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% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat 324: Lecture 21\\
Correlation Analysis}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{April 21, 2005}
\maketitle \thispagestyle{empty}


\begin{enumerate}


\item The {\em sample correlation coefficient} $r$
is defined as $r = S_{xy}/\sqrt{S_{xx}S_{yy}}.$ In the previous
lecture, we showed that $0 \leq r^2 \leq 1$ so $-1 \leq r \leq 1$.

\item {\em Example 1.} Is there any linear relationship between the
mathematical achievement test scores ($x_i$) and calculus grades
($y_i$) for 10 college freshmen?\\
x$\|39,43,21,64,57,47,28,75,34,52$\\
y$\|65,78,52,82,92,89,73,98,56,75$
\begin{verbatim}
> cor(x,y)
[1] 0.8397859
\end{verbatim}


\item {\em Example 2.} Suppose that data $(x_j,y_j)$ satisfy a
nonlinear relationship $y_j = \sqrt{10^2 - x_j^2}$. Since this is a
highly nonlinear relationship, we should expect less correlation.
\begin{verbatim}
>x<--10:10
>y<-sqrt(10^2 -x^2)
> cor(x,y)
[1] 0
>summary(lm(y~x))
 Coefficients: Estimate
(Intercept) .... 7.392e+00
          x .... 4.997e-18
Multiple R-Squared: 9.504e-32
\end{verbatim}
\begin{figure}[h]
\centering
\renewcommand{\baselinestretch}{1}
\includegraphics[scale=0.8]{324.21.01.ps}
\caption{Complete lack of linear fit.}
\end{figure}
For this data set, $r=0$.

\item $r=\pm 1$ if and only if paired observations $(x_j,y_j)$ are on the
straight line. This can be seen from the fact that $r^2=1$ iff.
$SSE=\sum_{j=1}^n (y_j - \hat \beta_0 - \hat \beta_1 x_j)^2=0.$ This
is equivalent to all points $(x_j,y_j)$ passing through a line
$y=\hat \beta_0 + \hat \beta_1 x$. The correlation can be used to
measure if the paired observations show linear relationship.

\item {\em Population correlation coefficient} is defined as
$$\rho(X,Y) = \frac{{\bf Cov}(X,Y)}{\sqrt{{\bf Var}(X){\bf
Var}(Y)}}$$. The sample correlation is an estimate for the
population correlation.
%\begin{figure}[t]
%\centering
%\renewcommand{\baselinestretch}{1}
%\includegraphics[scale=0.26]{positivecor.ps}
%\includegraphics[scale=0.26]{negativecor.ps}
%\caption{Complete lack of linear fit.}
%\end{figure}

{\em Example 3.} Is there any relationship between the number of
times you miss submitting homeworks ($x_i$) and the final grade ($y_i$)?\\
x$\|03,04, 01,02,01,00,00,07,08,05$\\
y$\|70,75,80,75,85,95,90,70,50,60$
\begin{verbatim}
> cor(x,y)
[1] -0.8938321
\end{verbatim}

\item Assume $(X,Y)$ is distributed as a {\em bivariate normal}
whose joint density is given by \bq f(x,y) =
\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp \Big[
-\Big(\frac{x-\mu_1}{\sigma_1}\Big)^2\\ -
2\rho\frac{x-\mu_1}{\sigma_1}\frac{y-\mu_2}{\sigma_2}
-\Big(\frac{x-\mu_2}{\sigma_2}\Big)^2\Big]/(2(1-\rho^2)).\eq The
marginal distributions are $X \sim N(\mu_1,\sigma_1^2)$, $Y \sim
N(\mu_2,\sigma_2^2)$. Also $\rho(X,Y)=\rho$. If $\rho=0$, then the
joint density can be factored into the product of marginal
distributions showing the independence of $X$ and $Y$.



It can be shown that $Y|X=x$ is normally distributed with parameters
 $\mathbb{E}(Y|X=x) = \beta_0 + \beta_1 x$, where
$\beta_1 = \rho \frac{\sigma_2}{\sigma_1}$ and $\beta_0 = \mu_2 -
\rho \mu_1 \frac{\sigma_2}{\sigma_1}$, and $\mathbb{V} (Y|X=x) =
(1-\rho^2)\sigma_2^2$. This implies that $Y$ given $X=x$ can be
modeled using the linear regression.



 \item Testing $H_0:\rho =0$ vs. $H_1:\rho \neq 0$ is equivalent
 to $H_0:\beta_1 =0$ vs. $H_1:\beta_1 \neq 0$ in in bivariate normal data. So the appropriate test
 statistic for testing $H_0:\rho=0$ is $T= (\hat \beta_1-\beta_1)/S_{\hat
 \beta_1} \sim t_{n-2}$. Under $H_0$, the test
 statistic is equivalent to $T= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$.



\item Revisiting {\em Example 3.} Test for the dependence of the mathematical
achievement test scores ($x_i$) and calculus grades
($y_i$) for 10 college freshmen assuming $(x_i,y_i)$ are from bivariate normal.\\
x$\|39,43,21,64,57,47,28,75,34,52$\\
y$\|65,78,52,82,92,89,73,98,56,75$\\

{\em Solution.} There are two ways to solve this problem. The first
approach might be much simpler conceptually. The only thing you need
is to compute the correlation coefficient and follow the solution of
Exercise 12.67.
\begin{verbatim}
> r<-cor(x,y)
[1] 0.8397859
> r*sqrt(10-2)/sqrt(1-r^2)
[1] 4.375015
> qt(0.025,8)
[1] -2.306004
\end{verbatim}

An alternate method would be to perform a simple linear regression
analysis.
\begin{verbatim}
> summary(lm(y~x))
Coefficients: Estimate
   x          4.4
\end{verbatim}

\item Example 4. Exercise 12.67. A sample of 10,000 $(x_i,y_i)$ pairs resulted in
the sample correlation coefficient $r=0.022$. Test $H_0:\rho=0$ vs.
$H_1:\rho \neq 0$ at level $\alhpa=0.05$. Also test if $x_i$ and
$y_i$ are independent.\\
\begin{verbatim}
> r<-0.022
> r*sqrt(10000-2)/sqrt(1-r^2)
[1] 2.200313
> qnorm(0.025)
[1] -1.959964
\end{verbatim}

Since $t= 2.2$ we reject $H_0$.

\item In general, test $H_0:\rho=\rho_0 \mbox{ vs. } H_1:\rho
\neq \rho_0$ requires the Fisher transformation. It can be shown
that $\frac{1}{2}\ln \frac{1+r}{1-r} \sim N(\frac{1}{2}\ln
\frac{1+\rho}{1-\rho}, \frac{1}{n-3})$. So an appropriate test
statistic for testing $H_0:\rho =\rho_0$ is
$$Z = \frac{\frac{1}{2}\ln \frac{1+r}{1-r} -
\frac{1}{2}\ln \frac{1+\rho_0}{1-\rho_0}}{(n-3)^{-1/2}}.$$ and under
$H_0$, $Z \sim N(0,1)$. Constructing CI is based on the Fisher
transform.

\end{enumerate}
{\bf Assigned problems.} Exercise 12.60, 12.64.
  HW 6 due April 28th. Solve the assigned
problems in Lecture 19-22.
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