Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 5 Solution Created 2026-10-03 Updated 2026-10-14
Code each factor level by . A nonconstant factorial contrast indexed by a nonempty factor set has sign column . Flipping any coordinate in reverses that sign and pairs the treatment combinations, so exactly have each sign. Allocating the two signs to two experimental blocks therefore divides the full factorial design into equal experimental blocks. The contrast is confounded with experimental blocks because its column is constant within each experimental block: changing its coefficient and compensating by the two experimental block intercepts leaves every fitted value unchanged. It has no information separate from experimental block differences. Randomize the allocation of the two groups to days and the run order within days.
Write for temperature, pressure and catalyst, respectively, with low levels and catalyst one coded minus. In the supplied allocation, on day one and on day two. Thus the three-factor interaction is confounded with days. Every other nonconstant factorial column is orthogonal to the day contrast, because its product with is a nonconstant sign column on the full cube and consequently sums to zero.
If all interactions are negligible, the fitted model contains an intercept, one day contrast and the three main effects. The corrected total statistical degrees of freedom partition asThe three residual statistical degrees of freedom are the unused two-factor interaction columns , pooled as error under the stated negligible-interaction assumption. There is no independent pure-error replication of treatment combinations.
For four days with two runs each, all three main effects can be estimated. One regular allocation pairs each treatment combination with its complete opposite:The block-generating columns can be and ; their product is also constant in experimental blocks. Thus span the three experimental block contrasts, while and vary within experimental blocks. The main effects remain mutually orthogonal and separate from the experimental block effects; with an additive main-effect model there is one residual degree of freedom, represented by .
For the further request about , the distinction between regular confounding and unrestricted pairing is important. A regular four-block allocation cannot retain all three main effects and . Its three confounded columns form a rank-two defining subgroup. To preserve main effects, these columns must lie in . If the subgroup includes and a two-factor column, their product is a main-effect column, which is forbidden. Hence the only permissible nonidentity subgroup is , and it necessarily confounds . This proves impossibility in the usual regular factorial-blocking framework, rather than relying on a count of available statistical degrees of freedom.
The literal request does not explicitly restrict arrangements to regular experimental blocks. If arbitrary pairings are permitted, a nonregular construction can estimate together, assuming the other interactions are omitted from the model. An explicit allocation isEvery treatment combination occurs exactly once. Fit one unrestricted intercept per day and the four coefficients . By estimability from within-block differences, subtracting the second run from the first in each experimental block removes all day effects and yields the coefficient matrixThus all four coefficients are uniquely determined from the four day differences. In this broader design class the answer is yes, under the reduced four-effect model. The resulting fit is saturated: four experimental block intercepts plus four treatment coefficients use all eight observations, leaving no residual estimate of error variance. Replication or external error information would be needed for inference. This does not contradict the impossibility for regular orthogonal confounding, and it does not claim simultaneous unbiased estimation when arbitrary additional interactions are retained.