1. Total scores on the GMAT are normally distributed and historically have a pop

Question

1. Total scores on the GMAT are normally distributed and historically have a population standard deviation of 113. The Graduate Management Admission Council (GMAC), who administers the test, claims that the mean total score is 529. Suppose a random sample of 8 students took the test, and their scores are given below: (hint: assume s = 113) 699, 560, 414, 570, 521, 663, 727, 416 a. Find a point estimate for the population mean b. Construct a 95% confidence interval for the true mean score for the population. c. Does this interval contain the value reported by GMAC? d. How many students should be surveyed to estimate the mean score within 25 points with 90% confidence? e. How many students should be surveyed to estimate the mean score within 25 points with 95% confidence? f. How many students should be surveyed to estimate the mean score within 25 points with 99% confidence?

Explanation / Answer

a)

The sample mean is, using technology,

X = sample mean =    571.25   [ANSWER]

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b)      

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    571.25          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    113          
n = sample size =    8          
              
Thus,              
              
Lower bound =    492.9464339          
Upper bound =    649.5535661          
              
Thus, the confidence interval is              
              
(   492.9464339   ,   649.5535661   ) [ANSWER]

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c)

YES, 529 IS INSIDE THIS INTERVAL.

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d)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    113  
E = margin of error =    25  
      
Thus,      
      
n =    55.27533498  
      
Rounding up,      
      
n =    56   [ANSWER]

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