Chapter 8 Homework
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Question 4
Assume that a sample is used to estimate a population proportion $p$. Find the $99.9 \%$ confidence interval for a sample of size 322 with $50.9 \%$ successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
\[
< p<
\]

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 ：We are given that the sample proportion (\(\hat{p}\)) is 50.9% or 0.509, the sample size (n) is 322, and we are asked to find the 99.9% confidence interval for a population proportion.

Step 2 ：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 Z is the Z-score corresponding to the desired level of confidence.

Step 3 ：The Z-score for a 99.9% confidence interval can be found from a standard normal distribution table or calculated using a statistical function. In this case, the Z-score (Z) is approximately 3.291.

Step 4 ：Substituting the given values into the formula, we get the margin of error as approximately 0.092.

Step 5 ：Subtracting and adding the margin of error from the sample proportion, we get the lower and upper bounds of the confidence interval as approximately 0.417 and 0.601 respectively.

Step 6 ：\(\boxed{\text{Final Answer: The 99.9% confidence interval for a sample of size 322 with 50.9% successes is given by the inequality } 0.417 < p < 0.601. \text{ In other words, we are 99.9% confident that the true population proportion lies between 41.7% and 60.1%.}}\)