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\title{Stat 312: Lecture 22\\
Inference on Correlation}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
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\section*{Concepts}
\begin{enumerate}
\item Assume that $X$ and $Y$ are two random variables whose joint
probability density function $f$ is the {\em bivariate normal}
such that $X \sim N(\mu_X,\sigma_X^2)$, $Y \sim
N(\mu_Y,\sigma_Y^2)$ and $\rho(X,Y)=\rho$. In this situation, we
can show that $\mathbb{E}(Y|X=x) = \beta_0 + \beta_1 x$, where
$\beta_1 = \rho \sigma_Y/\sigma_X$. This is the regression model
we studied!

 \item Testing $H_0:\rho =0$ vs. $H_1:\rho \neq 0$ is equivalent
 to $H_0:\beta_1 =0$ vs. $H_1:\beta_1 \neq 0$. So the appropriate test
 statistic for testing $H_0:\rho=0$ is $T= \hat \beta_1/S_{\hat
 \beta_1} \sim t_{n-2}$ (see Lecture 19, Concept 3). This test
 statistic is equivalent to $T= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$.

 \item Testing for independence is equivalent to testing
$\rho=0$ in bivariate data.
\end{enumerate}

\section*{In-class problems}
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 0.05. Also est if $x_i$ and
$y_i$ are independent.\\
 {\em Solution.}
\begin{verbatim}
> r<-0.022
> r*sqrt(10000-2)/sqrt(1-r^2)
[1] 2.200313
> pt(2.2,9998)
[1] 0.9860852
\end{verbatim}
Using Concept 2, $t= 2.2$. $P$-value =
$P(|T|>2.2)=2(1-0.986)=0.028$. So we reject $H_0$. Since
independence is equivalent to zero correlation in bivariate data,
we can use the same test procedure and end up with the same
result.\\

{\em Example.} 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
\end{verbatim}

An alternate method would be to do a simple linear regression
analysis.
\begin{verbatim}
> summary(lm(y~x))
Coefficients:
Estimate  t value Pr(>|t|)
   x          4.4  0.0024
\end{verbatim}

\section*{Self-study problems}
Please read P.533. Example 12.16.


\section*{Homework 7} Due April 29. 11:00am.\\
12.30., 12.36., 12.58., 12.60.,12.66.


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