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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 16\\
Other two sample tests}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
%\date{Feb 20, 2003}
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\maketitle \thispagestyle{empty} \subsection*{Concepts}
\begin{enumerate}
\item 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 $X_i \sim \mbox{Bernoulli}(p_X)$ and $Y_j \sim
\mbox{Bernoulli}(p_Y)$. Let $\hat p_X = \sum_{i=1}^n X_i/n$ and
$\hat p_Y = \sum_{j=1}^m Y_j/m$. $${\bf Var}(\hat p_X - \hat p_Y)
= \frac{p_X(1-p_X)}{n} + \frac{p_Y(1-p_Y)}{m}.$$
\item Difference between population proportions: for large $n$ and $m$, use a $Z$-statistic for testing
$H_0:p_X = p_Y$:
$$Z=\frac{\hat p_X -\hat p_Y - \mathbb{E}(\hat p_X -\hat p_Y)}
{\sqrt{{\bf Var}(\hat p_X - \hat p_Y)}} \sim N(0,1).$$ Since
$p_X$ and $p_Y$ are unknown, we estimate them from the samples.
\end{enumerate}
\subsection*{In-class problems}
{\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 9.13. 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 \\
  Number of major malfunctions & 42 & 294 \\ \hline
\end{tabular}\\

{\em Solution.} Let $p_U$ and $p_N$ be the proportions of births
with major malfunctions among Marijuana users and non users. The
hypothesis we are testing is $$H_0: p_U = p_N \mbox{ vs. } H_0:
p_U > p_N.$$ Under $H_0: p_U=p_N = p$, $\hat p =
(42+294)/(1246+11,178)=0.027$. Then the estimate for the variance
in Concept 2 under $H_0$ would be $\hat p(1- \hat
p)(1/n+1/m)=0.0048^2$. Also $\hat p_U =42/1246=0.034, \hat
p_N=294/11178=0.026$. 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 reject $H_0$ at 0.05
level.\\
{\em Example 2.} There are two coins. You threw 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*{Self-study problems}
Example 9.9, 9.10., Read p.379-381., Example 9.11., 9.12.\\ {\em
Note:} Starting lecture 17, we will jump to Chapter 12 and study
linear regression. If you use R, you can do your assignments real
quick.
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