Question

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 $29,000 and $47,600. 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 $500? (b) How large a sample should be taken if the desired margin of error is $300? (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.

Answer 1

the planning value for the population sd is:-

- - (mar = min) – ( 1760 – 2900) – 1656

z critical value for 95% confidence level, both tailed test be:-

** = 1.96

a).margin of error (E) = 500

the needed sample size be:-

1.96 * 4650 500 500 = E ) 333

b).margin of error (E) = 300

the needed sample size be:-

$=\left ( \frac{z^**\sigma}{E} \right )^2=\left ( \frac{1.96*4650}{300} \right )^2\approx\mathbf{923}$

c).margin of error (E) = 100

the needed sample size be:-

-639) - (1.96.9656) * ss01

d).NO, i will not recommend trying to obtain $100 margin of error.Because, it will be tome consuming ,tedious, and costly.

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