Question

Exercise 8.27 (Algorithmic)} Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $50,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation (to the nearest whole number)? 3750 How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)? a. $500? 216 b. $200? 1351 5403 c. $100? d. Would you recommend trying to obtain the $100 margin of error? No, sample requires too much time and expense e. When the desired confidence level is fixed, what happens to the sample size as the margin of error is decreased? It increases

Answer 1

Maximum = 50000

Minimum = 30000

Range = maximum - minimum = 50000 - 30000= 20000

Using Range rule of thumb

Standard deviation = Range / 4 = 20000 / 4 = 5000

(a)

Z_$\alpha$/2 = 1.96

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [1.96 * 5000 / 500]²

n = 384.16

Sample size = n = 385

(b)

n = [1.96 * 5000 / 200]²

n = 2401

Sample size = n = 2401

(c)

n = [1.96 * 5000 / 100]²

n = 9604

Sample size = n = 9604