Re: st: Wald Statistic.

[Richard Williams <Richard.A.Williams.5@nd.edu>]

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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