Thanks all for your answers.
When I wrote 'of type Pearson chi-squared' I didn't want to mean that
it was specifically chi-squared, but that it was of the type that
could be obtained as an option when performing a plain frequency
analysis, without having to carry out regressions.
Steve's proposal makes me a little bit nervous: I was taught that
using O.L.S. regression for a binary response is inadequate, but I
suppose there are exceptions.
Angel Rodriguez-Laso
2008/10/2 Steven Samuels <[email protected]>:
>>>
>>>
>>
> 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 <[email protected]>:
>>>
>>> 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 <[email protected]>:
>>>
>>> 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 <[email protected]>:
>>>
>>> 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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