Question

You may need to use the appropriate appendix table or technology to answer this question. The Economic Policy Institute periodically issues reports on workers' wages. Suppose 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. (Round your answers to four decimal places.) (a) What is the probability that a sample of 40 male graduates will provide a sample mean within $1.00 of the population mean, $37.39? (b) What is the probability that a sample of 40 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? part (a), because the standard error is higher part (a), because the standard error is lower part (b), because the standard error is higher part (b), because the standard error is lower (d) What is the probability that a sample of 110 female graduates will provide a sample mean more than $0.60 below the population mean, $27.83?

Answer 1

information, we have @ n. From the given Given that, X-Me +1.00, M 37. 39,5 4.60 na 40 The formula for Standard z-score is. 122 -M P(-1.00 (X- M L1.60 CO) = P 14.60 .: -e 126 101)= v =P ( 0 1243 <Z2 073). = P(-1.37 2Z<1:37) = P(Z 21037) - P(Z 2-1.37) from the standard normal table", the area to the left of 2 = 1.37 is 0.0853 and the area to the left of z = 1.3:1 PS 6.9147 p(-60 LZ-M < 10) = P(2.21.37) - P(Za-1:37) 0.9147 - 0-0853 = 0.8994 Hence, the probability that a sample of Ho male gradualy will provide a sample mean within $1.00 a the population mean $37.39 is 10.8294

6 from the givers information, we have ME + 1.00 112 27.83,5-410 P(-10 2 ā-Malo) a pl T -M 40 - 110 a 4:10 luo " = p ( 3 < 27 0.6483) 0-64830-03 = P(-1.54 < Z 21:54) = P(22154)-P(24-1054) from the "standard normal tale", the area to the left of z = -1.54 0.0618 and the area to the left of z = losu is 0.9382. & P(ahocZ-N210) = P(Z <1.54) - P(ZL11.5U). - 0.9382 -0.0618 - 0.8164 " The probability that a sample of 40 female gradity will Provide a sample mean within $10 of the population mean $ 97.33 510-87641 - The higher probability of obtaining a sample estimate within $10 of the population mean is obtained as shown below: The probability in part @ for a sample of 40 male graduates will provide a sample mean within 8.100 of the population mean $ 37 39 is 0.8294. The probability in part 6 for a sample of the female

a 10 of the gradudes will provide a sample population mean $ 27.83 is mean within 0.8264. Hence, the probability for females is higher than the probability et males for a sample meren within $ 1.00 of the population means This indicates that probability of part 0 is greater than part @. Past 6 , because the standard awe is higher. di The probability that a sample at no female graduates will Provide a sample mean more than $ 0-60 below the population mean is ostained below! from the given information , we have -A = 06 , M = 94.83,6= 4.10, no 110 The required probability is, P(&M 70.60) a ir plānu 2 0.60) laplz 21:60] from the standard normal table", the area to the left of 7 160 is 0.9452 - Planne >0.60) = 1-P(231.60) = 1-0.9052 = 0.0548 Hence, The Probability that a sample of 110 female graduates will provide a sample mean more than $ 0.60 below the popuktion mean is 10-0548] . * please rate a