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.
Q.What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?
The way I set it up is z=(-.30)/(2.05/square root 120))= -1.60
(.30) /(2.05/square root 120))= 1.60
=.8904 which says not correct?
Please help!
2 answers
you did it correctly. you just have to take the negative z score of -1.6 and convert it to a percentage that's to the left of that. to 4 decimal places it's .0548
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3 answers