Question

we would like to construct a 90% confidence interval estimate of the annual starting salaries for college graduates with a degree in finance. based on past studies, the standard deviation for the salary population is 16500 dollars. what is the minimum sample size needed to estimate this population mean with a margin of error no larger than $2000? a. 96 b. 185 c. 215 d. 112

Answer 1

To estimate the population mean with a margin of error no larger than $2000, we need to find the minimum sample size using the formula N = (Z * σ / E)², where N is the sample size, Z is the Z-score corresponding to the desired confidence level (Z = 1.645 for 90% confidence), σ is the standard deviation, and E is the margin of error (E = $2000). Plugging in the values, the minimum sample size needed is 15. So the correct answer is Option D.

Step-by-step explanation:

To estimate the population mean with a margin of error no larger than $2000, we need to find the minimum sample size using the formula:

N = (Z * σ / E)²

where N is the sample size, Z is the Z-score corresponding to the desired confidence level (Z = 1.645 for 90% confidence), σ is the standard deviation, and E is the margin of error (E = $2000).

Plugging in the values:

N = (1.645 * 16500 / 2000)² = 14.2

Rounding up to the nearest whole number, the minimum sample size needed is 15. Therefore, the correct answer is d. 112.