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\begin{document}

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\date{May 16, 2003}

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

% 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

\begin{enumerate}

\item[1.] Let $X_1,\cdots,X_n$ be a random sample from Bernoulli distribution with parameter $p$.
\begin{itemize}
\item[(a)] Find the maximum likelihood estimator of $(1-p)^2$ (20pts).

\item[(c)] Is your estimator in (a) unbiased? If it is biased, find an unbiased estimator of $(1-p)^2$(20pts).
\end{itemize}
{\em Solution.} (a) First find the MLE for $p$. Note that the Bernoulli probability function is $P(X_i=x)=p^x(1-p)^{1-x}$ for $x=0,1$ (5pts). The likelihood function is then $L(x_1,\cdots,x_n; p)= p^{\sum_{i=1}^n x_i}(1-p)^{\sum_{i=1}^n 1-x_i$ (5pts). By solving $\partial \log L/\partial p=0$, we get $\hat p=\bar x$ (5pts). From the invariance principle, $\widehat{(1-p)^2} = (1-\bar x)^2$ (5pts). (b) $\mathbb{E}(1-\bar X)^2 = 1-2\mathbb{E}\bar X + \mathbb{E} {\bar X}^2$. Note that $\mathbb{E} {\bar X}^2 = {\bf Var} \bar X - (\mathb{E} \bar X)^2 = p(1-p)/n - p^2$. So $\mathbb{E}(1-\bar X)^2= 1 -2p + p(1-p)/n - p^2$. Hence it is biased (5pts). We only need to find an unbiased estimator of $p^2$. See Midterm I, where it is given by $\bar X - S^2$, where $S$ is the sample variance. So the unbiased estimator of $(1-p)^2$ is $1-2\bar X + \bar X - S^2= 1-\bar X - S^2$ (15pts).

\item[2.] Gross sales before and after a training program is given by
\begin{verbatim}
Salesperson :  1  2   3  4   5  6
Sales before: 90 83 105 97 110 78
Sales after : 97 80 110 93 123 84
\end{verbatim}
Determine if the training program is effective at $\alpha=0.1$. Carefully do your analysis stating all the relevant assumptions (30 pts).\\
{\em Solution.} This problem can be also solved using regression analysis. The points will be given in three categories. {\bf Model specification} (10 pts). We assume that the gross sale $X_i$ of $i$-th sales person before the training to follow normal, i.e. $X_i \sim N(\mu_b,\sigma_b^2)$. We also assume that the gross sale $Y_i$ of $i$-th sales person after the training to follow normal, i.e. $X_i \sim N(\mu_a,\sigma_a^2)$. {\bf Inference setup} (10 pts): the null hypothesis of interest would be $H_0: \mu_a = \mu_b$ vs. $H_1: \mu_a > \mu_b$. The appropriate test statistic would be the one sample $t$-statistic (this is not a two sample $t$-test since $X_i$ and $Y_i$ are paired and correlated). Note that the sales difference $W_i=X_i - Y_i \sim N(\mu_a-\mu_b,\sigma^2)$ for some $\sigma$ (figure out the relationship between $\sigma, \sigma_a$ and $\sigma_b$). So the $t$-statistic is $T= \bar W/(S/\sqrt{6}),$ where $S$ is the sample variance of $W$ and it is distributed as $T \sim t_5$. {\bf Actual computation} (10 pts). You need to compute $S$ by yourself and actual R output either in $\tt{pt}$ or $\tt{qt}$ will be given.

\item[3.] Consider a linear model $Y=\beta_0 + \beta_1 x + \epsilon$. There is a reason to believe that $\beta_1=0$. Following sample data are given: $(x_1,y_1), \cdots, (x_n,y_n)$. We assume $\epsilon$ to follow zero mean normal distribution.
 \begin{itemize}
\item[(a)] Estimate $\beta_0$ by minimizing the sum of the residuals. No point given for not showing the minimization steps (10pts).
\item[(b)] Determine if the estimator in (a) is an unbiased maximum likelihood estimator. Please derive everything ( 20pts).
\item[(c)] When $n=2$, show that the estimator in (a) is the minimum variance unbiased estimator (MVUE) among the estimators of the form $c_1Y_1+c_2Y_2$ (10pts).
\end{itemize}
{\em Solution.} This problem tests your total understanding of linear model and estimation. Since $\beta_1=0$, we are dealing with a little dumber version of linear model $Y=\beta_0 + \epsilon$. (a) The the sum of squared residuals $SSE =\sum_{i=1}^n (y_i-\beta_0)^2$ (3pts). Solving $\partial SSE/\partial \beta_0 = 0$, we get $\beta_0=\bar y$ (7pts). (b) The estimator is $\hat \beta_0=\bar  Y$ (3pts). Do not mix up little $y_i$ and big $Y_i$. It will cost you 2 points. Then it can be shown that $\mathbb{E} \hat \beta_0 = \beta_0$ (see Lecture 20 derivations) (7pts). We need to check if $\hat \beta_0$ is MLE. Note that $Y_i \sim N(\beta_0,\sigma^2)$ for some $\sigma$. The maximum likelihood estimator of $\beta_0$ is $\bar Y$ (see Lecture 4 normal distribution example). Hence $\hat \beta_0$ is the unbiased MLE (10pts). (c) To make $c_1Y_1 + c_2Y_2$ unbiased, we need $\mathbb{E}(c_1Y_1+c_2Y_2) = c_1\beta_0 + c_2\beta_0=\beta_0$. So $c_2=1-c_1$. Now minimize the variance ${\bf Var}(c_1Y_1 + c_2Y_2) = c_1^2{\bf Var}Y_1 + c_2^2{\bf Var} Y_2 = (c_1^2+(1-c_1)^2)\sigma^2$ for some $\sigma$. By differentiating it w.r.t. $c_1$, we get $c_1=1/2$. Hence, $\bar Y$ is MVUE when $n=2$.

\item[4.] Data for the average blood lead level ($\tt{blood}$) of children age 6 months to 5 years and the amount of lead used in gasoline production ($\tt{gasoline}$) for ten 6-months periods are available.\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...

  $\tt{blood}$ &48&  59  &79 & 80&  95&  95 & 97& 102& 102& 107\\
\hline
  $\tt{gasoline}$ & 9.3 &11.0 &12.8 &14.1& 13.6 &13.8 &14.6 &14.6 &16.0 &18.2\\

  \hline
\end{tabular}\\
R output for analyzing this data is given below.
\begin{verbatim}
>blood
[1]  48  59  79  80  95  95  97 102 102 107
>gasoline
[1]  9.3 11.0 12.8 14.1 13.6 13.8 14.6 14.6 16.0 18.2
>summary(lm(blood~gasoline))
Call:
lm(formula = blood ~ gasoline)
Residuals:
     Min       1Q   Median       3Q      Max
-11.5112  -6.6140  -0.2788   7.6404  10.0596
Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  -14.312     16.101  -0.889 0.399984
gasoline       7.298      1.150   6.344 0.000222 ***
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Residual standard error: 8.508 on 8 degrees of freedom
Multiple R-Squared: 0.8342,     Adjusted R-squared: 0.8135
F-statistic: 40.25 on 1 and 8 DF,  p-value: 0.0002220
\end{verbatim}
\begin{itemize}
\item[(a)] Is there any linear relationship between blood lead level and the amount of lead used in gasoline production? What is the statistical significance of your conclusion? (20 pts).
\item[(b)] What is the correlation coefficient between two variables? (10pts).
\end{itemize}
{\em Solution.} (a) We test if the intercept slop is zero, i.e. $H_0:\beta_1=0$ (see Lecture 22 Concept 2). The $P$-value is extremely small so we conclude that there is a linear relationship with $99\%$ significance. (b) The coefficient of determination $r^2=0.8442$. If you take the squre root, $r=\pm \sqrt{0.844}$. From the negative regrssion slope, we see that the correlation must be negative. So $\rho= -  \sqrt{0.844}$ is the correlation.

\item[5.] Suppose that 20 out of 50 females and 10 out of 40 males are
depressed in a sample. Determine whether the frequency of
depression is related to sex (30 pts). \\
{\em Solution.} This was an in-class example we studied. Let us formulate this problem in terms of categorical data.
{\bf Model specification} (10 pts). Let $p_{11}$ and $p_{21}$ be the respective proportions of depression in females and males. $p_{12}$ and $p_{22}$ are the respective proportions of non-depression in females and males. We are interested
 in testing $H_0: p_{11}=p_{21}, p_{12}=p_{22}$. 4 pts will be deducted if you do not clearly specify what these $p_{ij}$'s are. {\bf Inference setup} (10 pts). Under $H_0$, we compute the expected number of elements in each cell and compute the $\chi^2$ test statistic value. {\bf Actual computation} (10pts). Then the cutoff value $\chi^2_{\alpha,1}$ for $\alpha$-level test can
be computed from $N(0,1)$. Note that $\alpha=P(\chi^2_1 > \chi^2_{\alpha,1}) = P(Z^2>
\chi^2_{\alpha,1})=2P(Z > \chi_{\alpha,1}).$ So
$\chi^2_{\alpha,1}=Z^2_{\alpha/2}$. For $\alpha=0.05$,
$\chi^2_{\alpha,1}=1.96^2$. For this problem, $\tt{pnorm}$ or $\tt{qnorm}$ values will be given.

\end{enumerate}

\end{document}