Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 4 Solution Created 2026-10-03 Updated 2026-10-14
A randomized complete block design groups similar experimental units into experimental blocks and places every treatment once in each experimental block, independently randomizing treatment allocation within experimental blocks. Blocking removes background between-block variation from within-block treatment comparisons. A balanced incomplete block design has treatments and experimental blocks of size , with each treatment occurring in experimental blocks and every distinct treatment pair together in experimental blocks. Counting treatment occurrences and co-occurrences gives and .
A Latin square of order places each of treatment symbols once in every row and every column of a array. This controls two nuisance directions. A Graeco-Latin square superposes two mutually orthogonal Latin squares: each square is Latin, and each ordered pair of symbols from the two squares occurs exactly once. The two symbol factors can then be fitted together with row and column effects. Random permutations of row labels, column labels and treatment symbols provide a suitable randomized allocation without destroying these balances.
For the operator experiment, fit the additive normal linear modelHere account for the two fabric directions, and is the operator effect. Additivity, independent equal-variance errors and the normal assumption supply the exact F-test; there is only one observation per cell, so an unrestricted interaction model is not available.
In the analysis of variance for a Latin square, the three centered factor spaces each have dimension . Together with the intercept the model has rank ten, leaving residual statistical degrees of freedom. The corrected total has . The completed calculation isUnder the null hypothesis , the operator F-statistic isThe upper five-percent critical value for is , so the operator effect is not significant at five percent in this model.
The decision is independent of the order of fitting rows, columns and operators. To prove the required orthogonality, let be any centered row coefficients and any centered operator coefficients. Their observation-space inner product is , since every operator occurs once in each row. The same calculation applies to columns and operators, and all row-column pairs occur once as well. Thus the three centered spaces are mutually orthogonal; their projections commute and sequential sums of squares equal adjusted sums of squares in every fitting order.
With machines included, the layout is a Graeco-Latin square: the machine symbols are Latin, and all sixteen operator-machine pairs occur exactly once. Fitin addition to the earlier constraints. The machine space contributes three new statistical degrees of freedom and is orthogonal to the previous factor spaces, so the operator sum of squares remains . The residual sum of squares and statistical degrees of freedom becomeHenceThis exceeds but not . Therefore the operator effect is now significant at five percent, but not at one percent. Adjustment for machines has removed substantial residual variation without altering the operator numerator. This illustrates why a balanced nuisance factor can change precision and significance even when it does not change the estimated treatment contrasts.