Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,000 and $50,000. Assume that a 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 $500? Round your answer to next whole number.

$230?

$140?

Answer 1

Given that,

a) Population standard deviation = = high value - low value / 4 ( range rule thumb )

Population standard deviation = = 50000 - 10000 / 4 = 10000

b)

Margin of error = E = 500

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [1.96 * 10000 / 500]2

Sample size = n = 1537

#

Margin of error = E = 230

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [1.96 * 10000 / 230]2

Sample size = n = 7262

#

Margin of error = E = 140

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [1.96 * 10000 / 140]2

Sample size = n = 19400

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