Re: st: Weighted Averages

[Steven Samuels <sjsamuels@gmail.com>]

You are welcome, Christopher.   You will find communication easier if  
you use standard terms.  In survey usage, "N" is population size; "n"  
is sample size. (In some data sets, weights have been normalized to  
sum to "n", a practice I don't like.)  Also,  the proper term for  
Stata's default variance estimate in regression is "linearized".  Its  
publication (Woodruff, 1971) preceded White's publication by over 10  
years. And White's estimate, I believe, applied only to a standard  
generated model (mean + error term), not to a finite population  
sampling design with weights (not to mention clusters and strata).

 Steve

Woodruff, RS. 1971. A simple method for approximating the variance of  
a complicated estimate. Journal of the American Statistical  
Association : 411-414.



On Jan 17, 2011, at 1:34 PM, Christopher Steiner wrote:

Thanks Steve!

I noticed that with binary data, the discrepancy I received was near
the sqrt(DEFF), which I didn't realize Stata was accounting for.
Also, I did not realize that the formula I posted was the frequency
weight formula, so thanks for pointing that out.

In this particular application, the average of the weights is 1, so
summing them is equivalent to N.

This is my first time with survey data, so I'm learning fast.  I'm,
confident now that Stata's doing things "correctly," and did a little
reading of the survey book last night.

Thanks,
Christopher Paul Steiner

On Mon, Jan 17, 2011 at 8:04 AM, Steven Samuels <sjsamuels@gmail.com>  
wrote:
> 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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--
Christopher Paul Steiner
Third Year Grad Student, Ph.D. Economics
University of California, San Diego
University of Illinois Alumnus, BS Mathematics & Economics
cpsteiner@gmail.com | cpsteiner@alumni.illinois.edu | c1steiner@ucsd.edu
(Note the number "1" instead of the "p" in the UCSD email address.)
<3!

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