Suppose an urn contains N balls that are green and red. You want to estimate p = proportion of green balls. You could:
1) Sample n balls with replacement and calculate phat1 = sample proportion of green balls. 2) Sample n balls without replacement and calculate phat2 = sample proportion of green balls. But, comparing the variances of the estimators: Var(phat1) = p(1-p)/n Var(phat2) = p(1-p)/n * (N-n)/(N-1) Thus, it is always the case that sampling without replacement leads to an estimator with a smaller variance. This may surprise some folks. Jason . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
