This is my homework problem. Just wondering if I solved it correctly.
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is:
a. $500
b. $200
c. $100
My answers:
Planning value: (range/4) = (15,000/4) = 3,750
a. [(1.96²)(3750²)] / (500²) = 217
b. same as above but replaced 500² with 200² = 1351
c. same as above but replaced 200² with 100² = 5403
Am I correct? Thank you for any input!
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is:
a. $500
b. $200
c. $100
My answers:
Planning value: (range/4) = (15,000/4) = 3,750
a. [(1.96²)(3750²)] / (500²) = 217
b. same as above but replaced 500² with 200² = 1351
c. same as above but replaced 200² with 100² = 5403
Am I correct? Thank you for any input!
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the approach and calculations are correct. Only comment is that sometimes I see as estimate of the std dev from the range to be range/6 ( assuming +/- 3 std dev). But your assumption of range/4 leads to a larger planning estimate for the std dev, and thus is more conservative (leads to larger sample size needed) - so I think range/4 is to be preferred here.