BusinessWeek conducted a survey of graduates from 30 top MBA programs (BusinessW

Question

BusinessWeek conducted a survey of graduates from 30 top MBA programs (BusinessWeek, September 22, 2003). On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is $168,000 and $117,000, respectively. Assume the standard deviation for the male graduates is $40,000, and for the female graduates it is $25,000. 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? 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? 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? Why? What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4000 below the population mean?

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 168000-10000=   158000      
x2 = upper bound = 168000+10000=   178000      
u = mean =    168000      
n = sample size =    40      
s = standard deviation =    40000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.58113883      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.58113883      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.056923149      
P(z < z2) =    0.943076851      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.886153702   [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 117000-10000=   107000      
x2 = upper bound = 117000+10000=   127000      
u = mean =    117000      
n = sample size =    40      
s = standard deviation =    25000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.529822128      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.529822128      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.005706018      
P(z < z2) =    0.994293982      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.988587964      
  
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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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