Question 1
In a sample of 350 adults, 305 had children. Cónstruct a $90 \%$ confidence interval for the true population proportion of adults with children.
Give your answers as decimals, to three places
\[
< p<
\]

Step 1 ：First, calculate the sample proportion (p̂), which is the number of adults with children divided by the total number of adults. In this case, p̂ = \(\frac{305}{350} = 0.871\)

Step 2 ：Next, find the standard error of the proportion. The formula for this is \(\sqrt{\frac{p̂ * (1 - p̂)}{n}}\), where n is the total number of adults. Substituting the values, we get \(\sqrt{\frac{0.871 * (1 - 0.871)}{350}} = 0.018\)

Step 3 ：Then, find the z-score for a 90% confidence interval. The z-score for a 90% confidence interval is 1.645

Step 4 ：Calculate the margin of error by multiplying the z-score by the standard error. This gives us \(1.645 * 0.018 = 0.029\)

Step 5 ：Finally, calculate the confidence interval by adding and subtracting the margin of error from the sample proportion. This gives us \((0.871 - 0.029, 0.871 + 0.029) = (0.842, 0.901)\)

Step 6 ：Final Answer: The $90 \%$ confidence interval for the true population proportion of adults with children is \(\boxed{(0.842, 0.901)}\)