You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be $95 \%$ confident that you estimate is within $1 \%$ of the true population proportion. How large of a sample size is required? Do not round mid-calculation. \[ n= \]

Step 1 ：We are given that we want to estimate a population proportion with no preliminary estimation. We want to be 95% confident that our estimate is within 1% of the true population proportion. We need to find out how large of a sample size is required.

Step 2 ：We can use the formula for the sample size needed to estimate a population proportion with a certain level of confidence and margin of error. The formula is: \[n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}\] where: n is the sample size, Z is the z-score for the desired confidence level, p is the estimated population proportion, and E is the desired margin of error.

Step 3 ：Since we don't have a preliminary estimation for the population proportion, we can use p = 0.5, which maximizes the product p*(1-p) and thus gives us the largest possible sample size. This is a conservative approach that ensures our sample size will be large enough, regardless of the true population proportion.

Step 4 ：The z-score for a 95% confidence level is approximately 1.96, and the desired margin of error is 1% or 0.01.

Step 5 ：Substituting these values into the formula, we can calculate the required sample size: Z = 1.96, p = 0.5, E = 0.01.

Step 6 ：By calculating, we get n = 9604.

Step 7 ：So, the required sample size to estimate the population proportion with a 95% confidence level and a margin of error of 1% is \(\boxed{9604}\).