A poll of 600 frequent airline travelers found that 60 reall

A poll of 600 frequent airline travelers found that 60% really like the food on their flights. What is the 95% confidence interval for the true proportion of airline travelers who really like the food on their flights?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.6          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.02          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.56080072          
upper bound = p^ + z(alpha/2) * sp =    0.63919928          
              
Thus, the confidence interval is              
              
(   0.56080072   ,   0.63919928   ) [ANSWER]