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% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat 324: Lecture 18\\
Two Sample Tests}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{April 12, 2005}
\maketitle \thispagestyle{empty}


\begin{enumerate}


\section*{Concepts}

\begin{enumrate}



\item {\em Example 1.} A study was conducted to compare the reaction
times of men and women to a stimulus. 12 men and 9 women were
employed in the experiment. Let $X_i$ be the reaction time of the
$i$-th man and $Y_j$ be the reaction time of the $j$-th woman. There
is a reason to believe that $X_i \sim N(\mu_X, 0.18)$ and $Y_j \sim
N(\mu_Y, 0.14)$. The following measurements were taken: $\bar x =
3.6$ seconds, $\bar y= 3.8$ seconds. Is there any difference between
the reaction times of men and women? Test this at $\alpha =0.1$.


\begin{verbatim}
> z=(3.6-3.8)/sqrt(0.18/12+0.14/9)
> z
[1] -1.144155
\end{verbatim}

\item Let $X_1,\cdots,X_n$ and $Y_1,\cdots,Y_m$ be two independent
samples with mean $\mu_X$, $\mu_Y$ and variance $\sigma_X^2$,
$\sigma_Y^2$ respectively. $$\mathbb{E}(\bar X - \bar Y) = \mu_X
-\mu_Y. $$
$${\bf Var} (\bar X -\bar Y) = \sigma_X^2/n +
\sigma_Y^2/m.$$


Let $X_1,\cdots,X_n$ and $Y_1,\cdots,Y_m$ be two independent samples
with $X_i \sim N(\mu_X,\sigma_X^2)$ and $Y_j \sim N(\mu_Y,
\sigma_Y^2)$. Then $$Z=\frac{\bar X - \bar Y}{\sqrt{\sigma_X^2/n +
\sigma_Y^2/m}} \sim N(0,1).$$


\item {\em Example 2.} A study was conducted to compare the reaction
times of men and women to a stimulus. 50 men and 65 women were
employed in the experiment. Let $X_i$ be the reaction time of the
$i$-th man and $Y_j$ be the reaction time of the $j$-th woman. The
following measurements were taken: $\bar x = 3.6$ seconds, $\bar y=
3.8$ seconds, $s_X^2=0.18$, $s_Y^2=0.14$. Is there any difference
between the reaction times of men and women?
\begin{verbatim}
> z=(3.6-3.8)/sqrt(0.18/50+0.14/65)
> z
[1] -2.63664
> qnorm(0.05)
[1] -1.64
\end{verbatim}

What is the main reason you are getting the different results in
Example 1 and 2? Increasing the sample size enable us to distinguish
the small difference that was difficult to do so in the small
sample.


\item Assume $n$ and $m$ to be large. The test statistic for
testing
$$H_0: \mu_X = \mu_Y \mbox{ vs. } H_1: \mu_X \neq \mu_Y$$
$$Z=\frac{\bar X - \bar Y}{\sqrt{s_X^2/n + s_Y^2/m}} \sim N(0,1).$$


\item {\em Example 3.} A study was conducted to compare the weights of
cats and dogs.\\
Weights of cats: 20, 21, 35, 13, 21, 10.\\
 Weights of dogs: 31, 10,
20, 40.
\\Assuming normality and equal variance for both cats and
dogs, determine if there is any difference between the weights of
cats and dogs at $\alpha = 0.1$.
\begin{verbatim}
> x<-c(20,21,35,13,21,10)
> y<-c(31,10,20,40)
> Sp<-sqrt((5*var(x)+3*var(y))/8)
[1] 10.52824
> t=(mean(x)-mean(y))/(Sp*sqrt(1/5+1/3))
> t
[1] -0.6827026
> qt(0.05,8)
[1] -1.859548
\end{verbatim}

\item Let $X_1,\cdots,X_n$ and $Y_1,\cdots,Y_m$ be two independent
samples from normal distributions with the same population variance,
i.e. $$X_i \sim N(\mu_X, \sigma^2) \mbox{ and } Y_j \sim
N(\mu_Y,\sigma^2).$$ The pooled estimator of $\sigma^2$ is
$$S_p^2 = \frac{(n-1)S_X^2 + (m-1)S_Y^2}{n+m-2}.$$
It can be shown that $\mathbb{E} S_p^2 = \sigma^2$. The proper test
statistic for testing
$$H_0: \mu_X = \mu_Y \mbox{ vs. } H_1: \mu_X \neq \mu_Y$$
is
$$T=\frac{\bar X - \bar Y -(\mu_X -\mu_Y)}{S_p\sqrt{1/n + 1/m}} \sim t_{n+m-2}.$$
Reject $H_0$ if $|T| > t_{\alpha/2,n+m-2}$.


Note. {\tt R} package has a built-in command {\tt t.test} that can
be used to perform two sample $t$ tests.

%If you use R, it is very easy to do two sample hypothesis testing.
%\begin{verbatim}
%>t.test(x,y,alternative="two.sided",
%var.equal=TRUE,conf.level=0.9)
%        Two Sample t-test
%data:x and y t = -0.7725, df = 8, p-value = 0.462 alternative
%hypothesis: true difference in means is not equal to 0 90 percent
%confidence interval:
% -17.88739   7.38739
\end{enumerate}
{\bf Assigned problems.} Exercise 9.7., Exercise 9.22 under the
assumption of equal variance. HW 5 due April 14th. Solve the
assigned problems in Lecture 15-18.



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