\documentclass[fleqn]{book}
%\usepackage{times}
%\usepackage{chicago}
%\input dbp-book.sty

\input 312HW.sty
\def\betatilde{\widetilde\beta}
\def\Ztilde{\widetilde Z}
\def\Ytilde{\widetilde Y}

\def\yhat{\widehat y}
\def\rank{\text{rank}}
\def\xbar{\bar x}
\def\ybar{\bar y}

\def\Btilde{\widetilde B}
\def\bhat{\widehat b}
\def\ghat{\widehat g}
%\pagewidth{5.5in}


\voffset-1in
\hoffset0.5in
\topmargin0pt
\textheight 600pt
\textheight 650pt

\def\taa{widetilde\aa}
\def\corr{\text{corr}}

\begin{document}

%knit("homework4.Rnw",quiet=T)
<<echo=F>>=
BC.coeff <- function(out,digits=1){
	coeff <- summary(out)$coeff[,1:2]

	dd <- dimnames(coeff)
	dd[[1]] <- gsub("(p|t)\\K[a-z]*","",dd[[1]],perl=T)
	dd[[2]] <- c("Est","StdErr")
	dimnames(coeff) <- dd
	print(out$call)
	print( t(coeff),digits )
}
@

\HW{4}{Monday 3 October}

\problem[cath]
The catheter data set is taken from a well known text book.
If you happen to know the book please do not just repeat the analysis
it presents.
<<>>=
cath <- read.table("catheter.txt",header=T)
outHW <- lm(distance ~ height + weight,cath)
outH <- lm(distance ~ height, cath)
outW <-lm(distance ~ weight, cath)
cath[1:3,]
@
The summary information for each fit (\verb!outHW!, \verb!outH!, \verb!outW!) 
seems to suggest that height by itself is a good predictor of distance,
that weight by itself is a good predictor of distance, but when both predictors are used then neither 
is particularly useful. (The stars in the summary table suggest `significance'.)

\ppart (15 points)
Add more variables to the cath data.frame:  height and weight centered to zero means (call them hcen
and wcen); and things like \verb!wres <-lm(wcen~ hcen, cath)$res!.
Explain why some of the coeffs and  std. errors are the same and some are different for the models:
\begin{verbatim}
distance ~ height
distance ~ hcen
distance ~ hcen + wres
distance ~ height+wres
distance ~ hcen+wcen
distance ~ height+weight
\end{verbatim}


\ppart (5 points)
Explains why \verb!summary(outHW)! is misleading regarding the value of \verb!height! and \verb!weight! 
as predictors.

%<<>>=
%source("cath.R")
%@

\ppart (5 points)
Explain how the output from \verb!cor(cath)! is relevant to the problem.

\problem[reparam] 
The handout two\_factors.pdf showed how to calculate several least squares fits using the Box-Cox data:
<<>>=
BC <- read.table("../Handouts/boxcox.data", header=T,sep="\t")
BC$rate <- 1/BC$time # transformation suggested by BHH page 235
BC$Htreatment <- C(BC$treatment,helmert)
out5 <- lm(rate ~ -1 + treatment,BC)
out7 <- lm(rate ~ Htreatment,BC)
@
In class  I showed (page~9  of the handout) how to transform results from one 
parametrization into results for a different parametrization, using \verb!out5! and \verb!out6!
as an example. For this homework problem I want you to recreate the shortened summary
<<echo=F>>=
BC.coeff(out5,3)
@
\noindent
using only the information contained in \verb!out7!, which is essentially the same 
as the~\verb!out9! generated by: 
<<>>=
C7 <- contrasts(BC$Htreatment)
dummyT <- outer(BC$treat,levels(BC$treat),"==")+0 
X7 <- cbind(1, dummyT %*% C7)
out9 <- lm(BC$rate ~ -1+X7)
@
Display all the \Rlang/ code that you use.

\ppart (5 points)
Show that \verb!X7! is equal to \verb!dummyT %*% K7! where \verb!K7 <- cbind(1,C7)!.

\ppart (10 points)
If $\ghat$ is the vector of coefficients from \verb!out7! and $\bhat$
is the vector of coefficients from \verb!out5!, show that $\bhat=K_7 \ghat$.

\ppart (10 points)
Use (ii) and \verb!out7! to recreate the shortened summary for \verb!out5!.

\References
\end{document}