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

[John Fox]

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 at stat.math.ethz.ch 
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Denis Chabot
> Sent: Wednesday, October 05, 2005 8:22 AM
> To: r-help at 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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