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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: Final Exam Solutions}

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


\begin{enumerate}


\item We wish to fit paired sample $(x_1,y_1)$, $\cdots$, $(x_n,y_n)$ with a linear regression model $Y=c x + \epsilon$. We assume that $\epsilon$ follows a normal distribution with zero mean and variance $\sigma^2$.
 \begin{itemize}
\item[(a)] Estimate $c$ by minimizing the sum of the squared residuals (5pts).

\item[(b)] Write the log-likelihood function as a function of $c$ and $\sigma$ (5pts).
\item[(c)] Find the likelihood estimator of $c$ by differentiating the log-likelihood function in (b). Derive everything (5pts).
\item[(d)] Either prove or disprove unbiasedness of the estimator you computed in (c) (5pts, no point given if (c) is incorrect).
\item[(e)] Compute the variance of the estimator you computed in (a) (5pts, no point given if (a) is incorrect).
 
\item[(f)] If the sample correlation coefficient of the above paired data is $0.5$, 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 (5pts).

\end{itemize}
{\em Solution.} (a) The sum of squared residuals $SSE =\sum_{i=1}^n (y_i-cx_i)^2$ (2pts). Letting $\partial SSE/\partial c = 0$, we get $\sum_{i=1}^n x_i(y_i - cx_i)=0$. Solving this, $\hat c = \sum_{i=1}^n x_i y_i/\sum_{i=1}^n x_i^2$ (3pts). (b) Note that $\mathbb{E} Y_i = cx_i$ and ${\bf Var} Y_i = \sigma^2$ for some $\sigma$. 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 
$$\log L(c,\sigma)= const. -n\log \sigma + \frac{1}{2\sigma^2}\sum_{i=1}^n(y_i-cx_i)^2.$$ (c)
Letting $\partial \log L/\partial c=0$, we get $\sum_{i=1}^n x_i(y_i - cx_i)=0$ so we get the same answer. (d) $\mathbb{E} \hat c = \sum_{i=1}^n x_i\mathbb{E} Y_i/\sum_{i=1}^n x_i^2$ =c. (e) $\hat c = \sum_{i=1}^n x_iY_i/\sum_{i=1}^n x_i^2$. ${\bf Var} \hat c = \sum_{i=1}^n x_i^2 {\bf Var} Y_i/[\sum_{i=1}^n x_i^2]^2 = \sigma^2/\sum_{i=1}^n x_i^2$. (f) The sample correlation is given by $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$, 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 $\bar z=\bar w =0$. So the sample correlation for $(z_i,w_i)$ would be $S_{zw}/\sqrt{S_{zz}S_{ww}}$ where $S_{zw}=\sum_{i=1}^n(z_i-\bar z)(w_i-\bar w)= \sum_{i=1}^n z_iw_i=S_{xy}$. Also $S_{zz}=S_{xx}$ and $S_{ww}=S_{yy}$. So the sample correlation is the same. 



\item When you throw a coin 10 times, you observed 3 heads. Test if the coin is biased at $\alpha=0.2$. Clearly state appropriate parameters, hypotheses, a test statistic. You may use the following R output (15pts).\\
\begin{verbatim}
> pbinom(0:5,10,0.5)
 [1] 0.00 0.01 0.06 0.17 0.38 0.62
\end{verbatim}

{\em Solution} Let $p$ be the probability of getting head.
Then $H_0: p=0.5$ vs. $H_1: p \neq 0.5$. Assign $X_i=1$ if $i$-th
tossing yields head and $X_i=0$ otherwise.  Note that $P(X_i=1)=p, 
P(X_i=0)=1-p$. The point
estimator for $p$ would be $\hat p=\sum_{i=1}^{10} X_i/10=\bar X$.
We will take $\sum_{i=1}^{10} X_i$ rather than $\bar X$ as the
test statistic because it will be much easier to figure out the
distribution of the test statistic. Note
that $\sum_{i=1}^{10} X_i \sim \mbox{Binomial}(10,p)$. Under $H_0$, $\sum_{i=1}^{10} X_i
\sim \mbox{Binomial}(10,0.5)$. We reject $H_0$ if
$\sum_{i=1}^{10} x_i$ is either too small or too large. 
Note that $P(\sum_{i=1}^{10} X_i \leq 3) = P(\sum_{i=1}^{10} X_i \geq 7) = 0.17$ from symmetry. So the $P$-value
is 0.34. We do not reject the null hypothesis at $\alpha=0.2$. 

\item A manufacturer of automatic washers offers a particular model in one of three colors: white, green and blue. Of the 100 washers sold, 40 were white, 35 green, 25 blue. In each of the subsequent questions, clearly state appropriate parameters, hypotheses, a test statistic and its distribution under the null assumption.


\begin{itemize}
\item[(a)] Would you conclude that customers have no preference? Test it at $\alpha=0.05$ (10 pts).

\item[(b)] Would you conlude that customers have a preference for white color? Test it at $\alpha=0.05$ (10 pts). 
\end{itemize}
{\em Solution.}  Let $p_w, p_r,p_b$ be the proportion of customers who prefer white, green and blue respectively (a) If there is no preference, we expect $H_0:p_w=p_r=p_b=1/3$. Under the null, the expected numbers of washers customers choose are $33.3$ for each color. The chi-square statistic value is $\chi^2  = (40-33.3)^2/33.3 + (35-33.3)^2/33.3 + (25-33.3)^2/33.3 = 3.50$. There are three categories. So we compare it with the cutoff value 4.60 from $\chi^2_2$.  Do not reject $H_0$. (b) If there is no preference for white, we expect $p_w=1/3=0.333$. So the hypotheses of interest would be $H_0:p_w=1/3 \mbox{ vs. } H_1:p_w > 1/3$. The test statistic is based on $z$-statistic of large sample proportion. $z$-value is $z=(0.4 - 0.333)/\sqrt{0.333 0.666/1000}=4.50$. Since the rejection region is $z > 1.64$, we reject $H_0$ and conclude that the customers prefer white color. 

\item A quality control engineer has measured the numbers of defectives per day from a certain production process for 50 days and recorded below. Test if the number of defectives follows a binomial distribution at $\alpha=0.05$. (20pts).

\begin{center}
\begin{tabular}{|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  number of defects & frequencies\\
  \hline
0 &10\\ \hline
1 &24\\ \hline
2 &10\\ \hline
3 &6\\ \hline
\end{tabular}\\
\end{center}
{\em Solution.} We test if data follow Binomial$(3,p)$. First we estimate $p$ by matching the sample mean and the population mean, which gives the maximum likelihood estimation of $p$. $\bar x = (0\cdot 10 + 1 \cdot 24 + 2 \cdot 10 + 3 \cdot 6)/50=1.24=3p$. $\hat p=0.41$. Let's see if data follow $P(X=x)={3 \choose x}\hat p^x(1- \hat p)^{3-x}$. The null hypothesis we need to test is $p_x=P(X=x)$, $H_0: p_0=(1-\hat p)^3=0.20,p_1=3\hat p(1-\hat p)^2=0.41,p_2=3\hat p^2(1-\hat p)=0.30,p_3=\hat p^3=0.07$. The expected numbers of defects are 10.1, 20.7, 15, 3.5. Then $\chi^2=(10-10.1)^2/10.1 + (24-20.7)^2/20.7 + (10-15)^2/15 + (6-3.5)^2/3.5=3.97.$ The $95\%$ cut-off value for $\chi^2_{4-1-1}$ is 5.99 so we do not reject $H_0$. 

%\item Aptitude tests should produce scores with a lrage amount of variations to distinguish between persons with a low aptitude and pwersons with high aptitude. The standard test used by a certain industry produce scores with a mean of 100 points and a standard devation of 10 points. A new test is developed and was given to 5 prospective employees and produced a mean of 110 points and a sample standard devation of 12 points. Are scores from the new test significant more variable than scores from the standard? Test your claim at $\alpha=0.05$ (10pts).\\

%{\em Solution.} The hypotheses of interest is whether $H_0: \sigma^2 =10$ vs. $H_1: \sigma^2 >10$. Note that

 
\item Problem on the coefficient of determination.
\begin{itemize}
\item[(a)] The following it the R output of a regression analysis based on linear model $Y=\beta_0 + \beta_1 x + \epsilon$ for 11 paired data $(x_i,y_i)$. Compute the coefficient of determination and the coefficient of correlation (5pts).

\item[(b)] It is shown during the lecture that the coefficient of determination is always between 0 and 1. Prove it (15pts).\\
{\em Solution.} (a) Note that $t$-value $=3=r\sqrt{11-2}/\sqrt{1-r^2}$. Solving this we get $r^2=0.5$. The correlation is then $r= \sqrt{0.5}$. (b) The coffined of determination is given by $r^2=1-SSE/SST$, where $SSE=\sum_{i=1}^n (y_i - \hat \beta_0 - \hat \beta_1x_i)^2$, while $SST= S_{yy}= \sum_{i=1}^n (y_i - \bar y)^2$. It can be shown that $SSE \leq SST$ (see lecture note) so $0 \leq r^2 \leq 1$. 



\end{itemize}


\end{enumerate}
\end{document}