testing non-linear component in mgcv:gam
In this case the models being compared are really identical, and the P-value is meaningless numerical noise. If your main focus is hypothesis testing and you really need near exact p-values, then do this sort of testing using unpenalized models. i.e. don't have mgcv::gam estimate the EDF of the smooth, just use `s(...,fx=TRUE)' to estimate it unpenalized. This gives horrible point estimates and excellent p-values. best, Simon
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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