The Economic Policy Institute periodically issues reports on wages of entry leve

Question

The Economic Policy Institute periodically issues reports on wages of entry level workers. The Institute reported that entry level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05. Use z-table. Round your answers to four decimal places. a. What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68? (Round z value in intermediate calculations to 2 decimal places.) b. What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80? (Round z value in intermediate calculations to 2 decimal places.) 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 $.50 of the population mean? Part (b) Do we have a higher probability of obtaining a sample estimate within $.50 of the population mean? Yes Why? Because the standard error for female graduates is smaller than d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean? (Round z value in intermediate calculations to 2 decimal places.) :the standard error for male graduates

Explanation / Answer

1)
we need to find p(21.68 - 0.50 < x < 21.68 + 0.50)

P(21.18 < x < 22.18)

= P((21.18 - 21.68) / ( 2.30 / sqrt(50) < z < (22.18 - 21.68) / ( 2.30 / sqrt(50))
= P(-1.537 < z < 1.537)

P(21.18 < x < 22.18) = P(-1.537 < z < 1.537) = 0.8761

2)

we need to find p(18.80 - 0.50 < x < 18.80 + 0.50)

P(18.3 < x < 19.3)

= P((18.3 - 18.80) / ( 2.05 / sqrt(50) < z < (19.3 - 18.80) / ( 2.05/ sqrt(50))
= P(-1.724 < z < 1.724)

P(18.3 < x < 19.3) = P(-1.724 < z < 1.724) = 0.9164

d)
P(x < 19.1)

z= (x - mean) / (s/sqrt(n))
= (19.1 - 18.80) / ( 2.05/ sqrt(50)
= 1.035
P(x < 19.1) = P ( z < 1.035) = 0.8504

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