To construct a 99% confidence interval for the true population proportion of people with kids, we follow these steps:

1. Identify the sample proportion ($\hat{p}$):

The sample proportion ($\hat{p}$) is calculated as: $\hat{p} = \frac{\text{number of people with kids}}{\text{total sample size}} = \frac{308}{400} = 0.77$

2. Determine the critical value for a 99% confidence interval:

For a 99% confidence interval, the critical value ($z_{\alpha/2}$) corresponds to the Z-value where the area under the standard normal curve between -$z_{\alpha/2}$ and $z_{\alpha/2}$ is 99%. This value is approximately: $z_{\alpha/2} = 2.576$

3. Compute the standard error (SE):

The standard error (SE) of the sample proportion is calculated using the formula: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.77(1 - 0.77)}{400}} = \sqrt{\frac{0.77 \times 0.23}{400}} = \sqrt{\frac{0.1771}{400}} \approx 0.0210$

4. Calculate the margin of error (ME):

The margin of error is given by: $ME = z_{\alpha/2} \times SE = 2.576 \times 0.0210 \approx 0.0541$

5. Construct the confidence interval:

The confidence interval is: $\hat{p} \pm ME = 0.77 \pm 0.0541$ Thus, the confidence interval is: $(0.7700 - 0.0541, 0.7700 + 0.0541) = (0.7159, 0.8241)$

Final Answer:

The 99% confidence interval for the true population proportion is: $(0.7159, 0.8241)$

Would you like further details or have any questions?

Here are five follow-up questions you could explore:

1. How would the confidence interval change if the sample size increased?
2. What is the interpretation of this confidence interval in context?
3. How would a 95% confidence interval differ from a 99% one in this case?
4. How does the sample proportion ($\hat{p}$) affect the width of the interval?
5. How does the standard error influence the margin of error?

Tip: A larger sample size reduces the margin of error, leading to a narrower confidence interval, all else being equal.