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\begin{document}

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%\date{January 23, 2003}

% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat312: Midterm II}

% For single author (just remove % characters)
\author{Moo K. Chung, Yulin Zhang\\
mchung@stat.wisc.edu, yulin@stat.wisc.edu}
% For two authors (default example)
\maketitle
\thispagestyle{empty}
\begin{center}
\small
Answer all questions clearly and {\bf circle your final answer}.\\ 
Your answers should be correct up to the {\bf second decimal places}.\\
One page note and a calculator are allowed.\\
No textbooks, scrap papers or hand-held computers are allowed.\\ 
If you are introducing new variables, why don't you share the \\
secrete of what those variables mean by defining them clearly.\\
This exam booklet consists of {\bf three problems and one page R output}.\\
\vspace{1cm}
\large
{\bf Name:}\underline{\hspace{10.5cm}}\\
{\bf Student ID:}\underline{\hspace{10cm}}\\

\normalsize
\vspace{1cm}
Pledge: On my honor, I have neither given nor received unauthorized\\
aid on this examination.\\
\vspace{1cm}
\large
{\bf Signature:}\underline{\hspace{10cm}}\\
\vspace{2cm}

\begin{tabular}{|r|l|c|}\hline
Question & Points}\\
\hline
1 & \\
\hline
2 & \\
\hline
3 & \\
\hline
%4 & \\
%\hline
Total & \\
\hline
\end{tabular}
\end{center}

\newpage

Please use the following R output in answering the appropriate
exam problems.

\begin{verbatim}

       > pnorm(c(1.0,1.1,1.2,1.3,1.4,1.5,1.6))
       [1] 0.84 0.86 0.88 0.90 0.92 0.93 0.95

       > pt(c(1.0,1.1,1.2,1.3,1.4,1.5,1.6),9)
       [1] 0.83 0.85 0.87 0.89 0.90 0.92 0.93

       > pt(c(1.0,1.1,1.2,1.3,1.4,1.5,1.6),10)
       [1] 0.83 0.85 0.87 0.89 0.90 0.92 0.93

\end{verbatim}

\newpage
\begin{enumerate}
\item A mixture of pulverized fuel ash and Portland cement should
have a compressive strength of more than 1000KN/m$^3$. The
mixture will not be used unless experimental evidence indicates
that the strength specification has been met. Suppose compressive
strength for specimens of this mixture is normally distributed
with $\sigma^2 = 98^2$. The sample mean of compressive strength
for 10 specimens is
$\bar x = 1050$.\\
(1) What are the appropriate null and alternate hypothesis (5 points)? \\
(2) What would be the appropriate test statistic for the
hypothesis testing?\\
What is the probability distribution of your
test statistic when $H_0$ is true (5 points each) ?\\
\newpage
(3) Do hypothesis testing at $85\%$ level. Has the strength
specification
been met (10 points)?\\
(4) Compute the $P$-value (5 points).

%{\em Solution.} (1) Let $\mu$ be the population mean of
%compressive strength. Then $H_0: \mu = 1000$ and $H_1: \mu >
%1000$. (2) Since the population variance is known, the test
%statistic should be the $Z$ statistic $Z=(\bar
%X-\mu)/(\sigma/\sqrt{n})$. Under $H_0$, $Z \sim N(0,1)$. (3)
%$z=50/(98/\sqrt{10}) = 1.61.$ From R output, $85\%$ level
%rejection region is $z > 1.1$. So we reject $H_0$. (4) P-value $=
%P(Z>1.61) = 1-P(Z \leq 1.61) = 0.05$.

\newpage
\item A farmer want to know if there is any difference in
weight gain from  young male and female pigs when both sexes are
subjected to identical feeding program. He measured the weight
gain for 7 female pigs ($x_i$) and 5 male pigs ($y_i$) and
observed $\bar x =9.8, \bar y = 9.2, s_x = 0.62, s_y = 0.86$.
$s_x$ and $ s_y$ are the sample variance of female pigs
and male pigs respectively.\\
(1) State the appropriate null and alternate hypothesis (5
points).\\
(2) Compute the pooled sample variance. (5 points).
\newpage
(3) What is the probability distribution of the test statistic
under $H_0$?\\ What are the statistical assumptions on measurements
to get this distribution? (5 points each).\\
(4) Compute the $P$-value. Based on the $P$-value, do hypothesis
testing at $\alpha = 0.05$ (5 points each).\\
%{\em Solution.} (1) Let $\mu_X$ and $\mu_Y$ be the weight gains
%for female and male pigs respectively. Then  $H_0: \mu_X = \mu_Y
%\mbox{ vs. } H_1: \mu_X \neq \mu_Y$. (2) The pooled variance is
%$$s_p^2 =\frac{(n-1)s_x^2 + (m-1)s_y^2}{n+m-2} =0.53.$$ (3)
%%$$T=\frac{\bar X - \bar Y}{S_p\sqrt{1/n + 1/m}} \sim t_{10}.$$
%We need to assume that $x_i$ come from $N(\mu_X,\sigma^2)$ and
%$y_j$ come from  $N(\mu_Y,\sigma^2)$. Further, $x_i$ and $y_j$
%should be independent observations. (4) $P$-value $= P(|T|>
%1.41)=2(1-P(1.41))=2(1-0.9)=0.2$. Since our $P$-value is bigger
%than the given $\alpha$, we do not reject $H_0$.

\newpage
\item A study found that of 238 individuals with coronary heart
disease (CHD), 145 smoked cigarettes; of 476 in the same age
group but {\bf without} CHD, 192 smoked cigarettes. The
researchers want to know if the proportion of cigarettes smokers
is the same in the population of
individulas with CHD as in the population with no CHD.\\
(1) State the null and alternate hypothesis (5 points).\\
(2) What's your test statistic and its distribution under the null
hypothesis (5 points) ?
\newpage
(3) Perform the hypothesis test using a significance level
$\alpha=0.14$. (10 points).\\
(4) Calculate a 86$\%$ confidence interval for the difference
between the two proportions (10 points).\\
% {\em Solution.} (1) Let
%$p_C$ and $p_N$ be the propotions of smokers in CHD group and non
%CHD group respectively. Then $H_0: p_C = p_N$ vs. $H_1:p_C \neq
%p_N$. (2) The point estimator for $p_C -p_N$ is $\hat p_C -\hat
%p_N =X/n -Y/m$, where $X,Y$ are the total number of smokers in
%CHD group and non CHD group respectively and $n = 238, m = 476$.
%Then under $H_0: p_C=p_N=p$, the test statistic is
%$$Z=\frac{X/n -Y/m}{\sqrt{p(1-p)(1/n + 1/m)}} \sim N(0,1).$$
%Since $p$ is unknown, you estimate it by pooling the samples:
%$\hat p =(x+y)/(m+n)$. (3) $\hat p = (145+192)/(238+476) = 0.472$
%and $z=5.20$. We reject $H_0$ if $|z|$ is large. Since
%$\alpha=0.14$, the cutoff value should be
%$z_{\alpha/2}=z_{0.07}=1.5$ from the R output. (4) Although
%$\tt{qnorm}$ values are not given, we can estimate $z_{0.07}$
%from the R output. Note that $P(Z > z_{0.07})=0.07$.S So
%$z_{0.07} =1.5$. A $100(1-\alpha)\%$ confidence interval is
%$$\hat p_C - \hat p_N \pm z_{\alpha/2}\sqrt{\hat p_C(1-\hat
%p_C)/238 + \hat p_N (1-\hat p_N)/476}.$$ With $\hat p_C = 145/238$
%and $\hat p_N =192/476$, we get $[0.15,0.26]$.
\end{document}