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Nov 18, 2018 at 2:04 comment added Heteroskedastic Jim @baxx there's nothing stopping you from interpreting the exponentiated coefficient but it's only true when B = 0. And as in my example, that can be grossly irrelevant.
Nov 18, 2018 at 1:47 comment added Heteroskedastic Jim @baxx something like that. It depends on the value of B. You effectively tie them together by setting up an interaction.
Nov 18, 2018 at 1:38 comment added baxx From this then, it's not possible to only give the odds ratio of $A$ in this context? By which I mean, reporting $\exp(0.756)$ is incorrect because of the interaction ( I can't report this as being adjusted for B and AxB ). And in the case of $\exp(\beta_1 + \beta_3 \times B )$ it's not an odds ratio only for $A$, because it contains B. Does that make sense? Thanks
Nov 17, 2018 at 22:45 history edited Heteroskedastic Jim CC BY-SA 4.0
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Nov 17, 2018 at 22:38 history answered Heteroskedastic Jim CC BY-SA 4.0