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 (to the nearest whole number)?

How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)?

a. $500?
b. $200?
c. $100?

Answer 1

217

1341

5403

No

Explanation:

Given:

c = 95%

E = 500,200,100

RANGE = 45000 — 30000 = 15000

The standard deviation is approximately one forth of the range:

σ=RANGE/4

=15000/4

=3750

Formula sample size:

n = ((z_
`\alpha`/2*σ)/E)^2

For confidence level 1—a = 0.95, determine z_
`\alpha`/2 = z_0.025 using table 1 (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

z_
`\alpha`/2 = 1.96

The sample size is then (round up!):

a.n = ((z_
`\alpha`/2*σ)/E)^2 =217

b. n = ((z_
`\alpha`/2*σ)/E)^2 =1351

c. n = ((z_
`\alpha`/2*σ)/E)^2 =5403

d. It is not recommendable to try to obtain the $100 margin of error, because it will cost a lot of time and money to obtain such a large random sample.

Answer 2

a) 217

b) 1351

c) 5403

Explanation:

Given that:

confidence interval (c) = 0.95

`\alpha =1-0.95=0.05\\(\alpha )/(2) =(0.05)/(2)=0.025`

The Z score of
`(\alpha )/(2)` is from the z table is given as:

`Z_{(\alpha )/(2) }=Z_(0.025)=1.96`

Range = $45000 - $30000 = $15000

The standard deviation (σ) is given as:

`\sigma=(Range)/(4) =(15000)/(4)=3750`

Sample size (n) is given as:

`n=(\frac{Z_{(\alpha)/(2) }\sigma}{E} )^2`

a) E = $500

`n=(\frac{Z_{(\alpha)/(2) }\sigma}{E} )^2= ((1.96*3750)/(500) )^2` ≈ 217

b)
`n=(\frac{Z_{(\alpha)/(2) }\sigma}{E} )^2= ((1.96*3750)/(200) )^2` ≈ 1351

c)
`n=(\frac{Z_{(\alpha)/(2) }\sigma}{E} )^2= ((1.96*3750)/(100) )^2` ≈ 5403