Out of 600 people sampled, 276 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.Give your answers as decimals, to three places < p <

To construct a 90% confidence interval for the true population proportion of people with kids, follow these steps:

1. Calculate the sample proportion (p̂): $\hat{p} = \frac{\text{number of people with kids}}{\text{total number of people sampled}} = \frac{276}{600} = 0.460$

2. Find the standard error (SE) of the sample proportion: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.460 \times (1 - 0.460)}{600}} = \sqrt{\frac{0.460 \times 0.540}{600}} = \sqrt{\frac{0.2484}{600}} = \sqrt{0.000414} \approx 0.020$

3. Determine the z-score for a 90% confidence interval: For a 90% confidence interval, the z-score (z*) is approximately 1.645.

4. Calculate the margin of error (ME): $ME = z^* \times SE = 1.645 \times 0.020 = 0.033$

5. Construct the confidence interval: $\hat{p} - ME < p < \hat{p} + ME$ $0.460 - 0.033 < p < 0.460 + 0.033$ $0.427 < p < 0.493$

So, the 90% confidence interval for the true population proportion of people with kids is $0.427 < p < 0.493$.