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 21.6, and the variance is known to be 153.76. How large of a sample would be required in order to estimate the mean per capita income at the 98th percentile with a margin of error of 1.5?

Answer 1

To estimate the mean income with a 1.5 margin of error at the 98th percentile, a sample size of approximately 371 is required.

Step-by-step explanation:

The student is interested in determining the sample size required to estimate the mean per capita income with a given margin of error and confidence level. To find the sample size (n), the formula n = (Z*σ/E)^2 is used, where Z is the z-score corresponding to the confidence level, σ is the population standard deviation, and E is the margin of error. Since the variance is given as 153.76, the standard deviation σ is the square root of the variance, which is 12.4 (in thousands of dollars). For a 98th percentile confidence level, the z-score (Z) is approximately 2.33.

Using the formula:

n = (Z*σ/E)^2
n = (2.33*12.4/1.5)^2
n ≈ (28.872/1.5)^2
n ≈ (19.248)^2
n ≈ 370.14

Therefore, a sample size of approximately 371 is required to achieve the desired margin of error at the 98th percentile confidence level.