(a) In a set of two-way classified data according to k levels of factor A and r levels of factor B, there is one observation in…

(a) Number of error contrasts

The model is y_ij = μ + αᵢ + βⱼ + e_ij (i = 1,…,k; j = 1,…,r), with one observation per cell, so kr observations. A linear function L = Σ l_ij y_ij is an error contrast if E(L) = 0 for all values of μ, αᵢ, βⱼ, i.e. it depends on the errors alone.

E(L) = μΣΣl_ij + Σᵢ αᵢ(Σⱼ l_ij) + Σⱼ βⱼ(Σᵢ l_ij) vanishes for all parameter values only if Σⱼ l_ij = 0 for each i (k conditions) and Σᵢ l_ij = 0 for each j (r conditions) (ΣΣl_ij = 0, which makes L a contrast, then follows). These k + r conditions are not independent: the sum of the k row totals equals the sum of the r column totals. Exactly k + r − 1 are independent, so the coefficient vectors (l_ij) form a space of dimension kr − (k + r − 1) = (k − 1)(r − 1) = (r − 1)(k − 1).

Explicit set: L_ij = y_ij − y_i1 − y_1j + y_11, i = 2,…,k, j = 2,…,r. Each has E(L_ij) = 0 (every μ, α and β term cancels), and they are linearly independent because only L_ij contains y_ij. There are (k − 1)(r − 1) of them, and the dimension count shows that no more can be independent.

Check: the kr − 1 independent contrasts split into (k − 1) for A, (r − 1) for B and (k − 1)(r − 1) for error, since (k − 1) + (r − 1) + (k − 1)(r − 1) = kr − 1.

(b) Two-stage sampling

Technique: the population consists of N first-stage units (fsu), e.g. villages, each containing M second-stage units (ssu), e.g. households. n fsu are drawn by simple random sampling without replacement (SRSWOR) and, from each selected fsu, m ssu are drawn likewise. A list of ssu is needed only for the selected fsu, which cuts cost (other examples: schools and students, blocks and farms).

Notation (equal-sized fsu): Y_ij is the value of the jth ssu in the ith fsu, Ȳᵢ the fsu mean, Ȳ = ΣȲᵢ/N, S_b² = Σ(Ȳᵢ − Ȳ)²/(N − 1), Sᵢ² = Σⱼ(Y_ij − Ȳᵢ)²/(M − 1), S_w² = ΣSᵢ²/N. Estimator: ȳ̄ = (1/n)Σᵢ∈s ȳᵢ, where ȳᵢ is the mean of the m ssu in fsu i. Since E₂(ȳᵢ) = Ȳᵢ, E(ȳ̄) = Ȳ.

V(ȳ̄) = V₁E₂(ȳ̄) + E₁V₂(ȳ̄). E₂(ȳ̄) = (1/n)Σᵢ∈s Ȳᵢ is an SRSWOR mean of the Ȳᵢ, so V₁ = (1/n − 1/N)S_b². V₂(ȳ̄) = (1/n²)Σᵢ∈s(1/m − 1/M)Sᵢ², so E₁V₂ = (1/n)(1/m − 1/M)S_w². V(ȳ̄) = (1/n − 1/N)S_b² + (1/n)(1/m − 1/M)S_w²

(i) Stratified random sampling: if every fsu is taken (n = N), the fsu act as strata, the first term vanishes and V = (1/N)(1/m − 1/M)S_w² = (1/N²)Σᵢ(1/m − 1/M)Sᵢ², the usual stratified variance with N equal strata of size M. (ii) Cluster sampling: if all M ssu of each selected fsu are enumerated (m = M), the second term vanishes and V = (1/n − 1/N)S_b², the usual cluster-sampling variance with clusters of size M.

(c)(i) Multiple correlation of X₃ on X₁, X₂

As the Y's are uncorrelated with unit variance, Var(Xᵢ) = 2, Cov(X₁, X₂) = Var(Y₂) = 1, Cov(X₁, X₃) = Var(Y₁) = 1, Cov(X₂, X₃) = Var(Y₃) = 1. Hence r₁₂ = r₁₃ = r₂₃ = 1/2. R²₃.₁₂ = (r₁₃² + r₂₃² − 2r₁₂r₁₃r₂₃)/(1 − r₁₂²) = (1/4 + 1/4 − 1/4)/(3/4) = 1/3. Check: σ′Σ₁₂⁻¹σ/σ₃₃ with σ = (1, 1)′ and Σ₁₂ = [[2, 1], [1, 2]] gives (2/3)/2 = 1/3. R₃.₁₂ = 1/√3 ≈ 0.577

(c)(ii) First principal component

Σ = 6I + 3·11′ with 1 = (1, 1, 1)′, and |Σ − λI| = 0 gives (λ − 15)(λ − 6)² = 0. For the vector 1, Σ1 = 15·1; for any x orthogonal to 1, Σx = 6x. So λ₁ = 15, λ₂ = λ₃ = 6, and 15 + 6 + 6 = 27 = trace Σ. The normalised eigenvector for 15 is (1, 1, 1)′/√3, so first principal component Y₁ = (X₁ + X₂ + X₃)/√3, with Var(Y₁) = 15. Proportion of total variability explained = 15/27 = 5/9 ≈ 0.556 (55.6%). The other two eigenvalues are equal, so the remaining components are not unique.