Question

1. In a large Midwestern university, a random sample of 100 entering freshman in 2009 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened the next year. In 2011, a random sample of 100 entering freshman found that 10 had finished in the bottom third of their high school class. Let p1 be the proportion of admitted freshman who had finished in the bottom third of their high school class in 2009. Let p2 be the proportion of admitted freshman who had finished in the bottom third of their high school class in 2011.

The two sample proportions are 20% and 10% respectively which untested would seem like the proportion finishing in the bottom third of class has decreased. However, comparing the sample proportions does not build any uncertainty into it. This is why we perform an inference procedure.

To explore the suspicion above, conduct a significance test at level 0.05; calculator output is shown. Answer the following questions.

1. Give the test decision: do not reject (reject or do not reject)
2. Evidence ________does not favor_______(favors or does not favor) that college admissions standards for the bottom third of the class have changed from 2009 to 2011.
3. Is this result borderline? In other words, if you changed the decision level to another usual level such as 0.01 or 0.10, would it change the decision you made here? (yes or no)
4. A confidence interval for the true difference in proportions of freshman who had finished in the bottom third of class in years of 2009 and 2011 is (0.018, 0.182). This supports your decision in the significance test because the bounds are _______=_______in sign.

Answer 1

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 >   0          
                  
sample #1   ----->              
first sample size,     n1=   100          
number of successes, sample 1 =     x1=   20          
proportion success of sample 1 , p̂1=   x1/n1=   0.2000          
                  
sample #2   ----->              
second sample size,     n2 =    100          
number of successes, sample 2 =     x2 =    10          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.1000          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2000   -   0.1000   =   0.1000
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.1500          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.05050          
Z-statistic = (p̂1 - p̂2)/SE = (   0.100   /   0.0505   ) =   1.9803
                  
z-critical value , Z* =        1.6449   [excel function =NORMSINV(α)]      
p-value =        0.0238   [excel function =NORMSDIST(-z)]      
decision :    p-value<α,Reject null hypothesis               

=================

a) reject

b) Evidence______(favors ) that college admissions standards for the bottom third of the class have changed from 2009 to 2011.

c) YES

d) positive in sign