To solve this problem, we need to calculate the sample size necessary to obtain a 99.5% confidence interval with a margin of error of 0.08, based on the estimate that 14% of the operations result in complications.

Here is the formula for determining the required sample size for a proportion:

$n = \left(\frac{Z_{\alpha/2}^2 \cdot p \cdot (1 - p)}{E^2}\right)$

Where:

• $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level.
• $p$ is the estimated proportion (0.14 in this case).
• $E$ is the margin of error (0.08 in this case).
• $n$ is the sample size we need to find.

Step 1: Find the Z-score for a 99.5% confidence interval

A 99.5% confidence interval corresponds to a Z-score of approximately 2.807 (since half of 0.5% lies in each tail of the normal distribution).

Step 2: Plug the values into the formula

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

First, calculate $2.807^2$, $p \cdot (1 - p)$, and then divide by the square of the margin of error.

Let me calculate that for you:The sample size required to obtain a 99.5% confidence interval with a margin of error of 0.08, using the estimate that 14% of operations result in complications, is 149.

Would you like further details or have any questions?

Here are five related questions for deeper understanding:

1. How does the margin of error affect the required sample size?
2. Why is the Z-score for a 99.5% confidence interval higher than for a 95% interval?
3. How would the sample size change if the estimated proportion of complications were different?
4. What impact does decreasing the confidence level have on the sample size?
5. What happens to the sample size if we want to reduce the margin of error?

Tip: The higher the confidence level and the smaller the margin of error, the larger the required sample size will be.