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. 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 prob- ability of obtaining a sample estimate within $.50 of the population mean? Why? d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean? Note: In some of the questions you are required to "show"the sampling distribution. This means to state the mean and standard deviation of the estimator and to comment on the shape.

Answer 1

Solution:

a. We are given that the entry-level wage for male college graduate were $21.68 per hour with standard deviation $2.30

We have to find:

P(21.68 -0.50 < i < 21.68 +0.50)

The sampling distribution of the mean is:

2.30 Mi = 21.68,01 = Oz = 750 = 0. == 0.325269

Now using the z-score formula, we have:

21.18 - 21.68 P(21.68–0.50 S 7 S 21.68+0.50) = P( 0.325269 -szs 22.18 - 21.68 0.325269

= P(-1.537 < < 1.537)

= P(Z < 1.537) - P(Z < -1.537)

Now using the standard normal table, we have:

P(21.68 -0.50 < T < 21.68 +0.50) = P(Z < 1.537) - P(Z < -1.537) = 0.9379 - 0.0621 = 0.8758

Therefore, the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean 21.68 is 0.8758

b. We are given that the entry-level wage for female college graduate were $18.80 per hour with standard deviation $2.05

We have to find:

P(18.80 – 0.50 < i < 18.80 +0.50)

The sampling distribution of the mean is:

di = 18.80,05 = 2.05 == 0.28991378 50

Now using the z-score formula, we have:

18.3 – 18.80 P(18.80–0.50 S T S 18.80+0.50) = P( 0.28991378 IA 19.3 - 18.80 0.28991378

= P(-1.725 <2<1.725)

= P(Z < 1.725) - P(Z < -1.725)

Now using the standard normal table, we have:

$\large \boldsymbol{P(18.80-0.50 \leq \bar{x} \leq 18.80+0.50)=P(z \leq 1.725)-P(z \leq -1.725)=0.9577-0.0423=0.9154}$

Therefore, the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean 18.80 is 0.9154

c. Part b has a higher probability of obtaining a sample estimate within $0.50 of the population mean because the standard deviation is smaller in part b.

d. We are given that the entry-level wage for female college graduate were $18.80 per hour with standard deviation $2.05

We have to find:

$\large P( \bar{x}>18.80-0.30)$

The sampling distribution of the mean is:

$\large \mu_{\bar{x}}=18.80, \sigma_{\bar{x}}=\frac{2.05}{\sqrt{120}}=0.18713854$

Now using the z-score formula, we have:

18.50 – 18.80 P(ī > 18.50) = P(2>, 0.18713854

   $\large =P( z > -1.603)$

  

Now using the standard normal table, we have:

Pã> 18.80 – 0.30) = P(Z > -1.603) = 0.9455

Therefore, the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below population mean 18.80 is 0.9455