(a) What is the role of properties of completeness and sufficiency in Statistical Inference ? Explain. In U (0, θ), find out…

(a) Role of completeness and sufficiency; UMVUE of θ in U(0, θ)

A statistic T is sufficient for θ if the conditional distribution of the sample given T does not depend on θ; it keeps all the information about θ in the data, so the sample can be reduced to T without loss. T is complete if Eθ[g(T)] = 0 for all θ implies g(T) = 0 almost surely; this makes an unbiased estimator that is a function of T unique. Role in inference: by the Rao-Blackwell theorem, if T is sufficient and W is unbiased for θ, then E(W | T) is unbiased with variance not larger than that of W. By the Lehmann-Scheffé theorem, if T is complete and sufficient and h(T) is unbiased for θ, then h(T) is the UMVUE. So sufficiency gives data reduction and completeness gives uniqueness and minimum variance. UMVUE in U(0, θ): let X₁, …, Xₙ be a sample. Step 1: The joint density is θ^(−n) for 0 < xᵢ < θ, which equals θ^(−n) 1{x₍ₙ₎ < θ}. By the factorisation criterion, T = X₍ₙ₎ = max Xᵢ is sufficient. Step 2: T has density n t^(n−1)/θⁿ for 0 < t < θ. Step 3: Completeness. If Eθ g(T) = 0 for all θ, then ∫₀^θ g(t) t^(n−1) dt = 0 for all θ. Differentiating in θ gives g(θ) θ^(n−1) = 0, so g = 0. Thus T is complete. Step 4: E(T) = ∫₀^θ t · n t^(n−1)/θⁿ dt = nθ/(n + 1), so ((n + 1)/n) T is unbiased for θ. Step 5: By Lehmann-Scheffé, the UMVUE of θ is θ̂ = ((n + 1)/n) X₍ₙ₎.

(b) Chi-square test of equal probability of male and female births

H₀: p = 1/2 (the number of boys in a family of four is Binomial(4, 1/2)); H₁: p ≠ 1/2. Step 1: Probabilities are 1/16, 4/16, 6/16, 4/16, 1/16, so the expected frequencies for 400 families are 25, 100, 150, 100, 25. Step 2: χ² = Σ(O − E)²/E = 81/25 + 121/100 + 25/150 + 324/100 + 49/25 = 3·24 + 1·21 + 0·167 + 3·24 + 1·96 = 9·82. Step 3: No parameter is estimated (p is specified), so the degrees of freedom are 5 − 1 = 4 and the critical value is 9·488. Step 4: Since 9·82 > 9·488, H₀ is rejected at 5%. The data are not consistent with equal probability of male and female births.

(c) Likelihood ratio test for the mean of N(θ, σ²), σ² unknown

Definition: for H₀: θ ∈ Θ₀ against H₁: θ ∈ Θ − Θ₀, the likelihood ratio is λ(x) = sup over Θ₀ of L / sup over Θ of L. The LR test rejects H₀ when λ ≤ k, where k is chosen so that the size is α. Step 1: Over the whole space the MLEs are θ̂ = x̄ and σ̂² = Σ(xᵢ − x̄)²/n. Under H₀ the MLE is σ̂₀² = Σ(xᵢ − θ₀)²/n. Step 2: λ = (σ̂²/σ̂₀²)^(n/2) = [Σ(xᵢ − x̄)²/Σ(xᵢ − θ₀)²]^(n/2). Step 3: Since Σ(xᵢ − θ₀)² = Σ(xᵢ − x̄)² + n(x̄ − θ₀)², λ^(2/n) = 1/[1 + t²/(n − 1)], where t = √n(x̄ − θ₀)/s and s² = Σ(xᵢ − x̄)²/(n − 1). Step 4: λ ≤ k holds exactly when |t| ≥ c. Under H₀, t follows Student's t with n − 1 degrees of freedom, so c = t(α/2, n−1). The LR test rejects H₀ when |t| > t(α/2, n−1); it is the usual two-sided t-test.