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 \(\boxed{4148}\)