Question

The Economic Policy Institute periodically issues reports on wages of entry level workers. The institute reported that entry level wages for male college student graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2020. 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 probability of obtaining a sample estimate within $.50 of the population mean? a or b?

d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean? Answer

Answer 1

= sample mean of entry level wages for male college graduates

=sample mean of entry level wages for female college graduates

= entry level wages for male college student graduates =21.68

=entry level wages for female college student graduates=18.80

=2.30

=2.05

a) we have to find probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68

i.e,bound on error =.50

i.e,

we know confidence interval =

therefore ,bound on error =

                        is the z critical value needed for level of confidence.

bound on error =     .50=                      ( is the standard error of mean)

                          

                             =1.5371 , says sample mean of sample size 50 is within 1.5371 standard deviation

required probability= = 2*0.4382 =0.8764 (from z table., shown below)

b)we have to find the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80

bound on error =.50

n=50

.50 =

   .50 =

   =1.72

=2*0.4573 =0.9146

finding area from z table

c) in part b

we can see it has a probability of 0.9146 where part a has a probability of 0.876

d) the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean

i.e probability that   

since we dont seperate tables for normal distribution, convert them into z

   for =.30 , z=-98.85

     for =18.80,z=0

(area to the left of 0)

here is a rough graph of this