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From |
Steven Samuels <[email protected]> |

To |
[email protected] |

Subject |
Re: st: Weighted Averages |

Date |
Mon, 17 Jan 2011 14:07:46 -0500 |

Steve

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

Christopher:After looking more closely at your formulas, typos aside, I thinkthat youwere 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 withreplacementand your weights were frequency weights, not probability weights.The sumof probability weights is an estimate of N. Dividing a variance by Nwouldordinarily make the standard error of the mean much too small. Ifyours aresometimes larger than the linearized variance estimates, youprobably alsomade other mistakes in the formula or your calculations. Steve [email protected] On Jan 16, 2011, at 3:46 PM, Steven Samuels wrote: Christopher:The variance formula you present has little relation to the trueformula,whether for sampling with or without replacement. See for examplepage 230of Sharon Lohr. 2009. Sampling: Design and Analysis. Boston, MA:CengageBrooks/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 * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

-- Christopher Paul Steiner Third Year Grad Student, Ph.D. Economics University of California, San Diego University of Illinois Alumnus, BS Mathematics & Economics [email protected] | [email protected] | [email protected] (Note the number "1" instead of the "p" in the UCSD email address.) <3! * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: Weighted Averages***From:*Christopher Steiner <[email protected]>

**Re: st: Weighted Averages***From:*Steven Samuels <[email protected]>

**Re: st: Weighted Averages***From:*Steven Samuels <[email protected]>

**Re: st: Weighted Averages***From:*Christopher Steiner <[email protected]>

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