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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 $p$ given that $\hat{p}=0.85$ and $n=100$.
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 $= \pm \bar{i}$
The $90 \%$ confidence interval is to
i
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Step 1 ：The point estimate is simply the sample proportion, which is 0.85.

Step 2 ：The margin of error can be calculated using the formula for the confidence interval of a proportion, which is \( \hat{p} \pm Z \sqrt{(\hat{p}(1-\hat{p}))/n} \), where Z is the Z-score corresponding to the desired level of confidence. For a 90% confidence interval, the Z-score is approximately 1.645.

Step 3 ：The confidence interval is then the point estimate plus and minus the margin of error.

Step 4 ：Let's calculate the margin of error and the confidence interval. Given values are \( \hat{p} = 0.85 \), \( n = 100 \), and \( Z = 1.645 \).

Step 5 ：The margin of error is calculated as \( 0.05873824882476495 \).

Step 6 ：The confidence interval is calculated as \( (0.791261751175235, 0.908738248824765) \).

Step 7 ：The point estimate for p is 0.85. The margin of error is approximately 0.059, and the 90% confidence interval for p is approximately (0.791, 0.909).

Step 8 ：Final Answer: The point estimate is \( \boxed{0.85} \). The margin of error is \( \pm \boxed{0.059} \). The 90% confidence interval is \( \boxed{(0.791, 0.909)} \).