Suppose we want to find a $99 \%$ confidence interval for a population proportion. Determine the sample size required to produce a margin of error of at most 2.5 percentage points if we have reason to believe that the population proportion is at least 0.42. Enter your answer as a whole number.

Step 1 ：We are given that the margin of error (E) is 0.025, the estimated population proportion (p) is 0.42, and the Z-score (Z) corresponding to a 99% confidence level is 2.576.

Step 2 ：We want to find the sample size (n) required to produce a margin of error of at most 2.5 percentage points. The formula for this is \(n = \frac{{Z^2p(1-p)}}{E^2}\).

Step 3 ：Substituting the given values into the formula, we get \(n = \frac{{(2.576)^2 * 0.42 * (1-0.42)}}{(0.025)^2}\).

Step 4 ：Calculating the above expression, we find that the required sample size is \(\boxed{2587}\).