\documentclass[12pt]{article} 
%\usepackage{amsbsy} % for \boldsymbol and \pmb 
\usepackage{graphicx} % To include pdf files!
\usepackage{amsmath}
\usepackage{amsbsy}
\usepackage{amsfonts}
%\usepackage[colorlinks=true, pdfstartview=FitV, linkcolor=blue, citecolor=blue, urlcolor=blue]{hyperref} % For links
\usepackage{fullpage}           % Good for US Letter paper

\topmargin=-0.75in
\textheight=9.5in
\usepackage{fancyhdr} 
\renewcommand{\headrulewidth}{0pt}  % Otherwise there's a rule under the header
\setlength{\headheight}{15.2pt}
\fancyhf{}
\pagestyle{fancy} 

\cfoot{Page \thepage {} of 3}
% \pagestyle{empty} % No page numbers


\begin{document}
\enlargethispage*{1000 pt} 

\begin{flushright}
Name \underline{\hspace{60mm}} \\ 
$\,$ \\
Student Number \underline{\hspace{60mm}}
\end{flushright}   
\vspace{2mm}

\begin{center}   
{\Large \textbf{STA 431 Quiz 6}}\\
\vspace{1 mm}
\end{center}

\noindent
In homework, you examined the reliability of a long test with equivalent items, meaning that the variances of the error terms were all the same. The following model is more realistic. Independently for $i = 1, \ldots, n$, let 
\begin{equation*}
    W_i = \nu_i + \lambda_i X + e_i,
\end{equation*}
where $E(X)=\mu$, $Var(X)=\phi$, $E(e_i)=0$, and $Var(e_i)= \omega_i$. The $e_i$ are independent of one another, and they are independent of $X$. See the footnote\footnote{In this model, $X$ is the true latent quantity being measured by a test or exam. The test has $n$ questions, or ``items." The observable variables $W_1, \ldots, W_n$ are the number of points the student gets on each item. These numbers are not just the truth plus a piece of random noise. Each item has its own slope and intercept, and also the error terms have different variances. This is reasonable. Some test questions are worth more marks (that's $\nu_i$), and some are better than others.} for the intuition behind the model, if you are interested.

The parameters $\nu_i$, $\lambda_i$ and $\omega_i$ are fixed constants, and we need to make them reasonably well-behaved. Assume that as $n \rightarrow \infty$, 

\begin{center}
\begin{tabular}{ccc}
${\displaystyle \overline{\lambda}_n = \frac{1}{n}\sum_{i=1}^n\lambda_i \rightarrow \lambda \neq 0}$ & 
 and  &
${\displaystyle \overline{\omega}_n = \frac{1}{n}\sum_{i=1}^n\omega_i \rightarrow \omega}$
\end{tabular}
\end{center}
These are ordinary limits.

\vspace{3mm}

\begin{enumerate} 

\item (1 point) Score on the entire test will be the average $\overline{W}_n = \frac{1}{n} \sum_{i=1}^n W_i$. To make Question~\ref{bigee} easier, simplify $\overline{W}_n$. Use the formula for $W_i$ above.  

\vspace{50mm}

\item (1 point) Calculate the variance of $\overline{e}_n = \frac{1}{n}\sum_{i=1}^n e_i$.

\vspace{50mm}

\newpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\item (6 points) \label{bigee} Calculate $\rho_n^2$, the reliability of $\overline{W}_n$. Show your work and \emph{simplify}. Simplification matters this time. \textbf{Circle your final answer.}  


\newpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

~ \vspace{100mm}


\item (2 points) Again denoting the reliability of $\overline{W}_n$ by $\rho_n^2$, calculate $\displaystyle \lim_{n \rightarrow \infty} \rho_n^2$. Show some of the steps.



\end{enumerate} % End of quiz questions


\end{document}