Paper I — Q8
(a)(i) What are principal components ? Show that the principal components are uncorrelated. (10 marks) (a)(ii) Obtain the…
What are principal components ? Show that the principal components are uncorrelated. 10 marks
Obtain the principal components and the amount of variation explained by each principal component associated with the following dispersion matrix : Σ = 4 & 2 & 1 2 & 3 & 1 1 & 1 & 2 Comment on the results. 10 marks
For the given data, the yield of the treatment B in the second block is missing and is denoted as 'y'. Estimate the missing value, and analyse the data by assuming the level of significance = 0·05. [Given that F(3, 4) = 6·59; and F(2, 3) = 9·55] 20 marks
Distinguish between Sampling and Non-sampling Errors. What are their sources ? How these errors can be controlled ? 10 marks
हिंदी में प्रश्न पढ़ें
मुख्य घटक क्या हैं ? दर्शाइए कि मुख्य घटक असहसंबंधित हैं। (10 अंक)
निम्नलिखित प्रकीर्णन आव्यूह से संबंधित मुख्य घटकों को प्राप्त कीजिए तथा प्रत्येक मुख्य घटक द्वारा स्पष्ट की गई परिवर्तन की मात्रा प्राप्त कीजिए : Σ = 4 & 2 & 1 2 & 3 & 1 1 & 1 & 2 परिणामों पर टिप्पणी कीजिए। (10 अंक)
दिए गए आंकड़ों के लिए, दूसरे खंड में उपचार B की उपज लुप्त है और इसे 'y' से दर्शाया गया है। लुप्त मान का आकलन कीजिए, और आंकड़ों का सार्थकता स्तर 0·05 पर विश्लेषण कीजिए। [दिया गया है F(3, 4) = 6·59; और F(2, 3) = 9·55] (20 अंक)
प्रतिचयन और अप्रतिचयन त्रुटियों के बीच अंतर कीजिए। उनके स्रोत क्या हैं ? इन त्रुटियों को कैसे नियंत्रित किया जा सकता है ? (10 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) A 3x3 matrix labeled Sigma (dispersion matrix) with the following values: Row 1: 4, 2, 1 Row 2: 2, 3, 1 Row 3: 1, 1, 2
(b) Table with header 'Block' spanning three columns labeled I, II, III. The rows are labeled by 'Treatments' on the left. The data is as follows: Row A: 22, 24, 22 Row B: 20, y, 18 Row C: 21, 25, 20 Below the table, the text reads: [Given that F(3, 4) = 6.59; and F(2, 3) = 9.55]
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)
(i) Principal Components and Uncorrelatedness
Definition: Principal Components (PCs) are a set of new variables, Y₁, Y₂, …, Yₚ, which are linear combinations of the original random variables X₁, X₂, …, Xₚ. They are defined such that the first component explains the maximum variance, the second explains the maximum remaining variance, and so on. Mathematically, Yᵢ = aᵢ₁X₁ + … + aᵢₚXₚ, where the coefficients aᵢⱼ are the elements of the i-th eigenvector of the dispersion matrix Σ.
Proof of Uncorrelatedness: Let Yᵢ and Yⱼ be two distinct principal components (i ≠ j) corresponding to eigenvalues λᵢ and λⱼ. The covariance between them is: Cov(Yᵢ, Yⱼ) = aᵢ' Σ aⱼ Since aⱼ is an eigenvector of Σ, we know Σ aⱼ = λⱼ aⱼ. Substituting this into the equation: Cov(Yᵢ, Yⱼ) = aᵢ' (λⱼ aⱼ) = λⱼ (aᵢ' aⱼ) By the properties of symmetric matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal, meaning aᵢ' aⱼ = 0. Therefore, Cov(Yᵢ, Yⱼ) = 0. This proves that the principal components are uncorrelated.
(ii) Calculation of Principal Components
1. Eigenvalues: We solve the characteristic equation |Σ - λ I| = 0:
| 4-λ & 2 & 1 2 & 3-λ & 1 1 & 1 & 2-λ | = 0 Expanding the determinant yields the cubic equation: λ³ - 9λ² + 20λ - 13 = 0. The roots (eigenvalues) are: λ₁ = 6.049 λ₂ = 1.643 λ₃ = 1.308 (Note: The sum of eigenvalues equals the trace of the matrix: 6.049 + 1.643 + 1.308 = 9.0).
2. Variation Explained: Total Variance = 9.
- PC1: (6.049 / 9) × 100 = 67.2%
- PC2: (1.643 / 9) × 100 = 18.3%
- PC3: (1.308 / 9) × 100 = 14.5%
3. Eigenvectors and components: Solving (Σ − λI)a = 0 and normalising (a'a = 1): Y₁ = 0.737X₁ + 0.591X₂ + 0.328X₃ (λ₁ = 6.049) Y₂ = 0.591X₁ − 0.328X₂ − 0.737X₃ (λ₂ = 1.643) Y₃ = 0.328X₁ − 0.737X₂ + 0.591X₃ (λ₃ = 1.308) The eigenvectors are mutually orthogonal, as the proof in (i) requires.
Comment: The first principal component accounts for 67.2% of the total variation, and the first two together for 85.5%. Y₁ has all positive loadings, so it is a weighted overall "size" index dominated by X₁ and X₂; Y₂ contrasts X₁ with X₃. Since λ₂ and λ₃ are close, Y₂ and Y₃ are not sharply separated, so a one-dimensional summary through Y₁ is reasonable, but retaining two components gives a fuller picture.
(b) Missing Value Estimation and ANOVA
1. Estimation of Missing Value (y): For a Randomized Block Design (RBD) with t treatments and b blocks, the formula for a missing observation y is: y = (t T'_B + b B'_II - G')/((t-1)(b-1)) Where: t = 3 (Treatments A, B, C) b = 3 (Blocks I, II, III) T'_B = 20 + 18 = 38 (Treatment B total excluding y) B'_II = 24 + 25 = 49 (Block II total excluding y) G' = 22+24+22+20+18+21+25+20 = 172 (Grand total excluding y)
y = (3(38) + 3(49) - 172)/((3-1)(3-1)) = (114 + 147 - 172)/4 = 89/4 = 22.25
2. ANOVA Table: The completed Grand Total G = 172 + 22.25 = 194.25. Degrees of Freedom (d.f.) are adjusted: Total d.f. = 9-1-1=7. One d.f. is lost to the estimation of y, so Error d.f. = 7 - 2 - 2 = 3.
- Treatments: d.f. — 2; Sum of Squares (SS) — 10.7917; Mean Square (MS) — 5.3958; F-ratio — 5.25
- Blocks: d.f. — 2; Sum of Squares (SS) — 22.6250; Mean Square (MS) — 11.3125; F-ratio — 11.01
- Error: d.f. — 3; Sum of Squares (SS) — 3.0833; Mean Square (MS) — 1.0278; F-ratio —
- Total: d.f. — 7; Sum of Squares (SS) — 36.5000; Mean Square (MS) — ; F-ratio —
3. Conclusion: (The treatment sum of squares from the filled-in table is slightly biased upward. The bias is [B′ − (t−1)y]²/[t(t−1)] = (49 − 44.5)²/6 = 3.375, so the corrected treatment SS is 7.417 and F = 3.61, which leads to the same conclusion. F(3, 4) is not needed for a 3 × 3 layout.)
- Treatments: F_calc = 5.25. The critical value F(2, 3) = 9.55. Since 5.25 < 9.55, the difference between treatments is not significant.
- Blocks: F_calc = 11.01. The critical value F(2, 3) = 9.55. Since 11.01 > 9.55, the difference between blocks is significant.
(c) Sampling vs. Non-sampling Errors
1. Distinction and Sources:
- Definition: Sampling Error — Chance fluctuation between sample statistic and population parameter.; Non-sampling Error — Bias arising from factors other than sampling.
- Sources: Sampling Error — Random selection, sample size, sampling design.; Non-sampling Error — Specification, coverage, non-response, measurement, processing.
- Control: Sampling Error — Can be reduced by increasing sample size (n) or using stratification.; Non-sampling Error — Controlled by training, call-backs, and data validation.
- Quantification: Sampling Error — Can be estimated (Standard Error).; Non-sampling Error — Generally cannot be quantified.
2. Control Measures: Sampling errors are controlled mathematically by increasing the sample size or using more efficient designs like stratified random sampling. Non-sampling errors are controlled through administrative measures, such as rigorous training of enumerators, conducting follow-up interviews for non-respondents, and implementing data validation checks to minimize measurement and processing errors.
How this answer will be evaluated
Framework: null
What earns full marks: Rigorous derivation for (a), precise ANOVA for (b), and clear distinction for (c).
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Define principal components, prove uncorrelatedness, and compute components for the given matrix. 20 marks
explain— definition/context → points in order → small example → short close
Must cover
- Define principal components as linear combinations of variables.
- Show covariance between components is zero using eigenvector properties.
- Compute eigenvalues of the given 3x3 dispersion matrix.
- Calculate percentage of variation explained by each component.
Loses marks
- Fails to show the uncorrelatedness proof.
- Computes eigenvalues but does not calculate the percentage of variation.
Earns more
- Correctly computes eigenvectors (e.g., Y1 = 0.737X1 + 0.591X2 + 0.328X3) for the eigenvalues.
- Provides a clear comment on the results (e.g., dominance of first component).
Extra mark
- Explicitly states the characteristic equation used for eigenvalues.
- (b) Estimate the missing value 'y' and perform ANOVA on the data. 20 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- RBD missing value y = (tT′+bB′−G′)/((t−1)(b−1)) = 22.25
- ANOVA with adjusted df (total 7, error 3): trt F = 5.25 ns, block F = 11.01 significant vs F(2,3)=9.55
- F(3,4) unused (df mismatch)
Loses marks
- Uses incorrect degrees of freedom in the ANOVA table.
- Fails to compare calculated F with the provided critical values.
Earns more
- Correctly adjusts degrees of freedom for the missing observation.
- Clearly states the conclusion regarding the significance of treatments.
Extra mark
- Explicitly writes the formula for the missing value before substitution.
- (c) Distinguish between sampling and non-sampling errors, their sources, and control. 10 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- Sampling error (chance fluctuation; random selection/size/design) vs non-sampling (specification, coverage, non-response, measurement, processing)
- Control: larger n / stratification and SE quantification vs training, call-backs, validation
Loses marks
- Confuses the sources of the two types of errors.
- Fails to address how the errors can be controlled.
Earns more
- Uses paired headings or point-by-point contrast to distinguish the two types.
- Mentions that sampling error can be quantified but non-sampling cannot.
Extra mark
- Provides a specific example of a non-sampling error source (e.g., non-response).
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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