Question

(Use this problem for questions 16-20.) In a recent market research report, based upon responses from 5,000 randomly-chosen recent college graduates working in Washington, DC, the average starting salaries for college graduates employed in Washington, DC, is $45,000 with a standard deviation of $7,500. (Each question here is worth two points.) 16. With a 68% confidence level, what is the range of starting salaries in Washington, DC for recent college graduates? Show your work 17. With a 95%% confidence level, what is the range of starting salaries in Washington, DC for recent college graduates? Show your work 18. With a 99%% confidence level, what is the range of starting salaries in Washington, DC for recent college graduates? Show your work 19. At a significance level of 5%, what is the confidence interval of starting salaries in Washington, DC for recent college graduates? Explain. 20. If the market research firm had chosen to use only 500 random responses (instead of 5,000) to calculate average starting salaries for recent college graduates working in Washington, DC, how would the confidence interval in question 19 change? a. It would narrow b. It would expand reducing sample size reduces precision, increases CI) c. It would not change

Answer 1

h>>30, distribution so, it com be affroximal-cel (CLT) to Normal ñ = 45000, 5=7500, h = 5000 88% (5 = *t for ; (using Empirical Pales = 45000 75 VSTV = 45000 + 106.07 = |(444893.93 , $ 45106.07) 17. 95% = † 1.96 use -NORMS INV () -0.05 17500 i 45000 $ 1.96 45000 207.89 -($ 44792.11, $45207.89 )

18. 99% (r = x 2.58 ; NORMS INVB) = 2.58 When a 0.01 / 7500 = 4.5000+2:58 (1960) -4-5000+ 2.58 = 45000 I 273.65 =ll $ 44726,35, $ 45273.65) 19 Q.17 95% CI = { $ 44 792011, $ 45207689) starting salary would be the lower bound of that gutenne 1. 44792-177 as n decrease E will lucrease as a result the confidence Interval width will expond