[R] testing non-linear component in mgcv:gam
John Fox
jfox at mcmaster.ca
Wed Oct 5 23:01:58 CEST 2005
Dear Denis,
You got me: I would have thought from ?summary.gam that this would be the
same as the adjusted R^2 for a linear model. Note, however, that the
percentage of deviance explained checks with the R^2 from the linear model,
as expected.
Maybe you should address this question to the package author.
Regards,
John
--------------------------------
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
--------------------------------
> -----Original Message-----
> From: Denis Chabot [mailto:chabotd at globetrotter.net]
> Sent: Wednesday, October 05, 2005 3:33 PM
> To: John Fox
> Cc: R list
> Subject: Re: [R] testing non-linear component in mgcv:gam
>
> Thank you everyone for your help, but my introduction to GAM
> is turning my brain to mush. I thought the one part of the
> output I understood the best was r-sq (adj), but now even
> this is becoming foggy.
>
> In my original message I mentioned a gam fit that turns out
> to be a linear fit. By curiosity I analysed it with a linear
> predictor only with mgcv package, and then as a linear model.
> The output was identical in both, but the r-sq (adj) was 0.55
> in mgcv and 0.26 in lm. In lm I hope that my interpretation
> that 26% of the variance in y is explained by the linear
> relationship with x is valid. Then what does r2 mean in mgcv?
>
> Denis
> > summary.gam(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
>
>
> > summary(sed.true.lin)
>
> Call:
> lm(formula = wm.sed ~ Temp, weights = N.sets)
>
> Residuals:
> Min 1Q Median 3Q Max
> -0.6138 -0.1312 -0.0325 0.1089 1.1449
>
> 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
>
> Residual standard error: 0.3061 on 35 degrees of freedom
> Multiple R-Squared: 0.285, Adjusted R-squared: 0.2646
> F-statistic: 13.95 on 1 and 35 DF, p-value: 0.000666
>
>
> Le 05-10-05 à 09:45, John Fox a écrit :
>
> > 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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> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide!
> >> http://www.R-project.org/posting-guide.html
> >>
> >
> >
>
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