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\begin{document}

% Don't want date printed
\date{May 11, 2005}

% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat324: Final Exam Solutions}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
% For two authors (default example)
\maketitle

You may use the following {\tt R} output for the answering exam
problems.
\begin{verbatim}
> qnorm(c(0.005,0.01,0.025,0.05,0.1))
[1] -2.575829 -2.326348 -1.959964 -1.644854 -1.281552
> qt(0.025,5:11)
[1] -2.57 -2.44 -2.36 -2.30 -2.26 -2.22 -2.20
> qt(0.05,5:11)
[1] -2.01 -1.94 -1.89 -1.85 -1.83 -1.81 -1.79
> qt(0.01,5:11)
[1] -3.36 -3.14 -2.99 -2.89 -2.82 -2.76 -2.71
> qf(0.025,1,5:11)
[1] 0.001084778 0.001067109 0.001054611 0.001045308 0.001038115
0.001032389 [7] 0.001027723
> qf(0.05,1,5:11)
[1] 0.004344768 0.004273760 0.004223537 0.004186155 0.004157256
0.004134252 [7] 0.004115506
> qf(0.01,1,5:11)
[1] 0.0001735012 0.0001706780 0.0001686808 0.0001671942 0.0001660447
[6] 0.0001651297 0.0001643840
\end{verbatim}
\begin{enumerate}

\item Two players $A$ and $B$ flip a biased coin alternately and the first player
to obtain a head wins. The probability of obtaining a head is $p >
0$ at each toss. Suppose $A$ flips the coin first.\\
(a) Find the probability that $A$ wins the game. Derive your result [10pts].\\
(b) Find the expected number of tosses $A$ must make till $A$
wins. Derive your result [10pts].\\
 {\it Solution.} (a) Let $X$ be the number of tosses $A$ makes until
he gets head.
$$P(A \mbox{ wins }) = \sum_{i=1}^{\infty} P(X=i)= p + (1-p)^2p + (1-p)^4p + \dots\\
= \frac{p}{1-(1-p)^2}.$$ (b) $$\mathbb{E} X = p + 2(1-p)^2p +
3(1-p)^4p + \cdots = \sum_{x=1}^{\infty} x(1-p)^{2(x-1)}p.$$ let
$r=(1-p)^2$. Then $\mathbb{E} X = p\sum_{x=1}^{\infty} x r^{x-1} =
p\frac{d}{dr}\sum_{x=1}^{\infty} r^{x}=
p\frac{d}{dr}\Big(\frac{r}{1-r}\Big)=\frac{p}{(1-r)^2}
=\frac{p}{(1-(1-p)^2)^2}.$



\item (a) Suppose that $X$ and $Y$ are Bernoulli random variables with parameter $0< p < 1$ such that $\mbox{cov}(X,Y) = 0$.
Prove that $X$ and $Y$ are independent [10pts].\\
(b) Let $X_1,\cdots,X_n$ be a random sample from the Bernoulli
distribution with parameter $p$. Find an unbiased estimator for
$(1-p)^2$ [10pts].


{\em Solution.} (a) Note ${\bf Cov}(X,Y)=\mathbb{E}(XY)
-\mathbb{E}X\mathbb{E}Y = P(X=1,Y=1) -P(X=1)P(Y=1) =0$. Hence
$P(X=1,Y=1)=P(X=1)P(Y=1)=p^2$. Then it follows
 $P(X=1,Y=0)=P(X=1)-P(X=1,Y=1)= p(1-p)=P(X=1)P(Y=0)$
 and similarly for other cases. So $X$ and $Y$ are independent.\\
(b) Note that $\mathbb{E} S^2 = p(1-p)$ and $\mathbb{E} \bar X = p$.
Hence $\mathbb{E} (1-\bar X-S^2) = p^2 -2p+1 =(1-p)^2$.


\item 47 out of 102 doctors knew the generic name
for the drug methadone. Compute a $98\%$ confidence interval for the
propotion of all doctors who {\bf do not know} the generic name for
methadone. [10 pts].


{\em Solution.} {\bf Problem setup} [5pts]. Let $X_i$ be 1 if the
$i$-th doctor knows the name and 0 otherwise. Then $X_i \sim
Bernoulli(p)$. The unbiased point estimate for $p$ is $\bar X$.
Since the sample size is large, we use the central limit theorem and
the statistic is
$$Z = \frac{\hat p - p}{\sqrt{p(1-p)/n}} \sim N(0,1).$$
{\bf Computation} [5pts]. This requires solving a quadratic equation
so we will approximate this by substituting $p$ in the denominator
with estimate $\hat p=47/102$. Then a CI for $p$ becomes $\hat p \pm
z_{\alpha/2}\sqrt{\hat p(1-\hat p)/n}$. So CI is $[0.35, 0.58]$. The
proportion of doctors who do not know is $q=1-p$. Hence a CI for $q$
is $[1-0.58, 1-0.35]=[0.42, 0.65]$.



\item We wish to fit observations $(x_1,y_1)$, $\cdots$, $(x_n,y_n)$ with a linear  model $Y=c x + \epsilon$.
We assume that $\epsilon \sim N(0,\sigma^2)$.
 \begin{itemize}
\item[(a)] Estimate $c$ by the least squares estimation. Determine if it is unbiased [10pts].

\item[(b)] Estimate $c$ by the maximum likelihood method and compute the variance of the estimator [15pts].


\item[(c)] If the sample correlation coefficient of the above paired data is denoted by $r$,
what is the sample correlation correlation coefficient of data
$(x_1 -\bar x,y_1 -\bar y), \cdots, (x_n -\bar x, y_n -\bar y)$?
 $\bar x$ and $\bar y$ are the respective sample means of $x_i$'s and $y_i$'s.
  Prove your result [10pts].
\item[(d)] Prove the sample correlation $r$ is a number between -1 and 1 [10pts].
\end{itemize}
{\em Solution.} (a) The sum of squared residuals $SSE =\sum_{i=1}^n
(y_i-cx_i)^2$ (2pts). Letting $\frac{\partial SSE}{\partial c} = 0$,
we get $\sum_{i=1}^n x_i(y_i - cx_i)=0$. Solving this, $\hat c =
\frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}$ (3pts). $\mathbb{E}
\hat c = \sum_{i=1}^n x_i\mathbb{E} Y_i/\sum_{i=1}^n x_i^2$ =c
(5pts).


\\ (b) Note
that $\mathbb{E} Y_i = cx_i$ and ${\bf Var} Y_i = \sigma^2$. So $Y_i
\sim N(cx_i, \sigma^2)$. Then the likelihood function is given by
$$L(c,\sigma)=\frac{const.}{\sigma^n}\exp \Big(\frac{\sum_{i=1}^n(y_i-cx_i)^2}{2\sigma^2}\Big).$$ The log-likelihood function is then
$$\log L(c,\sigma)= const. -n\log \sigma + \frac{1}{2\sigma^2}\sum_{i=1}^n(y_i-cx_i)^2.$$
Letting $\frac{\partial \log L}{\partial c}=0$, we get $\sum_{i=1}^n
x_i(y_i - cx_i)=0$ so we get the same answer as in (a).  $\hat c =
\frac{\sum_{i=1}^n x_iY_i}{\sum_{i=1}^n x_i^2}$. ${\bf Var} \hat c =
\frac{\sum_{i=1}^n x_i^2 {\bf Var} Y_i}{[\sum_{i=1}^n x_i^2]^2} =
\frac{\sigma^2}{\sum_{i=1}^n x_i^2}$.\\
(c) The sample correlation is given by
$r_{x,y}=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$, where $S_{xy} =
\sum_{i=1}^n (x_i-\bar x)(y_i -\bar y)$. Let $(z_i,w_i)=(x_i-\bar
x,y_i -\bar y)$. Note that $\bar z=\bar w =0$. Also $S_{zz}=S_{xx}$
and $S_{ww}=S_{yy}$. $S_{zw}=\sum_{i=1}^n(z_i-\bar z)(w_i-\bar w)=
\sum_{i=1}^n z_iw_i=S_{xy}$. So the sample correlation is the same,
i.e. $r_{z,w} = r_{x,y}$.\\
(d) We show that $0 \leq r^2 \leq 1$.   First we show that
$$SSE=\min_{\beta_0,\beta_1} \sum_{j=1}^n(y_j-\beta_0-\beta_1
x_j)^2 \leq \min_{\beta_0} \sum_{j=1}^n(y_j-\beta_0)^2 \leq
\sum_{j=1}^n(y_j-\bar y)^2\leq SST.$$ Hence $0\leq r^2 =
1-\frac{SSE}{SST}\leq 1$ and the proof follows.


\item Three different types of boxes underwent compression strength
test. The following data are compression strength (lb). Assume the
normality of data.

\begin{array}{ccccc}
\mbox{Type I   }  &65 & 79 & 72 & 68 \\
\mbox{Type II} & 78 & 77 & 68 & 77 \\
\mbox{Type III} & 60 & 62 & 74 & & \\
\end{array}

The following {\tt R} output may be used to answer the following
questions; however, you are not required to use it.
\begin{verbatim}
> type1<-c(65,79,72,68)
> type2<-c(78,77,68,77)
> type3<-c(60,62,74)
> z<-c(type1,type2,type3)
> w<-c(0,0,0,0,1,1,1,1,2,2,2)
> summary(lm(z~w))

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)   73.158      3.092  23.663 2.05e-09 ***
w             -2.474      2.564  -0.965     0.36
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

Residual standard error: 6.738 on 9 degrees of freedom Multiple
R-Squared: 0.09376,    Adjusted R-squared: -0.006933 F-statistic:
0.9312 on 1 and 9 DF,  p-value: 0.3598
\end{verbatim}

(a) Test if the true compression strength for box types I and III
are equal at $\alpha =0.1$ Assume {\bf equal} variance for different
box types [10pts].\\
(b) Test if the true compression strength for all box types I, II
and III are equal at $\alpha =0.1$. Assume {\bf equal} variance for
different box types. Clearly write down the null and alternate
hypotheses, the test statistic, and the statistical model you are
using
[20pts].\\
{\em Solution.} (a) Let $X_i \sim N(\mu_X, \sigma_X^2)$ be the
compression strength for type I and $Y_i \sim N(\mu_Y,\sigma_Y^2)$
be the compression strength for type II. The pooled estimator of
$\sigma^2$ is
$$S_p^2 = \frac{(n-1)S_X^2 + (m-1)S_Y^2}{n+m-2} = 44.93,$$
where $S_X^2 = 36.67$ and $S_Y^2 = 57.33$. 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 $|T| > t_{\alpha/2,n+m-2}$. Since $t=0.165$ and
$t_{0.05,5} = 2.015$, we do not reject $H_0$ at $\alpha=0.1$.\\
(b) {\bf Problem Setup.} Combine $x_i$ and $y_i$ into a single
variable $z_i$ with $1 \leq i \leq 11$. Introduce a dummy variable
$w_i =0$ for type I, $w_i=1$ for type II, and $w_i=2$ for type III.
Then we set up a linear model $Z_i = \beta_1 + \beta_2w_i +
\epsilon_i,$ where $\epsilon_i \sim N(0,\sigma^2)$. Then we test
$H_0: \beta_2 = 0 \mbox{ vs. } H_1:\beta_2 \neq 0$ [15pts]. {\bf
Computation.} The test statistic is $T=\frac{\hat \beta_2}{S_{\hat
\beta}} \sim t_{11-2}$. Since $t=-0.965$ and $t_{0.05,9} =1.83$, we
do not reject $H_0$ at $\alpha = 0.1$ [5pts].





\end{enumerate}

\end{document}