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More than 200,000 people worldwide take the GMAT examination each year as they a

ID: 3135103 • Letter: M

Question

More than 200,000 people worldwide take the GMAT examination each year as they apply for MBA programs. Their scores vary Normally with mean about and standard deviation about . One hundred students go through a rigorous training program designed to raise their GMAT scores.
Test the following hypotheses about the training program

in each of the following situations:
(a) The students' average score is . Is this result significant at the 5% level?
(b) The average score is . Is this result significant at the 5% level?

The test statistic corresponding to the first situation is , and the test statistic corresponding to the second situation is .
Both results are significant at . The test statistic corresponding to the first situation is , and the test statistic corresponding to the second situation is .
Both results are not significant at . The test statistic corresponding to the first situation is , and the test statistic corresponding to the second situation is .
The first result is not significant and the second result is, at .

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   525  
Ha:    u   >   525  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    541.4          
uo = hypothesized mean =    525          
n = sample size =    100          
s = standard deviation =    100          
              
Thus, z = (X - uo) * sqrt(n) / s =    1.64          
              
Also, the p value is              
              
p =    0.050502583          
              
As P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

IT IS NOT SIGNIFICANT. [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   525  
Ha:    u   >   525  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    541.5          
uo = hypothesized mean =    525          
n = sample size =    100          
s = standard deviation =    100          
              
Thus, z = (X - uo) * sqrt(n) / s =    1.65          
              
Also, the p value is              
              
p =    0.049471468          
              
As P < 0.05, we   REJECT THE NULL HYPOTHESIS.          
IT IS SIGNIFICANT. [ANSWER]      

**************************************

Hence,

OPTION C: The test statistic corresponding to the first situation is z = 1.64, and the test statistic corresponding to the second situation is z = 1.65.
The first result is not significant and the second result is, at alpha = 0.05. [ANSWER, C]

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