(a) Let X₁, X₂, ..., Xₙ be a random sample from Poisson distribution with mean λ > 0. Define a statistic W = (1 − 1/n)^T, T =…

(a)(i) The joint pmf is p(x₁,...,xₙ; λ) = ∏ᵢ e^(−λ) λ^(xᵢ)/xᵢ! = e^(−nλ) λ^(Σxᵢ)/∏xᵢ!.

By the Neyman factorization theorem, this factors as c(λ) λ^(T) h(x), so T = Σᵢ Xᵢ is sufficient for λ.

For completeness, T ~ Poisson(nλ), since the sum of independent Poisson variables is Poisson. If E_λ[h(T)] = 0 for all λ > 0, then Σₜ h(t) e^(−nλ)(nλ)^t/t! = 0. Put z = nλ > 0. Then Σₜ h(t) z^t/t! = 0 for all z > 0. A power series identically zero must have all coefficients zero, so h(t) = 0 for every t. Hence T is complete sufficient.

(ii) As stated, T itself is not unbiased for e^(−λ), because E(T) = nλ. The intended unbiased statistic is W = (1 − 1/n)^T. Using the Poisson pgf, E(W) = E[(1 − 1/n)^T] = Σₜ (1 − 1/n)^t e^(−nλ)(nλ)^t/t! = e^(−nλ) exp[nλ(1 − 1/n)] = e^(−nλ) e^(nλ−λ) = e^(−λ). Thus W is unbiased for e^(−λ).

(iii) Since T is complete sufficient and W is a function of T unbiased for e^(−λ), by the Lehmann–Scheffé theorem W is the UMVUE of e^(−λ).

For X ~ Poisson(λ), log f = x log λ − λ − log x!, so the score is U = x/λ − 1. Hence I₁(λ) = Var(X/λ − 1) = Var(X)/λ² = λ/λ² = 1/λ. For n observations, I_n(λ) = n/λ. For g(λ) = e^(−λ), g′(λ) = −e^(−λ). Therefore the Cramér–Rao lower bound is CRLB = [g′(λ)]²/I_n(λ) = e^(−2λ)/(n/λ) = λ e^(−2λ)/n.

Now let a = 1 − 1/n. Then E(W²) = E[a^(2T)] = exp[nλ(a² − 1)] = exp[nλ((1 − 1/n)² − 1)] = exp(−2λ + λ/n). So Var(W) = e^(−2λ+λ/n) − e^(−2λ) = e^(−2λ)(e^(λ/n) − 1). Since e^(λ/n) − 1 > λ/n for λ > 0, Var(W) > λ e^(−2λ)/n = CRLB. Thus W is UMVUE but does not attain the CRLB.

(b)(i) f_Y(y) = ∫_−∞^∞ f(x,y) dx = y^(3/2)e^(−y)/√(2π) ∫_−∞^∞ e^(−yx²/2) dx. Since ∫_−∞^∞ e^(−yx²/2) dx = √(2π/y), f_Y(y) = y e^(−y), y > 0. Thus Y ~ Gamma(shape = 2, rate = 1).

The conditional density is f_X|Y(x|y) = f(x,y)/f_Y(y) = √(y)/√(2π) e^(−yx²/2), so X|Y = y ~ N(0, 1/y).

(ii) For Y ~ Gamma(2,1): E(Y) = 2, V(Y) = 2. For X|Y = y ~ N(0, 1/y): E(X|Y) = 0, V(X|Y) = 1/Y. Equivalently, conditional on Y = y, V(X|Y=y) = 1/y.

(iii) By the law of total expectation, E(X) = E[E(X|Y)] = E(0) = 0. By the law of total variance, V(X) = E[V(X|Y)] + V[E(X|Y)] = E(1/Y) + 0. For Y ~ Gamma(2,1), E(1/Y) = ∫₀^∞ (1/y) y e^(−y) dy = ∫₀^∞ e^(−y) dy = 1. Therefore V(X) = 1.

(c) Use the Kruskal–Wallis test, since three independent samples are compared for equality of effectiveness. H₀: the three methods are equally effective. H₁: not all methods are equally effective.

The 18 scores in increasing order are: 60(B), 65(C), 67(B), 68(A), 72(B), 77(C), 78(C), 79(A), 80(A), 82(B), 83(A), 84(B), 85(A), 86(B), 88(C), 90(A), 91(B), 93(C).

Ranks by method: Method A: 4, 8, 9, 11, 13, 16; R₁ = 61. Method B: 1, 3, 5, 10, 12, 14, 17; R₂ = 62. Method C: 2, 6, 7, 15, 18; R₃ = 48.

Here N = 18, k = 3, n₁ = 6, n₂ = 7, n₃ = 5. The Kruskal–Wallis statistic is H = [12/(N(N+1))] Σⱼ Rⱼ²/nⱼ − 3(N+1) = (12/342)[61²/6 + 62²/7 + 48²/5] − 57 = 0.1968.

Under H₀, H is approximately χ² with k − 1 = 2 degrees of freedom. At 5% level, χ²₀.₀₅,₂ = 5.991. Since 0.1968 < 5.991, we fail to reject H₀.

Conclusion: At the 0.05 level of significance, there is no sufficient evidence to conclude that the three methods are not equally effective.