这是一份nottingham诺丁汉大学MATH4068作业代写的成功案例

$$
\begin{aligned}
\Lambda &=\frac{\left|\mathbf{A E A}^{\prime}\right|}{\left|\mathbf{A}(\mathbf{E}+\mathbf{H}) \mathbf{A}^{\prime}\right|}=\frac{3.058 \times 10^{8}}{3.092 \times 10^{8}} \
&=.9889>\Lambda_{.05,2,1,18}=.703 .
\end{aligned}
$$
For the $B C$ interaction, we have
$$
\begin{aligned}
\Lambda &=\frac{\left|\mathbf{B E B}^{\prime}\right|}{\left|\mathbf{B}(\mathbf{E}+\mathbf{H}) \mathbf{B}^{\prime}\right|}=\frac{4.053 \times 10^{6}}{4.170 \times 10^{6}} \
&=.9718>\Lambda_{.05,2,1,18}=.703 .
\end{aligned}
$$
For $A B C$, we obtain
$$
\begin{aligned}
\Lambda &=\frac{\mid \text { GEG }^{\prime} \mid}{\left|\mathbf{G}(\mathbf{E}+\mathbf{H}) \mathbf{G}^{\prime}\right|}=\frac{2.643 \times 10^{12}}{2.927 \times 10^{12}} \
&=.9029>\Lambda_{.05,4,1,18}=.551 .
\end{aligned}
$$
In summary, factors $A, B$, and $C$ and the $A B$ interaction are significant.

MATH4068 COURSE NOTES ：

Factor A
$$
T^{2}=N\left(\mathbf{A} \overline{\mathbf{y}}{\ldots}\right)^{\prime}\left(\mathbf{A}{\mathrm{pl}^{\mathrm{A}}} \mathbf{A}^{\prime}\right)^{-1}\left(\mathbf{A} \mathbf{y}{\ldots}\right) $$ is distributed as $T{a-1, v_{E}}^{2}$.
Factor B
$$
T^{2}=N\left(\mathbf{B}{\ldots}\right)^{\prime}\left(\mathbf{B S}{\mathrm{pl}} \mathbf{B}^{\prime}\right)^{-1}\left(\mathbf{B}{\mathbf{y}}^{\ldots}\right) $$ is distributed as $T{b-1, v_{E}}^{2}$.
AB Interaction
$$
T^{2}=N\left(\mathbf{G}{\ldots}\right)^{\prime}\left(\mathbf{G S}{\rho l} \mathbf{G}^{\prime}\right)^{-1}\left(\mathbf{G}{\ldots}\right) $$ is distributed as $T{(a-1)(b-1), v E^{*}}^{2}$.