We want to obtain a sample to estimate a population mean. Based on previous evidence, researchers believe the population standard deviation is approximately $\sigma=58.1$. We would like to be $99 \%$ confident that the estimate is within 2.5 of the true population mean. How large of a sample size is required? \[ n= \]

Step 1 ：We are given that the population standard deviation, \(\sigma\), is 58.1, the desired margin of error, \(E\), is 2.5, and the confidence level is 99%.

Step 2 ：We need to find the z-score, \(Z_{\alpha/2}\), for a 99% confidence level. This can be found from a standard normal distribution table or calculated using statistical software. The z-score for a 99% confidence level is approximately 2.576.

Step 3 ：We can now substitute these values into the formula for the sample size in a confidence interval estimation for a population mean: \[n = \left(\frac{Z_{\alpha/2} * \sigma}{E}\right)^2\]

Step 4 ：Substituting the given values, we get: \[n = \left(\frac{2.576 * 58.1}{2.5}\right)^2\]

Step 5 ：Calculating the above expression, we find that the required sample size is approximately 3583.6.

Step 6 ：Since we can't have a fraction of a sample, we'll round up to the nearest whole number. Therefore, the required sample size is \(\boxed{3584}\).