Construct a 90 percent confidence interval for the proportio

Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.)

The 90% confidence interval is from, to?

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.365853659          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.075224017          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.123732496          
lower bound = p^ - z(alpha/2) * sp =   0.242121162          
upper bound = p^ + z(alpha/2) * sp =    0.489586155          
              
Thus, the confidence interval is              
              
(   0.242121162   ,   0.489586155   ) [ANSWER]