Using R for Introductory Statistics : John Verzani

Two extensions of the linear model 339> curve(g(x, Y=53.903, k=1.393, t0=1.958), add=TRUE)Finally, so we can compare, we find the AIC:> AIC(res.g)[1] 1559Next, we fit the Richards model. First, we try to use the same values, to see if that willwork (not shown).> curve(f(x, Y=53.903, k=1.393, t0=1.958, m=1),add=TRUE)> legend(4,20, legend=c("logisticgrowth","Richards"),lty=l:2)It is not a great fit, but we try these as starting points for the algorithm anyway:> res.f=nls(size ~ f(age, Y, k, t0, m),data=urchin.growth,+ start=c(Y=53, k=1.393, t0=1.958, m=1))Error in numericDeriv(form[[3]], names(ind), env) :Missing value or an Infinity produced whenevaluating the modelThis is one of the error messages that can occur when the initial guess isn’t good or themodel doesn’t fit well.Using a little hindsight, we think that the problem might be t o and k. For this model, afew exploratory graphs indicate that we should have t≥t 0 for a growth model, as thegraphs decay until t0. So,we should start with t 0 res.f=nls(size ~ f(age, Y, k, t0, m),+ start=c(Y=53, k=.5, t0=0, m=1), data=urchin.growth)> res.fNonlinear regression modelmodel: size ~ f(age, Y, k, t0, m)data: urchin.growthY k t0 m57.2649 0.7843 −0.8587 6.0636residual sum-of-squares: 6922> curve(f(x, Y=57.26, k=0.78, t0=-0.8587, m = 6.0636),add=TRUE, lty=2)Now we have convergence. The residual sum-of-squares, 6,922, is less than the 7,922 forthe logistic model. This is a good thing, but if we add parameters this is often the case. †We compare models here with AIC.> AlC(res.f)[1] 1548