Final answer:
To construct a confidence interval for the true population proportion of adults with children, we can use the formula CI = p' ± Z * √[(p' * (1 - p')) / n]. In this case, the 99% confidence interval is (0.648, 0.712).
Step-by-step explanation:
To construct a confidence interval for the true population proportion, we can use the formula:
CI = p' ± Z * √[(p' * (1 - p')) / n]
Where:
- CI is the confidence interval
- p' is the sample proportion
- Z is the z-score corresponding to the desired confidence level
- n is the sample size
In this case, the sample proportion is 401/590 = 0.680, and the sample size is 590. The z-score for a 99% confidence level is approximately 2.576. Plugging in these values:
CI = 0.680 ± 2.576 * √[(0.680 * (1 - 0.680)) / 590] = (0.648, 0.712)
Therefore, the 99% confidence interval for the true population proportion of adults with children is (0.648, 0.712).