You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $\sigma=64.3$. You would like to be $99 \%$ confident that your estimate is within 3.75 of the true population mean. How large of a sample size is required? Do not round mid-calculation. Give your answer in whole people. \[ \mathrm{n}= \] Submit Question Jump to Answer
Solution
Step 1 ：We are given that the population standard deviation \(\sigma\) is 64.3, the desired margin of error \(E\) is 3.75, and the confidence level is 99%.
Step 2 ：The Z-score corresponding to a 99% confidence level is approximately 2.576. This value can be found using a standard normal distribution table or a statistical calculator.
Step 3 ：We can use the formula for the sample size needed to estimate a population mean with a certain level of confidence: \(n = (Z*\sigma/E)^2\).
Step 4 ：Substituting the given values into the formula, we get \(n = (2.576*64.3/3.75)^2\).
Step 5 ：Calculating the above expression, we find that \(n\) is approximately 1950.6.
Step 6 ：Since we can't have a fraction of a person, we'll round up to the nearest whole number. Therefore, the required sample size is \(\boxed{1951}\).
