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

From: Simon Wood <sw283_at_maths.bath.ac.uk>
Date: Mon 24 Oct 2005 - 23:41:19 EST

Looks like a bug in mgcv::summary.gam when the model is strictly parametric... I'll take a look and fix it. thanks, Simon

>
> 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@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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> >>
> >
> >
>
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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 Mon Oct 24 23:54:21 2005

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