Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for $p$, the margin of error, and the confidence interval. Assume the results come from a random sample.

A $90 \%$ confidence interval for the proportion of the population in Category A given that $23 \%$ of a sample of 400 are in Category A.
Round your answer for the point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places.

Point estimate $=\mathbf{i}$.
Margin of error $= \pm \mathbf{i}$
The $90 \%$ confidence interval is
to

Step 1 ：Given that 23% of a sample of 400 are in Category A, we can calculate the point estimate for the proportion p as 0.23.

Step 2 ：Next, we calculate the standard error of the proportion using the formula \(\sqrt{p(1-p)/n}\), where p is the point estimate and n is the sample size. Substituting the given values, we get the standard error as approximately 0.021.

Step 3 ：We then calculate the margin of error by multiplying the standard error by the z-score for the desired level of confidence. For a 90% confidence level, the z-score is approximately 1.645. This gives us a margin of error of approximately \(\pm 0.035\).

Step 4 ：Finally, we calculate the 90% confidence interval by adding and subtracting the margin of error from the point estimate. This gives us a confidence interval of \((0.195, 0.265)\).

Step 5 ：\(\boxed{\text{The point estimate for } p \text{ is } 0.23. \text{ The margin of error is } \pm 0.035. \text{ The 90% confidence interval is } (0.195, 0.265)}\)