A survey of an urban university showed that 750 of 1100 stud

A survey of an urban university showed that 750 of 1100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.681818182          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.014043509          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.036173681          
lower bound = p^ - z(alpha/2) * sp =   0.645644501          
upper bound = p^ + z(alpha/2) * sp =    0.717991863          
              
Thus, the confidence interval is              
              
(   0.645644501   ,   0.717991863   ) [ANSWER]