hH YiX2l + Â3 £ Yi*3l 2
Due to residual (RSS) Total
£ of £ y2
n - 3
Recall the identity
TSS has, as usual, n — 1 df and RSS has n — 3 df for reasons already discussed. ESS has 2 df since it is a function of p2 and fa. Therefore, following the ANOVA procedure discussed in Section 5.9, we can set up Table 8.1.
Now it can be shown6 that, under the assumption of normal distribution for ui and the null hypothesis p2 = p3 = 0, the variable
is distributed as the F distribution with 2 and n — 3 df.
What use can be made of the preceding F ratio? It can be proved7 that under the assumption that the ui ~ N(0, a2),
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