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Re: st: Wald Statistic.


From   Richard Williams <Richard.A.Williams.5@nd.edu>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Wald Statistic.
Date   Mon, 07 Mar 2005 17:58:43 -0500

At 05:20 PM 3/7/2005 -0500, you wrote:
Hi!!!
I am doing random effects logistics regression.  Stata, by default, uses a
Wald chi2 test of overall significant.  I have been doing some research,
and a Wald statistic is suppossed to be a test of individual parameters,
dividing the parameter by its standard error and squaring it.  How does it
work for an overall model fit?  How are the degrees of freedom calculated?
The data manual for xt procedures does not have the details.
Wald tests are not limited to individual parameters. For example, the command

test x1 x2 x3

would do a Wald test of whether all three variables had zero effects. The algebra is a little more complicated than it is when testing a single parameter, but it can be done. Similarly a command like

test x1=x2

does a Wald test. That doesn't fully answer your question, but it may at least clear up a misconception about what a Wald test can do. An advantage of Wald tests is that they only require the estimation of a single model (calculations are then done using the parameter estimates and the variances and covariances of the estimates). Likelihood ratio tests require the estimation of two models. Most people seem to think that an LR test will be superior to the corresponding Wald test (although it often matters little), but there are situations in which LR is inappropriate and Wald is better.

Incidentally, Stata has this way of surprising you and sometimes reporting Fs when you expect chi-squares, or Wald chi-squares when you expect Likelihood Ratio chi-squares. One example of this is explained at

http://www.stata.com/support/faqs/stat/chi2.html

and I suspect something similar applies to your problem.


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Richard Williams, Notre Dame Dept of Sociology
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