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\subsection*{pg. 110 \#5)}

According to Dr. Horvath, what they mean by $\mathbf{Y} = \boldsymbol{\theta} + \boldsymbol{\varepsilon}$ is the following:
\[
	\begin{array}{l}
	Y_1 = \theta_1 + \varepsilon_1\\
	Y_2 = \theta_2 + \varepsilon_2\\
	Y_3 = \theta_3 + \varepsilon_3\\
	Y_4 = \theta_4 + \varepsilon_4
	\end{array}
\]
But with the restriction $\theta_1+\theta_2+\theta_3+\theta_4=0$ we can write it as:
\[
	\begin{array}{l}
	Y_1 = \theta_1 + \varepsilon_1\\
	Y_2 = \theta_2 + \varepsilon_2\\
	Y_3 = \theta_3 + \varepsilon_3\\
	Y_4 = -(\theta_1+\theta_2+\theta_3) + \varepsilon_4
	\end{array}
\]
So we can write the model as $\mathbf{Y} = \mathbf{X} \boldsymbol{\theta} + \boldsymbol{\varepsilon}$, where:
\[
\mathbf{X}= \left(
	\begin{array}{ccc}
	1&0&0 \\
	0&1&0 \\
	0&0&1 \\
	-1&-1&-1
	\end{array}
\right)
\qquad \mathrm{ and } \qquad
\boldsymbol{\theta}= \left(
	\begin{array}{c}
	\theta_1\\
	\theta_2\\
	\theta_3\\
	\end{array}
\right)
\]
Now with $\mathbf{A}=(1,0,-1)$ you can find the $F$ statistic for the test and you'll get the answer in the book.  I didn't use the Lagrange multiplier hint (or at least I didn't realize I was using it if I did), but doing it this way requires inverting a 3x3 matrix and I don't know if he'll ask one that requires that.

\subsection*{pg. 118 \#4)}

For this one, to set it up like a linear model, you can look at the model:
\[
	\begin{array}{l}
	Y_1 = \mu + \varepsilon_1\\
	Y_2 = \mu + \varepsilon_2\\
	\qquad \vdots \\
	Y_n = \mu + \varepsilon_3\\
	Y_{n+1} = \lambda + \varepsilon_{n+1}
	\end{array}
\]
Which we can write as $\mathbf{Y} = \mathbf{X} \boldsymbol{\beta} + \boldsymbol{\varepsilon}$, where:
\[
\mathbf{X}= \left(
	\begin{array}{ccc}
	1&0 \\
	1&0 \\
	\vdots & \vdots \\
	1&0 \\
	0&1
	\end{array}
\right)
\qquad \mathrm{ and } \qquad
\boldsymbol{\beta}= \left(
	\begin{array}{c}
	\mu \\
	\lambda
	\end{array}
\right)
\]
So we can test $H: \mu = \lambda$ using $\mathbf{A} = (1,-1) $.  We get:
\[
(\mathbf{X}^T\mathbf{X})^{-1} = 
	\left(
	\begin{array}{cc}
	1/n&0 \\
	 0&1
	\end{array}
	\right)
\quad \mathrm{ so } \quad
\mathbf{\hat{\beta}} = 
	\left(
	\begin{array}{c}
	\bar{Y}_n \\
	Y_{n+1}
	\end{array}
	\right)
\quad \mathrm{ and } \quad
A(\mathbf{X}^T\mathbf{X})^{-1}A^T= \frac{1}{n}+1
\]
And
\[
RSS/(n-1) = \frac{1}{n-1} 
\left[  
	\left( 
		\begin{array}{c}
		 Y_1 \\
		\vdots \\
		Y_n \\
		 Y_{n+1}
		\end{array}	
	\right) - 
	\left(
		\begin{array}{c}
		\bar{Y}_1 \\
		\vdots \\
		\bar{Y}_n \\
		 Y_{n+1}
		\end{array}
	\right)  
\right]^T
\left[  
	\left( 
		\begin{array}{c}
		 Y_1 \\
		\vdots \\
		Y_n \\
		 Y_{n+1}
		\end{array}	
	\right) - 
	\left(
		\begin{array}{c}
		\bar{Y}_1 \\
		\vdots \\
		\bar{Y}_n \\
		 Y_{n+1}
		\end{array}
	\right)
\right] = S_n^2
\]
So I get:
\[
F=\frac{(\bar{Y}_n - Y_{n+1})^2}{S_n^2 \left(1 + \frac{1}{n} \right)}
\]

Now if you take the square root, it kind of looks like the answer in the back of the book.  If you can get it into a t-distribution, let me know.  It seems like it wouldn't be so bad, but I'm a little burned out.

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