Re: st: vce(analytic) for proportions-- what is it exactly?

[Steve Samuels <sjsamuels@gmail.com>]

Tamer,

You are looking at the wrong commands and, therefore, the wrong -help-.

If you have survey data, you must first run -svyset- and then use
Stata's survey commands, such as -svy: proportion-. The default vce
option for -svyset- is vce(linearized), also known as Taylor
linearized.. Formulas are contained in Stata's Survey Manual or in
any sampling text. (My favorite is Sharon Lohr, 2009. Sampling: Design
and Analysis. Boston, MA: Cengage Brooks/Cole.)

What are standard errors based on vce(analytic) in (non-survey)
-proportion-? In the absence of weights or clusters, they are the
formulas found in introductory texts: se = sqrt of
P(1-P)/n. For a complex survey design (multiple stages, strata,
weights), this formula would be quite wrong.


Steve

Steven Samuels
Consulting Statistician
18 Cantine's Island
Saugerties NY 12477 USA



On Jan 16, 2013, at 3:52 PM, Tamer Farag wrote:

Hello All,

I am calculating survey weighted proportions using the svy: proportion command.  The Stata help file states that the default for calculating variance is vce(analytic).  Link below of the stata help file for proportion:

http://www.stata.com/help.cgi?proportion 

Specifically, it states:  


----+ SE/Cluster +-------------------------------------------------------
vce(vcetype) specifies the type of standard error reported, which
       includes types that are derived from asymptotic theory, that allow
       for intragroup correlation, and that use bootstrap or jackknife
       methods; see [R]  ( http://www.stata.com/help.cgi?vce_option )vce_option ( http://www.stata.com/help.cgi?vce_option ).
vce(analytic), the default, uses the analytically derived variance
       estimator associated with the sample proportion.

But just what is an "analytically derived variance estimator", exactly?  Is there any reason not to use it?

Many thanks,
Tamer



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