Since 1973 Gallup has asked Americans how much confidence th

Since 1973 Gallup has asked Americans how much confidence they have in a variety of us institutions one question asked of those polled whether they have a great deal of confidence in banks. In 2007 a random sample of 1008 adults Americans, 413 said yes; and in 2013 of a random sample of 1529 adult Americans, 398 said yes. For the two years, find and interpret the 95% confidence interval for the difference between the portages of adult Americans who had a great deal of confidence in banks.

Solution

          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.409722222      
p2 = x2/n2 =    0.26030085      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.019127434      
          
For the   95%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.025      
z(alpha/2) =    1.959963985      
          
lower bound = p1^ - p2^ - z(alpha/2) * sd =    0.11193229      
upper bound = p1^ - p2^ + z(alpha/2) * sd =    0.186910454      
          
Thus, the confidence interval is          
          
(   0.11193229   ,   0.186910454) [CONFIDENCE INTERVAL]

Thus, we are 95% confident that the true difference in the proportions of adult Americans who had a great deal of confidence in banks is between 0.1119 and 0.1869.

Thus, this shows that the population proportion of 2007 is significantly greater than the proportion in 2013, at 0.025 significance.