In a simple random sample of 116 Americans 65 reported suffe

In a simple random sample of 116 Americans, 65 reported suffering from seasonal allergies. Find a 82% confidence interval for the population proportion

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.560344828          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.04608449          
              
Now, for the critical z,              
alpha/2 = (1-0.82)/2 =    0.09          
Thus, z(alpha/2) =    1.340755034          
Thus,              
Margin of error = z(alpha/2)*sp =    0.061788011          
lower bound = p^ - z(alpha/2) * sp =   0.498556816          
upper bound = p^ + z(alpha/2) * sp =    0.622132839          
              
Thus, the confidence interval is              
              
(   0.498556816   ,   0.622132839   ) [ANSWER]