A local political candidate conducts a telephone survey of 2

A local political candidate conducts a telephone survey of 250 voters concerning their

opinion of local issues. Seventy

-

five people believe the city needs a new library.

Construct a 95% confidence

interval for the proportion of the population that holds the same opinion

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.3          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.028982753          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.056805153          
lower bound = p^ - z(alpha/2) * sp =   0.243194847          
upper bound = p^ + z(alpha/2) * sp =    0.356805153          
              
Thus, the confidence interval is              
              
(   0.243194847   ,   0.356805153   ) [ANSWER]