Business Week conducted a survey of graduates from 30 top MBA programs (Business
ID: 3355721 • Letter: B
Question
Business Week conducted a survey of graduates from 30 top MBA programs (Business Week, September 22, 2003). The survey found that the average annual salary for male and female graduates 10 years after graduation was $168,000 and $117,000, respectively. Assume the population standard deviation for the male graduates is $35,000, and for the female graduates it is $30,000. When calculating values for z, round to two decimal places.
A) What is the probability that a simple random sample of 40 male graduates will provide a sample mean within $10,000 of the population mean, $168,000 (to 4 decimals)?
B) What is the probability that a simple random sample of 40 female graduates will provide a sample mean within $10,000 of the population mean, $117,000 (to 4 decimals)?
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 $10,000 of the population mean? D)What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean (to 4 decimals)?
Explanation / Answer
z = (x - Mean)/(SD/sqrt N)
(a) z = (158000 - 168000)/(40000/sqrt 40) = -1.5811 and z = (178000 - 168000)/(40000/sqrt 40) = 1.5811
P(158000 < x < 178000) = P(-1.5811 < z < 1.5811) = 0.8861 [Using z Tables]
(b) z = (107000 - 117000)/(25000/sqrt 40) = -2.5298 and z = (127000 - 117000)/(25000/sqrt 40) = 2.5298
P(107000 < x < 117000) = P(-2.5298 < z < 2.5298) = 0.9886 [Using z Tables]
c)
We have a higher probability in part B, as there is less variation in the salary of females.
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d)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 168000-4000= 164000
u = mean = 168000
n = sample size = 100
s = standard deviation = 40000
Thus,
z = (x - u) * sqrt(n) / s = -1
Thus, using a table/technology, the left tailed area of this is
P(z < -1 ) = 0.158655254 [ANSWER]
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