Suppose a survey client requires a margin of error of approx

Suppose a survey client requires a margin of error of approximately 2%. Using the quick method (that is, you have no estimate for p), approximately how many people must be polled?

Solution

18.

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.025  
As there is no previous estimate for p, we set p = 0.5.      
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
E =    0.02  
p =    0.5  
      
Thus,      
      
n =    2400.911763  
      
Rounding up,      
      
n =    2401   [ANSWER]