Re: st: Test for trend in surveys

[Steven Samuels <sjhsamuels@earthlink.net>]

> >
There is, to my knowledge, no such thing as test for trend of type  
Pearson chi-squared.  I suspect that �ngel is referring to the  
Cochran-Armitage test one degree-of-freedom chi square test for trend  
(A. Agresti, 2002, Categorical Data Analysis, 2nd Ed. Wiley Books,  
Section 5.3.5).

Let Y be the 0-1 binary outcome variable and X be the variable which  
contains category scores. One survey-enabled approach is Phil's  
suggestion: use -svy: logit-.

However -svy: reg- will produce a result closer to that of the  
Cochran-Armitage test. Why? The Cochran-Armitage test statistic is  
formally equivalent to an O.L.S. regression of Y on X, with a  
standard error  for beta which substitutes the total variance for the  
residual variance. The statistic is (beta/se)^2.  The total variance  
is equal to P(1-P), where P is the overall sample proportion. In  
other words, the standard error is computed under the null hypothesis  
of equal proportions.

The -svy: reg- command will estimate the same regression coefficient,  
but with a standard error that is robust to heterogeneity in  
proportions.  In both survey-enabled commands, t = (b/se) has a t  
distribution with degrees of freedom (d.f.) based on the survey  
design; t^2  has an F(1, d.f.) distribution.


-Steve


> > On Sep 30, 2008, at 6:39 AM, Philip Ryan wrote:
> >
> > Well, the z statistic testing the coefficient on the exposure  
> > variable is as
> > valid and as useful a summary (test) statistic as the chi-square  
> > statistic
> > produced by a test of trend in tables. If you prefer chi-squares,  
> > you could
> > just square the z statistic to get the chi-square on 1 df. And if  
> > you prefer
> > likelihood ratio chi-squares to the Wald z (or Wald chi-square)  
> > then the
> > modelling approach can deliver that also.
> >
> > Phil
> >
> > Quoting �ngel Rodr�guez Laso <angelrlaso@gmail.com>:
> >
> > Thanks to Philip and Neil for their advice.
> >
> > Philip's proposal is absolutely compatible with survey data, but I  
> > was
> > interested in a summary statistic of the type of Pearson chi-squared.
> >
> > To this respect, Neil puts forward a test (nptrend) that would be
> > perfect if it allowed complex survey specifications. I believe strata
> > and clusters are not important because the formula for the standard
> > error of this nonparametric test (see Stata Reference Manual K-Q page
> > 338) should not be affected by these specifications. But nptrend does
> > not accept weights as an option, what I think makes it unsuitable for
> > complex survey analyses.
> >
> > Angel Rodriguez Laso
> >
> > 2008/9/29 Philip Ryan <philip.ryan@adelaide.edu.au>:
> >
> > For a 2 x k table [with a k-category "exposure" variable] just set  
> > up a
> > logistic
> > dose-response model:
> >
> > svyset <whatever>
> > svy: logistic <binary outcome var> <exposure var>
> >
> > and check the coefficient of <exposure var>, along with its  
> > confidence
> > interval
> > and P-value.
> >
> > If you prefer a risk metric rather than odds, then use svy:  
> > glm..... with
> > appropriate link and error specifications.
> >
> > Phil
> >
> >
> > Quoting �ngel Rodr�guez Laso <angelrlaso@gmail.com>:
> >
> > Dear Statalisters,
> >
> > Is there a way to carry out a test for trend in a two-way table in
> > survey analysis in Stata?
> >
> > Many thanks.
> >
> > Angel Rodriguez Laso
> > *
> > *--
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