Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 329 with 48.9% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

99.5% C.I. = 

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.


To find the 99.5% confidence interval for a population proportion \( p \), we use the following formula for the confidence interval (CI):

\[
CI = \left( \hat{p} - Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \ \hat{p} + Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)
\]

Where:
- \( \hat{p} \) is the sample proportion.
- \( Z_{\alpha/2} \) is the critical value corresponding to the 99.5% confidence level.
- \( n \) is the sample size.

### Step 1: Identify the values
- Sample size \( n = 329 \)
- Sample proportion \( \hat{p} = 48.9\% = 0.489 \)
- Confidence level = 99.5%, which means \( \alpha = 1 - 0.995 = 0.005 \), so \( \alpha/2 = 0.0025 \).

The critical value \( Z_{\alpha/2} \) for a 99.5% confidence interval (using standard normal distribution tables) is approximately **2.807**.

### Step 2: Calculate the margin of error
The margin of error (ME) is given by:
\[
ME = Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Substitute the values:
\[
ME = 2.807 \times \sqrt{\frac{0.489 \times (1 - 0.489)}{329}}
= 2.807 \times \sqrt{\frac{0.489 \times 0.511}{329}}
= 2.807 \times \sqrt{\frac{0.249459}{329}}
= 2.807 \times \sqrt{0.0007585}
= 2.807 \times 0.02754
\]
\[
ME \approx 0.07734
\]

### Step 3: Find the confidence interval
Now, calculate the confidence interval:
\[
CI = ( \hat{p} - ME, \ \hat{p} + ME )
= ( 0.489 - 0.07734, \ 0.489 + 0.07734 )
= ( 0.41166, \ 0.56634 )
\]

### Final Answer:
The 99.5% confidence interval is:
\[
(0.412, 0.566)
\]

Would you like more details or have any questions? Here are 5 related questions to extend this:

1. What is the interpretation of a 99.5% confidence interval in statistics?
2. How does increasing the confidence level affect the width of the confidence interval?
3. What happens if the sample size is increased to 500 with the same sample proportion?
4. How would the confidence interval change if we wanted a 95% confidence level instead?
5. What is the formula for finding the critical value \( Z_{\alpha/2} \) from a standard normal distribution?

**Tip:** The width of a confidence interval depends on both the sample size and the confidence level—the higher the confidence level, the wider the interval.

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 329 with 48.9% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Learn how to calculate the 99.5% confidence interval for a population proportion using a sample size of 329 with 48.9% successes. This step-by-step guide covers key statistical concepts such as critical values, confidence intervals, and the margin of error. Suitable for high school and college-level students.

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