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.

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