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% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat 324: Lecture 24\\
Testing equality of regression parameters in two groups}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{May 3, 2005}
\maketitle \thispagestyle{empty}


\begin{enumerate}

\item As an application of multiple regression, we show how to
perform a test on equality of regression parameters in two groups.
Suppose there are $m=43$ autistic and $n=32$ normal subjects. The
data can be found at\\ {\tt
http://www.stat.wisc.edu/$\sim$mchung/teaching/data/autism/}. The
fractional anisotropy (FA) measure $y_i$ is regressed over age $x_i$
and plotted in Figure 1 for each group. Let
$$Y_i = \beta_{a0} + \beta_{a1} x_i + \epsilon_i$$
be the linear model for the autism group and
$$Y_i = \beta_{c0} + \beta_{c1} x_i + \epsilon_i$$
be the linear model for the control group.

\begin{figure}[h]
\centering
\includegraphics[width=80mm]{324.24.01.ps}
\includegraphics[width=80mm]{324.24.02.ps}
\caption{Regression lines for each group.}\label{fig:hist}
\end{figure}

\begin{verbatim}
> summary(lm(aut$FA~aut$age))
Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.5031720  0.0104110  48.331  < 2e-16 ***
aut$age     0.0041600  0.0007205   5.773 9.08e-07 ***

> summary(lm(con$FA~con$age))
Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.5165714  0.0076978  67.107  < 2e-16 ***
con$age     0.0032801  0.0004563   7.189 5.33e-08 ***
\end{verbatim}
Then we test if $H_0: \beta_{a1}=\beta_{c1}$ based on the two sample
$t$ test with unequal variance. We present a different method using
the multiple regression framework.

\item We combine both the autistic subjects data with the normal control
data in such a way that $x_i, i=1,\cdots, m$ are ages for the
autistic subjects and $x_i, i=m+1,m+n$ are ages for the normal
controls. We also combine FA measurers $y_i$ together. Then we can
combine two independent linear models into a single big linear
model:
$$Y_i = \beta_0 + \beta_1 x_i + \beta_2 z_i + \beta_3 x_iz_i +
\epsilon_i, i =1,\codts,m+n$$ where $z_i$ is a dummy variable taking
values $0$ if the $i$-th subject is autistic and 1 otherwise. For
autistic subjects,
$$Y_i=\beta_0 + \beta_1 x_i + \epsilon_i, i=1,\cdots,m$$
For normal subjects,
$$Y_i=(\beta_0 +\beta_2) + (\beta_1 + \beta_3)x_i + \epsilon_i, i=1,\cdots,n$$
Hence we have $\beta_{a1}=\beta_1$ and $\beta_{c1}=\beta_1+\beta_3$
and testing if the regression slopes of the both groups are equal is
equivalent to testing the following:
$$H_0:\beta_3 =0 \mbox{ vs. } H_1:\beta_3 \neq 0}.$$
The null hypothesis corresponds to the reduced model

$$H_0: Y_i = \beta_0 + \beta_1 x_i + \beta_2 z_i + \epsilon_i$$
and the alternate hypothesis corresponds to the full model
$$H_1: Y_i = \beta_0 + \beta_1 x_i + \beta_2 z_i + \beta_3x_iz_i+
\epsilon_i, \beta_3 \neq 0.$$

The fit of model is measured by the sum of squared errors (SSE). Let
$SSE_0$ and $SSE_1$ be the SSEs for the null and the alternate model
respectively. Then
$$SSE_0 = \sum_{i=1}^{m+n} (y_i-\hat \beta_0 - \hat \beta_1
x_i-\hat \beta_2 z_i)^2,$$ where $\hat \beta_i$ are the least
squares estimates of $\beta_i$.

$$SSE_1 = \sum_{i=1}^{m+n} (y_i-\hat \beta_0 - \hat \beta_1
x_i- \hat \beta_2 z_i -\hat \beta_3x_iz_i)^2,$$ where $\hat \beta_i$
are the least squares estimates of $\beta_i$. Then under $H_0$, the
test statistic is

$$F= \frac{(SSE_0-SSE_1)/1}{SSE_0/(m+n-1-3) } \sim F_{1,m+n-1-3}$$

\begin{verbatim}
> summary(lm(FA~age+group))
Residual standard error: 0.02007 on 72 degrees of freedom Multiple
R-Squared: 0.5126,     Adjusted R-squared: 0.4991 F-statistic: 37.86
on 2 and 72 DF,  p-value: 5.805e-12

> summary(lm(FA~age+group+age*group))
Residual standard error: 0.02007 on 71 degrees of freedom Multiple
R-Squared: 0.5195,     Adjusted R-squared: 0.4992 F-statistic: 25.59
on 3 and 71 DF,  p-value: 2.485e-11

> (0.5195-0.5126)/0.5195*(43+32-1-3)
[1] 0.9430221
> qf(0.95,1,43+32-1-3)
[1] 3.97581
\end{verbatim}
In our example, the resulting $F$ value is 0.943. Hence we can not
reject $H_0$ at level $\alpha=0.05$ (95$\%$ significance) and
conclude that there is no difference in regression slope between the
two groups.


\end{enumerate}
{\bf Assigned problems.} Exercise 13.36., Exercise 13.40. HW 7 due
last day of class. Solve the assigned problems in Lecture 23-24 and
Exercise 13.25 from Lecture 22. (generate a similar computer output
using R).
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