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 probability of obtaining a sample estimate within $.50 of the population mean?
Do we have a higher probability of obtaining a sample estimate within $.50 of the population mean?
d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to your Z scores.
I need in answers of the questions pls
B.2.05/square root of (50)=.2899
.5/.2899=1.724732667
Under the z score table lookup +/- 1.72
=.9573-.0427
=.9146
To answer these questions, we can use the normal distribution and the concept of sampling distribution.
a. To find the probability that a sample of 50 male graduates will provide a sample mean within $0.50 of the population mean, we need to calculate the sampling distribution's standard deviation. The formula for the standard deviation of the sampling distribution is given by:
Standard Error = (Population Standard Deviation) / sqrt(sample size)
Standard Error = 2.30 / sqrt(50) ≈ 0.325
Next, we use the standard normal distribution (Z-distribution) to find the probability. We want to find the probability that the sample mean falls within $0.50 of the population mean, so we consider a range of $21.68 ± $0.50.
Z1 = ($21.68 - $0.50 - $21.68) / 0.325 ≈ -1.54
Z2 = ($21.68 + $0.50 - $21.68) / 0.325 ≈ 1.54
Using a standard normal distribution table or a calculator, we can find the cumulative probability between Z1 and Z2. This probability represents the likelihood that a sample of 50 male graduates will provide a sample mean within $0.50 of the population mean.
b. We follow the same steps as in part (a) to find the probability for female graduates.
Standard Error = 2.05 / sqrt(50) ≈ 0.290
Z1 = ($18.80 - $0.50 - $18.80) / 0.290 ≈ -1.72
Z2 = ($18.80 + $0.50 - $18.80) / 0.290 ≈ 1.72
We calculate the cumulative probability between Z1 and Z2 using the standard normal distribution.
c. To determine which case has a higher probability of obtaining a sample estimate within $0.50 of the population mean, compare the probabilities calculated in parts (a) and (b). Whichever probability is higher indicates a higher probability of obtaining a sample estimate within $0.50 of the population mean.
d. To find the probability that a sample of 120 female graduates will provide a sample mean more than $0.30 below the population mean, we can use a similar approach.
Standard Error = 2.05 / sqrt(120) ≈ 0.187
Z = ($18.80 - $0.30 - $18.80) / 0.187 ≈ -1.60
We can find the cumulative probability for Z < -1.60 using the standard normal distribution to determine the probability of getting a sample mean more than $0.30 below the population mean.