Assume you plan to construct a 95 confidence interval The nu

Assume you plan to construct a 95% confidence interval. The numbers of online applications from simple random sample of college applications for 2003 and for the current year are given below Find (a) the margin of error F, and (b) the 95% confidence interval.

Solution

If we are trying to get the confidence interval for the DIFFERENCE in the two proportions, then:

a)

Formulating the hypotheses          

Ho: p1 - p2   =   0  
Ha: p1 - p2   =/=   0  

Here, we see that pdo =    0   , the hypothesized population proportion difference.  
          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.351351351      
p2 = x2/n2 =    0.571428571      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.12208977      
      
  
          
For the   95%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.025      
z(alpha/2) =    1.959963985      

Thus,

Margin of error = z*(alpha/2)*sd = 0.239291552 [ANSWER]

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

b)

Also,
          
lower bound = p1^ - p2^ - z(alpha/2) * sd =    -0.459368772      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    0.019214332      
          
Thus, the confidence interval is          
          
(   -0.459368772   ,   0.019214332 ) [ANSWER]