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To solve this problem, we need to calculate the sample size required for a 99.5% confidence interval, given that the estimated proportion of complications \( p = 0.14 \) and the margin of error \( E = 0.08 \).

The formula for the sample size \( n \) when estimating a population proportion is:

\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]

Where:
- \( Z \) is the Z-value corresponding to the desired confidence level (for 99.5%, \( Z \approx 2.807 \)).
- \( p = 0.14 \) is the estimated proportion.
- \( E = 0.08 \) is the margin of error.

Let's plug in the values and calculate the sample size:

\[
n = \left( \frac{2.807^2 \cdot 0.14 \cdot (1 - 0.14)}{0.08^2} \right)
\]

Now I will compute this.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can compute the sample size manually by following the formula I provided:

\[
n = \left( \frac{2.807^2 \cdot 0.14 \cdot (1 - 0.14)}{0.08^2} \right)
\]

If you want more help with it, feel free to ask!

Would you like me to explain the process in more detail, or do you have any questions?

Here are 5 related questions you could explore:
1. How does the confidence level affect the sample size in a survey?
2. What are the implications of a higher margin of error in statistical studies?
3. How is the Z-value for a confidence interval determined?
4. Can a higher sample size reduce the margin of error, and by how much?
5. What role does the estimated proportion \( p \) play in calculating sample size?

**Tip:** Always round the sample size up to the next whole number when dealing with statistical calculations.

A medical researcher wants to construct a 99.5% confidence interval for the proportion of knee replacement surgeries that result in complications. An article in a medical journal suggested that approximately 14% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.08?

This problem focuses on determining the necessary sample size to estimate the proportion of knee surgeries resulting in complications with a 99.5% confidence interval. The estimated complication rate is 14%, and the margin of error is set at 0.08. Learn how to apply the sample size formula n = (Z^2 * p * (1 - p)) / E^2 to calculate the appropriate sample size.

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