1. **Problem Statement:** An economist wants to estimate the mean number of school children (in thousands) who read above their grade level in a major city in Florida. A sample of size $n=45$ is taken, with sample mean $\bar{x}=11.2$ and population standard deviation $\sigma=3.7$. The population distribution is skewed to the right. 2. **Goal:** Calculate the margin of error for the population mean. 3. **Key Formula for Margin of Error (ME):** $$ ME = z^* \times \frac{\sigma}{\sqrt{n}} $$ where: - $z^*$ is the critical value from the standard normal distribution corresponding to the desired confidence level, - $\sigma$ is the population standard deviation, - $n$ is the sample size. 4. **Important Considerations:** - Since the population standard deviation $\sigma$ is known, the $z$-distribution is appropriate. - The sample size $n=45$ is greater than 30, which is generally considered large enough for the Central Limit Theorem to apply, making the sampling distribution of the sample mean approximately normal even if the population is skewed. - Therefore, using the $z$-distribution to calculate the margin of error is appropriate here. 5. **Summary:** Use the $z$-distribution method for margin of error calculation because: - Population standard deviation is known. - Sample size is sufficiently large ($n=45$). - The Central Limit Theorem justifies normal approximation despite skewness. 6. **Final Answer:** The most appropriate method to calculate the margin of error for the population mean in this scenario is to use the $z$-distribution formula: $$ ME = z^* \times \frac{3.7}{\sqrt{45}} $$ where $z^*$ depends on the chosen confidence level.