Paper I — Q7
(a) A very big population is divided into two strata. The allocation of units of stratified random sample of size n for the two…
A very big population is divided into two strata. The allocation of units of stratified random sample of size n for the two strata under Neyman allocation are n'_1 and n'_2, and under other type of allocation are n_1 and n_2. Define r = n'_1/n'_2 and μ = n_1/(rn_2). Then prove that the efficiency of stratified random sampling with respect to stratified random sampling under Neyman allocation is given by e = μ(r+1)²/((μr + 1)(μ + r)). 20 marks
A bank has 40000 clients in its computer files, divided into 4000 branches, each managing exactly 10 clients. To estimate the proportion of clients for whom the bank has granted loan, a simple random sample of 40 branches is selected. From the selected sample, for each branch i, a list of clients (A_i) having a loan is prepared; i = 1, 2, ..., 40. The data observed from the selected sample are Σ(i=1 to 40) A_i = 200 and Σ(i=1 to 40) A_i² = 1156. What type of sampling is this? 3 marks
State the expression of the parameter to estimate and obtain its unbiased estimate. 6 marks
Estimate the variance of the unbiased estimator obtained in part (ii). 6 marks
Verify whether the following BIBD are possible: (1) v = b = 22, r = k = 7, λ = 2; (2) v = 10, b = 18, r = 9, k = 5, λ = 4. Given that the design is resolvable.
Given below is the incidence matrix (N) of a block design. Find the degrees of freedom associated with the adjusted treatment sum of squares and the degrees of freedom for the error sum of squares.
हिंदी में प्रश्न पढ़ें
एक बहुत बड़ी समष्टि को दो स्तरों में विभाजित किया गया है। नेमन नियतन के अनुसार, दो स्तरों के लिए, आमाप n के स्तरीत यादृच्छिक प्रतिदर्श की इकाइयों के नियतन n'_1 और n'_2 हैं और दूसरे प्रकार की नियतन विधि के अनुसार n_1 तथा n_2 हैं। r = n'_1/n'_2 तथा μ = n_1/(rn_2) को परिभाषित कीजिए। तब सिद्ध कीजिए कि नेमन नियतन के अंतर्गत स्तरीत यादृच्छिक प्रतिचयन के सापेक्ष स्तरीत यादृच्छिक प्रतिचयन की दक्षता e = μ(r+1)²/((μr + 1)(μ + r)) है। (20 अंक)
एक बैंक में इसके कम्प्यूटर की फाइलों में 40000 ग्राहक हैं, जिनको 4000 शाखाओं में बाँटा गया है, प्रत्येक शाखा ठीक 10 ग्राहकों का प्रबंध करती है। जिन ग्राहकों को बैंक का ऋण दिया गया है, उनके अनुपात का आकलन करने के लिए 40 शाखाओं का एक सरल यादृच्छिक प्रतिदर्श चुना गया है। चुने गये प्रतिदर्श में से, प्रत्येक शाखा i के लिए ग्राहकों, जिन्होंने ऋण लिया है, उनकी एक सूची (A_i) तैयार की गई है; i = 1, 2, ..., 40 है। चयनित प्रतिदर्श से प्रेक्षित आँकड़े Σ(i=1 से 40) A_i = 200 और Σ(i=1 से 40) A_i² = 1156 प्राप्त हुए हैं। यह किस प्रकार का प्रतिचयन है? (3 अंक)
प्राचल जिसका आकलन करना है, उसके लिए व्यंजक (एक्सप्रेशन) लिखिए और उसका अनभिनत आकलक ज्ञात कीजिए। (6 अंक)
भाग (ii) में प्राप्त अनभिनत आकलक के विचरण का आकलन कीजिए। (6 अंक)
सत्यापित कीजिए कि क्या नीचे दिये गये BIBD संभव हैं: (1) v = b = 22, r = k = 7, λ = 2; (2) v = 10, b = 18, r = 9, k = 5, λ = 4। दिया गया है कि अभिकल्पना विभोज्य है।
नीचे एक खंडक अभिकल्पना का आपतन आव्यूह (N) दिया गया है। समायोजित उपचार वर्गों के योग से संबद्ध स्वातंत्र्य-कोटियाँ और त्रुटि वर्गों के योग के लिए स्वातंत्र्य-कोटियाँ ज्ञात कीजिए।
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.
(b) Summation from i=1 to 40 of A_i = 200 and summation from i=1 to 40 of A_i squared = 1156.
(c) Incidence matrix N of a block design, given as a 6x4 matrix with rows: row 1 = [1, 0, 0, 0]; row 2 = [1, 0, 1, 0]; row 3 = [1, 0, 1, 0]; row 4 = [0, 1, 0, 1]; row 5 = [0, 1, 0, 1]; row 6 = [0, 0, 0, 1]. The question asks to find the degrees of freedom associated with the adjusted treatment sum of squares and the degrees of freedom for the error sum of squares.
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) Efficiency Relative to Neyman Allocation
Let the two strata have sizes N₁, N₂, weights W₁ = N₁/N, W₂ = N₂/N and variances S₁², S₂². For a stratified random sample of size n, the variance of the unbiased estimator of the population mean is
V = W₁²S₁²/n₁ + W₂²S₂²/n₂ (the population is very large, so finite population corrections are ignored).
Put A = W₁²S₁², B = W₂²S₂². Under Neyman allocation n′₁ ∝ W₁S₁ and n′₂ ∝ W₂S₂, so
r = n′₁/n′₂ = W₁S₁/(W₂S₂), hence r² = A/B, i.e. B = A/r².
Neyman allocation gives n′₁ = nr/(r+1), n′₂ = n/(r+1). Therefore
V_N = A/n′₁ + B/n′₂ = (r+1)/n (A/r + B)
= (r+1)/n (A/r + A/r²) = A(r+1)²/(n r²).
For the other allocation, μ = n₁/(r n₂), so n₁ = μ r n/(μ r + 1), n₂ = n/(μ r + 1). Hence
V = A/n₁ + B/n₂ = (μ r + 1)/n (A/(μ r) + B)
= (μ r + 1)/n (A/(μ r) + A/r²)
= A(μ r + 1)(μ + r)/(n μ r²).
Efficiency of the other allocation relative to Neyman allocation is
e = V_N/V = [A(r+1)²/(n r²)] / [A(μ r + 1)(μ + r)/(n μ r²)]
= μ(r+1)²/((μ r + 1)(μ + r)).
Hence proved.
(b)(i) Type of Sampling
This is single-stage cluster sampling with equal clusters. The branches are clusters, each of size 10. A simple random sample of 40 branches is selected, and all clients in the selected branches are examined.
(b)(ii) Parameter and Its Unbiased Estimate
Let Aᵢ be the number of loan clients in branch i, i = 1, ..., 4000. The parameter is the population proportion
P = (Σᵢ₌₁⁴⁰⁰⁰ Aᵢ)/(4000 × 10) = Ā/10,
where Ā is the population mean number of loan clients per branch. The unbiased estimator is
P̂ = (Σᵢ₌₁⁴⁰ Aᵢ)/(40 × 10) = 200/400 = 1/2 = 0.5.
(b)(iii) Estimated Variance of P̂
Here N = 4000, n = 40, M = 10. Sample mean per branch is ā = 200/40 = 5. Sample variance of Aᵢ is
s_A² = [Σ Aᵢ² - nā²]/(n - 1)
= (1156 - 40 × 25)/39 = 156/39 = 4.
Finite population correction f = n/N = 40/4000 = 0.01. Hence
Var(P̂) = (1 - f) s_A²/(n M²)
= (0.99 × 4)/(40 × 100) = 3.96/4000 = 0.00099.
Thus the estimated variance is 0.00099 (estimated standard error ≈ 0.0315).
(c)(i) Are the Resolvable BIBDs Possible?
For a BIBD, necessary conditions are vr = bk and λ(v - 1) = r(k - 1). For a resolvable BIBD, v/k must be an integer.
(1) v = b = 22, r = k = 7, λ = 2. The two equations hold: 22×7 = 22×7 and 2×21 = 7×6 = 42. But v/k = 22/7 is not an integer. Hence it is not possible as a resolvable BIBD. (Also, by Bruck-Ryser-Chowla, since v is even, k - λ = 5 is not a perfect square, so no symmetric BIBD exists.)
(2) v = 10, b = 18, r = 9, k = 5, λ = 4. Here vr = 10×9 = 90 = 18×5, and λ(v - 1) = 4×9 = 36 = 9×4. Also v/k = 10/5 = 2, and b = r(v/k) = 9×2 = 18. However, a resolvable design with v = 2k and λ = k - 1 exists only if a Hadamard matrix of order v = 10 exists. Hadamard matrices exist only for orders 1, 2 or multiples of 4; 10 is not such an order. Hence this is not possible as a resolvable BIBD. The same conclusion follows from Bose's result: a resolvable BIBD satisfies b ≥ v + r − 1; here b = 18 = v + r − 1, so the design would be affine resolvable, which requires k²/v = 25/10 to be an integer. It is not, so the design is impossible.
(c)(ii) Degrees of Freedom from the Incidence Matrix
The incidence matrix N is 6×4, so v = 6 treatments and b = 4 blocks. Total observations n = sum of entries = 1 + 2 + 2 + 2 + 2 + 1 = 10. The incidence graph has two connected components: treatments 1,2,3 with blocks 1,3, and treatments 4,5,6 with blocks 2,4. Thus c = 2.
Adjusted treatment degrees of freedom = v - c = 6 - 2 = 4.
Error degrees of freedom = n - b - v + c = 10 - 4 - 6 + 2 = 2.
How this answer will be evaluated
Framework: UPSC Statistics Paper 1
What earns full marks: Rigorous derivation in (a); correct identification and calculation in (b); precise condition checking in (c).
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) (a) Prove the efficiency of a general allocation relative to Neyman allocation is μ(r+1)²/[(μr+1)(μ+r)] (20 marks) 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- V(ȳ_st)=W₁²S₁²/n₁+W₂²S₂²/n₂ with fpc ignored for a very large population
- Neyman allocation nᵢ′ ∝ WᵢSᵢ gives r=W₁S₁/(W₂S₂), so B=A/r²
- V_N=A(r+1)²/(nr²) using n₁′=nr/(r+1), n₂′=n/(r+1)
- Other allocation n₁=μrn/(μr+1), n₂=n/(μr+1) gives V=A(μr+1)(μ+r)/(nμr²), and e=V_N/V simplifies to the result
Loses marks
- Inverting the ratio (V/V_N)
- Algebraic errors in expressing n₁, n₂ via μ and r
Earns more
- Clear definition of efficiency as the variance ratio
- Checks e=1 when μ=1
Extra mark
- Notes e ≤ 1 by AM–GM, confirming Neyman optimality
- (b)(i) (b)(i) Type of sampling (3 marks) 3 marks
define— precise definition → the distinguishing feature → one example
Must cover
- Identifies single-stage cluster sampling
- Branches are clusters of equal size M=10
- Clusters selected by SRSWOR
- All units in selected clusters are enumerated
Loses marks
- Calling it stratified or two-stage sampling
- Treating clients as the sampling units
Earns more
- Contrasts with stratified sampling
- States N=4000, n=40
Extra mark
- Mentions the intra-cluster correlation's effect on efficiency
- (b)(ii) (b)(ii) Expression for the parameter and its unbiased estimate (6 marks) 6 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Parameter P=ΣAᵢ/(NM) over all 4000 branches
- Estimator P̂=ΣAᵢ/(nM) over sampled branches
- Justifies unbiasedness from SRSWOR of clusters
- P̂=200/400=0.5
Loses marks
- Dividing by 40 instead of 400
- Using a ratio estimator without justification
Earns more
- Expresses P as the mean cluster total divided by M
- Correct notation
Extra mark
- Estimates the total number of loan clients as 20,000
- (b)(iii) (b)(iii) Estimate the variance of P̂ (6 marks) 6 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Sample mean ā=5 and s²=[ΣAᵢ²−nā²]/(n−1)=156/39=4
- f=n/N=0.01
- v(P̂)=(1−f)s²/(nM²)
- v(P̂)=0.99×4/4000=0.00099
Loses marks
- Using the binomial p(1−p)/n formula for clients
- Omitting M² in the denominator
Earns more
- Standard error ≈ 0.0315
- Uses the between-cluster variance correctly
Extra mark
- Computes the design effect relative to SRS of 400 clients
- (c)(i) (c)(i) Verify whether the resolvable BIBDs (22,22,7,7,2) and (10,18,9,5,4) are possible (8 marks) 8 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Checks vr=bk and λ(v−1)=r(k−1) for both designs
- Design 1: v/k=22/7 not an integer, so not resolvable (and Bruck–Ryser–Chowla rules out the symmetric design since v is even and k−λ=5 is not a square)
- Design 2: parameter conditions hold and v/k=2, but b=v+r−1 makes it affine resolvable
- Design 2: k²/v=2.5 is not an integer (equivalently no Hadamard matrix of order 10), so impossible
Loses marks
- Concluding possible merely because the basic equalities hold
- Ignoring the resolvability condition
Earns more
- States Bose's inequality b ≥ v+r−1 for resolvable designs
- Clear verdict for each design
Extra mark
- Notes a non-resolvable 2-(10,5,4) design does exist
- (c)(ii) (c)(ii) Degrees of freedom for adjusted treatment SS and error SS from the given incidence matrix (7 marks) 7 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- v=6 treatments, b=4 blocks, n=10 plots
- Identifies the design as disconnected with c=2 components
- Adjusted treatment df=v−c=4
- Error df=n−b−v+c=2
Loses marks
- Using v−1=5 for treatments
- Counting n as 24 (cells) instead of 10
Earns more
- Lists the components {t₁,t₂,t₃;b₁,b₃} and {t₄,t₅,t₆;b₂,b₄}
- Checks total df 9=3+4+2
Extra mark
- Explains via rank of the C-matrix = v−c
What "Prove" is asking you to do
Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.
Structure that answers it
Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved
Where marks are lost
Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Statistics 2024 Paper I
- Q4 (a) Find the most powerful test of size α(= 0·05) for testing H₀: μ = 0 vs. H₁: μ = 1, gi…
- Q5 (a) How will you justify the usage of the principle of least squares in estimating the pa…
- Q6 (a) (X, Y) has bivariate normal distribution BN(μ₁, μ₂, σ₁², σ₂², ρ). (i) Show that X and…
- Q7 (a) A very big population is divided into two strata. The allocation of units of stratifi…
- Q8 (a) A 2²-factorial design was used to develop the yield of a crop. Two factors A and B we…