Exampleslinmod.Rnw

\documentclass{article}

\begin{document}
\SweaveOpts{concordance=TRUE}

\title{Linear modelling critical examples}

\section{Hypothesis testing using C matrix}

<<>>=
data(ToothGrowth)

fm1<-lm(len~dose+supp,ToothGrowth)
#summary(fm1)

## see below:
(varcov.mat<-round(vcov(fm1),digits=3))
@

Test: $H_0: \beta_0=2\beta_1, \beta_2=0$. 

<<>>=
y<-ToothGrowth$len
X<-model.matrix(fm1)
XT.X <- t (X)%*%X
XT.y <- t(X)%*%y
G<-solve(XT.X)

## recovers Vcov matrix:
4.24^2*G

beta.hat <- G%*%XT.y

C<-matrix(c(1,-2,0,0,0,1),byrow=T,ncol=3)
## this changes:
c<-matrix(c(0,0),byrow=T,ncol=1)

(yT.y <- t(y)%*%y)
(bT.XT.y<-t(beta.hat)%*%t(X)%*%y) 

## by definition of S_r:
Sr <- yT.y - bT.XT.y
n<-60
p<-3  ## num parameters
sigma.hat.2 <- (1/(n-p)) * Sr
#sqrt(sigma.hat.2)

num<-t(C %*% beta.hat - c) %*% 
     ginv(C %*% G %*% t(C) ) %*% 
     (C %*% beta.hat - c )
q<-2 ## num of hypotheses
(F<-num/(sigma.hat.2*q))
1-pf(F,q,n-p)
@

In the exam exercise we are given:

<<>>=
given<-solve(C%*%solve(XT.X)%*%t(C))/4.24^2
@

So, the middle term is:

<<>>=
given*4.24^2
@

Cross-check:

<<>>=
ginv(C %*% G %*% t(C) )
@

So we can do:

<<>>=
num<-t(C %*% beta.hat - c) %*% 
     ginv(C %*% G %*% t(C) ) %*% 
     (C %*% beta.hat - c )

num2<-t(C %*% beta.hat - c) %*% 
     (given*4.24^2)%*% 
     (C %*% beta.hat - c )

q<-2 ## num of hypotheses
(F<-num/(sigma.hat.2*q))
1-pf(F,q,n-p)
@

\end{document}