Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $42,000 and $60,600. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. (Round your answers up to the nearest whole number.)

What is the planning value for the population standard deviation?

(a)

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

(b)

How large a sample should be taken if the desired margin of error is $300?

(c)

How large a sample should be taken if the desired margin of error is $100?

(d)

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

Answer 1

Solution :

Given that,

Population standard deviation = = 60600 - 42000 / 4 = 4650

a) Margin of error = E = 500

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [1.96 * 4650 / 500]2

n = 332.25

Sample size = n = 333

b) Margin of error = E = 300

sample size = n = [Z/2* / E] 2

n = [1.96 * 4650 / 300]2

n = 922.94

Sample size = n = 923

c) Margin of error = E = 100

sample size = n = [Z/2* / E] 2

n = [1.96 * 4650 / 100]2

n = 8306.49

Sample size = n = 8307

d) No, the sample size would probably be too time consuming and costly.