A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a $2 \%$ margin of error at a $99 \%$ confidence level, what size of sample is needed? Give your answer in whole people.
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Step 1 ：We are given a political candidate who wants to conduct a poll to determine what percentage of people support her. She wants a 2% margin of error at a 99% confidence level. We are asked to find the sample size needed.

Step 2 ：We use the formula for the sample size in a proportion, which is \(n = \frac{Z^2 * P * (1-P)}{E^2}\), where n is the sample size, Z is the Z-score (which corresponds to the desired confidence level), P is the estimated proportion of the population (we'll use 0.5 since we don't have any prior information), and E is the margin of error.

Step 3 ：The Z-score for a 99% confidence level is approximately 2.576. This value can be found in a Z-table or using a statistical calculator.

Step 4 ：Substituting the given values into the formula, we get \(n = \frac{(2.576)^2 * 0.5 * (1-0.5)}{(0.02)^2}\)

Step 5 ：Solving the above expression, we find that the sample size needed is 4148.

Step 6 ：Thus, the candidate needs to survey at least 4148 people to achieve a 2% margin of error at a 99% confidence level.

Step 7 ：Final Answer: The required sample size is \(\boxed{4148}\)