= Solution
A <regression factor> is a <categorical predictor> represented by level indicators and contrasts. Treating year as a <regression factor> permits arbitrary differences between the three yearly means, reflecting conditions such as weather, rather than imposing a linear time trend.
Let $Y_{ijk}$ be the yield for seed variety $i\in\{1,2\}$, fertiliser level $j\in\{L,H\}$, and year $k\in\{2004,2005,2006\}$. A treatment-coded <two-factor normal linear model> with additive year effects is
$$
Y_{ijk}=\beta_0+a\mathbf1_{\{i=2\}}+b\mathbf1_{\{j=H\}}+c_k+d\mathbf1_{\{i=2,j=H\}}+\varepsilon_{ijk},\qquad c_{2004}=0,
$$
with independent $N(0,\sigma^2)$ errors. The unrestricted coefficients are $\beta_0,a,b,c_{2005},c_{2006},d$: six parameters for twelve observations. Equivalently use baseline-zero variety, fertiliser, and interaction arrays. There are no year-treatment interactions in this fit.
Dividing each sum of squares by its <statistical degrees of freedom> gives the completed <analysis of variance>:
$$
\begin{array}{l|rrrrr}
\text{source}&\mathrm{df}&\mathrm{SS}&\mathrm{MS}&F&p\\\hline
\text{seed}&1&5.3868&5.3868&11.3748&0.014993\\
\text{fertiliser}&1&0.0768&0.0768&0.1622&0.701122\\
\text{year}&2&6.2883&3.14415&6.6392&0.030146\\
\text{seed:fertiliser}&1&16.0083&16.0083&33.803&0.001137\\
\text{residual}&6&2.8414&0.4735667&&
\end{array}
$$
In particular, the missing interaction statistic is $16.0083/(2.8414/6)\simeq33.803$. Its <null hypothesis> is $H_0:d=0$, against $H_1:d\ne0$: the change in mean yield from high fertiliser is the same for both varieties. Under the <null hypothesis> and the independent homoscedastic <normal linear model>, this <nested-model F-test> has distribution $F_{1,6}$. The printed $p$-value 0.001137 strongly rejects no interaction.
\b[Do not drop fertiliser just because its main-effect ANOVA row has $p=0.7011$.] In this balanced factorial design that row measures a fertiliser effect averaged over varieties; large opposite effects can cancel. The strongly supported interaction requires retaining its associated main effects by <regression model hierarchy>. Seed and year also have evidence of effects in their ANOVA rows, so this table alone gives no compelling simplification of the fitted terms.
The <interaction contrast> is $-4.6200$, with <standard error> 0.7946. The fitted high-minus-low fertiliser effect is $+2.4700$ for variety 1 but $2.4700-4.6200=-2.1500$ for variety 2. The fitted variety-2-minus-variety-1 contrast is $+3.6500$ under low fertiliser and $3.6500-4.6200=-0.9700$ under high fertiliser. Thus \b[variety 2 performs best under low fertiliser, while high fertiliser benefits variety 1 and reduces variety 2's fitted yield.] The marginal fertiliser effect is only $(2.4700-2.1500)/2=0.1600$.
The four fitted means in the baseline year are $5.5050,7.9750,9.1550,7.0050$ for $(1,L),(1,H),(2,L),(2,H)$ respectively. Add $1.5775$ in 2005 and $0.0875$ in 2006 to every fitted mean. The 2005 coefficient has $p=0.017646$ under a $t_6$ test of a zero contrast with 2004; the analogous 2006 contrast has $p=0.863216$. The latter is lack of evidence for a difference, not proof that the yearly means coincide. <Standard errors> for the combined contrasts require the coefficient <covariance matrix>. The experiment is small: <residual standard error> is 0.6882 on only six <statistical degrees of freedom>. If the same fields were reused, possible within-field dependence and field allocation would need checking before treating the fitted error independence as established.
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