Re: st: Weighted Averages

[Steven Samuels <sjsamuels@gmail.com>]

Christopher:

After looking more closely at your formulas, typos aside,  I think  
that you were trying to estimate the variance of the  mean as:

(Estimated Population Variance)/(sum of weights)

This would be true only if you had a simple random sample with  
replacement and your weights were frequency weights, not probability  
weights.  The sum of probability weights is an estimate of N. Dividing  
a variance by N would ordinarily make the standard error of the mean  
much too small. If yours are sometimes larger than the linearized  
variance estimates, you probably also made other mistakes in the  
formula or your calculations.

Steve
sjsamuels@gmail.com

On Jan 16, 2011, at 3:46 PM, Steven Samuels wrote:

Christopher:

The variance formula you present has little relation to the true  
formula, whether for sampling with or without replacement.  See for  
example page 230 of  Sharon Lohr. 2009. Sampling: Design and Analysis.  
Boston, MA: Cengage Brooks/Cole.


On Jan 15, 2011, at 8:02 PM, Christopher Steiner wrote:

Hello everyone:

I am computing some basic summary statistics with weighted means from
a weighted, but otherwise simple design survey.  When I use the
following commands:

svyset [pweight=weight2]
svy: reg fcost_1

I get a weighted average of "fcost_1" that matches my hand
calculation.  I also receive White "robust" standard errors, which is
fine.  However, when I do a hand calculation of regular standard
errors using the formula:

sigma^2 = [sum(weights*(x-xbar))/sum(weights)] * (N/N-1)

and then divide by sum(weights) to get the standard error, I often
receive *larger* standard errors than the robust estimate.  Is this a
function of the pweights?  Around 10% of the values are also missing,
so is it a function of this?  Or am I doing something incorrectly?

Thank so much,
Christopher Paul Steiner

--
Christopher Paul Steiner
Third Year Grad Student, Ph.D. Economics
University of California, San Diego
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