A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a $1 \%$ margin of error at a $90 \%$ confidence level, what size of sample is needed? Give your answer in whole people.

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 $1 \%$ margin of error at a $90 \%$ 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 desired margin of error.

Step 3 ：For a $90 \%$ confidence level, the Z-score is approximately 1.645. The margin of error is $1 \%$, or 0.01. Let's plug these values into the formula and calculate the sample size.

Step 4 ：Substituting the given values into the formula, we get: \(n = \frac{(1.645)^2 * 0.5 * (1-0.5)}{(0.01)^2}\)

Step 5 ：Solving the above expression, we find that the required sample size is approximately 6766.

Step 6 ：\(\boxed{6766}\) is the required sample size.