Paper I — Q7
(a) Consider the following data given for a BIBD with v = b = 4, r = k = 3, λ = 2 and N = 12 : Block : 1 2 3 4 Treatment 1 : 73…
Consider the following data given for a BIBD with v = b = 4, r = k = 3, λ = 2 and N = 12 : Block : 1 2 3 4 Treatment 1 : 73 74 – 71 Treatment 2 : – 75 67 72 Treatment 3 : 73 75 68 – Treatment 4 : 75 – 72 75 Analyse the design. [Given that : F₃,₅ (0·05) = 5·41] 15
The data matrix of a random sample of size n = 3 from a bivariate normal population BVN (μ₁, μ₂, σ₁², σ₂², ρ) is X = [6 10; 10 6; 8 2]. Test the null hypothesis H₀ : μ = μ₀ against H₁ : μ ≠ μ₀, where μ₀' = (8, 5), at 10% level of significance. [You are given : F₀.₁₀; ₂, ₁ = 49·5, F₀.₁₀; ₁, ₂ = 8·53]
Suppose n₁ = 11 and n₂ = 12, observations are made on two random vectors X₁ and X₂ which are assumed to have bivariate normal distribution with a common covariance matrix Σ, but possibly different mean vectors μ₁ and μ₂. The sample mean vectors and pooled covariance matrix are X̄₁ = (-1, -1)', X̄₂ = (2, 1)', S_pooled = (7 -1; -1 5). Obtain Mahalanobis sample distance D² and Fisher's linear discriminant function. Assign the observation X₀ = (0, 1)' to either population Π₁ or Π₂. 10+10=20
A sample of size n is drawn with equal probability and without replacement from a population with size N. Let Ŷ_N = Σᵣ₌₁ⁿ aᵣ yᵣ be any linear estimate of the population mean Ȳ_N, where aᵣ are constants and yᵣ denotes the value of the unit included in the sample at the rᵗʰ draw.
Show that Ŷ_N is an unbiased estimate of Ȳ_N if and only if Σᵣ₌₁ⁿ aᵣ = 1
Under above condition V(Ŷ_N) = (S²/N)[NΣᵣ₌₁ⁿ aᵣ² - 1]
If aᵣ = 1/n, for what value of n may this variance of the sample mean in simple random sampling without replacement be exactly half the variance of the mean of a random sample of the same size taken with replacement ? 15 marks
हिंदी में प्रश्न पढ़ें
किसी बी.आई.बी.डी. (BIBD), जहाँ v = b = 4, r = k = 3, λ = 2 और N = 12, के लिए दिए गए निम्नलिखित आँकड़ों पर विचार कीजिए : खण्ड : 1 2 3 4 उपचार 1 : 73 74 – 71 उपचार 2 : – 75 67 72 उपचार 3 : 73 75 68 – उपचार 4 : 75 – 72 75 अभिकल्पना का विश्लेषण कीजिए । [दिया गया है : F₃,₅ (0·05) = 5·41] 15
एक द्विचर प्रसामान्य समष्टि BVN (μ₁, μ₂, σ₁², σ₂², ρ) से लिए गए आमाप n = 3 के एक यादृच्छिक प्रतिदर्श का न्यास मैट्रिक्स X = [6 10; 10 6; 8 2] है। वैकल्पिक परिकल्पना H₁ : μ ≠ μ₀ के विरुद्ध निराकरणीय परिकल्पना H₀ : μ = μ₀, का परीक्षण 10% सार्थकता-स्तर पर कीजिए, जहाँ μ₀' = (8, 5) है। [आपको दिया गया है : F₀.₁₀; ₂, ₁ = 49·5, F₀.₁₀; ₁, ₂ = 8·53]
मान लीजिए कि दो यादृच्छिक सदिशों X₁ और X₂, जो एक समान सहप्रसरण आव्यूह Σ, किन्तु सम्भवतः भिन्न माध्य सदिशों μ₁ और μ₂ के साथ द्विचर प्रसामान्य बंटन का अनुसरण करते माने जाते हैं, पर n₁ = 11 और n₂ = 12 प्रेक्षण बनाए जाते हैं। प्रतिदर्श माध्य सदिश और संयुक्त सहप्रसरण आव्यूह हैं : X̄₁ = (-1, -1)', X̄₂ = (2, 1)', Sसंयुक्त = (7 -1; -1 5)। महालनोबिस प्रतिदर्श दूरी D² और फिशर के रैखिक विभिक्तकर फलन को प्राप्त कीजिए। प्रेक्षण X₀ = (0, 1)' को या तो समष्टि Π₁ या Π₂ को निर्दिष्ट कीजिए। 10+10=20
N आकार की समष्टि से n आकार का एक प्रतिदर्श समान प्रायिकता एवं प्रतिस्थापन रहित के साथ चुना गया । मान लीजिए कि Ŷ_N = Σᵣ₌₁ⁿ aᵣ yᵣ समष्टि माध्य Ȳ_N का कोई रैखिक आकल है, जहाँ aᵣ अचर हैं और yᵣ rवें ड्रा पर प्रतिदर्श में सम्मिलित इकाई का मान है ।
दर्शाइए कि Ŷ_N, Ȳ_N का एक अनभिनत आकल है यदि और केवल यदि Σᵣ₌₁ⁿ aᵣ = 1
उपर्युक्त प्रतिबंध के अंतर्गत V(Ŷ_N) = (S²/N)[NΣᵣ₌₁ⁿ aᵣ² - 1]
यदि aᵣ = 1/n, तो n के किस मान के लिए प्रतिस्थापन रहित सरल यादृच्छिक प्रतिचयन में प्रतिदर्शी माध्य का यह प्रसरण उसी आकार के प्रतिस्थापन सहित लिए गए यादृच्छिक प्रतिदर्श के माध्य के प्रसरण का बिल्कुल आधा होगा ? 15 अंक
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) Table with header row 'Treatment' and 'Block' (subdivided into columns 1, 2, 3, 4). Rows are: 1: 73, 74, -, 71 2: -, 75, 67, 72 3: 73, 75, 68, - 4: 75, -, 72, 75
(b) Vector X-bar-1 is a column vector (-1, -1). Vector X-bar-2 is a column vector (2, 1). Matrix S_pooled is a 2x2 matrix with rows: [7, -1], [-1, 5]. Vector X-0 is a column vector (0, 1).
(c) Table with 4 columns and 11 rows (including header). The table is split into two halves, each with 2 columns.
Left half: Header: Days | Time Row 1: 1 | 23 Row 2: 2 | 25 Row 3: 3 | 12 Row 4: 4 | 07 Row 5: 5 | 17 Row 6: 6 | 16 Row 7: 7 | 13 Row 8: 8 | 26 Row 9: 9 | 27 Row 10: 10 | 12
Right half: Header: Days | Time Row 1: 11 | 14 Row 2: 12 | 14 Row 3: 13 | 16 Row 4: 14 | 19 Row 5: 15 | 23 Row 6: 16 | 24 Row 7: 17 | 12 Row 8: 18 | 18 Row 9: 19 | 11 Row 10: 20 | 08
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Analysis of the BIBD
Model: y_ij = μ + τ_i + β_j + e_ij; H₀: τ₁ = τ₂ = τ₃ = τ₄ (intra-block analysis). Totals: G = 870; treatments T = (218, 214, 216, 222); blocks B = (221, 224, 207, 218). CF = G²/N = 870²/12 = 63075. SS(total) = Σy² − CF = 63156 − 63075 = 81. SS(blocks, unadjusted) = ΣB²/k − CF = 189390/3 − 63075 = 55. Adjusted treatment totals Q_j = kT_j − (totals of the blocks containing treatment j): Q = (−9, −7, −4, 20), with ΣQ = 0. SS(treatments, adjusted) = ΣQ_j²/(λkv) = 546/24 = 22.75. SS(error) = 81 − 55 − 22.75 = 3.25.
Source: d.f.; SS; MS
- Blocks (unadjusted): 3; 55; 18.33
- Treatments (adjusted): 3; 22.75; 7.583
- Error: 5; 3.25; 0.65
- Total: 11; 81
F = 7.583/0.65 = 11.67 > F₃,₅(0.05) = 5.41, so H₀ is rejected: the treatments differ significantly (p ≈ 0.011).
Adjusted effects: τ̂_j = Q_j/(λv) = (−1.125, −0.875, −0.5, 2.5). SE of a difference = √(2kMSE/(λv)) = √0.4875 = 0.698, so CD(5%) = 2.571 × 0.698 = 1.79. Treatment 4 exceeds each other treatment by 3.0 to 3.6, more than the CD, while treatments 1, 2, 3 differ among themselves by at most 0.63.
(b)(i) Hotelling's T² test for the mean vector
H₀: μ = (8, 5)′ against H₁: μ ≠ (8, 5)′; n = 3, p = 2. x̄ = (8, 6)′; deviations (−2, 4), (2, 0), (0, −4) give S = [[4, −4], [−4, 16]], |S| = 48, S⁻¹ = (1/48)[[16, 4], [4, 4]]. x̄ − μ₀ = (0, 1)′, so T² = n(x̄ − μ₀)′S⁻¹(x̄ − μ₀) = 3 × 4/48 = 0.25. F = (n − p)T²/[p(n − 1)] = 0.25/4 = 0.0625, which follows an F distribution with (p, n − p) = (2, 1) d.f. under H₀. Critical value F₀.₁₀;₂,₁ = 49.5. Since 0.0625 < 49.5, H₀ is not rejected at the 10% level: the data are consistent with μ = (8, 5)′. (F₀.₁₀;₁,₂ = 8.53 is not needed.)
(b)(ii) Mahalanobis D² and Fisher's discriminant function
d = x̄₁ − x̄₂ = (−3, −2)′; |S| = 35 − 1 = 34, S⁻¹ = (1/34)[[5, 1], [1, 7]], S⁻¹d = (−1/2, −1/2)′. D² = d′S⁻¹d = 1.5 + 1 = 2.5. Fisher's function: ŷ = d′S⁻¹x = −(x₁ + x₂)/2. Mean scores: ȳ₁ = 1, ȳ₂ = −1.5, midpoint m = −0.25. Rule (equal priors and costs): assign x to Π₁ if ŷ ≥ m, else to Π₂. For x₀ = (0, 1)′: ŷ₀ = −0.5 < −0.25, so x₀ is assigned to Π₂. Check: (x₀ − x̄₁)′S⁻¹(x₀ − x̄₁) = 37/34 ≈ 1.09 exceeds (x₀ − x̄₂)′S⁻¹(x₀ − x̄₂) = 20/34 ≈ 0.59, so x₀ is nearer Π₂.
(c) Linear estimate under SRSWOR
(i) At every draw each unit is equally likely, so E(y_r) = Ȳ_N for all r and E(Ŷ_N) = Ȳ_NΣa_r. This equals Ȳ_N for every population only if Σa_r = 1; conversely Σa_r = 1 gives E(Ŷ_N) = Ȳ_N.
(ii) Let σ² = (N − 1)S²/N. Then V(y_r) = σ² and, for r ≠ s, Cov(y_r, y_s) = −σ²/(N − 1) = −S²/N (for n = N, Σy_r = NȲ_N is constant, so Nσ² + N(N − 1)Cov = 0). Hence V(Ŷ_N) = (N − 1)S²Σa_r²/N − (S²/N)[(Σa_r)² − Σa_r²]. With Σa_r = 1 this is (S²/N)[(N − 1)Σa_r² − 1 + Σa_r²] = (S²/N)[NΣa_r² − 1].
(iii) For a_r = 1/n, Σa_r² = 1/n and V(ȳ) = (S²/N)(N/n − 1) = (N − n)S²/(Nn). With replacement, V(ȳ) = σ²/n = (N − 1)S²/(Nn). Setting (N − n) = ½(N − 1): n = (N + 1)/2 (N odd), e.g. N = 5, n = 3.
How this answer will be evaluated
Framework: UPSC Statistics Paper 1
What earns full marks: Rigorous derivations, correct ANOVA, and clear statistical interpretation
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Analyse the given BIBD (v = b = 4, r = k = 3, λ = 2, N = 12) with an ANOVA of treatments adjusted for blocks and the F-test against F₃,₅(0.05) = 5.41. 15 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- CF = 63075, SS(total) = 81 and SS(blocks) = 55
- Adjusted treatment totals Q = (−9, −7, −4, 20), SS(treatments, adjusted) = ΣQ²/(λkv) = 22.75, SS(error) = 3.25, with d.f. 3, 3, 5
- F = 11.67 > 5.41, so the treatment differences are significant
Loses marks
- Using the unadjusted treatment sum of squares
- F-test without the critical value or the conclusion
Earns more
- ANOVA table with correct d.f. and mean squares (7.583 and 0.65)
- Adjustment of treatments for blocks explained (intra-block analysis)
Extra mark
- Adjusted effects with CD = 1.79 identifying treatment 4
- (b)(i) Test H₀: μ = (8, 5)′ against H₁: μ ≠ (8, 5)′ at the 10% level with Hotelling's T² for the given bivariate sample of size 3. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- x̄ = (8, 6)′ and S = [[4, −4], [−4, 16]] with divisor n − 1
- T² = 0.25 and F = (n − p)T²/[p(n − 1)] = 0.0625 with (2, 1) d.f.
- Comparison with F₀.₁₀;₂,₁ = 49.5 and H₀ not rejected
Loses marks
- Using (1, 2) d.f. and the value 8.53
- Dividing the sum of products by n instead of n − 1
Earns more
- S⁻¹ = (1/48)[[16, 4], [4, 4]]
- Null and alternative hypotheses stated
Extra mark
- Equivalent critical value T² = 198
- (b)(ii) Obtain the Mahalanobis sample distance D² and Fisher's linear discriminant function, and assign x₀ = (0, 1)′ to Π₁ or Π₂. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- D² = (x̄₁ − x̄₂)′S⁻¹(x̄₁ − x̄₂) = 2.5
- Fisher's function ŷ = (x̄₁ − x̄₂)′S⁻¹x = −(x₁ + x₂)/2 with midpoint −0.25
- ŷ₀ = −0.5 < −0.25, so x₀ is assigned to Π₂
Loses marks
- Assigning x₀ to Π₁ by reversing the rule
- Using S instead of S⁻¹ in D²
Earns more
- S⁻¹ = (1/34)[[5, 1], [1, 7]]
- Cross-check with distances 37/34 and 20/34
Extra mark
- Stating the equal priors and costs assumption of the rule
- (c) Show that Σa_r y_r is unbiased for the population mean iff Σa_r = 1, that its variance is (S²/N)[NΣa_r² − 1], and find the n for which, with a_r = 1/n, the SRSWOR variance is half the SRSWR variance. 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- E(y_r) = Ȳ_N at each draw, so E(Ŷ_N) = Ȳ_NΣa_r and the if-and-only-if condition Σa_r = 1
- V(y_r) = σ² and Cov(y_r, y_s) = −S²/N leading to (S²/N)[NΣa_r² − 1]
- n = (N + 1)/2 from N − n = (N − 1)/2
Loses marks
- Proving only the 'if' part of (i)
- Using S² in place of σ² = (N − 1)S²/N for sampling with replacement
Earns more
- Justification of the covariance −σ²/(N − 1)
- Variance with replacement σ²/n = (N − 1)S²/(Nn)
Extra mark
- Numerical illustration N = 5, n = 3
What "Analyse" is asking you to do
Break the subject into its working parts and show how they act on each other. The marks are in the interconnections — which factor drives which, and what the resulting structure explains — not in the inventory of factors.
Structure that answers it
Define the whole → separate it into its parts → show which part drives which → what that interaction produces → what the structure implies
Where marks are lost
A flat list of causes with no account of which drives which. An answer of neatly separated headings, each self-contained, scores as description.
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