Out of 600 people sampled, 174 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 four places
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Step 1 ：First, calculate the sample proportion (p̂), which is the number of people with kids divided by the total number of people sampled. In this case, we have 174 people with kids out of a sample of 600, so p̂ = 174 / 600 = 0.29.

Step 2 ：Next, calculate the standard error of the proportion. The formula for this is the square root of [(p̂ * (1 - p̂)) / n], where n is the total number of people sampled. Substituting the values we have, the standard error (SE) = √[(0.29 * (1 - 0.29)) / 600] = 0.0185.

Step 3 ：Then, calculate the 90% confidence interval using the formula: p̂ ± Z * SE, where Z is the Z-score for a 90% confidence interval. For a one-tailed test, Z = 1.645. Substituting the values we have, the confidence interval = 0.29 ± 1.645 * 0.0185.

Step 4 ：Finally, round the confidence interval to four decimal places. The lower limit of the confidence interval is 0.2595 and the upper limit is 0.3205.

Step 5 ：\(\boxed{[0.2595, 0.3205]}\) is the 90% confidence interval for the true population proportion of people with kids.