Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $22.1, and the variance is known to be $141.61. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error at most $0.64?

Answer 1

The question requires calculating the sample size needed to estimate the mean per capita income with an 85% confidence level and a margin of error of $0.64, using a known variance. The formula to be applied involves the variance, margin of error, and the z-score for the specified confidence interval.

Step-by-step explanation:

The student is tasked with determining how large a sample would be required to estimate the mean per capita income in a major city in California with an error margin of at most $0.64 and at the 85% confidence level, given a variance of $141.61 and an assumed mean income of $22.1 thousand. The calculation requires the use of the formula for the sample size in estimation problems, which is based on the variance, the error bound (margin of error), and the z-score corresponding to the desired level of confidence.

To solve this, we use the formula: sample size (n) = (z^2 * variance) / (margin of error)^2. The z-score for an 85% confidence level can be found using a z-table or statistical software. Once the z-score is identified, we can plug in the known variance and the square of the desired margin of error to find the required sample size. This problem is a typical example of statistical inference in econometrics, where population parameters are estimated based on sample data.