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.