An answer is what exactly you have said (comparing a linear and a non-linear model). e.g.
model1<-lm(yv~xv)
model2<-lm(yv~xv+I(xv^2)) #Even if we restrict ourselves to the inclusion of a quadratic term, there are many curves we can describe, depending upon the signs of the linear and quadratic terms
anova(model1,model2)
Analysis of Variance Table
Model 1: yv ~ xv
Model 2: yv ~ xv + I(xv^2)
Res.Df RSS Df Sum of Sq F Pr(>F)
1 16 91.057
2 15 68.143 1 22.915 5.0441 0.0402 *
The more complicated curved model is a significant improvement over the linear model (p=0.04) so, in that case, we accept that there is evidence of curvature in the data.
gd047
2010-06-10 12:24:40