Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 36 / 1

Write $1$ for the all-ones vector and setThe centering assumption makes $P_0{P}_X=0$. Thus the full normal linear model has fitted-value orthogonal projection $P_0+P_X$, and its residual sum of squares is $R=Y^{\mathsf{T}}(I-P_0-P_X)Y$. Assume $\nu >0$, as is needed to estimate the unknown error variance from residuals.

For the null that all slopes vanish, the reduced fitted value is $\bar{Y}1$, with residual sum of squares $R_0=Y^{\mathsf{T}}(I-P_0)Y$. The F-test isReject for an upper-tail value exceeding the $1-\alpha$ quantile of this F-distribution. The required theorem is Cochran's theorem: for an isotropic normal distribution, projections onto orthogonal subspaces are independent, and the squared norm on a rank-$r$ subspace divided by ${\sigma }^2$ has a chi-squared distribution with $r$ degrees of freedom. Here the null makes the projected mean in $\mathrm{col}(X)$ zero, so $(R_0-R)/{\sigma }^2\sim {\chi }_p^2$, independently of $R/{\sigma }^2\sim {\chi }_{\nu }^2$.

Let $p_1,p_2$ be the numbers of columns in the two predictor blocks. To test the first block while retaining the second as nuisance, formThe partial F-test for nested linear models isAgain reject in the upper tail. The difference of the nested fitted-value projections has rank $p_1$, is orthogonal to the full residual projection, and annihilates the null mean, which gives the same independent chi-squared argument. If a block has no columns, the corresponding test is vacuous.

The phrase orthogonal statistical parameters refers to zero cross-block Fisher information, not to the numerical parameter vectors being perpendicular. In this model the cross-block information is $X_1^{\mathsf{T}}{X}_2/{\sigma }^2$, so the orthogonal coefficient blocks in a centered normal linear model condition isThen the two fitted subspaces are orthogonal, the block least-squares estimators have zero covariance and are independent because they are jointly Gaussian, and each block's extra sum of squares is unchanged by including the other block. Centering also makes both blocks orthogonal to the intercept.

Both fits are the same additive two-way analysis of variance model: a common intercept, five country effects represented by four contrasts, and eight item effects represented by seven contrasts. The three missing cells are omitted, leaving $37$ observations and model rank $1+4+7=12$, hence $25$ residual degrees of freedom. The residual sum of squares is $659.44$ in either ordering, giving

These are sequential sums of squares, so each row measures the improvement when that factor is added after the preceding factors. The country-first sum $1115.56$ describes its unadjusted contribution before accounting for item; its displayed $F=10.573$ and $p=3.73\times 10^{-5}$ are not the country-adjusted-for-item test. When country is added after item, its extra sum is $1616.74$, givingThus the additive normal linear model gives strong evidence of country differences after accounting for the different products. Conversely, item added after country has extra sum $16910.20$, soThe printed zero for this probability is numerical formatting, not a probability literally equal to zero. Product differences are much larger than residual variation, as expected from their different scales.

The explanation for the order dependence is that missing cells can destroy factor orthogonality in two-way ANOVA. A complete equally replicated crossed design would make centered country and item indicator columns orthogonal. Here the missing observations change the item mix across countries and the country mix across items, so the two factor subspaces are not orthogonal. The two sequential partitions still have the same total explained sum: $1115.56+16910.20=16409.02+1616.74=18025.76$ to the displayed precision. For inference about either factor conditional on the other, compare the full model with the reduced model that omits that factor, rather than interpreting its first-entry sum as adjusted evidence.

The omnibus country test does not identify the particular country contrasts. As an extension, fitting the supplied price table gives the item-adjusted UK-minus-US contrast about $21.33$ pounds, with standard error $2.824$ and an ordinary Gaussian-model $95\%$ confidence interval approximately $[15.51,27.14]$ pounds. This is a derived model contrast, with Student t-distribution reference on $25$ degrees of freedom; simultaneous exploration of many country pairs requires control for multiple hypothesis testing.

Check residuals against fitted values, product and country, together with a normal Q-Q plot and influential observations. The homoscedasticity assumption on absolute prices may be doubtful because expensive products can have larger absolute variation. A model for logarithmic prices gives multiplicative country effects and can make the scale assumption more plausible. Country-by-item interaction terms would allow discounts to differ by product, but there is at most one observed value per cell: an unrestricted interaction saturates the $37$ observed means and leaves no pure-error degrees of freedom, so it cannot be tested against independent measurement noise from these data alone. Additional products or genuine cell replication would support that extension. The selected products and non-random missing entries also limit claims about a whole country's consumer prices; the conditional inference is for the stated additive model and observed price collection.