= Solution
This is an <analysis of covariance>: achievement is the response, aptitude is a quantitative <covariate>, and teaching method is a categorical explanatory variable. The usual <normal linear model> assumes independent errors with common <variance> $\sigma^2$.
`read.table` reads the data with a header row; `attach` makes its columns directly accessible, and the two printed vectors confirm their ordering. `rep(1:3, each=7)` supplies the three group labels, and `factor` makes them categorical. The contrast option selects <treatment coding> for unordered factors, with method A as the first-level reference. The <polynomial> option would apply to ordered factors and has no role here.
The formula using `aptitude * Group` includes both main effects and their <interaction>. Writing $I_B,I_C$ for the method indicators, its fitted mean has the form
$$
E(Y_i)=\beta_0+\beta_1x_i+\tau_BI_{B,i}+\tau_CI_{C,i}+\gamma_Bx_iI_{B,i}+\gamma_Cx_iI_{C,i}.
$$
Thus A, B and C have separate intercepts and slopes. There are six coefficients and $21-6=15$ residual <statistical degrees of freedom>. The <analysis of variance> table uses <sequential sums of squares>: adding aptitude to an intercept-only model removes $36.57548$ from the <residual sum of squares>; subsequently adding the two group indicators removes $16.93200$; adding the two slope <interactions> removes only $0.66714$. The remaining sum of squares is $9.63490$, giving mean square $9.63490/15=0.64233$.
Each displayed <F-test> divides the added term's sum of squares by its <statistical degrees of freedom> and then by this full-model <residual mean square>. In particular the <interaction> test is
$$
F=\frac{0.66714/2}{9.63490/15}=0.51932,\qquad p=0.60524.
$$
There is no evidence that the slopes differ, so the common-slope model is preferable among the fitted models. The small group $p$-value concerns differences after including aptitude, whereas the first aptitude row is sequential and does not adjust for Group. It should not be interpreted as the <covariate> test after treatment adjustment.
The additive formula gives the parallel-lines model
$$
E(Y_i)=\beta_0+\beta_1x_i+\tau_BI_{B,i}+\tau_CI_{C,i}.
$$
It has four coefficients and $17$ residual <statistical degrees of freedom>. Removing the <interaction> raises the <residual sum of squares> to $9.63490+0.66714=10.30204$, but lowers the <residual mean square> to $10.30204/17=0.60600$. The sequential aptitude and group sums of squares are unchanged. The adjusted group test is now $F=(16.93200/2)/0.60600=13.97024$ on $(2,17)$ <statistical degrees of freedom>, with $p=0.0002579$: after allowing for initial aptitude, the methods are not all equally effective.
`summary(..., cor=F)` displays the fitted coefficients, their <standard errors>, the individual $t$ statistics, and two-sided $p$-values; it omits the coefficient-correlation <matrix>. The intercept $2.8367$ is the fitted A mean at aptitude zero, which is outside the observed range and is mainly a parametrization device. The common slope $0.7429$ predicts about $0.74$ extra achievement units per aptitude unit, holding method fixed. Its adjusted $t$ test is $5.2267$ on $17$ <statistical degrees of freedom>, with $p$ about $0.0001$. The method coefficients are adjusted differences from A, not the B and C intercepts themselves. The three fitted lines are
$$
\boxed{\widehat y_A=2.8367+0.7429x,\quad\widehat y_B=5.0245+0.7429x,\quad\widehat y_C=4.6979+0.7429x.}
$$
The B-minus-A estimate is $2.1878$ with <standard error> $0.4545$ and $t=4.8139$; the C-minus-A estimate is $1.8612$ with <standard error> $0.4240$ and $t=4.3901$. Both comparisons are strongly significant.
The B-minus-C comparison needs its own <treatment contrast>; its significance cannot be read off either comparison with A. Reconstructing the coefficient <covariance> from the design gives
$$
\widehat\tau_B-\widehat\tau_C=0.32653,\qquad\operatorname{se}=0.42831,\qquad t=0.7624,\quad p\simeq0.4563.
$$
Thus there is no detectable difference between B and C. This is consistent with pooling their effects in a further reduced model, but does not establish exact equality.
The <residual standard error> is $\sqrt{0.60600}=0.7785$. The five-number residual summary shows deviations on both sides of zero, with the largest magnitude about $1.74$; it does not by itself verify constant <variance> or normality. The <coefficient of determination> $R^2=0.8386$ means about $84\%$ of the achievement variation about its overall mean is explained. The overall $F=29.43$ compares all three nonintercept coefficients with zero, on $(3,17)$ <statistical degrees of freedom>; its $p$-value is $5.889\times10^{-7}$. Printed coefficient $p$-values rounded to `0.0000` are small, not literally zero.
The following original sketch shows the separate-slope model and the preferred parallel-line fit.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-43-teaching-methods.png]
{title=Observed teaching scores and separate-slope versus common-slope fitted models}
\b[At the same initial aptitude, B and C give higher predicted achievement than A, with no convincing difference between B and C.] Interpreting this adjusted association as a causal teaching effect additionally relies on the experimental allocation and error assumptions.
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