having never been married. (a) Construct a $99 \%$ confidence interval to estimate the population proportion of adults who have never been married. Please write your answer in this format $(x, y)$ without spaces, where $x$ and $y$ represent real numbers rounded to 2 decimal places. (b) Which of the following values are reasonable estimates of the proportion of U.S. residents aged 15 and over who have never been married? 0.19 0.05 0.25 0.36

Step 1 ：Calculate the sample proportion (p̂) as the number of successes divided by the total number of observations. So, p̂ = \(\frac{42}{130} = 0.3231\).

Step 2 ：Calculate the standard error (SE) for a proportion as \(\sqrt{\frac{p̂ * (1 - p̂)}{n}}\), where n is the total number of observations. So, SE = \(\sqrt{\frac{0.3231 * (1 - 0.3231)}{130}} = 0.0426\).

Step 3 ：A 99% confidence interval is calculated as p̂ ± Z * SE, where Z is the Z-score that corresponds to the desired level of confidence. For a 99% confidence interval, Z = 2.576.

Step 4 ：So, the 99% confidence interval is \(0.3231 ± 2.576 * 0.0426 = (0.3231 - 0.1097, 0.3231 + 0.1097) = (0.2134, 0.4328)\).

Step 5 ：\(\boxed{\text{So, the 99% confidence interval for the population proportion of adults who have never been married is (0.21, 0.43).}}\)

Step 6 ：The values 0.19, 0.25, and 0.36 are all within the 99% confidence interval we calculated, so they are all reasonable estimates of the proportion of U.S. residents aged 15 and over who have never been married. The value 0.05 is not within the confidence interval, so it is not a reasonable estimate.