Question

8. -/4.16 points ASWSBE14 8.E.027. My Notes + Ask Your Teacher You may need to use the appropriate appendix table or technology to answer this question Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $42,000 and $57,000. 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 $5007 (b) How large a sample should be taken if the desired margin of error is $200? (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 This answer has not been graded yet.

Answer 1

$\dpi{100} \mathbf{Solution}$

given : mar = 57000, min = 42000

Planning value for population standard deviation () = max - min 57000 - 42000 -= 3750

$\dpi{100} \mathbf{formula:}\;sample\;size\;(n)=\left [ \frac{z_{\frac{\alpha }{2}}*\sigma}{E} \right ]^{2}$

$\dpi{100} \mathbf{for \;95\%\; confidence \;level, \;z_{\frac{\alpha }{2}}=1.96}$

a for margin of error E) = 500

sample sise (n) = [1.96 * 37501 400

sample size (n) = 337.6

sample size (n) 338

b) for margin of error E) = 200

[1.96 * 37507 sample size (n) = 200

sample size (n) = 1350.6

sample size (n) 1351

c) for margin of error E) = 100

[1.96 * 37507 sample size (n) = 100

sample size (n) = 5402.25

sample size (n) 5403

d) No, the sample size would probably be too time consuming and costly