[R] testing non-linear component in mgcv:gam
Denis Chabot
chabotd at globetrotter.net
Wed Oct 5 16:08:44 CEST 2005
Fair enough, Andy. I thought I was getting both predictive ability
and confirmation that the phenomenon I was studying was not linear. I
have two projects, in one prediction is the goal and I don't really
need to test linearity. In the second I needed to confirm a cycle was
taking place and I thought my procedure did this. There was no
theoretical reason to anticipate the shape of the cycle, so GAM was
an appealing methodology.
Denis
Le 05-10-05 à 09:38, Liaw, Andy a écrit :
> I think you probably should state more clearly the goal of your
> analysis. In such situation, estimation and hypothesis testing are
> quite different. The procedure that gives you the `best' estimate is
> not necessarily the `best' for testing linearity of components. If
> your
> goal is estimation/prediction, why test linearity of components?
> If the
> goal is testing linearity, then you probably need to find the
> procedure
> that gives you a good test, rather than relying on what gam() gives
> you.
>
> Just my $0.02...
>
> Andy
>
>
>> From: Denis Chabot
>>
>> 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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