Statistics 2021 Paper I 50 marks Prove

Q4

(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 = Σᵢ₌₁ⁿ Xᵢ (i) Show that T is complete sufficient statistic. (ii) Show that T is unbiased for e^(−λ). (iii) Show that even though T is UMVUE, it does not attain the CRLB for g(λ) = e^(−λ). (20 marks) (b) Let f(x, y) = \frac{e^{\frac{-yx^2}{2}} y^{3/2} e^{-y}}{\sqrt{2\pi}}, -\infty < x < \infty, y > 0. (i) Obtain the marginal distribution of Y and conditional distribution of X given Y. (ii) Find E(Y), V(Y), E(X|Y), V(X|Y). (iii) Use (ii) to find E(X), V(X). (5+5+5 marks) (c) A company's trainees are randomly assigned to groups which are through a certain industrial inspection procedure by three different methods. At the end of the instructing period they are tested for inspection performance quality. The following are their scores : Method A : 80 83 79 85 90 68 Method B : 82 84 60 72 86 67 91 Method C : 93 65 77 78 88 Using the appropriate non-parametric test, determine at 0·05 level of significance whether the three methods are equally effective. (15 marks)

हिंदी में प्रश्न पढ़ें

(a) माना X₁, X₂, ..., Xₙ प्वासों बंटन, जिसका माध्य λ > 0, से लिया गया एक यादृच्छिक प्रतिदर्श है । एक प्रतिदर्शज परिभाषित है W = (1 − 1/n)^T, T = Σᵢ₌₁ⁿ Xᵢ (i) दर्शाइए कि T पूर्ण पर्याप्त प्रतिदर्शज है । (ii) दर्शाइए कि T, e^(−λ) के लिए अनभिनत है । (iii) भले ही T, UMVUE (यू.एम.वी.यू.ई.) है, दर्शाइए कि यह g(λ) = e^(−λ) के लिए CRLB (सी.आर.एल.बी.) प्राप्त नहीं करता है । (20 अंक) (b) माना f(x, y) = [e^(−yx²/2) y^(3/2) e^(−y)] / √(2π) , −∞ < x < ∞, y > 0. (i) Y का उपांत बंटन और Y के दिए होने पर X का सप्रतिबंध बंटन प्राप्त कीजिए । (ii) E(Y), V(Y), E(X|Y), V(X|Y) ज्ञात कीजिए । (iii) (ii) का उपयोग करते हुए E(X), V(X) ज्ञात कीजिए । (5+5+5 अंक) (c) तीन विभिन्न तरीकों से एक निश्चित औद्योगिक निरीक्षण प्रक्रिया द्वारा एक कंपनी के प्रशिक्षार्थियों को यादृच्छया समूहों में नियत किया गया । प्रशिक्षण अवधि की समाप्ति पर निरीक्षण प्रदर्शन गुणवत्ता के लिए उनका परीक्षण किया गया । उनके स्कोर निम्न हैं : रीति A : 80 83 79 85 90 68 रीति B : 82 84 60 72 86 67 91 रीति C : 93 65 77 78 88 उपयुक्त अप्राचलिक परीक्षण का उपयोग करते हुए, 0·05 सार्थकता स्तर पर निर्धारित कीजिए कि क्या तीनों रीतियाँ समान रूप से प्रभावी हैं । (15 अंक)

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Directive word: Prove

This question asks you to prove. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

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How this answer will be evaluated

Approach

Prove all required results systematically, spending approximately 40% of time on part (a) given its 20 marks, 30% on part (b) for 15 marks, and 30% on part (c) for 15 marks. Structure as: (a) establish completeness via exponential family, sufficiency via factorization, unbiasedness via expectation calculation, and CRLB non-attainment via variance comparison; (b) integrate to obtain marginal Gamma distribution, derive conditional Normal, then apply law of total expectation/variance; (c) state Kruskal-Wallis test assumptions, compute ranks, calculate H-statistic, and compare with χ² critical value.

Key points expected

  • Part (a)(i): Apply factorization theorem to show T is sufficient; use completeness property of Poisson exponential family with natural parameter space containing an open set
  • Part (a)(ii): Calculate E[W] = E[(1-1/n)^T] using Poisson MGF to verify unbiasedness for e^(-λ)
  • Part (a)(iii): Compute Var(W), derive CRLB for g(λ)=e^(-λ), and explicitly show strict inequality Var(W) > CRLB
  • Part (b): Identify Y ~ Gamma(5/2, 1) marginal; X|Y ~ N(0, 1/Y); apply E(X)=E[E(X|Y)] and V(X)=E[V(X|Y)]+V[E(X|Y)]
  • Part (c): Apply Kruskal-Wallis H-test: pool and rank all 18 observations, compute rank sums per method, calculate H = [12/N(N+1)]Σ(Ri²/ni) - 3(N+1), compare with χ²₂,₀.₀₅ = 5.991

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly identifies Poisson exponential family form for (a), recognizes Gamma kernel in (b) integrand, and properly states Kruskal-Wallis null hypothesis with independent samples assumption for (c)States basic sufficiency/unbiasedness definitions but misidentifies distribution families or omits key assumptions like independence in (c)Confuses completeness with sufficiency, fails to recognize Gamma/Normal structures, or applies wrong test (e.g., ANOVA) for (c)
Method choice20%10Uses factorization theorem and exponential family completeness for (a); employs law of total expectation/variance elegantly in (b); selects Kruskal-Wallis over ANOVA given small samples and no normality assurance in (c)Applies correct broad methods but with inefficient approaches (e.g., direct integration without MGF for Poisson expectation)Attempts Lehmann-Scheffé without verifying completeness, uses conditional expectation incorrectly, or applies parametric tests to (c)
Computation accuracy20%10Exact calculation: E[(1-1/n)^T] = e^(-λ) verified via Σ(1-1/n)^k e^(-λ)λ^k/k!; Gamma(5/2,1) moments correct; Kruskal-Wallis H-statistic computed precisely with proper ranking of tiesCorrect final expressions with minor arithmetic errors in series summation or integration constantsMajor errors: incorrect MGF usage, wrong Gamma shape/rate parameters, miscalculated rank sums, or wrong degrees of freedom
Interpretation20%10Explains why UMVUE non-attainment of CRLB occurs (non-linear function of natural parameter); interprets hierarchical structure in (b) as Normal-Gamma mixture; concludes practical significance for industrial training in (c)States results without explaining the theoretical significance of CRLB violation or mixture structureNo interpretation of why results matter; fails to connect (b)(ii) to (b)(iii); omits conclusion on method effectiveness in (c)
Final answer & units20%10Clear boxed final answers: explicit variance inequality for (a)(iii), numerical E(X)=0, V(X)=2/3 for (b), and H=2.78 < 5.991 leading to 'fail to reject H₀' for (c) with α=0.05 statedCorrect answers present but poorly organized or missing final decision statement in (c)Missing final answers, incorrect decision (reject H₀), or no significance level mentioned for hypothesis test

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