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\centerline{\Large \bf HOMEWORK\#2} \vskip6pt
\centerline{\sc Due on 12PM (noon) May 29} \vskip24pt

Please email your homework (scanned handwritten solution or typed solution) to my email address with subject "HW 2 of NCKU course"
\begin{enumerate}
%\item KNNL Problem 2.53

\item A clay tile company is interested in studying the effect of cooling temperature on strength.
The company has five ovens which produce the tiles, four tiles were baked in each oven
and then randomly assigned to one of the four cooling temperatures. The data are
shown below.

\begin{table}[h]
\begin{center}
\begin{tabular}{c|ccccc|c}
Cooling & \multicolumn{5}{c}{Oven}\\ 
Temp    & 1& 2& 3 & 4 & 5 & mean \\ \hline
5$^{\circ}$   &     3 & 10 &  7    &   4   & 3      &5.4       \\
10$^{\circ}$  &     3 &  8 & 12    &    2  &  4     &5.8       \\
15$^{\circ}$  &     9 & 13 &  15   &     3 &   10   &10       \\
20$^{\circ}$  &     7 & 12 &   9   &      8&     13 &9.8 \\\hline   
Mean& 5.50 & 10.75 & 10.75& 4.25&7.50& 7.75\\\hline                      
\end{tabular}
\end{center}
\end{table}

\begin{enumerate}
 \item Which type of design was employed? Describe how the fundamental principles of
  experimental design were followed in this design.
 \item If $MS_E= 6.275$, compute the F-statistic to determine if there is a difference
among the four cooling temperatures (use $\alpha$= 5\%).
 \item Estimate the relative effiency, and interpret your result.
 \item Suppose the company believes there is a jump in the strength at 12.5$^{\circ}$
but otherwise cooling temperature has no effect, that is, 5$^{\circ}$  and 10$^{\circ}$
are not different, neither are 15$^{\circ}$ and 20$^{\circ}$ 
, but these two groups of temperatures have different effects. Find
a set of orthogonal contrasts that would allow you to test this.
 \item Test these contrasts.

\end{enumerate}





\item A $2^{5-2}$ design is defined by $\mathbf{D} = \mathbf{AC}, \ \mathbf{E} = \mathbf{BC}$.

\begin{itemize}
\item Find its defining words and resolution. 

% Its defining words are $\mathbf{ACD}$ and $\mathbf{BCE}$, and its defining contrast subgroup is $I = \mathbf{ACD} = \mathbf{BCE} = \mathbf{ABDE}$. The design is of resolution III.

\item In the course of the analysis of this experiment, it is thought that factor $\mathbf{E}$ and all  interactions involving $\mathbf{E}$ are negligible. In addition to estimating the four main effects, there are still three degrees of freedom left. What two-factor interactions can be estimated with these three degrees of freedom? Here an effect is considered to be estimatible, if it does aliased with any same-order or lower-order interaction effects.

% To determine which two-factor interactions can be estimated with the remaining three degrees of freedom, note that
%\[
%I = 134 = 235 = 1245
%\]
%implies that
%\[
%12 = 234 = 135 = 45,
%\]
%\[
%23 = 124 = 5 = 1345,
%\]
%\[
%24 = 123 = 345 = 15.
%\]
% As factor 5 is inert, we see from above that the two-factor interactions which can be estimated with these three degrees of freedom are 12, 23, and 24. Indeed, these are the only two-factor interactions not aliased with the four main effects of 1, 2, 3, and 4 that can be estimated with the remaining three degrees of freedom.}

\end{itemize}


\item Consider regression model $y_i\sim N(\beta_2+\beta_1 x_i, \sigma^2)$ for $i=1,...,n_1$ and $y_i\sim N(\beta_1+\beta_2 x_i, \sigma^2)$ for $i=n_1+1,...,n_1+n_2$, and all $y_i$'s are independent. Furthermore, we assume that $\sum_{i=1}^{n_1+n_2} x_i=0$.
\begin{itemize}
\item Express the model in a matrix form, i.e.,
rewritten the model as $Y=X\beta+\epsilon$, where $Y=(y_1,\dots,y_{n_1+n_2})'$, and $\beta$ is the  parameter vector.
\item Find out the OLS estimation $(b_1, b_2)$ for $\beta_1$ and $\beta_2$. (Please provide an answer which no longer involves matrix operation, otherwise only partial credits will be granted.)
\item What is the variance of $b_1+b_2$?
\end{itemize}


\item You are interested in determining the linear relationship between the
flower opening size (nearest tenth of a mm) and the incidence of Fusarium head
blight (\% of infected spikes). You hire an undergraduate to collect the data and
she comes back from the field with the following table based on a collection of
29 wheat plants.


\begin{center}
  \begin{tabular}[h]{|c|c|c|c|c|c|c|c|c|}\hline
   Opening (mm)& 1.2 & 1.4& 1.7 &1.8 &1.9& 2.0 &2.2 &2.4\\\hline
   Mean incidence (\%)  & 5.3& 8.6& 10.2 &12.5& 14.4 &17.1& 18.2 &22.3\\\hline
   n &3 &5&2&5&5&6&5&3\\\hline
  \end{tabular}
\end{center}
\begin{itemize}
\item The individual measurements used to generate this table were thrown away. Can
you still do the inference here? If so, what would you do? Will you get the same
slope, intercept, and variance estimates? Explain.
\end{itemize}

\end{enumerate}

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