Paper II — Q5
(a) Apply the method of link relatives to the following data and calculate the seasonal indices : Price of Rice (in ₹ per 10…
Apply the method of link relatives to the following data and calculate the seasonal indices :
Price of Rice (in ₹ per 10 kg)
| Quarter | 2001 | 2002 | 2003 | 2004 |
|---|---|---|---|---|
| 1 | 75 | 86 | 90 | 100 |
| 2 | 60 | 65 | 72 | 78 |
| 3 | 54 | 63 | 66 | 72 |
| 4 | 59 | 80 | 82 | 93 |
10 marks
Derive the means and variances of the sampling distributions of the OLS estimates of α and β in the two-variable linear model Y = α + βX + u. 10 marks
Consider, in the usual notations, the equation y = Y₁β + X₁γ + u, where y is an (n × 1) vector, Y₁ is an (n × (g-1)) matrix, X₁ is an (n × k) matrix. Derive the equations for the two-stage least square method of estimation. 10 marks
If the survivorship function l(x) in life table is linear between x and x+1, and complete expectations of life at ages 40 and 41 for a particular group of persons are 21·39 years and 20·91 years respectively and l(40) = 41176, find the number of persons that attain the age 41. 10 marks
Compute the T-scores corresponding to test score x for the following frequency distribution :
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f | 2 | 3 | 8 | 6 | 1 |
(Cumulative Normal Distribution Table is given in Page No. 9)
10 marks
हिंदी में प्रश्न पढ़ें
निम्नलिखित आँकड़ों पर शृंखलित आपेक्षिक विधि का प्रयोग कीजिए और ऋतुनिष्ठ सूचकांकों की गणना कीजिए :
चावल का मूल्य (₹ में प्रति 10 किलोग्राम)
| क्वार्टर | 2001 | 2002 | 2003 | 2004 |
|---|---|---|---|---|
| 1 | 75 | 86 | 90 | 100 |
| 2 | 60 | 65 | 72 | 78 |
| 3 | 54 | 63 | 66 | 72 |
| 4 | 59 | 80 | 82 | 93 |
10
द्विचर रैखिक निदर्श Y = α + βX + u में α और β के ओ. एल. एस. आकलनों के प्रतिचयन बंटनों के मध्यों और प्रसरणों को व्युत्पन्न कीजिए।
10
प्रचलित संकेतों में, समीकरण y = Y₁β + X₁γ + u पर विचार कीजिए, जहाँ y एक (n × 1) सदिश है, Y₁ एक (n × (g-1)) आव्यूह है, X₁ एक (n × k) आव्यूह है। आकलन की द्विचरण न्यूनतम वर्ग विधि के लिए समीकरणों को व्युत्पन्न कीजिए।
10
यदि वय सारणी में उत्तरजीविता फलन l(x), x और x+1 के बीच रैखिक है तथा किसी विशेष व्यक्तियों के समूह के लिए आयु 40 और 41 पर जीवन की पूर्ण प्रत्याशा क्रमशः 21·39 वर्ष और 20·91 वर्ष हैं, और l(40) = 41176, तब आयु 41 तक पहुँचने वाले व्यक्तियों की संख्या ज्ञात कीजिए।
10
निम्नलिखित बारंबारता बंटन के लिए परीक्षण स्कोर x के संगत (T-समांकी) का परिकलन कीजिए :
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f | 2 | 3 | 8 | 6 | 1 |
(संचयी प्रसामान्य बंटन सारणी पृष्ठ सं० 9 में दी गई है)
10
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) Seasonal indices by link relatives
Link relative = (price of a quarter ÷ price of the preceding quarter) × 100; 2001 Q1 has none.
- 2001: –, 80.00, 90.00, 109.26
- 2002: 145.76, 75.58, 96.92, 126.98
- 2003: 112.50, 80.00, 91.67, 124.24
- 2004: 121.95, 78.00, 92.31, 129.17
Averages: Q1 = 126.74 (three values), Q2 = 78.40, Q3 = 92.72, Q4 = 122.41.
Chain relatives (Q1 = 100): Q2 = 78.40; Q3 = 78.40 × 92.72/100 = 72.69; Q4 = 72.69 × 122.41/100 = 88.98; next Q1 = 88.98 × 126.74/100 = 112.78 (unrounded values carried).
Correction d = (112.78 − 100)/4 = 3.19. Corrected chain relatives: 100, 75.20, 66.30, 79.40; mean = 80.23.
Seasonal index = corrected value ÷ 80.23 × 100: Q1 = 124.65, Q2 = 93.74, Q3 = 82.65, Q4 = 98.97 (sum 400). Rice is dearest in Q1 and cheapest in Q3.
(b) Means and variances of the OLS estimates
Assume X fixed, E(uᵢ) = 0, Var(uᵢ) = σ², Cov(uᵢ, uⱼ) = 0. Let xᵢ = Xᵢ − X̄, Sxx = Σxᵢ², kᵢ = xᵢ/Sxx, so Σkᵢ = 0, ΣkᵢXᵢ = 1, Σkᵢ² = 1/Sxx.
β̂ = ΣxᵢYᵢ/Sxx = ΣkᵢYᵢ = β + Σkᵢuᵢ.
- E(β̂) = β
- Var(β̂) = σ²Σkᵢ² = σ²/Sxx
α̂ = Ȳ − β̂X̄ = α + Σ(1/n − X̄kᵢ)uᵢ.
- E(α̂) = α
- Var(α̂) = σ²[1/n − 2X̄Σkᵢ/n + X̄²Σkᵢ²] = σ²(1/n + X̄²/Sxx) = σ²ΣXᵢ²/(nSxx)
Also Cov(α̂, β̂) = −X̄σ²/Sxx. If uᵢ ~ N(0, σ²), α̂ and β̂ are exactly normal with these means and variances.
(c) Two-stage least squares equations
Let X (n × K) contain all exogenous variables of the system, including X₁, with K ≥ g − 1 + k (order condition). Y₁ is correlated with u, so OLS of y on [Y₁ X₁] is inconsistent. Let P = X(X′X)⁻¹X′; it is symmetric and idempotent, and PX₁ = X₁.
Stage 1: regress Y₁ on X: Π̂ = (X′X)⁻¹X′Y₁, giving Ŷ₁ = XΠ̂ = PY₁.
Stage 2: regress y on Ŷ₁ and X₁. The normal equations are Ŷ₁′Ŷ₁β̂ + Ŷ₁′X₁γ̂ = Ŷ₁′y X₁′Ŷ₁β̂ + X₁′X₁γ̂ = X₁′y. Since Ŷ₁′Ŷ₁ = Y₁′PY₁ and Ŷ₁′X₁ = Y₁′X₁, they become Y₁′PY₁β̂ + Y₁′X₁γ̂ = Y₁′Py X₁′Y₁β̂ + X₁′X₁γ̂ = X₁′y.
With Z = [Y₁ X₁] and δ = (β′, γ′)′: δ̂ = (Z′PZ)⁻¹Z′Py. The estimated covariance is σ̂²(Z′PZ)⁻¹, with σ̂² = (y − Zδ̂)′(y − Zδ̂)/(n − g + 1 − k), using Y₁, not Ŷ₁, in the residuals.
(d) Persons attaining age 41
If l(x) is linear on (x, x + 1), the person-years lived in that year are L(x) = [l(x) + l(x+1)]/2. Since e°(x) = T(x)/l(x) and T(x) = L(x) + T(x+1), e°(x) = ½ + p[½ + e°(x+1)], where p = l(x+1)/l(x). So p = [e°(x) − ½]/[e°(x+1) + ½].
For x = 40: p = (21.39 − 0.5)/(20.91 + 0.5) = 20.89/21.41 = 0.97571. l(41) = 41176 × 20.89/21.41 = 40175.9 ≈ 40176.
About 40176 persons attain age 41 (about 1000 of the 41176 die in the year). Check: ½ + 0.97571 × 21.41 = 21.39.
(e) T-scores from the frequency distribution
Convert each score to its mid-point percentile rank PR = 100(cf below + f/2)/N, with N = 20; read z from the normal table with Φ(z) = PR/100; then T = 50 + 10z.
- x = 1: PR = 100(0 + 1)/20 = 5; z = −1.645; T = 33.6
- x = 2: PR = 100(2 + 1.5)/20 = 17.5; z = −0.93 (0.825 lies between Φ(0.93) = 0.8238 and Φ(0.94) = 0.8264); T = 40.7
- x = 3: PR = 100(5 + 4)/20 = 45; z = −0.13; T = 48.7
- x = 4: PR = 100(13 + 3)/20 = 80; z = 0.84; T = 58.4
- x = 5: PR = 100(19 + 0.5)/20 = 97.5; z = 1.96; T = 69.6
For PR below 50 the table is read by symmetry: z = −(z for 100 − PR).
How this answer will be evaluated
Framework: UPSC Statistics Paper 2
What earns full marks: All derivations complete with correct notation; all calculations accurate with proper interpretation; clean tables throughout
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Calculate the seasonal indices for the four quarters by the method of link relatives. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Link relatives (quarter ÷ preceding quarter × 100, 2001 Q1 omitted) and quarter averages 126.74, 78.40, 92.72, 122.41
- Chain relatives with Q1 = 100 and trend correction d = 3.19, giving 100, 75.20, 66.30, 79.40
- Seasonal indices 124.65, 93.74, 82.65, 98.97 as percentages of the mean 80.23 (sum 400)
Loses marks
- Omitting the trend correction, so the chain does not return to 100
- Indices that do not sum to about 400
Earns more
- Q1 average taken over three values, the other quarters over four
- Formula for the correction d = (next Q1 chain − 100)/4 shown
Extra mark
- Reads the pattern: rice dearest in Q1, cheapest in Q3
- (b) Derive the means and variances of the sampling distributions of the OLS estimates of α and β in Y = α + βX + u. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Assumptions stated (X fixed, E(u) = 0, constant variance σ², uncorrelated errors) and β̂ = β + Σkᵢuᵢ with kᵢ = xᵢ/Sxx
- E(β̂) = β and Var(β̂) = σ²/Sxx
- E(α̂) = α and Var(α̂) = σ²(1/n + X̄²/Sxx) = σ²ΣXᵢ²/(nSxx)
Loses marks
- Dropping the X̄²/Sxx term from Var(α̂)
- Claiming unbiasedness without E(u) = 0 and fixed X
Earns more
- α̂ written as α + Σ(1/n − X̄kᵢ)uᵢ
- Cov(α̂, β̂) = −X̄σ²/Sxx
Extra mark
- States that normal errors make α̂ and β̂ exactly normal
- (c) Derive the equations of two-stage least squares estimation for y = Y₁β + X₁γ + u. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Stage 1: regress Y₁ on all exogenous variables X (not on X₁ alone), Ŷ₁ = X(X′X)⁻¹X′Y₁ = PY₁
- Stage 2: regress y on Ŷ₁ and X₁ and write the normal equations, reduced using PX₁ = X₁
- Estimator δ̂ = (Z′PZ)⁻¹Z′Py with Z = [Y₁ X₁], δ = (β′, γ′)′
Loses marks
- Regressing Y₁ on X₁ only in stage 1
- Treating 2SLS as one OLS regression of y on Y₁ and X₁
Earns more
- Order condition K ≥ g − 1 + k and the reason OLS is inconsistent
- P symmetric and idempotent, so Ŷ₁′Ŷ₁ = Y₁′PY₁
Extra mark
- Covariance estimate σ̂²(Z′PZ)⁻¹ with residuals computed from Y₁, not Ŷ₁
- (d) Find the number of persons attaining age 41 when l(x) is linear between x and x + 1, e°(40) = 21.39, e°(41) = 20.91 and l(40) = 41176. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Linearity gives L(x) = [l(x) + l(x+1)]/2 and e°(x) = ½ + p[½ + e°(x+1)] with p = l(x+1)/l(x)
- p = (21.39 − 0.5)/(20.91 + 0.5) = 20.89/21.41 = 0.97571
- l(41) = 41176 × 20.89/21.41 ≈ 40176 persons
Loses marks
- Omitting the ½ terms (curtate instead of complete expectation)
- Final answer not given as a number of persons
Earns more
- Uses T(x) = L(x) + T(x+1) to link e°(x) and e°(x+1)
- Checks that the result reproduces e°(40) = 21.39
- (e) Compute the T-scores for the test scores x = 1 to 5 of the frequency distribution, using the normal table. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Mid-point percentile ranks PR = 100(cf below + f/2)/N: 5, 17.5, 45, 80, 97.5
- z from the normal table: −1.645, −0.93, −0.13, 0.84, 1.96
- T = 50 + 10z: 33.6, 40.7, 48.7, 58.4, 69.6
Loses marks
- Using the cumulative percentage without the f/2 correction
- Using the linear z = (x − x̄)/s instead of the normal table
Earns more
- N = 20 and the formula for PR stated
- Symmetry used to read z for PR below 50
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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