= Solution
In the larger <normal linear model>, add laboratory intercept offsets $\gamma_BI_{Bi}+\gamma_CI_{Ci}$ to the mean in part (a). The <nested-model F-test> compares
$$
H_0:\gamma_B=\gamma_C=0
\qquad\text{against}\qquad
H_1:(\gamma_B,\gamma_C)\ne(0,0).
$$
The smaller and larger models fit four and six mean coefficients respectively. Under $H_0$, independent Gaussian errors with common <variance> give independent quantities $(\mathrm{RSS}_1-\mathrm{RSS}_2)/\sigma^2\sim\chi^2_2$ and $\mathrm{RSS}_2/\sigma^2\sim\chi^2_{24}$. Thus the <test statistic> is
$$
\boxed{F=\frac{(\mathrm{RSS}_1-\mathrm{RSS}_2)/2}{\mathrm{RSS}_2/24}
=\frac{818.72/2}{5103.0/24}=1.9253,\qquad F\mid H_0\sim F_{2,24}.}
$$
The <p-value> is the upper-tail probability $\Pr(F_{2,24}\ge1.9253)=0.1677$. At the conventional 5% <significance level>, do not reject the common-intercept restriction. \b[Prefer the simpler model `tumour1` on this comparison.] This does not prove the intercepts coincide, nor address the curvature identified in part (b); the exact <F-test> calibration assumes the underlying error and mean assumptions. The one-decimal RSS values give $818.7$ when subtracted, whereas the table's difference uses unrounded values.
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