To construct a 99% confidence interval for the proportion $p$ of adult residents who are parents, we will follow these steps:

1. Determine the sample proportion $\hat{p}$: $\hat{p} = \frac{x}{n} = \frac{208}{400} = 0.52$

2. Calculate the standard error (SE): $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.52(1 - 0.52)}{400}} = \sqrt{\frac{0.52 \times 0.48}{400}} = \sqrt{\frac{0.2496}{400}} \approx 0.025$

3. Determine the critical value for a 99% confidence interval: For a 99% confidence level, the critical value $z$ (from the standard normal distribution) is approximately 2.576.

4. Calculate the margin of error (ME): $ME = z \times SE = 2.576 \times 0.025 \approx 0.0644$

5. Construct the confidence interval: $\text{Lower limit} = \hat{p} - ME = 0.52 - 0.0644 \approx 0.4556$ $\text{Upper limit} = \hat{p} + ME = 0.52 + 0.0644 \approx 0.5844$

Thus, the 99% confidence interval for the proportion $p$ of adult residents who are parents is approximately:

$(0.456, 0.584)$

Summary:

• Lower Limit: 0.456
• Upper Limit: 0.584

If you have any questions or need further details, feel free to ask! Here are some related questions to consider:

1. What assumptions are made when calculating the confidence interval for proportions?
2. How would the confidence interval change with a larger sample size?
3. What is the impact of the confidence level on the width of the confidence interval?
4. How can this method be applied to different populations or categories?
5. What are the limitations of using confidence intervals in statistical analysis?

Tip: When interpreting confidence intervals, remember that they provide a range of plausible values for the population parameter, not a guarantee that the true parameter lies within that range.