(a) For a two variable linear regression model Yᵢ = a + bXᵢ + eᵢ, where E(eᵢ) = 0, Var(eᵢ) = σ²ₑ, Cov(eᵢ, eⱼ) = 0 for i ≠ j…

(a) For given X-values, set xᵢ = Xᵢ - X̄ and Sₓₓ = ∑(Xᵢ - X̄)² = ∑xᵢ². The least-squares normal equations are

nâ + b̂∑Xᵢ = ∑Yᵢ,

â∑Xᵢ + b̂∑Xᵢ² = ∑XᵢYᵢ.

Solving them gives

b̂ = ∑(Xᵢ - X̄)(Yᵢ - Ȳ)/Sₓₓ = ∑xᵢYᵢ/Sₓₓ,

and

â = Ȳ - b̂X̄.

Substitute Yᵢ = a + bXᵢ + eᵢ. Since ∑xᵢ = 0,

∑xᵢYᵢ = a∑xᵢ + b∑xᵢXᵢ + ∑xᵢeᵢ = bSₓₓ + ∑xᵢeᵢ.

Hence

b̂ - b = ∑xᵢeᵢ/Sₓₓ.

Therefore

Var(b̂) = E[(∑xᵢeᵢ)²]/Sₓₓ² = σ²ₑ∑xᵢ²/Sₓₓ² = σ²ₑ/Sₓₓ.

For the intercept, Ȳ = a + bX̄ + ē, so

â - a = ē - X̄(b̂ - b) = ∑[1/n - X̄xᵢ/Sₓₓ]eᵢ.

Let cᵢ = 1/n - X̄xᵢ/Sₓₓ. Then

Var(â) = σ²ₑ∑cᵢ².

Now

∑cᵢ² = ∑[1/n² - 2X̄xᵢ/(nSₓₓ) + X̄²xᵢ²/Sₓₓ²] = 1/n - 0 + X̄²/Sₓₓ = 1/n + X̄²/Sₓₓ.

Thus

Var(â) = σ²ₑ(1/n + X̄²/Sₓₓ) = σ²ₑ∑Xᵢ²/(nSₓₓ).

For the covariance, write b̂ - b = ∑dᵢeᵢ with dᵢ = xᵢ/Sₓₓ. Then

Cov(â, b̂) = Cov(∑cᵢeᵢ, ∑dᵢeᵢ) = σ²ₑ∑cᵢdᵢ.

But

∑cᵢdᵢ = ∑[(1/n - X̄xᵢ/Sₓₓ)(xᵢ/Sₓₓ)] = (1/(nSₓₓ))∑xᵢ - (X̄/Sₓₓ²)∑xᵢ² = 0 - X̄/Sₓₓ = -X̄/Sₓₓ.

Hence

Var(b̂) = σ²ₑ/Sₓₓ, Var(â) = σ²ₑ(1/n + X̄²/Sₓₓ), Cov(â, b̂) = -σ²ₑX̄/Sₓₓ.

The result requires Sₓₓ > 0 and the stated uncorrelated, homoscedastic error assumptions.

(b) Let Y = (Y₁, Y₂)' and define the linear transformation

Y₁ = X₁ + X₂ + X₃, Y₂ = X₂ - X₃.

So Y = AX, where

A = [[1, 1, 1], [0, 1, -1]].

Since X ~ N₃(μ, Σ), any linear transformation is normal:

Y ~ N₂(Aμ, AΣA').

Now μ = (1, 2, 1)', so

Aμ = (1 + 2 + 1, 2 - 1)' = (4, 1)'.

Also Σ = [[9, 2, 2], [2, 3, 0], [2, 0, 2]].

Compute the variances:

Var(Y₁) = Var(X₁ + X₂ + X₃) = 9 + 3 + 2 + 2(2) + 2(2) + 2(0) = 22.

Var(Y₂) = Var(X₂ - X₃) = 3 + 2 - 2(0) = 5.

Cov(Y₁, Y₂) = Cov(X₁ + X₂ + X₃, X₂ - X₃) = Cov(X₁, X₂) - Cov(X₁, X₃) + Var(X₂) - Cov(X₂, X₃) + Cov(X₃, X₂) - Var(X₃) = 2 - 2 + 3 - 0 + 0 - 2 = 1.

Therefore

(Y₁, Y₂)' ~ N₂((4, 1)', [[22, 1], [1, 5]]).

Equivalently, Y₁ ~ N(4, 22), Y₂ ~ N(1, 5), and Cov(Y₁, Y₂) = 1.

(c) Let X = (X₁, X₂, ..., Xₙ)' ~ Nₙ(0, I). Put 1 = (1, 1, ..., 1)'. The sample mean is

X̄ = (1/n)1'X.

Define

B = I - (1/n)11'.

Then B is symmetric and idempotent, B² = B, and B1 = 0. Also,

∑ⱼ₌₁ⁿ(Xⱼ - X̄)² = X'X - nX̄² = X'X - (1/n)X'11'X = X'[I - (1/n)11']X = X'BX.

Hence

S² = (1/(n-1))∑(Xⱼ - X̄)² = X'BX/(n-1).

Now consider the scalar L = 1'X and the vector Z = BX. They are jointly normal because both are linear transforms of X. Their covariance is

Cov(L, Z) = Cov(1'X, BX) = 1' Var(X) B' = 1'B = (B1)' = 0.

Thus L and Z are independent, since zero covariance implies independence for jointly normal variables. Therefore X̄ = L/n is independent of Z. But S² = Z'Z/(n-1) is a function of Z. Hence

X̄ and S² are stochastically independent.

Equivalently, by the normal quadratic-form theorem, the idempotent matrices J/n and I - J/n have zero product, so X'(J/n)X and X'[I - J/n]X are independent; these determine X̄² and S², and the sign of X̄ is independent of S² as shown by the linear-form argument.

(d) Let p = 0.30 be the population proportion of defectives and q = 1 - p = 0.70. The phrase “within 2 percent of the true value” means a relative error:

|p̂ - p| ≤ 0.02p.

Thus the permissible absolute error is

e = 0.02p = 0.02 × 0.30 = 0.006.

For a large simple random sample,

p̂ ~ N(p, pq/n).

We require

P(|p̂ - p| ≤ e) = 0.955.

Standardize:

Z = (p̂ - p)/√(pq/n).

Then

P(|Z| ≤ z) = 0.955,

so

P(0 ≤ Z ≤ z) = 0.4775.

The given value gives z = 2.005. Now

z = e/√(pq/n) = e√n/√(pq).

Therefore

n = z²pq/e².

Substitute:

n = (2.005)²(0.30)(0.70)/(0.006)² = (4.020025)(0.21)/0.000036 = 0.84420525/0.000036 = 23450.145833.

Rounding up to the next integer,

n = 23451.

Since the population is very large, the finite-population correction is negligible. The normal approximation is valid because np and nq are large.

If “within 2 percent” were instead interpreted as an absolute margin of 2 percentage points, the corresponding size would be 2111.

(e) The size and shape of plots and blocks strongly affect the precision and validity of field experiments because they determine how much soil heterogeneity, border effect, competition, and management variation enters the experimental error.

For plot size, larger plots generally reduce border effects, interplot competition, and the relative influence of soil micro-variation. They give more representative yields, especially for crops with spreading roots or mobile pests. However, increasing plot size increases total variance per plot and cost, and it may reduce the number of replications for a fixed area. Very small plots are economical and allow more replication, but they suffer more from border effects, damage, missing plants, and treatment interference. The optimum plot size is usually studied by uniformity trials; variance per unit area decreases with plot size at a diminishing rate, so beyond a point the gain in precision is small.

For plot shape, long narrow rectangular plots are often more efficient than square plots because they sample soil heterogeneity better and reduce the correlation between adjacent plots. Their orientation should be chosen with respect to the fertility gradient, usually so that each plot is as representative as possible and block-to-block variation is minimized. Square or compact plots reduce perimeter and border effects, so the best shape depends on the crop and the pattern of soil variation.

For blocks, blocks should be compact and homogeneous so that variation within a block is small relative to variation between blocks. If a fertility gradient is known, blocks are usually laid out perpendicular to that gradient. Large blocks contain more heterogeneity and inflate experimental error; very small blocks may not provide enough error degrees of freedom and can create practical difficulties. The number and shape of plots within a block should balance precision, error degrees of freedom, and field layout. Thus careful choice of plot and block size and shape reduces experimental error and makes treatment comparisons more reliable.