In a study by von zur Muhlen et al A16 155 subjects with syn

In a study by von zur Muhlen et al. (A­16), 155 subjects with syncope or near syncope were studied. Syncope is the temporary loss of consciousness due to a sudden decline in blood flow to the brain. Of these subjects, 78 also reported having cardiovascular disease. Construct a 99 percent confidence interval for the population proportion of subjects with syncope or near syncope who also have cardiovascular disease.

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.503225806          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.040160131          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.399780165          
upper bound = p^ + z(alpha/2) * sp =    0.606671448          
              
Thus, the confidence interval is              
              
(   0.399780165   ,   0.606671448   ) [ANSWER]