Assume that a sample is used to estimate a population proportion p. Find the $90 \%$ confidence interval for a sample of size 167 with 100 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
\[
90 \% \text { C.I. }=
\]
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
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Step 1 ：The problem is asking for a 90% confidence interval for a population proportion. The formula for a confidence interval for a population proportion is given by: \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired level of confidence, and \(n\) is the sample size.

Step 2 ：In this case, \(\hat{p} = \frac{100}{167}\), \(Z\) is the Z-score corresponding to a 90% confidence level, and \(n = 167\).

Step 3 ：The Z-score for a 90% confidence level is approximately 1.645. This value can be found in a standard Z-table or using a calculator that can compute inverse normal probabilities.

Step 4 ：We plug these values into the formula and compute the confidence interval: \(\hat{p} = 0.5988023952095808\), \(Z = 1.645\), \(n = 167\), and \(se = 0.0379282403674407\).

Step 5 ：We calculate the lower and upper bounds of the 90% confidence interval: \(ci_{lower} = 0.5364104398051408\) and \(ci_{upper} = 0.6611943506140208\).

Step 6 ：\(\boxed{\text{Final Answer: The 90% confidence interval for the population proportion is }(0.536, 0.661)}\)