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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 18\\
Other two sample tests}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{November 11, 2004}
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\maketitle \thispagestyle{empty} \subsection*{Concepts}
\begin{enumerate}
\item {\em Paired data.} For a given paired sample $(X_1,Y_1), \cdots, (X_n,Y_n)$
with $\mathbb{E}X_i =\mu_X$ and $\mathbb{E}Y_i =\mu_Y$, a test
statistic for testing $H_0: \mu_X =\mu_Y$ can be based on one
sample test.
\item Let $X_i \sim \mbox{Bernoulli}(p_1)$ and $Y_j \sim
\mbox{Bernoulli}(p_2)$. Let $\hat p_1 = \sum_{i=1}^n X_i/n$ and
$\hat p_2 = \sum_{j=1}^m Y_j/m$. $${\bf Var}(\hat p_1 - \hat p_2) =
\frac{p_1(1-p_1)}{n} + \frac{p_2(1-p_2)}{m}.$$
\item Suppose $p_1$ and $p_2$ denote the proportion of individuals in population 1 and 2
respectively. For sufficiently large $n$ and $m$, we use a
$Z$-statistic for testing $$H_0:p_1 = p_2 \mbox{ vs. } H_1:p_1 \neq
p_2:$$
$$Z=\frac{\hat p_1 -\hat p_2 - \mathbb{E}(\hat p_1 -\hat p_2)}
{\sqrt{{\bf Var}(\hat p_1 - \hat p_2)}}.$$
\end{enumerate}
\subsection*{Examples}
{\em Example 1.} 10 students took two midterm exams.\\
Student\hspace{0.35cm}  $\|$  01  02  03  04 05 06 07 08 09 10\\
Midterm 1$\|$ 80 75 60 90 99 60 55 85 65 70\\
Midterm 2$\|$ 70 60 70 72 95 66 60 80 70 60\\
Is the first exam easier than the second exam? Test it at level
0.05.\\
{\em Solution.} Let $X_i$ and $Y_i$ be the first and the second
midterm scores for the $i$-th student. Perform the $t$-test on
$H_0: \mu_Y-\mu_X =0$ based on observations $Y_i-X_i$.

\begin{verbatim}
> x<-c(80, 75, 60, 90, 99, 60, 55,
85, 65, 70)
> y<-c(70, 60, 70, 72, 95, 66, 60,
80, 70, 60)
> t.test(y-x, conf.level=0.95)
 One Sample t-test
data:  y - x t = -1.1739, df = 9,
p-value = 0.2706 alternative
hypothesis: true mean is not equal
to 0 95 percent confidence
interval:
 -10.537282   3.337282
\end{verbatim}
Example 2. Is it harmful to use Marijuana when mothers are pregnant?\\
\begin{tabular}{|c|c|c|}
\hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
   & User & Nonuser \\
   \hline
  Sample size & 1246 & 11,178 \\
  Birth defects & 42 & 294 \\ \hline
\end{tabular}\\

{\em Solution.} Let $p_1$ and $p_2$ be the birth defect proportions
among Marijuana users and non users. The hypothesis we are testing
is $$H_0: p_1 = p_2 \mbox{ vs. } H_0: p_1
> p_2.$$
From data $\hat p_1 =42/1246=0.034, \hat p_2=294/11178=0.026$. Under
$H_0: p_1=p_2 = p$, $\hat p = (42+294)/(1246+11,178)=0.027$. Then
the estimate for the variance in Concept 2 under $H_0$ would be
$\hat p(1- \hat p)(1/n+1/m)=0.0048^2$. Then $z = 1.53$. We reject
$H_0$ if $z$ is larger so the $P$-value is $P(Z > 1.53) =
1-P(Z<1.53)=1-\tt{pnorm}(1.53) =0.06.$ So we do not reject $H_0$ at
0.05 level but we reject $H_0$ at 0.1 level.


{\em Example 3.} There are two coins. You toss the first coin 100
times and observed 40 heads. When you threw the second coin 110
times, you observed 50 heads. Test if the two coins give the
same number of heads.\\
{\em Solution.} Let $p_1$ and $p_2$ be the probabilities of
getting heads for the first and the second coins respectively.
$$H_0: P_1 = p_2 \mbox{ vs. } H_1:p_1 \neq p_2.$$
Let $X$ be the number of heads when you threw the first coin
$n=100$ times and $Y$ be the number of heads when you threw the
second coin $m=110$ times. $\hat p_1 -\hat p_2 =X/n -Y/m$. Then
the test statistic is $Z=\frac{X/n -Y/m}{\sqrt{p(1-p)(1/n +
1/m)}} \sim N(0,1),$ where $p=p_1=p_2$. Since $p$ is unknown, you
estimate it by pooling the samples: $\hat p =(x+y)/(m+n)$. $\hat
p = (40+50)/(100+110) = 0.43$ and $z=-0.60$. Since we reject
$H_0$ if $|z|$ is large, the $P$-value would be $P(|Z|> 0.6) =
2P(Z>0.6) = 2(1-P(Z \leq 0.6))=2(1-0.73)=0.54$. So we do not
reject $H_0$ at any level smaller than 0.5.

\subsection*{Review problems}
Example 9.9, 9.11.\\ {\em Note:} Starting lecture 19, we will jump
to Chapter 12 and study linear regressions.
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