[R] anova.lm and F-test
peter dalgaard
pdalgd at gmail.com
Mon Jul 9 16:24:27 CEST 2012
On Jul 9, 2012, at 15:40 , Suresh Krishna wrote:
>
> Hello,
>
> Why does anova.lm sometimes return a p-value and at other times not ? Is it because it recognizes nested models from non-nested ones ?
>
>> x<-seq(1,100,1)
>> y<-3*x+rnorm(100)
>> anova(lm(y~x),lm(y~x+I(x^2)),test="F")
> Analysis of Variance Table
>
> Model 1: y ~ x
> Model 2: y ~ x + I(x^2)
> Res.Df RSS Df Sum of Sq F Pr(>F)
> 1 98 90.449
> 2 97 90.288 1 0.16117 0.1732 0.6782
>
>> anova(lm(y~x),lm(y~I(x^2)+I(x^3)),test="F")
> Analysis of Variance Table
>
> Model 1: y ~ x
> Model 2: y ~ I(x^2) + I(x^3)
> Res.Df RSS Df Sum of Sq F Pr(>F)
> 1 98 90.4
> 2 97 7345.7 1 -7255.3
>
You have Df and Sum of Sq with opposite sign, so more parameters with a worse fit. The models are not nested, so the F test makes no sense.
I'd say that the real question is why anova.lm doesn't protest loudly when detecting this? One possible answer is that it also misses other non-nested cases where the signs do not clash, and warning only in some of the incorrect cases could lead the naive user to believe that the other ones are OK. E.g. this F test is equally meaningless
> anova(lm(y~I(x^4)),lm(y~I(x^2)+I(x^3)),test="F")
Analysis of Variance Table
Model 1: y ~ I(x^4)
Model 2: y ~ I(x^2) + I(x^3)
Res.Df RSS Df Sum of Sq F Pr(>F)
1 98 186639
2 97 7101 1 179538 2452.4 < 2.2e-16 ***
(Non-nestedness could in principle be determined by checking whether cbind(model.matrix(m1), model.matrix(m2)) has higher rank that both of its constituents, but numerical rank determination is a bit error-prone and slow, so this was not implemented).
--
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
More information about the R-help
mailing list