In a survey of 653 lesbian, gay; bisexual, or transgender adults in a certain country, 448 said that they consider themselves bisexual Construct a $90 \%$ confidence interval for the population proportion. Interpret the results. A. $90 \%$ confidence interval for the population proportion is ( $[, \square$ ) (Round to three decimal places as needed)

Step 1 ：First, we need to calculate the sample proportion \(\hat{p}\), which is the ratio of the number of bisexual adults to the total number of adults surveyed. In this case, \(\hat{p} = \frac{448}{653} = 0.686\).

Step 2 ：Next, we need to find the z-score corresponding to a $90\%$ confidence level. The z-score for a $90\%$ confidence level is approximately 1.645.

Step 3 ：We then calculate the standard error using the formula \(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(n\) is the total number of adults surveyed. Substituting the given values, we get the standard error as approximately 0.018.

Step 4 ：Finally, we can calculate the confidence interval using the formula \(\hat{p} \pm z \times se\). Substituting the calculated values, we get the lower and upper bounds of the confidence interval as approximately 0.656 and 0.716, respectively.

Step 5 ：This means that we are $90\%$ confident that the true population proportion of bisexual adults in this country is between $65.6\%$ and $71.6\%$.

Step 6 ：The $90 \%$ confidence interval for the population proportion is \(\boxed{(0.656, 0.716)}\).