Question

You are to estimate the average starting income offer for 2019 graduates with a major in marketing. Based on a pilot study, your estimate of the standard deviation is $6,000. Precision is important, so your allowable error is $500. Furthermore, you need a high level of confidence; you simply do not want to make a mistake. So the level of confidence associated with your estimate has been established at 98%. What is the required size for your simple random sample?

Answer 1

Answer

given that

margin of error(E) = 500

standard deviation (sd) = 6000

z critical = 2.33

sample size n = ((z*sd)/E)^2

= ((2.33*6000)/500)^2

= 27.96^2

= 781.76

= 782 (rounded to nearest whole number)