Question 1132051: BusinessWeek conducted a survey of graduates from 30 top MBA programs (BusinessWeek,
September 22, 2003). The sample of 1800 respondents had 987 female graduates. On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is $1.68mn and $1.17mn, respectively. Assume the standard deviation for the male graduates is $0.4 mn, and for the female graduates it is $0.25 mn
Question 1 :Construct the 95% confidence interval for proportion of males in the population
Question 2:Assuming a normal distribution, what is the probability that a female graduate will earn more than $ 1.75mn 10 years after graduation
Question 3:If the top 10% of the earners among males have to be identified, what should the annual salary cut off be?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Question 1
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1800 - 987 = 813
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sample proportion of males is 813/1800 is approximately 0.45
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sample standard error is square root(0.45*(1-0.45)/1800) is approximately 0.01
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alpha(a) = 1 -(95/100) = 0.05
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critical probability(p*) is 1 -(a/2) = 0.975
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critical value is the z-score associated with p* which is 1.96
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Note the sample size is sufficiently large, we can use the normal distribution
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margin of error(ME) = critical value * standard error
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ME = 1.96 * 0.01 is approximately 0.02
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95% confidence limit for the male proportion is 0.45 + or - 0.02, which is (0.43, 0.47)
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Question 2
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probability(P) ( X > 1.75 ) = 1 - P(X < 1.75 )
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z-score(1.75) = (1.75 - 1.17)/0.25 = 2.32
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P associated with a z-score of 2.32 is 0.9898
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P (X > 1.75) = 1 - 0.9898 = 0.0102 is approximately 0.01
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Question 3
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1 - 0.10 = 0.90
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z-score associated with a probability of 0.90 is 1.29
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1.29 = (X - 1.68)/0.4
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X -1.68 = 0.516
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X = 2.196 is approximately 2.20
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The annual salary cut off for males is $ 2.20mn
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