这是一份PSU宾夕法尼亚州立大学STAT 502作业代写的成功案
The estimates use the $c_{i k}$ and $C_{k}$ values in Table 10.4:
$$
\begin{aligned}
&\hat{a}{1}^{}=75 /\left[(-5)^{2}+(-3)^{2}+\cdots+(5)^{2}\right]=75 / 70=1.071 \ &\hat{a}{3}^{}=-31 /\left[(-5)^{2}+(7)^{2}+\cdots+(5)^{2}\right]=-31 / 180=-.172
\end{aligned}
$$
The estimate of $\mu$ (from Table $10.1$ ) is $\bar{X}{. .}=14.3$, and the estimation formula is $$ \hat{\mu}{i}=14.3+1.071 c_{i 1}-.172 c_{i 3} .
$$
From this equation, we estimate $\mu_{1}$ by inserting $c_{11}=-5$ and $c_{13}=-5$ to obtain
$$
\hat{\mu}_{1}=14.3+1.071(-5)-.172(-5)=9.8
$$
STAT 502 COURSE NOTES ：
$$
S S_{a \text { linear }(b)}=3,698.0+6,160.5+5,304.5-15,000=163.0 \text {. }
$$
We divide this by its degrees of freedom to obtain
$$
M S_{a \text { linear }(b)}=163 / 2=81.5 .
$$
Finally, we divide this by $M S_{w}$ to obtain the $F$ ratio
$$
F_{(2,108)}=81.5 / 40=2.04
$$