Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $41,000 and $55,200. 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 $600?

(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

What is the planning value for the population standard deviation?

s=range/4=55200-41000/4= 3550

population standard deviation=$3550

population standard deviation=$3550

(a)

How large a sample should be taken if the desired margin of error is $600?

n=(Z*sigma/MOE)^2

n=(1.96*3550/600)^2

n=134.4827

n=135

(rounded to next integer)

n=135

(b)

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

n=(Z*sigma/MOE)^2

n=(1.96*3550/300)^2

n= 537.9307

n=538

n=538

(c)

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

n=(Z*sigma/MOE)^2

n=(1.96*3550/100)^2

n=4841.376

n=4842

4842

(d)

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

Yes since it results is larger sample size

as sample size increases ,margin of error decreases and sampling error decreases