Econ Stats Ch 7 Questions

22. The Wall Street Journal reported that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is if = $2400. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? b. What is the advantage of a larger sample size when attempting to estimate the population mean?

NORM.DIST(16842,16642,2400/SQRT(30),TRUE) - NORM.DIST(16442,16642,2400/SQRT(30),TRUE) = .3519 (and just change the 30 as you do the next ones)

q28 A population proportion is .40. A random sample of size 200 will be taken and the sample proportion p— will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within ±.03 of the population proportion? b. What is the probability that the sample proportion will be within ± .05 of the population proportion?

Using Excel: NORM.DIST(.43,0.40,SQRT(.4*.6/200),TRUE) - NORM.DIST(.37,0.40,SQRT(.4*.6/200),TRUE) = .6135

23. The Economic Policy Institute periodically issues reports on worker's wages. The institute reported that mean wages for male college graduates were $37.39 per hour and for female college graduates were $27.83 per hour in 2017. Assume the standard deviation for male graduates is $4.60, and for female graduates it is $4.10. a. What is the probability that a sample of 50 male graduates will provide a sample mean within $1.00 of the population mean, $37.39? 7.6 Sampling Distribution of 331 b. What is the probability that a sample of 50 female graduates will provide a sample mean within $1.00 of the population mean, $27.83? c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within $1.00 of the population mean? Why? d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.60 below the population mean, 27.83?

a. NORM.DIST(38.39,37.39,4.6/SQRT(50),TRUE) - NORM.DIST(36.39,37.39,4.6/SQRT(50),TRUE) = .8758 b. Using Excel: NORM.DIST(28.83,27.83,4.1/SQRT(50),TRUE) - NORM.DIST(26.83,27.83,4.1/SQRT(50),TRUE) = .9154 c. In part (b) we have a higher probability of obtaining a sample mean within $1.00 of the population mean because the standard error for female graduates (.5798) is smaller than the standard error for male graduates (.6505). d. Using Excel: NORM.DIST(27.23,27.83,4.1/SQRT(120),TRUE) = .0545

Q15. A population has a mean of 200 and a standard deviation of 50. Suppose a simple random sample of size 100 is selected and x is used to estimate a. What is the probability that the sample mean will be within ±5 of the population mean? b. What is the probability that the sample mean will be within -±10 of the population mean?

a. Using Excel: NORM.DIST(205,200,50/SQRT(100),TRUE) - NORM.DIST(195,200,50/SQRT(100),TRUE) = .6827 b.Using Excel: NORM.DIST(210,200,50/SQRT(100),TRUE) - NORM.DIST(190,200,50/SQRT(100),TRUE) = .9545 c

31. The president of Doerman Distributors, Inc. believes that 30% of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and p = .30. What is the sampling distribution of p for this study? b. What is the probability that the sample proportion p— will be between .20 and .40? c. What is the probability that the sample proportion will be between .25 and .35

b. Using Excel: NORM.DIST(.40,.30,SQRT(.3*.7/100),TRUE) - NORM.DIST(.20,.30,SQRT(.3*.7/100),TRUE) = .9709 c. Using Excel: NORM.DIST(.35,.30,SQRT(.3*.7/100),TRUE) - NORM.DIST(.25,.30,SQRT(.3*.7/100),TRUE) = .7248

Q36. The Grocery Manufacturers of America reported that 76% of consumers read the ingredients listed on a product's label. Assume the population proportion is p = .76 and a sample of 400 consumers is selected from the population. a. Show the sampling distribution of the sample proportion p, where T) is the proportion of the sampled consumers who read the ingredients listed on a product's label. b. What is the probability that the sample proportion will be within ± .03 of the population proportion? c. Answer part (b) for a sample of 750 consumers

b. Using Excel: NORM.DIST(.79,.76,SQRT(.76*.24/100),TRUE) - NORM.DIST(.73,.76,SQRT(.76*.24/100),TRUE) = .8399 c. Using Excel: NORM.DIST(.79,.76,SQRT(.76*.24/750),TRUE) - NORM.DIST(.73,.76,SQRT(.76*.24/750),TRUE) = .9456

Q20 20. U.S. Unemployment. Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study. a. Show the sampling distribution of x, the sample mean average for a sample of 50 unemployed individuals. b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1/2 week of the population mean?

b. Using Excel: NORM.DIST(18.5,17.5,4/SQRT(50),TRUE) - NORM.DIST(16.5,17.5,4/SQRT(50),TRUE) = .9229 c. Using Excel: NORM.DIST(18,17.5,4/SQRT(50),TRUE) - NORM.DIST(17,17.5,4/SQRT(50),TRUE) = .6232