A survey of households in a small town showed of 2083 sample

A survey of households in a small town showed of 2,083 sampled households, that in 848 households at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.407105137          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.01076459          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.027727747          
lower bound = p^ - z(alpha/2) * sp =   0.37937739          
upper bound = p^ + z(alpha/2) * sp =    0.434832884          
              
Thus, the confidence interval is              
              
(   0.37937739   ,   0.434832884   ) [ANSWER]