Re: Interesting problem with random sampling

[Jon Cryer]

"This may surprise some folks."

It certainly should surprise anyone who has studied statistics!

Jon Cryer

At 12:54 PM 1/28/2003 -0800, you 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.

Jason
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Jon Cryer, Professor Emeritus
Dept. of Statistics www.stat.uiowa.edu/~jcryer
and Actuarial Science office 319-335-0819
The University of Iowa home 319-351-4639
Iowa City, IA 52242 FAX 319-335-3017

"It ain't so much the things we don't know that get us into trouble.
It's the things we do know that just ain't so." --Artemus Ward

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