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? 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? 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? -Select your answer- Why? Because the standard error for female graduates is Select your answer the standard error for male graduates d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

Explanation / Answer

a.

Standard error of mean = SD / sqrt(n) = $2.30 / sqrt(50) = 0.3253

For a sample mean within $.50 of population mean $21.68, the mean should lie between $21.68 - 0.50 and $21.68 + 0.50 or between $21.18 and $22.18

Z value for $21.18 = (21.18 - 21.68) / 0.3253 = -1.54

Z value for $22.18 = (22.18 - 21.68) / 0.3253 = 1.54

So, P(21.18 < X < 22.18) = P(-1.54 < Z < 1.54) = P(Z < 1.54) - P(Z < -1.54)

= 0.9382 - 0.0618 = 0.8764

b.

Standard error of mean = SD / sqrt(n) = $2.05 / sqrt(50) = 0.29

For a sample mean within $.50 of population mean $18.80, the mean should lie between $18.80 - 0.50 and $18.80+ 0.50 or between $18.30 and $19.30

Z value for $18.30 = (18.30 - 18.80) / 0.29 = -1.72

Z value for $19.30 = (19.30 - 18.80) / 0.29 = 1.72

So, P(18.30 < X < 19.30) = P(-1.72 < Z < 1.72) = P(Z < 1.72) - P(Z < -1.72)

= 0.9573 - 0.0427 = 0.9146

c.

From part (a) and (b), we can see that sample of female graduates have a higher probability of obtaining a sample estimate within $.50 of the population mean.

Because the standard error for female graduates is lower than the standard error for male graduates.

d.

Standard error of mean = SD / sqrt(n) = $2.05 / sqrt(120) = 0.1871

For a sample mean more than $.30 below population mean $18.80, the mean should be more than $18.80 - 0.30 = $18.50

Z value for $18.50 = (18.50 - 18.80) / 0.1871 = -1.60

So, P(X > 18.50) = P(Z > -1.60) =

= 0.9452

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