Out of a sample of 400 families, 156 had children under the age of 18 living in the home. Based on this, construct a 99% confidence interval for the true population proportion of families with children under the age of 18 living in the home.Give your answers rounded to 4 decimal places.

To construct a 99% confidence interval for the true population proportion of families with children under the age of 18 living in the home, follow these steps:

1. Calculate the sample proportion (p̂): $\hat{p} = \frac{\text{number of families with children under 18}}{\text{total number of families}} = \frac{156}{400} = 0.39$

2. Determine the standard error (SE) of the sample proportion: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.39 \times (1 - 0.39)}{400}} = \sqrt{\frac{0.39 \times 0.61}{400}} = \sqrt{\frac{0.2379}{400}} = \sqrt{0.00059475} \approx 0.0244$

3. Find the critical value (z) for a 99% confidence interval:* The critical value for a 99% confidence interval is approximately 2.576 (this value can be found using a z-table or standard normal distribution table).

4. Calculate the margin of error (ME): $ME = z^* \times SE = 2.576 \times 0.0244 \approx 0.0628$

5. Construct the confidence interval: $\text{Lower bound} = \hat{p} - ME = 0.39 - 0.0628 = 0.3272$ $\text{Upper bound} = \hat{p} + ME = 0.39 + 0.0628 = 0.4528$

Therefore, the 99% confidence interval for the true population proportion of families with children under the age of 18 living in the home is $(0.3272, 0.4528)$.