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

Answer:

a) 217

b) 1351

c) 5403

Step-by-step explanation:

Given that:

confidence interval (c) = 0.95

$\alpha =1-0.95=0.05\\\frac{\alpha }{2} =\frac{0.05}{2}=0.025$

The Z score of $\frac{\alpha }{2}$ is from the z table is given as:

$Z_{\frac{\alpha }{2} }=Z_{0.025}=1.96$

Range = $45000 - $30000 = $15000

The standard deviation (σ) is given as:

$\sigma=\frac{Range}{4} =\frac{15000}{4}=3750$

Sample size (n) is given as:

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

a) E = $500

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

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

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

Answer 2

Answer:

217

1341

5403

No

Step-by-step 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.