To construct a 99% confidence interval for the true population proportion of adults with children, follow these steps:
1. Determine the sample proportion ([tex]\(\hat{p}\)[/tex]):
The sample proportion is calculated by dividing the number of adults with children ([tex]\(x\)[/tex]) by the total number of adults in the sample ([tex]\(n\)[/tex]).
[tex]\[
\hat{p} = \frac{x}{n} = \frac{324}{410} \approx 0.790
\][/tex]
2. Find the critical value ([tex]\(z\)[/tex]) for the 99% confidence level:
The critical value for a 99% confidence level can be found using the standard normal distribution. This value ([tex]\(z\)[/tex]) defines the number of standard errors to cover the central 99% of the distribution.
[tex]\[
z \approx 2.576
\][/tex]
3. Calculate the standard error (SE):
The standard error of the sample proportion is given by the formula:
[tex]\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = \sqrt{\frac{0.790 (1 - 0.790)}{410}} \approx 0.020
\][/tex]
4. Calculate the margin of error (ME):
The margin of error is the product of the critical value and the standard error:
[tex]\[
ME = z \cdot SE = 2.576 \cdot 0.020 \approx 0.052
\][/tex]
5. Determine the confidence interval:
The confidence interval is constructed by adding and subtracting the margin of error from the sample proportion.
[tex]\[
\text{Lower bound} = \hat{p} - ME = 0.790 - 0.052 \approx 0.738
\][/tex]
[tex]\[
\text{Upper bound} = \hat{p} + ME = 0.790 + 0.052 \approx 0.842
\][/tex]
Therefore, the 99% confidence interval for the true population proportion of adults with children is approximately [tex]\(0.738\)[/tex] to [tex]\(0.842\)[/tex].
