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Name \underline{\hspace{60mm}} \\ 
$\,$ \\
Student Number \underline{\hspace{60mm}}
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\begin{center}   
{\Large \textbf{STA 442/2101 F 2012 Quiz 5}}\\
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\begin{enumerate}
    \item (5 points) For this question, you may use two facts about the multinomial without proof:
        \begin{itemize}
            \item The unrestricted MLE is the vector of sample proportions $\overline{\mathbf{Y}}$.
            \item The likelihood function may be written 
                  $L(\boldsymbol{\theta})=\theta_1^{n\overline{y}_1} \, \theta_2^{n\overline{y}_2} \cdots \theta_c^{n\overline{y}_c}$.
        \end{itemize}
Denoting the \emph{restricted} MLE (restricted by $H_0$) as 
$\widehat{\boldsymbol{\theta}} = (\widehat{\theta}_1, \ldots \widehat{\theta}_c)$,  show that the likelihood ratio test statistic 
\begin{displaymath}
    G^2 = 
-2 \log \left(   
           \frac{\max_{\theta \in \Theta_0} L(\theta)}
                {\max_{\theta \in \Theta}   L(\theta)  }
           \right) = 
2n \sum_{j=1}^c \overline{y}_j\log 
    \left(\frac{\overline{y}_j}  {\widehat{\theta}_j}\right)
\end{displaymath} 

\newpage
    \item In a preference test of two (not three) detergents, eighty consumers out of 120 preferred Brand $A$ over Brand $B$.
     \begin{enumerate}
        \item (2 points)  Calculate the likelihood ratio test statistic for the null hypothesis of equal preference. Show some work. The final answer is a number. \textbf{Circle the number}.

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        \item (3 points) The critical $\chi^2$ value at $\alpha=0.05$ is 3.841. In plain, \textbf{non-statistical language}, what do you conclude from this test? You cannot get marks for this part without a reasonable answer to (a).
    \end{enumerate}

  \end{enumerate}

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