Re: [R] testing non-linear component in mgcv:gam

From: John Fox <jfox_at_mcmaster.ca>
Date: Wed 05 Oct 2005 - 23:45:03 EST


Dear Denis,

Take a closer look at the anova table: The models provide identical fits to the data. The differences in degrees of freedom and deviance between the two models are essentially zero, 5.5554e-10 and 2.353e-11 respectively.

I hope this helps,
 John



John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox

> -----Original Message-----
> From: r-help-bounces@stat.math.ethz.ch
> [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Denis Chabot
> Sent: Wednesday, October 05, 2005 8:22 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] testing non-linear component in mgcv:gam
>
> Hi,
>
> I need further help with my GAMs. Most models I test are very
> obviously non-linear. Yet, to be on the safe side, I report
> the significance of the smooth (default output of mgcv's
> summary.gam) and confirm it deviates significantly from linearity.
>
> I do the latter by fitting a second model where the same
> predictor is entered without the s(), and then use anova.gam
> to compare the two. I thought this was the equivalent of the
> default output of anova.gam using package gam instead of mgcv.
>
> I wonder if this procedure is correct because one of my
> models appears to be linear. In fact mgcv estimates df to be
> exactly 1.0 so I could have stopped there. However I
> inadvertently repeated the procedure outlined above. I would
> have thought in this case the anova.gam comparing the smooth
> and the linear fit would for sure have been not significant.
> To my surprise, P was 6.18e-09!
>
> Am I doing something wrong when I attempt to confirm the non-
> parametric part a smoother is significant? Here is my example
> case where the relationship does appear to be linear:
>
> library(mgcv)
> > This is mgcv 1.3-7
> Temp <- c(-1.38, -1.12, -0.88, -0.62, -0.38, -0.12, 0.12,
> 0.38, 0.62, 0.88, 1.12,
> 1.38, 1.62, 1.88, 2.12, 2.38, 2.62, 2.88, 3.12,
> 3.38, 3.62, 3.88,
> 4.12, 4.38, 4.62, 4.88, 5.12, 5.38, 5.62, 5.88,
> 6.12, 6.38, 6.62, 6.88,
> 7.12, 8.38, 13.62)
> N.sets <- c(2, 6, 3, 9, 26, 15, 34, 21, 30, 18, 28, 27, 27,
> 29, 31, 22, 26, 24, 23,
> 15, 25, 24, 27, 19, 26, 24, 22, 13, 10, 2, 5, 3,
> 1, 1, 1, 1, 1) wm.sed <- c(0.000000000, 0.016129032,
> 0.000000000, 0.062046512, 0.396459596, 0.189082949,
> 0.054757925, 0.142810440, 0.168005168,
> 0.180804428, 0.111439628, 0.128799505,
> 0.193707937, 0.105921610, 0.103497845,
> 0.028591837, 0.217894389, 0.020535469,
> 0.080389068, 0.105234450, 0.070213450,
> 0.050771363, 0.042074434, 0.102348837,
> 0.049748344, 0.019100478, 0.005203125,
> 0.101711864, 0.000000000, 0.000000000,
> 0.014808824, 0.000000000, 0.222000000,
> 0.167000000, 0.000000000, 0.000000000,
> 0.000000000)
>
> sed.gam <- gam(wm.sed~s(Temp),weight=N.sets)
> summary.gam(sed.gam)
> > Family: gaussian
> > Link function: identity
> >
> > Formula:
> > wm.sed ~ s(Temp)
> >
> > Parametric coefficients:
> > Estimate Std. Error t value Pr(>|t|)
> > (Intercept) 0.08403 0.01347 6.241 3.73e-07 ***
> > ---
> > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> >
> > Approximate significance of smooth terms:
> > edf Est.rank F p-value
> > s(Temp) 1 1 13.95 0.000666 ***
> > ---
> > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> >
> > R-sq.(adj) = 0.554 Deviance explained = 28.5%
> > GCV score = 0.09904 Scale est. = 0.093686 n = 37
>
> # testing non-linear contribution
> sed.lin <- gam(wm.sed~Temp,weight=N.sets)
> summary.gam(sed.lin)
> > Family: gaussian
> > Link function: identity
> >
> > Formula:
> > wm.sed ~ Temp
> >
> > Parametric coefficients:
> > Estimate Std. Error t value Pr(>|t|)
> > (Intercept) 0.162879 0.019847 8.207 1.14e-09 ***
> > Temp -0.023792 0.006369 -3.736 0.000666 ***
> > ---
> > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> >
> >
> > R-sq.(adj) = 0.554 Deviance explained = 28.5%
> > GCV score = 0.09904 Scale est. = 0.093686 n = 37
> anova.gam(sed.lin, sed.gam, test="F")
> > Analysis of Deviance Table
> >
> > Model 1: wm.sed ~ Temp
> > Model 2: wm.sed ~ s(Temp)
> > Resid. Df Resid. Dev Df Deviance F Pr(>F)
> > 1 3.5000e+01 3.279
> > 2 3.5000e+01 3.279 5.5554e-10 2.353e-11 0.4521 6.18e-09 ***
>
>
> Thanks in advance,
>
>
> Denis Chabot
>
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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Thu Oct 06 00:07:40 2005

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