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. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is
a. $500?
b. $200?
c. $100?
d. Would you recommend trying to obtain the $100 margin of error? Explain.

Answer 1

Answer:

we not recommend trying to obtain the $100 margin of error because the sample size is too larger.

Explanation:

given data

Annual starting salaries  = $30,000 and $45,000.

confidence interval = 95%

solution

we get here the planning value for the population standard deviation that is express as

planning value = (45000-30000) ÷ 4 =

planning value = 3750

and

for 95% confidence interval a is 0.05

we use here standard normal table and get

Z(0.025) = 1.96

so for error E = $500

n = (Z×S/E)²    

put here value

n = ( 1.96 × $\frac{3750}{500}$ )²

n = 216.09 = 217

and

now for error $200

n = ( 1.96 × $\frac{3750}{200}$ )²  

n = 1350.56  = 1351

and

for error E = $100

n = ( 1.96 × $\frac{3750}{100}$ )²  

n = 5402.25  =  5403

and

we not recommend trying to obtain the $100 margin of error because the sample size is too larger.

Answer 2

In the sampling, one should not recommend trying to obtain the $100 margin of error since the sample size is large.

How to explain the sampling?

Firstly, the planing value will be:

= (45000/30000) / 4

= 3750

Then the number will be:

= 1.96 × (3750/500)²

= 217

Also, for error 200, the desired margin will be:

= 1.96 × (3750/100)²

= 5403.

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