Question

Annual starting salaries for college graduates with degrees in
business administration are generally expected to be between $30,000 and
$45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired.

a. What is the planning value for the population standard deviation?

How large a sample should be taken if the desired margin of error is
b.$500?

c.$200?

d.$100?

e.Would you recommend trying to obtain the $100 margin of error? Explain.

Answer 1

a)
std. dev. = (45000 - 30000)/4 = 3750

b)
The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 500, σ = 3750

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 3750/500)^2
n = 216.09 i.e. 216

c)
n >= (zc *σ/E)^2
n = (1.96 * 3750/200)^2
n = 1350.56 i.e. 1351

d)
n >= (zc *σ/E)^2
n = (1.96 * 3750/100)^2
n = 5402.25 i.e. 5402

e)
No, because the sample size of 5402 is not practical to have.