Statistics 2022 Paper I 50 marks Discuss

Q4

(a) Consider Poisson distribution $$P_{\theta}(X = j) = \frac{e^{-\theta} \theta^{j}}{j!} = p_{j}, j = 0, 1, 2, ....$$ Let $f_{j}$ be the frequency for X = j and $E(f_{j}) = m_{j} = np_{j}$. Discuss how you obtain minimum chi-square estimate for $\theta$. Does minimum chi-square method necessarily yield a sufficient statistic even if it exists ? (20 marks) (b) (i) Let the joint probability density function of X and Y be $$f(x, y) = C . \exp \{-(4x^{2} + 9y^{2} - xy)\},$$ where C is a constant. Find E(X), V(X), E(Y), V(Y) and the correlation coefficient between X and Y. (10 marks) (ii) If $X_{1}, X_{2}, ..., X_{6}$ are independent random variables such that $$P(X_{i} = -1) = P(X_{i} = 1) = \frac{1}{2}, i = 1, 2, ..., 6,$$ then obtain the value of $$P\left[\sum_{i=1}^{6} X_{i} = 4\right].$$ (5 marks) (c) The following data present the time (in minutes), that a commuter had to wait to catch a bus to reach his destination : Use the sign-test at 0·05 level of significance to test the claim of the bus operators that commuters do not have to wait for more than 15 minutes before the bus is made available to them. [Given Z₍₀.₀₂₅₎ = 1·96, Z₍₀.₀₅₎ = 1·645] (15 marks)

हिंदी में प्रश्न पढ़ें

(a) प्वासों बंटन $$P_{\theta}(X = j) = \frac{e^{-\theta} \theta^{j}}{j!} = p_{j}, j = 0, 1, 2, ....$$ पर विचार कीजिए । मान लीजिए कि X = j की बारम्बारता $f_{j}$ है तथा $E(f_{j}) = m_{j} = np_{j}$ है । आप $\theta$ का न्यूनतम काई-वर्ग आकल कैसे प्राप्त करेंगे, इसकी विवेचना कीजिए । यदि पर्याप्त प्रतिदर्शज का अस्तित्व भी है तो क्या न्यूनतम काई-वर्ग विधि पर्याप्त प्रतिदर्शज अवश्य देगा ? (20 अंक) (b) (i) मान लीजिए कि X तथा Y का संयुक्त प्रायिकता घनत्व फलन $$f(x, y) = C . \exp \{-(4x^{2} + 9y^{2} - xy)\},$$ है, जहाँ C एक अचर है । E(X), V(X), E(Y), V(Y) और X और Y के बीच सहसंबंध गुणांक को ज्ञात कीजिए । (10 अंक) (ii) यदि स्वतंत्र यादृच्छिक चर $X_{1}, X_{2}, ..., X_{6}$ इस प्रकार हैं कि $$P(X_{i} = -1) = P(X_{i} = 1) = \frac{1}{2}, i = 1, 2, ..., 6$$ हैं, तब $$P\left[\sum_{i=1}^{6} X_{i} = 4\right]$$ का मान प्राप्त कीजिए । (5 अंक) (c) निम्नलिखित आँकड़े एक यात्री को उसके गंतव्य तक पहुँचने के लिए बस को पकड़ने के लिए किए गए प्रतीक्षा समय (मिनटों में) को दर्शाते हैं : साइन-परीक्षण का 0·05 सार्थकता स्तर पर उपयोग करते हुए बस संचालकों के द्वारा दावा कि यात्रियों को बस को पकड़ने के लिए 15 मिनट से अधिक प्रतीक्षा नहीं करनी पड़ती, का परीक्षण कीजिए। [दिया गया है Z₍₀.₀₂₅₎ = 1·96, Z₍₀.₀₅₎ = 1·645] (15 अंक)

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Directive word: Discuss

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

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

The directive 'discuss' in part (a) requires a balanced analytical treatment with derivation and critical evaluation, while parts (b) and (c) are primarily computational. Allocate approximately 40% of effort to part (a) given its 20 marks and theoretical depth, 30% to part (b) covering both (i) bivariate normal properties and (ii) probability calculation, and 30% to part (c) for the non-parametric test. Structure as: brief theoretical exposition for (a), systematic derivations for (b), and complete hypothesis testing procedure for (c).

Key points expected

  • For (a): Derivation of minimum chi-square estimator for Poisson parameter by minimizing Σ(f_j - np_j)²/(np_j) with respect to θ, leading to the estimating equation
  • For (a): Critical discussion that minimum chi-square method does NOT necessarily yield sufficient statistics—contrast with MLE which preserves sufficiency via factorization theorem; cite example where MCS estimator differs from sufficient statistic
  • For (b)(i): Recognition of bivariate normal form, completion of squares to identify μ_x = μ_y = 0, extraction of variances σ_x² = 9/35, σ_y² = 4/35 and covariance to find ρ = 1/6
  • For (b)(ii): Identification that ΣX_i follows distribution of (number of +1's) - (number of -1's), equivalent to 2×Binomial(6,½) - 6, yielding P(ΣX_i=4) = P(5 successes) = 6/64 = 3/32
  • For (c): Correct application of sign test with null hypothesis H₀: median ≤ 15 vs H₁: median > 15, counting positive signs (values > 15), using normal approximation with continuity correction, and proper conclusion based on Z = 1.645 critical value

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly sets up chi-square minimization for (a) with proper likelihood and constraint; identifies bivariate normal structure in (b)(i) with accurate matrix completion; establishes correct Binomial framework for (b)(ii); properly formulates sign test hypotheses and identifies test statistic for (c)Basic setup present but minor errors in constraint handling for (a), incomplete square completion for (b)(i), or incorrect direction of alternative hypothesis for (c)Fundamental setup errors: wrong objective function for chi-square, failure to recognize bivariate normal form, incorrect probability model for sum of Rademacher variables, or completely wrong test selection for (c)
Method choice20%10Selects appropriate calculus-based optimization for MCS in (a), matrix inversion/completion of squares for (b)(i), exact combinatorial counting for (b)(ii), and correct normal approximation with continuity correction for sign test in (c)Correct general methods but suboptimal choices: numerical instead of analytical optimization, unnecessary integration instead of pattern recognition, or omission of continuity correctionInappropriate methods: using MLE instead of MCS in (a), treating variables as independent in (b)(i), using CLT unnecessarily for exact calculation in (b)(ii), or using t-test instead of sign test for (c)
Computation accuracy20%10Flawless calculations: correct MCS estimating equation solution, precise variance-covariance extraction yielding ρ = 1/6, exact probability 3/32, and accurate Z-score with proper sign count leading to correct decisionMinor computational slips: arithmetic errors in solving equations, slight miscalculations in correlation coefficient, or small errors in binomial probabilityMajor computational failures: incorrect derivatives, wrong determinant calculations, probability values outside [0,1], or critical value misapplication causing wrong conclusion
Interpretation20%10For (a), clearly explains why MCS may lose sufficiency compared to MLE with concrete reasoning; for (b), interprets correlation magnitude and distributional properties; for (c), correctly interprets p-value in context of bus operator's claim with practical significanceAdequate interpretation but lacking depth: generic statements about sufficiency without comparison, mechanical reporting of statistics without context, or correct but minimal conclusionMisinterpretation of results: claiming MCS always preserves sufficiency, incorrect sign of correlation interpretation, or reversing null/alternative conclusion in hypothesis test
Final answer & units20%10All final answers explicitly stated with proper units where applicable: MCS estimator formula, numerical values for E(X), V(X), E(Y), V(Y), ρ, exact probability fraction, and clear reject/fail-to-reject decision with significance level for sign testMost answers present but some missing: estimator without final simplified form, correlation without numerical value, or decision without justificationMissing or incorrect final answers: no explicit estimator, incomplete probability distribution results, or no conclusion for hypothesis test despite calculations

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