Question

You may need to use the appropriate appendix table or technology to answer this 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.† Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05. (Round your answers to four decimal places.)

(a)

What is the probability that a sample of 60 male graduates will provide a sample mean within $0.50 of the population mean, $21.68?

(b)

What is the probability that a sample of 60 female graduates will provide a sample mean within $0.50 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 $0.50 of the population mean? Why?

part (a), because the standard error is lowerpart (b), because the standard error is lower    part (a), because the standard error is higherpart (b), because the standard error is higher

(d)

What is the probability that a sample of 130 female graduates will provide a sample mean less than the population mean by more than $0.30?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	21.68
std deviation   =σ=	2.3000

sample size       =n=	60
std error=σx̅=σ/√n=	0.2969

probability that a sample of 60 male graduates will provide a sample mean within $0.50 of the population mean, $21.68:

probability =

P(21.18<X<22.18)

=

P(-1.68<Z<1.68)=

0.9535-0.0465=

0.9070

b)

probability =

P(18.3<X<19.3)

=

P(-1.89<Z<1.89)=

0.9706-0.0294=

0.9412

c)

part (b), because the standard error is lower

d)

probability =

P(X<18.5)

=

P(Z<-1.67)=

0.0475