In article <[EMAIL PROTECTED]>, Jason Owen <[EMAIL PROTECTED]> wrote: >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. Why should it be surprising? You get more information from new balls than old ones. In fact, if you could mark a ball as having already been seen, you would do well to discard them from the sample. One way to make it clear is what happens if the sample size is N. Without replacement, you see all balls, and hence know the exact numbers of each. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University [EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558 . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
