When 650 college students were surveyed 495 said they owned

When 650 college students were surveyed, 495 said they owned a computer.
Construct a 98% confidence interval for the proportion of college students who own a
computer.

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.761538462          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.016714693          
              
Now, for the critical z,              
alpha/2 =   0.01          
Thus, z(alpha/2) =    2.326347874          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.72265427          
upper bound = p^ + z(alpha/2) * sp =    0.800422653          
              
Thus, the confidence interval is              
              
(   0.72265427   ,   0.800422653   ) [ANSWER]