\documentclass[11pt]{article}
\usepackage{headerfooter}
\usepackage{psfigOz}
\marginparwidth -.25in
\oddsidemargin -.25in
\evensidemargin  -.25in
\marginparsep 0pt
\topmargin -1.0in
\textwidth 6.75in
\textheight 10in
\parindent 0in
% Define a newcommand for solution
\newcounter{solution} 
\newcommand{\solution}{\addtocounter{solution}{1} {\bf Solution} \thesolution }
\newtheorem{solute}{Solution}[solution]
\newcounter{smallroman}
\newcounter{bigroman}
\newcounter{alphabet}
\newcounter{bigarabic}
\newcounter{project}
\pagestyle{empty}
\begin{document}
\section{Profiles}
Three vitamin E diet supplements with levels zero,
low, and high were compared for their effect on growth of guinea pigs
(Crowder and Hand 1990, pp. 21-29). 
Five guinea pigs received each supplement level
and their weight was recorded at the end of weeks
1, 3, 4, 5, 6, and 7. These weights are given in the table below.
The three mean vectors are

\begin{eqnarray*}
\overline{\bf x}^{\prime}_{1 (zero)} & = & (466.4,519.4,568.8,561.6,546.6,572.0) \\ 
\overline{\bf x}^{\prime}_{2 (low)} & = & (494.4,551.0,574.2,567.0,603.0,644.0) \\ 
\overline{\bf x}^{\prime}_{3 (high)} & = &(497.8,534.6,579.8,571.8,588.2,623.2)\\ 
\mbox{and the overall mean vector is}\\ 
\overline{y}^{\prime} & = & (486.2,535.0,574.3,566.8,579.3,613.1).
\end{eqnarray*}
A profile plot of these means is given below. 
There is a high degree of parallelism in the 
three profiles, with the possible exception of week 6 for group 1.
\protect
\begin{table}[h]
\centering 
\caption{Weight of Guinea Pigs under 3 Levels of Vitamin E 
Supplements}
\begin{tabular}{cccccccc}
Group & Animal & Week 1 &  Week 3 &  Week 4  & Week 5  & Week 6  & Week 
7 \\ \hline
1 & 1  & 455 & 460 & 510 & 504 & 436 & 466 \\
1 & 2 & 467 & 565 & 610 & 596 & 542 & 587\\
1 & 3 & 445 & 530 & 580 & 597 & 582 & 619\\
1 & 4 & 485 & 542 & 594 & 583 & 611 & 612\\
1 & 5 & 480 & 500 & 550 & 528 & 562 & 576\\
2 & 6 & 514 & 560 & 565 & 524 & 552 & 597\\
2 & 7 & 440 & 480 & 536 & 484 & 567 & 569\\
2 & 8 & 495 & 570 & 569 & 585 & 576 & 677\\
2 & 9 & 520 & 590 & 610 & 637 & 671 & 702\\
2 & 10 & 503 & 555 & 591 & 605 & 649 & 675\\
3 & 11 & 496 & 560 & 622 & 622 & 632 & 670\\
3 & 12 & 498 & 540 & 589 & 557 & 568 & 609\\
3 & 13 & 478 & 510 & 568 & 555 & 576 & 605\\
3 & 14 & 545 & 565 & 580 & 601 & 633 & 649\\
3 & 15 & 472 & 498 & 540 & 524 & 532 & 583\\ \hline
\end{tabular}
\end{table}
\begin{figure}[h]
\centerline{\psfig{height=2.5in,figure=guineapig.ps}}
% \centerline{\psfig{figure=guineapig.ps}}
\caption{\label{LinearGrowth} \small Growth profiles for zero, low, and high Vitamin E concentrations}
\end{figure}

\pagebreak

\section{Test for Parallel profiles}
Given the above groups we must use the extension to three groups of 
the Wilks \mbox{\boldmath $\Lambda$}. As derived previously, the 
between (hypothesis) matrix and the within (error) matrix becomes
\begin{equation}
\begin{array}{ccccccc}
{\bf B} & = & CBC^{\prime}& \mbox{and}& {\bf W} & = & CWC^{\prime} 
\nonunmber
\end{array}
\end{equation}
respectively.
Thus,
\begin{eqnarray*}
 \mbox{\boldmath $\Lambda$} & = & \frac{\mid  {\bf CWC}^{\prime} \mid}{ 
 \mid {\bf CBC}^{\prime} + {\bf CWC}^{\prime} \mid}\\
  & = & \frac{\mid  {\bf CWC}^{\prime} \mid}{ 
 \mid C\left ({\bf B + W}\right )C^{\prime} \mid}.
\end{eqnarray*}


Suppose we wish to test the profiles of the zero group and the low 
group.
\begin{eqnarray*}
T^{2} & = & \left(\overline{\bf x}_{1 (zero)} - \overline{\bf 
x}_{2 (low)}\right )^{\prime}C^{\prime}\left [ \left (\frac{1}{n_{1 (zero)}} + 
\frac{1}{n_{2 (low)}}\right ) C S_{pooled}C^{\prime}\right ]^{-1}C\left(\overline{\bf x}_{1 (zero)} - \overline{\bf 
x}_{2 (low)}\right ) \\
 & = & 26.4018
\end{eqnarray*}

To test for coincident profiles is the same as testing if the means 
are the same at each of the {\em p}-epochs.

\begin{eqnarray*}
T^{2} & = & C1^{\prime}\left(\overline{\bf x}_{1 (zero)} - \overline{\bf 
x}_{2 (low)}\right )\left [ \left (\frac{1}{n_{1 (zero)}} + 
\frac{1}{n_{2 (low)}}\right ) C1^{\prime} S_{pooled}C1\right ]^{-1}C1^{\prime}\left(\overline{\bf x}_{1 (zero)} - \overline{\bf 
x}_{2 (low)}\right )
\end{eqnarray*}

To test for flat profiles we have to get overall means for the 
groups being compared.

\begin{eqnarray*}
T^{2} & = & \left (n_{1(zero)}+n_{2 (low)}\right )\left(C\overline{\bf x}\right)^{\prime} \left [ \left (\frac{1}{n_{1 (zero)}} + 
\frac{1}{n_{2 (low)}}\right ) C S_{pooled}C^{\prime}\right ]^{-1}\left(C\overline{\bf x}\right) \\
 & = & 363.2166
\end{eqnarray*}

\end{document}
