Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,000 and $35,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired.
a. What is the planning value for the population standard deviation?
b. How large a sample should be taken if the desired margin of error is $400? Round your answer to next whole number.
$230?
$90?
a)
std dev,σ = Range/4 = (35000-10000)/4 = 6250
b)
i) Standard Deviation , σ =
6250
sampling error , E =
400
Confidence Level , CL= 95%
alpha = 1-CL = 5%
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 1.960
* 6250 / 400 ) ²
= 937.856
So,Sample Size needed=
938
---------------
ii)
Standard Deviation , σ =
6250
sampling error , E =
230
Confidence Level , CL= 95%
alpha = 1-CL = 5%
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 1.960
* 6250 / 230 ) ²
= 2836.616
So,Sample Size needed=
2837
------------------
iii)
Standard Deviation , σ =
6250
sampling error , E =
90
Confidence Level , CL= 95%
alpha = 1-CL = 5%
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 1.960
* 6250 / 90 ) ²
= 18525.554
So,Sample Size needed=
18526
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