Analysis of the Balanced Incomplete Block Design. Each of t treatments is applied a total of r times, distributed across b blocks, each made up

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1 1 Analysis of the Balanced Incomplete Bloc Design Each of t treatments is applied a total of r times, distributed across b blocs, each made up of plots. For = 1,...,t and = 1,...,b write n for the number of times treatment appears in bloc. Each n is either 0 or 1, and so n = and n = t. The word balanced comes from the fact that each pair of treatments appears together in λ blocs. Constraints on the parameters The total number of plots equals n = tr = b. The parameters are also connected by a relationship obtained by counting the total number of plots in all r blocs where treatment 1 appears. Each of those r blocs consists of plots, giving a total of r plots. Moreover, each of the r blocs contains treatment 1 together with the 1 other treatments that appear together with 1 somewhere. Thus <1> r 1 = 1 The linear model Define indicators B 1,...,B b to identify which bloc a plot belongs to, and indicators T 1,...,T t to identify the treatment applied to the plot. Notice that T B = n. The model asserts that the vector of yields Y has variance matrix σ 2 I n and vector of expected values in the subspace X spanned by B 1,...,B b,t 1,...,T t. That is, EY = β B + τ T, for unnown parameters β 1,...,τ t. The representation is overparametrized, because the spanning vectors are linearly dependent. For example, 1 = B = T. In particular, the τ are not identifiable unless they are subected to a constraint such as τ = 0, as explained later. Bloc and treatment subspaces The vectors {B / } form an orthonormal basis for a b-dimensional subspace X B of X. Denote by X T the part of X that is orthogonal to X B. In general, X T is not a subspce of spant 1,...,T t. Write H T for the linear map that is, a matrix, if you are thining of the elements of R n as column vectors and not as b t tables with some cells empty that proects R n onto X T. Write T for H T T, the component of T that is orthogonal to X B : T T = T component of T in X B = T Note that B B = T 1 n B <2> T T = T T because T T X T = T T 1 n 2 = r r and, for, <3> T T = T T = T T 1 n n = λ

2 2 Dimension of the treatment subspace <4> The subspace X T is spanned by the vectors H T B 1,...,H T B b,h T T 1,...,H T T t, that is, by T 1,..., T t because H T B = 0. The { T } are not linearly independent, because T = H T T = H T 1 = H T B = 0. However, the vectors T 2,..., T t are linearly independent. For suppose 0 = 2 θ T. Dot with T 1 to get 0 = θ λ That is, θ = 0. Dot with T, for 2toget 0=θ r r λ θ = θ r r + λ because θ = θ It follows that θ = 0 because, by <1>, <5> r r + λ = 0. Thus X T has dimension t 1. Proection matrix The proection H T onto X T equals M = T T To establish this fact, first notice that Mx = 0ifxis orthogonal to each T. That is, Mx = 0if x X T. To show that Mz = z for each z in X T it suffices to chec that M T = T for each, because each vector in X T is a linear combination of the spanning set { T }. T T T M T = = r r T λ = T from <4> and <5> That is, M T = T. It follows that H T = M, as asserted. T from <2> and <3> Least squares estimates The vector of fitted values must be a linear combination of the vectors spanning X, that is, Ŷ = β B + τ T Of course the coefficients are not uniquely determined by this equality, because the vectors are linearly dependent. We could replace the T by T plus a linear combination of the B s, then rewrite the representation as <6> Ŷ = β B + τ T where β is a linear combination of the β and the τ s. This representation has the advantage that the first sum gives the component of Ŷ in X B and the second sum gives the component in

3 3 the orthogonal subspace X T. In particular, β = 1 Y B = Y + = average yield in bloc. The τ will be uniquely determined if we add the constraint that τ = 0. Write Y + for the total yield from plots receiving treatment. Then, Y + = Y T = Ŷ T because Y Ŷ X T = β B T + τ T T = 1 = 1 Y + n + r r τ λ n Y + + τ λ τ τ by <5> Subect to τ = 0, these equations have the unique solution τ = Y + 1 n Y + That is, τ = Y T 1 n B Y = T Y Notice that n B Y is ust the sum of all yields over blocs in which treatment appears. Estimates of treatment differences The individual τ in the representation for EY are not uniquely determined, and hence they are not estimable, but we will see that their differences are functions of the expected values. With the nonunique choice of τ as above, E τ = EY T = τ T T because B T = 0 = r r τ λ τ = τ λ τ by <5> Consequently, E τ τ = τ τ. The differences are estimable because they equal the expectations of linear functions of the yields. Notice that the functions τ 1 t τ are estimable. The τ are uniquely determined only up to an additive constant. If they are constrained by τ = 0 then the τ are uniquely determined. Variances var τ τ = From <2> and <3>, and then <5>, 2 var Y T 2 T = σ 2 T T 2. T T 2 = T 2 + T 2 2 T T =2 r r +2 λ =2

4 It follows that var τ τ = 2σ 2 /. Notice the symmetry: all treatment differences can be estimated with the same accuracy because of the balance in the design. 4 A 2 15 factorial design in blocs of size 16. The following explanation is based on the discussion in Fisher s boo Design of Experiments, and on ideas in the solutions of Feng Liang and Andreas Schulz. The problem ased for a design such that no main effects or interactions are confounded. The solution amounts to finding a subgroup of size 2 11 of the group of all products of factors, such that no singletons or doublets are in the group. The group operation corresponds to pointwise multiplication of signed indicator vectors, that is, of vectors { +1 if factor i appears at high level F i = 1 if factor i appears at low level Notice that F i F i = 1, the constant vector of 1 s, for every i. As suggested in Section 45.1 of Fisher s boo, the subgroup G of factor combinations that are confounded with the blocs should be generated by S = S 0 S 1 S 2, the following collection of 11 generators, each consisting of a product of three factors: S 0 S 1 S 2 F 1 F 2 F 3 F 2 F 4 F 6 F 4 F 8 F 12 F 1 F 4 F 5 F 2 F 8 F 10 F 1 F 6 F 7 F 2 F 12 F 14 F 1 F 8 F 9 F 1 F 10 F 11 F 1 F 12 F 13 F 1 F 14 F 15 Notice the pattern. The 8 members of S 0 are obtained by writing F 1 followed by successive pairs, F 2 F 4,...,F 14 F 15. The set S 1 is obtained from S 0 by multiplying each subscript by 2 then discarding any products involving a subscript greater than 15. The set S 2 is obtained in similar fashion from S 1. To simplify the exposition I will refer to a factor whose subscript is an odd number as odd, to a factor whose subscript is an even number as even, to a factor whose subscript is twice an odd number as 2 odd, and so on. Notice that: A Products of one or two generators from S 0 contain two odd factors and at least one even factor. B Products of three or more generators from S 0 contain at least three odd factors. There are several facts that need to be established. i All different products of generators or 2 11 if we include the empty product as giving 1 give distinct factor combinations. For example, if F 1 F 3 F 4 F 6 = F 1 F 2 F 3 F 2 F 4 F 6 were in the list of generators then many of the products would correspond to the same factor combination. ii No singleton a product of the form F i belongs to G. iii No doublet a product of the form F i F, with i belongs to G. Proof of i It is sufficient to show that no generator can be expressed as a product of generators appearing earlier in the listing of S.

5 5 For generators in S 0 the property is obvious, because each new generator introduces a new odd factor. No generator in S 0 can appear as a product for a generator from S 1 S 2, because those generators contain no odd factors. No generator in S 1 is a product of earlier generators from S 1 because each generator introduces a new 2 odd factor. And so on. I have written the argument to suggest the extension to the more general case of n = 2 1 factors in blocs of size 2. Proof of ii From A, any product containing a generator from S 0 must contain at least two odd factors, which could not be cancelled out by generators in S 1 S 2. Thus no generators from S 0 appear in the product. A similar argument, with odd replaced by 2 odd, can be applied to any product of generators from S 1. Thus no generators from S 1 appear in the product. And so on. Proof of iii Suppose a product of generators gave a doublet. Consider how many generators could come from S 0. From B, three or more generators from S 0 would give three odd factors, which could not be cancelled out by generators from S 1 S 2. A single generator from S 0 would contribute two odd factors together with a single even factor that would have to be cancelled out by a product from S 1 S 2 that reduces a singleton, contradicting ii. A pair of generators from S 0 would contribute two odd factors, together with a pair of 2 odd factors that would have to be cancelled out by a product from S 1 S 2. That is, it would have to be possible to get a doublet of 2 odd factors using products from S 1 S 2. With no generators from S 0 we would be left with the tas of finding a product of generators from S 1 S 2 that reduced to a doublet. The last two possibilities both require construction of a doublet as a product from S 1 S 2. But that problem is essentially a copy of the problem of constructing a doublet from a product of generators from S 0 S 1 S 2, except that all subscripts get multiplied by 2. Repeat the argument to reduce the last problem to the impossible tas of getting a doublet as a product of generators from S 1 alone. Again I have written the argument to suggest the extension to the general case.