(a) (i) In stratified sampling under optimum allocation, how will you proceed to select units from different strata, if one or…

(a)(i) Optimum allocation when some nᵢ exceed Nᵢ

Compute the provisional optimum allocation nᵢ = n·NᵢSᵢ/ΣNᵢSᵢ (or nᵢ ∝ NᵢSᵢ/√Cᵢ if costs differ). For every stratum with nᵢ > Nᵢ, take all its units (nᵢ = Nᵢ), since it cannot supply more. Remove these strata, reduce the sample size to n′ = n − ΣNᵢ over the exhausted strata, and recompute the optimum allocation among the remaining strata with the same weights. Repeat until every nᵢ ≤ Nᵢ. Then take a complete enumeration in the exhausted strata and an SRSWOR of the revised size in the others.

(a)(ii) Estimate of the number of orchards

The unbiased estimator is Ŷ_st = ΣNᵢȳᵢ.

• A: Σy = 49, n = 15, contribution 275 × 49/15 = 898.33.
• B: Σy = 152, n = 10, ȳ = 15.2, contribution 146 × 15.2 = 2219.20.
• C: Σy = 189, n = 12, ȳ = 15.75, contribution 93 × 15.75 = 1464.75.
• D: Σy = 335, n = 11, contribution 62 × 335/11 = 1888.18.

Ŷ_st = 898.33 + 2219.20 + 1464.75 + 1888.18 = 6470.47, so the district has about 6470 orchards.

(b)(i) Second-order polynomial model: normal equations

Model: yᵢ = β₀ + β₁xᵢ + β₂xᵢ² + εᵢ. Assume fixed xᵢ (at least three distinct values), E(εᵢ) = 0, Var(εᵢ) = σ², uncorrelated errors, and normality for inference. Minimising Q = Σ(yᵢ − β₀ − β₁xᵢ − β₂xᵢ²)² gives nβ₀ + β₁Σx + β₂Σx² = Σy, β₀Σx + β₁Σx² + β₂Σx³ = Σxy, β₀Σx² + β₁Σx³ + β₂Σx⁴ = Σx²y, that is X′Xβ = X′y with X = [1, x, x²]. Interpretation: β₀ is the mean response at x = 0; β₁ is the slope at x = 0 (dy/dx = β₁ + 2β₂x); β₂ measures curvature, upward if positive and downward if negative; the turning point is x = −β₁/(2β₂).

(b)(ii) Why centre the predictor

With z = x − x̄ the model is y = γ₀ + γ₁z + γ₂z² + ε. Centring is recommended because x and x² are highly correlated when x is far from zero, which makes X′X ill-conditioned and the estimates unstable; centring makes Σz = 0, reduces this multicollinearity, and makes γ₀ the mean response at x̄ and γ₁ the slope at x̄. Fitted values are unchanged. Since 1, z = x − x̄ and z² = x² − 2x̄x + x̄² are linear combinations of 1, x, x² and vice versa, the two design matrices have the same column space, hence the same projection matrix: ŷᵢ = β̂₀ + β̂₁xᵢ + β̂₂xᵢ² = γ̂₀ + γ̂₁zᵢ + γ̂₂zᵢ², with the same residuals and SSE. The coefficients are related by β₂ = γ₂, β₁ = γ₁ − 2γ₂x̄, β₀ = γ₀ − γ₁x̄ + γ₂x̄².

(c) Split-plot designs

In a split-plot design, large main plots receive the levels of one factor, each main plot is divided into sub-plots, and the levels of a second factor are randomised to sub-plots within each main plot. Randomisation therefore occurs at two levels, with a main-plot error e₁ and a sub-plot error e₂.

Use it when one factor needs large units or is hard or costly to change (irrigation, tillage, machine settings), while another can be applied to small units, or when more precision is wanted on the sub-plot factor and the interaction.

e₁ is expected to be larger than e₂ because main plots are larger and more heterogeneous, so the variation among main plots includes soil or unit differences that are partly removed within a main plot, where adjacent sub-plots are more alike. Even when both are expressed per sub-plot, e₁ carries this extra variation, and it usually has fewer degrees of freedom. Main-plot comparisons are therefore less precise, sub-plot comparisons more precise.