Statistics 2025 Paper I 50 marks Derive

Q4

(a) Let X₁, X₂, ... be a sequence of random variables from Bernoulli distribution with mean θ, 0<θ<1. Derive SPRT for testing H₀ : θ = θ₀ versus H₁ : θ = θ₁ = 1 – θ₀, 0<θ₀<1. Also obtain expressions for OC function and ASN function. (20 marks) (b) A random sample of size n is taken from the exponential distribution with mean θ>0. Given that n₁ observations out of n observations are less than 'a'. Show that minimum Chi-square estimate and maximum likelihood estimate of θ are same. (15 marks) (c) The life of 6 items of brand-A and 6 items of brand-B are given below: A : 40 62 55 35 48 88 B : 50 70 65 30 45 92 Using Kolmogorov-Smirnov test, test whether the distribution of life of both the brands are same or not at 5% level of significance. [Given that D₍₆, ₆, ₀.₀₅₎ = 2/3] (15 marks)

हिंदी में प्रश्न पढ़ें

(a) मान लीजिए X₁, X₂, ... माध्य θ, 0<θ<1 वाले बर्नौली-बंटन से लिए गए यादृच्छिक चरों का एक अनुक्रम है । H₀ : θ = θ₀ विरुद्ध H₁ : θ = θ₁ = 1 – θ₀, 0<θ₀<1, के परीक्षण के लिए SPRT व्युत्पन्न कीजिए । इस परीक्षण के OC फलन तथा ASN फलन के व्यंजकों को भी प्राप्त कीजिए । (20 अंक) (b) माध्य θ>0 वाले चरघातांकी बंटन से आमाप n का एक यादृच्छिक प्रतिदर्श लिया गया है । दिया गया है कि, n प्रेक्षणों में से n₁ प्रेक्षण 'a' से छोटे हैं । दर्शाइए कि θ का न्यूनतम काई-वर्ग आकलक तथा अधिकतम संभाविता आकलक समान है । (15 अंक) (c) ब्रांड-A की 6 वस्तुओं और ब्रांड-B की 6 वस्तुओं के जीवनकाल नीचे दिये गये हैं : A : 40 62 55 35 48 88 B : 50 70 65 30 45 92 कोलमोगोरोव-स्मिरनोव परीक्षण का उपयोग करते हुए, 5% सार्थकता स्तर पर परीक्षण कीजिए कि दोनों ब्रांड का जीवन का बंटन समान है या नहीं । [दिया गया है कि D₍₆, ₆, ₀.₀₅₎ = ²/₃] (15 अंक)

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Directive word: Derive

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How this answer will be evaluated

Approach

Derive the SPRT for Bernoulli in part (a) with proper likelihood ratio development, spending ~40% time on this 20-mark component; for (b) prove equivalence of MCSE and MLE through differentiation of chi-square and likelihood functions (~30%); for (c) apply K-S test with correct empirical CDF construction and comparison against critical value D₍₆,₆,₀.₀₅₎ = 2/3 (~30%). Structure: state hypotheses → show derivations/computations → conclude with statistical decisions.

Key points expected

Part (a): Derive Wald's SPRT using likelihood ratio Λₙ = (θ₁/θ₀)^ΣXᵢ · ((1-θ₁)/(1-θ₀))^(n-ΣXᵢ) with continuation region A < Λₙ < B
Part (a): Obtain OC function L(θ) ≈ (A^(h(θ))-1)/(A^(h(θ))-B^(h(θ))) and ASN Eθ(N) ≈ L(θ)lnA + (1-L(θ))lnB / Eθ(Z) where Z = ln[f₁(X)/f₀(X)]
Part (b): Set up Pearson's chi-square with cells [0,a) and [a,∞), minimize Σ(Oᵢ-Eᵢ)²/Eᵢ to get MCSE, show same equation as MLE score function
Part (c): Construct ordered empirical distribution functions F₆(x) and G₆(x) for both brands, compute D₆,₆ = sup|F₆(x)-G₆(x)|
Part (c): Compare calculated D statistic with critical value 2/3, conclude whether to reject H₀ of identical distributions at 5% level

Evaluation rubric

Dimension	Weight	Max marks	Excellent	Average	Poor
Setup correctness	20%	10	Correctly specifies H₀, H₁, and parameter spaces for all three parts; for (a) identifies Bernoulli pmf and θ₁ = 1-θ₀ symmetry; for (b) defines exponential pdf and cell probabilities; for (c) properly states K-S null hypothesis and notes equal sample sizes	Basic hypotheses stated but some parameter constraints missing or incorrect pdf specification in one part; may confuse θ₀ and θ₁ roles in SPRT	Major errors in hypothesis specification, wrong distributions assumed, or completely omits setup for one or more sub-parts
Method choice	20%	10	Selects Wald's sequential probability ratio test framework for (a), Pearson's minimum chi-square method for (b), and two-sample K-S test for (c); correctly identifies that likelihood equations coincide in (b) and uses exact Dₙ,ₘ critical value	Correct general methods but may use approximate formulas for OC/ASN in (a) or incorrect minimization approach in (b); K-S procedure basically correct	Wrong test selection (e.g., t-test or Wilcoxon instead of K-S), uses fixed-sample test for sequential problem, or applies MLE directly without showing equivalence in (b)
Computation accuracy	20%	10	Accurate derivation of inequalities for SPRT continuation region; correct differentiation leading to identical estimating equations in (b); precise calculation of D = max\|F₆-G₆\| = 1/6 at x=55 or similar with correct supremum identification	Minor algebraic errors in likelihood ratio simplification or in solving estimating equations; D statistic computed with small arithmetic errors but correct concept	Major computational errors: wrong likelihood ratio, failed differentiation, incorrect empirical CDF values, or D statistic completely miscalculated
Interpretation	20%	10	Explains why SPRT boundaries depend on sum of successes; interprets MCSE-MLE equivalence as consequence of exponential family properties; correctly concludes 'do not reject H₀' since D₆,₆ = 1/6 < 2/3, noting brands have similar life distributions	Basic interpretation of results but limited insight into why methods work; conclusion stated without clear connection to critical value	Misinterprets SPRT as fixed-sample test, draws wrong conclusion about H₀ in K-S test, or fails to interpret the equivalence result in (b)
Final answer & units	20%	10	Presents complete expressions: SPRT decision boundaries in terms of ΣXᵢ, explicit OC and ASN formulas; clear statement that θ̂_MCSE = θ̂_MLE = a/[-ln(1-n₁/n)]; final K-S decision with D statistic value and correct comparison to 2/3	Final answers present but incomplete (e.g., OC function without simplified form) or missing explicit equality proof in (b); K-S conclusion correct but D value not stated	Missing final answers, incorrect boxed results, or no decision stated for hypothesis tests; completely wrong expressions for estimates

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