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Re: st: AW: Problem with logit


From   Hervé CACI <hcaci@wanadoo.fr>
To   <statalist@hsphsun2.harvard.edu>
Subject   Re: st: AW: Problem with logit
Date   Thu, 07 Aug 2003 22:27:19 +0200

le 7/08/03 12:04, Jann, Ben à ben.jann@soz.gess.ethz.ch a écrit :

> The model almost perfectly explains your data (only 1 failure and 2 successes
> are not "perfectly" determined). Because of that, the coefficients are huge
> (!). That the standard errors are huge as well, is probably due to very high
> collinearity between 'pa' and 'na' (what are 'pa' and 'na'?). Such a situation
> is likely to happen, if casenumbers are very low.

PA = Positive Affectivity, NA = Negative Affectivity.

PA & NA correlate at -0.31 in suicide attempters and +0.30 in controls. Both
coefficients are not significant. This is because suicide attempters are
depressed.
 
> (Furthermore, note that your logit estimation may be tremendously biased in
> general because of insufficient N)

Was may be a "sufficient N" ? I presume it depends on the number of
regressors and their distribution characteristics.
 
> The reason for the missing CIs in the first table, is evident if looking at
> the CIs in the second table. For example, exp(-17028.46) practically equals
> zero and exp(16912.24) practically equals infinity.
> (the CI's are so large becuase of the huge standard errors)

I understand.

> Try estimating a model including only one regressor. (or think about not using
> logistic regression at all)

It works fine. I also computed a set of repeated ANOVAs because I have a
retest at 6 months.

Thank you very much.
Hervé.
 
le 7/08/03 12:03, roger webb à roger.webb@man.ac.uk a écrit :

>With this type of result its best to tabulate the data and look at the
>individual cell values. Zero cell values will be the cause of the
>'successes and failures completely determined' message. Logistic
>regression isn't a valid method for your data set. Even without the
>zero cell values, your data are still far too sparse, especially for
>creating multivariate models. I suggest you forget it and run simple
>crosstabs using Fisher exact tests for significance testing.

I can't tabulate the values because PA and NA are "continuous" variables
(unit-weighted sums of items).

Thank you very much.
Hervé.


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