Suppose you have a box with a very large number of orange an

Suppose you have a box with a very large number of orange and blue beads. You want to estimate the proportion p of orange beads in the box and you want to be 93% confident that your point estimate, which is the sample proportion p

Solution

Want 93% CI , with Margin of Error = 0.02 Since no p-cap is given, and we want to account for worst scenerio, for maximum value of p*(1-p), we take p cap = 0.5 N = ( z(alpha/2 ) / ME ) ^2 * pcap * (1- pcap) = 2051.8635 hence we take N = 2052