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{\Large \textbf{STA 2101 Assignment 1 (Review)}}\footnote{This assignment was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistics, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/~brunner/oldclass/2101f19} {\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/2101f19}}}
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\noindent
The questions on this assignment are not to be handed in. They are practice for Quiz One on Friday September 20th. Please bring the R printout for Question~\ref{power} to the quiz, showing all input and output. Remember, \emph{this is not a group project}. You are expected to do the work yourself.

\begin{enumerate} 

\item In a political poll, a random sample of $n$ registered voters are to indicate which of two candidates they prefer. State a reasonable model for these data, in which the population proportion of registered voters favouring Candidate $A$ is denoted by $\theta$. Denote the observations $Y_1, \ldots, Y_n$.

\item  A medical researcher conducts a study using twenty-seven litters of cancer-prone mice. Two members are randomly selected from each litter, and all mice are subjected to daily doses of cigarette smoke. For each pair of mice, one is randomly assigned to Drug A and one to Drug B. Time (in weeks) until the first clinical sign of cancer is recorded. 
    \begin{enumerate}
        \item State a reasonable model for these data. Remember, a statistical model is a set of assertions that partly specify the probability distribution of the observable data. For simplicity, you may assume that the study continues until all the mice get cancer, and that log time until cancer has a normal distribution.
        \item What is the parameter space for your model?
    \end{enumerate}

\item  Suppose that volunteer patients undergoing elective surgery at a large hospital are randomly assigned to one of three different pain killing drugs, and one week after surgery they rate the amount of pain they have experienced on a scale from zero (no pain) to 100 (extreme pain). 
    \begin{enumerate}
        \item State a reasonable model for these data. For simplicity, you may assume normality.
        \item What is the parameter space? 
    \end{enumerate}

       \item Let $X_1, \ldots, X_n$ be a random sample (meaning independent and identically distributed) from a distribution with density $f(x) = \frac{\theta}{x^{\theta+1}}$ for $x>1$, where $\theta>0$. 
    \begin{enumerate}
        \item Find the maximum likelihood estimator of $\theta$. Show your work. The answer is a formula involving $X_1, \ldots, X_n$.
        \item Suppose you observe these data: \texttt{1.37, 2.89, 1.52, 1.77, 1.04, 2.71, 1.19, 1.13, 15.66, 1.43}. Calculate the maximum likelihood estimate. My answer is 1.469102.
    \end{enumerate} % See 2014 for good quiz questions, cut out of the 2016 assignment.

% \pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\item  Label each statement below True or False. Write ``T" or ``F" beside each statement. Assume the $\alpha=0.05$ significance level. 
    \begin{enumerate}
        \item \underline{\hspace{10mm}} The $p$-value is the probability that the null hypothesis is true. % F
        \item \underline{\hspace{10mm}} The $p$-value is the probability that the null hypothesis is false. % F
        \item \underline{\hspace{10mm}} In a study comparing a new drug to the current standard treatment, the null hypothesis is rejected. We conclude that the new drug is ineffective. % F
        \item \underline{\hspace{10mm}} If $p > .05$ we reject the null hypothesis at the .05 level. % F
        \item \underline{\hspace{10mm}} If $p < .05$ we reject the null hypothesis at the .05 level. % T
        \item  \underline{\hspace{10mm}} The greater the $p$-value, the stronger the evidence against the null hypothesis. % F
        \item \underline{\hspace{10mm}} In a study comparing a new drug to the current standard treatment, $p > .05$. We conclude that the new drug and the existing treatment are not equally effective. %F
        \item \underline{\hspace{10mm}} The 95\% confidence interval for $\beta_3$ is from $-0.26$ to $3.12$. This means $P\{-0.26 < \beta_3 < 3.12\} = 0.95$.  % F
    \end{enumerate}

        \item \label{t} Let $Y_1, \ldots, Y_n$ be a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$, so that $T = \frac{\sqrt{n}(\overline{Y}-\mu)}{S} \sim t(n-1)$. This is something you don't need to prove, for now. 
    \begin{enumerate}
        \item Derive a $(1-\alpha)100\%$ confidence interval for $\mu$. ``Derive" means show all the high school algebra. Use the symbol $t_{\alpha/2}$ for the number satisfying $Pr(T>t_{\alpha/2})= \alpha/2$.
        \item \label{ci} A random sample with $n=23$ yields $\overline{Y} = 2.57$ and a sample variance of $S^2=5.85$. Using the critical value $t_{0.025}=2.07$, give a 95\% confidence interval for $\mu$. The answer is a pair of numbers.
                
        \item Test $H_0: \mu=3$ at $\alpha=0.05$. 
            \begin{enumerate}
                \item Give the value of the $T$ statistic. The answer is a number.
                \item State whether you reject $H_0$, Yes or No.
                \item Can you conclude that $\mu$ is different from 3? Answer Yes or No.
                \item If the answer is Yes, state whether $\mu>3$ or $\mu<3$. Pick one.
            \end{enumerate}
        \item Show that using a $t$-test, $H_0:\mu=\mu_0$ is rejected at significance level $\alpha$ if and only the $(1-\alpha)100\%$ confidence interval for $\mu$ does not include $\mu_0$. The problem is easier if you start by writing the set of $T$ values for which $H_0$ is \emph{not} rejected.  
        \item In Question~\ref{ci}, does this mean $Pr\{1.53<\mu<3.61\}=0.95$? Answer Yes or No and briefly explain.
    \end{enumerate}

\pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

    \item Let $Y_1, \ldots, Y_n$ be a random sample from a distribution with mean $\mu$ and standard deviation $\sigma$. 
    \begin{enumerate}
        \item Show that the sample variance $S^2=\frac{\sum_{i=1}^n(Y_i-\overline{Y})^2}{n-1}$ is an unbiased estimator of $\sigma^2$.
        \item Denote the sample standard deviation by $S = \sqrt{S^2}$. Assume that the data come from a continuous distribution, so that $Var(S) > 0$. Using this fact, show that $S$ is a \emph{biased} estimator of $\sigma$.
    \end{enumerate}

\item \label{power} Let $Y_1, \ldots, Y_n$ as a random sample from a Bernoulli distribution with parameter $\theta$. That is, independently for $i=1, \ldots, n$, $P(y_i|\theta) = \theta^{y_i} (1-\theta)^{1-y_i}$ for $y_i=0$ or $y_i=1$, and zero otherwise. A large-sample test of $H_0: \theta = \theta_0$ uses the test statistic
\begin{displaymath}
    Z_1 = \frac{\sqrt{n}(\overline{Y}-\theta_0)}{\sqrt{\theta_0(1-\theta_0)}}
\end{displaymath}
When the null hypothesis is true, the distribution of $Z_1$ is approximately standard normal. What if the null hypothesis is false? In the following, you may use the following fact without proof: the Central Limit Theorem implies that  $\widehat{\theta}=\overline{Y}$ is approximately normal with mean $\theta$ and variance $\frac{\theta(1-\theta)}{n}$. 
    \begin{enumerate}
        \item Derive the approximate large-sample cumulative distribution function of $Z_1$ when $\theta$ is not necessarily equal to $\theta_0$. Derive means show the high school algebra. Use $\Phi(\cdot)$  to denote the cumulative distribution function of a standard normal. 
        \item Give an expression for the power of a two-sided, size $\alpha$ test of $H_0: \theta = \theta_0$. Let $z_{\alpha/2}$ denote the point with $\alpha/2$ of the standard normal above it. For example, $z_{0.025} = 1.96$. 
        \item Suppose we are testing $H_0: \theta = \frac{1}{2}$ at $\alpha = 0.05$, with $n=100$. If the true value of $\theta$ is 0.55, what is the power of the test? Your answer is a number between zero and one. I used R's \texttt{pnorm} function to calculate standard normal probabilities. 
        \item Suppose again that the true value of $\theta$ is 0.55. What is the smallest value of the sample size $n$ such that the power of the test will be at least 0.80? The answer is an integer.
    \end{enumerate}
Here are a few comments on this question. The use of \texttt{pnorm} tells you that you are \emph{not} estimating power by simulation. I used a \texttt{while} loop; look it up if you need to. My answer is greater than 700, but less than 800.

\textbf{Bring your R printout for this question to the quiz.} The printout must show all input an output.

\pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\item In the \emph{centered} linear regression model, sample means are subtracted from the explanatory variables, so that values above average are positive and values below average are negative. Here is a version with one explanatory variable. For $i=1, \ldots, n$, let $y_i = \beta_0 + \beta_1(x_i-\overline{x}) + \epsilon_i$, where
    \begin{itemize}
        \item[] $\beta_0$ and $\beta_1$ are unknown constants (parameters).
        \item[] $x_i$ are known, observed constants.
        \item[] $\epsilon_1, \ldots, \epsilon_n$ are unobservable random variables with $E(\epsilon_i)=0$, $Var(\epsilon_i)=\sigma^2$ and $Cov(\epsilon_i,\epsilon_j)=0$ for $i \neq j$. 
        \item[] $\sigma^2$ is an unknown constant (parameter).
        \item[] $y_1, \ldots, y_n$ are observable random variables.
    \end{itemize}
    \begin{enumerate}
        \item What is $E(y_i)$? $Var(y_i)$?
        \item Prove that $Cov(y_i,y_j)=0$. Use the following definition: \\
        $Cov(U,V) = E\{(U-E(U))(V-E(V)) \}$. 
        \item If $\epsilon_i$ and $\epsilon_j$ are independent (not just uncorrelated), then so are $y_i$ and $y_j$, because functions of independent random variables are independent. Proving this in full generality requires advanced definitions, but in this case the functions are so simple that we can get away with an elementary definition. Let $X_1$ and $X_2$ be independent random variables, meaning $P\{X_1 \leq x_1, X_2 \leq x_2 \} = P\{X_1 \leq x_1\}  P\{X_2 \leq x_2\}$ for all real $x_1$ and $x_2$. Let $Y_1=X_1+a$ and $Y_2=X_2+b$, where $a$ and $b$ are constants. Prove that $Y_1$ and $Y_2$ are independent.  
        \item In \emph{least squares estimation}, we observe random variables $y_1, \ldots, y_n$ whose distributions depend on a parameter $\theta$, which could be a vector. To estimate $\theta$, write the expected value of $y_i$ as a function of $\theta$, say $E_\theta(y_i)$, and then estimate $\theta$ by the value that gets the observed data values as close as possible to their expected values. To do this, minimize
\begin{displaymath}
Q =  \sum_{i=1}^n\left(y_i-E_\theta(y_i)\right)^2  .
\end{displaymath}
The value of $\theta$ that makes $Q$ as small as possible is the least squares estimate.

Using this framework, find the least squares estimates of $\beta_0$ and $\beta_1$ for the centered regression model. The answer is a pair of formulas. Show your work. 
        \item Because of the centering, it is possible to verify that the solution actually \emph{minimizes} the sum of squares $Q$, using only single-variable second derivative tests. Do this part too.
        \item How about a least squares estimate of $\sigma^2$?
        \item You know that the least squares estimators $\widehat{\beta}_0$ and $\widehat{\beta}_1$ must be unbiased, but show it by calculating their expected values for this particular case. 
        \item Calculate $\widehat{\beta}_0$ and $\widehat{\beta}_1$ for the following data. Your answer is a pair of numbers.
        
%        \begin{center}
~~~~~   \begin{tabular}{c|ccccc}
        $x$  &  8  &  7  &  7  &  9  &  4 \\ \hline
        $y$  &  9  & 13  &  9  &  8  &  6
        \end{tabular}
%        \end{center}
~~~~~ I get $\widehat{\beta}_1 = \frac{1}{2}$.

% \pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

        \item Going back to the general setting (not just the numerical example with $n=5$), suppose the $\epsilon_i$ are normally distributed.
            \begin{enumerate}
                \item What is the distribution of $y_i$?
                \item Write the log likelihood function. 
                \item Obtain the maximum likelihood estimates of $\beta_0$ and $\beta_1$; don't bother with $\sigma^2$. The answer is a pair of formulas. \emph{Don't do more work than you have to!} As soon as you realize that you have already solved this problem, stop and write down the answer. 
            \end{enumerate}
        \item Still for this centered model with a single explanatory variable, suppose we centered the $y_i$ values too. In this case what is the least squares estimate of $\beta_0$? Show your work. 
    \end{enumerate}

    \item Consider the centered \emph{multiple} regression model 
\begin{displaymath}    
    y_i = \beta_0 + \beta_1 (x_{i,1}-\overline{x}_1) + \cdots + \beta_{p-1} (x_{i,p-1}-\overline{x}_{p-1}) + \epsilon_i
\end{displaymath}    
with the usual details.
    \begin{enumerate}
        \item What is $E_{\boldsymbol{\beta}}(y_i)$?
        \item What is the least squares estimate of $\beta_0$? Show your work.
        \item For an ordinary uncentered regression model, what is the height of the least squares plane at the point where all $x$ variables are equal to their sample mean values?
    \end{enumerate}

\item Suppose that volunteer patients undergoing elective surgery at a large hospital are randomly assigned to one of three different pain killing drugs, and one week after surgery they rate the amount of pain they have experienced on a scale from zero (no pain) to 100 (extreme pain). Write a multiple regression model for these data; specify how the explanatory variables are defined.

% Cutting out the linear algebra part to make the first assignment more civilized. 
% Put it on Assignment 2.

\end{enumerate}


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# R for regression data
x = c(8, 7, 7, 9, 4); dx = x-7
y = c(9, 13,  9,  8,  6); dy = y-9
cbind(dx,dy)
lm(y~dx); sum(dx*dy)/sum(dx^2)
# beta0hat = 9, beta1hat = 0.5

# R for two-sample t-test
x = c(6,10,8,12); mean(x)
y = c(7, 4, 8, 7, 9); mean(y)
t.test(x,y,var.equal=T)

	Two Sample t-test

data:  x and y
t = 1.3528, df = 7, p-value = 0.2182
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1.495899  5.495899
sample estimates:
mean of x mean of y 
        9         7 

df = 1:10
critvalue = qt(0.975,df)
round(rbind(df,critvalue),3) # Round the whole thing to 3 digits

s2p = ((4-1)*var(x)+(5-1)*var(y))/7; s2p # 4.857143
se = sqrt(s2p*(1/4+1/5)); se # 1.478416
low = 2 - 2.365*se; low # -1.496454
high = 2 + 2.365*se; high # 5.496454

tstat = 2/se; tstat # 1.352799