Description Author(s) References See Also Examples
R code for Chapter 1 exercise solutions.
Simon Wood <simon@r-project.org>
Maintainer: Simon Wood <simon@r-project.org>
Wood, S.N. (2017) Generalized Additive Models: An Introduction with R, CRC
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 | library(gamair); library(mgcv)
## Q.8 Rubber
## a)
library(MASS)
m1 <- lm(loss~hard+tens+I(hard*tens)+I(hard^2)+I(tens^2)+
I(hard^2*tens)+I(tens^2*hard)+I(tens^3)+I(hard^3),Rubber)
plot(m1) ## residuals OK
summary(m1) ## p-values => drop I(tens^2*hard)
m2 <- update(m1,.~.-I(tens^2*hard))
summary(m2)
m3 <- update(m2,.~.-hard)
summary(m3)
m4 <- update(m3,.~.-1)
summary(m4)
m5 <- update(m4,.~.-I(hard^2))
summary(m5) ## p-values => keep all remaining
plot(m5) ## residuals OK
## b)
AIC(m1,m2,m3,m4,m5)
m6 <- step(m1)
## c)
m <- 40;attach(Rubber)
mt <- seq(min(tens),max(tens),length=m)
mh <- seq(min(hard),max(hard),length=m)
lp <- predict(m6,data.frame(hard=rep(mh,rep(m,m)),
tens=rep(mt,m)))
contour(mt,mh,matrix(lp,m,m),xlab="tens",ylab="hard")
points(tens,hard)
detach(Rubber)
## Q.9 warpbreaks
wm <- lm(breaks~wool*tension,warpbreaks)
par(mfrow=c(2,2))
plot(wm) # residuals OK
anova(wm)
## ... so there is evidence for a wool:tension interaction.
par(mfrow=c(1,1))
with(warpbreaks,interaction.plot(tension,wool,breaks))
## Q.10 cars
## a)
cm1 <- lm(dist ~ speed + I(speed^2),cars)
summary(cm1)
## Intercept has very high p-value, so drop it
cm2 <- lm(dist ~ speed + I(speed^2)-1,cars)
summary(cm2)
## both terms now significant, but try the alternative of
## dropping `speed'
cm3 <- lm(dist ~ I(speed^2),cars)
AIC(cm1,cm2,cm3)
plot(cm2)
# Clearly cm2, with speed and speed squared terms, is to be preferred,
# but note that variance seems to be increasing with mean a little:
# perhaps a GLM, better?
## b)
# In seconds, the answer is obtained as follows..
b <- coef(cm2)
5280/(b[1]*60^2)
# This is a long time, but would have a rather wide associated confidence
# interval.
## Q.11 QR
# The following is a version of the function that you should end up with.
fitlm <- function(y,X)
{ qrx <- qr(X) ## get QR decomposition
y <- qr.qty(qrx,y) ## form Q'y efficiently
R <- qr.R(qrx) ## extract R
p <- ncol(R);n <- length(y) ## get dimensions
f <- y[1:p]; r <- y[(p+1):n]## partition Q'y
beta <- backsolve(R,f) ## parameter estimates (a)
sig2 <- sum(r^2)/(n-p) ## resid variance estimate (c)
Ri <- backsolve(R,diag(ncol(R))) ## inverse of R matrix
Vb <- Ri%*%t(Ri)*sig2 ## covariance matrix
se <- diag(Vb)^.5 ## standard errors (c)
F.ratio <- f^2/sig2 ## sequential F-ratios
seq.p.val <- 1-pf(F.ratio,1,n-p) ## seq. p-values (e)
list(beta=beta,se=se,sig2=sig2,seq.p.val=seq.p.val,df=n-p)
} ## fitlm
# The following code uses the function to answer some of the question parts.
## get example X ...
X <- model.matrix(dist ~ speed + I(speed^2),cars)
cm <- fitlm(cars$dist,X) # used fitting function
cm$beta;cm$se # print estimates and s.e.s (a,c)
cm1<-lm(dist ~ speed + I(speed^2),cars) # equiv. lm call
summary(cm1) # check estimates and s.e.s (b,c)
t.ratio <- cm$beta/cm$se # form t-ratios
p.val <- pt(-abs(t.ratio),df=cm$df)*2
p.val # print evaluated p-values (d)
## print sequential ANOVA p-values, and check them (e)
cm$seq.p.val
anova(cm1)
## Q.12 InsectSprays
X <- model.matrix(~spray-1,InsectSprays)
X <- cbind(rep(1,nrow(X)),X) # redundant model matrix
C <- matrix(c(0,rep(1,6)),1,7) # constraints
qrc <- qr(t(C)) # QR decomp. of C'
## use fact that Q=[D:Z] and XQ = (Q'X')' to form XZ ...
XZ <- t(qr.qty(qrc,t(X)))[,2:7]
m1 <- lm(InsectSprays$count~XZ-1) # fit model
bz <- coef(m1) # estimates in constrained parameterization
## form b = Z b_z, using fact that Q=[D:Z], again
b <- c(0,bz)
b <- qr.qy(qrc,b)
sum(b[2:7])
## Q.13 trees
## a)
EV.func <- function(b,g,h)
{ mu <- b[1]*g^b[2]*h^b[3]
J <- cbind(g^b[2]*h^b[3],mu*log(g),mu*log(h))
list(mu=mu,J=J)
}
## b)
attach(trees)
b <- c(.002,2,1);b.old <- 100*b+100
while (sum(abs(b-b.old))>1e-7*sum(abs(b.old))) {
EV <- EV.func(b,Girth,Height)
z <- (Volume-EV$mu) + EV$J%*%b
b.old <- b
b <- coef(lm(z~EV$J-1))
}
b
## c)
sig2 <- sum((Volume - EV$mu)^2)/(nrow(trees)-3)
Vb <- solve(t(EV$J)%*%EV$J)*sig2
se <- diag(Vb)^.5;se
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