UPSC Statistics 2026 — Paper II

(a) A frailty/mixture problem: X | theta ~ Exponential(hazard theta) and theta ~ Weibull(shape alpha, scale lambda) with pdf f(theta) = (alpha/lambda)(theta/lambda)^(alpha-1) exp(-(theta/lambda)^alpha). Candidate must (i) derive the unconditional survival function S_X(x) by mixing (un-conditioning) over theta, and (ii) obtain the unconditional hazard rate h_X(x) and state its shape. Both sub-parts must be answered. (b) Design a single sampling plan by attributes, i.e. find the sample size n and acceptance number c, given AQL p1 = 0.05 with producer's risk alpha = 0.05 and LTPD p2 = 0.20 with consumer's risk beta = 0.10. The plan must be stated explicitly and its two risk conditions verified. (c) Describe how the choice of subgroup size n affects the X-bar, R and s control charts, treating the three stated ranges separately: n <= 5, 5 < n <= 10, and n > 10. The answer must cover both which dispersion chart is appropriate in each range and what happens to the limits and sensitivity of the X-bar chart. (d) For the given 3-state transition probability matrix P, use the Chapman-Kolmogorov equations (time-homogeneity) to evaluate the two-step probability P(X5 = 2 | X3 = 3) = p_32^(2) and the three-step probability P(X3 = 1 | X0 = 3) = p_31^(3). Both values must be computed. (e) Two parts: (i) list and explain the key features of IBM SPSS Statistics, and (ii) compare SPSS with R on the dimensions that matter to a statistician - cost/licensing, interface, extensibility, output, graphics, data handling and suitability for different tasks. Both parts must be answered.

• S(x | theta) = exp(-theta x)
• S_X(x) = E_theta[exp(-theta x)] (Laplace transform of the frailty distribution)
• h_X(x) = E[theta | X > x]
• h_X'(x) = -Var(theta | X > x) < 0
• Decreasing hazard rate (DFR) / decreasing failure rate
• h_X(0) = lambda*Gamma(1 + 1/alpha)
• alpha = 1 case: S_X(x) = 1/(1 + lambda x), Lomax/Pareto
• AQL p1 = 0.05, LTPD p2 = 0.20, alpha = 0.05, beta = 0.10
• Operating ratio R = p2/p1 = 4.0
• Cameron's table of Poisson unity values (np1 for P = 0.95, np2 for P = 0.10)
• c = 4
• n = 40
• Plan: sample 40, accept if defectives <= 4
• Verification alpha ~ 0.053, beta ~ 0.0996
• A2, A3, d2, c4, D3, D4, B3, B4 control-chart constants
• sigma-hat = R-bar/d2 and sigma-hat = s-bar/c4
• Relative efficiency of range versus standard deviation falls with n (about 0.955 at n = 5, 0.850 at n = 10)
• R chart for n <= 5, s chart for n > 10
• X-bar limit half-width = 3 sigma/sqrt(n)
• D3 = 0 for n <= 6, B3 = 0 for n <= 5
• Chapman-Kolmogorov equation p_ij^(m+n) = sum_k p_ik^(m) p_kj^(n)
• Time-homogeneous Markov chain
• Third row of P^2 = (0.40, 0.44, 0.16)
• P(X5 = 2 | X3 = 3) = 0.44
• P(X3 = 1 | X0 = 3) = 0.392
• IBM SPSS Statistics (Statistical Package for the Social Sciences)
• Data View / Variable View, variable and value labels, defined missing values
• SPSS Command Syntax and the Paste button; Output Viewer pivot tables
• Modules: Advanced Statistics, Custom Tables, Complex Samples, AMOS
• R is free open-source (GPL); CRAN; ggplot2; RStudio / R Markdown
• Proprietary licence versus open source; GUI versus script

(a(i)) Describe how a CUSUM control chart is constructed, specifically in its tabular (algorithmic) form, explaining and using the two design parameters named in the question: the reference value (allowance) K and the decision interval H. (a(ii)) A series-system reliability numerical (sub-marks 4 + 2 + 4 printed on the paper). All 6 chip types with the tabulated counts n_i and failure rates lambda_i must be fault-free, and all times to failure are exponential. Find (I) the system time-to-failure distribution (i.e. the system failure rate and reliability function R(t)), (II) the mean lifetime (MTTF), and (III) R(1000). (b) Three things are asked: (i) explain the minimax (and maximin) criterion in the theory of games; (ii) test the given 3 x 4 payoff matrix for stability, i.e. for the existence of a saddle point; and (iii) obtain the optimal (pure) strategies of both players and the value of the game. (c(i)) A basic deterministic EOQ numerical. With annual demand D = 10,000 units, ordering cost S = Rs 20 per order and holding cost C = 25 per cent of the Rs 25 acquisition cost, find (I) the economic order quantity and (II) the total inventory cost. Both sub-parts must be answered. (c(ii)) A 4 x 4 minimisation assignment problem. Using the Hungarian (Kuhn-Munkres) method, assign each of the 4 workers to exactly one of the 4 jobs so that total cost is minimum, and state the optimal assignment together with its total cost.

• Reference value / allowance / slack K = (delta/2) sigma = 0.5 sigma
• Decision interval H = h sigma, h = 4 or 5 (H = 5 sigma)
• C_i^+ = max[0, x_i - (mu_0 + K) + C_{i-1}^+]
• C_i^- = max[0, (mu_0 - K) - x_i + C_{i-1}^-]
• ARL_0 about 465 and ARL_1 about 10.4 for k = 0.5, h = 5
• V-mask: lead distance d, half-angle theta, H = d tan(theta)
• Series system: R_s(t) = product of component reliabilities
• lambda_s = sum n_i lambda_i = 1205.8496 per 10^6 hours
• lambda_s = 1.2058 x 10^-3 per hour
• System time-to-failure is exponential, R_s(t) = e^{-lambda_s t}
• MTTF = 829 hours
• R(1000) = 0.299 (about 29.9 per cent)
• Two-person zero-sum game, maximin and minimax criteria
• Row minima: -2, -1, -3; maximin = -1
• Column maxima: 1, 1, -1, 8; minimax = -1
• Saddle point at (A2, B3)
• Game is stable / strictly determined
• Optimal strategies A2 and B3; value of the game v = -1
• EOQ formula Q* = sqrt(2DS/C)
• C = 0.25 x 25 = Rs 6.25 per unit per year
• Q* = 252.98 ~ 253 units
• Total inventory cost = sqrt(2DSC) = Rs 1581.14
• At optimum ordering cost = carrying cost = Rs 790.57
• Hungarian (Kuhn-Munkres) assignment method
• Row reduction by row minima 2, 3, 1, 4
• Column reduction of column 1 by 3
• Minimum number of covering lines = order of matrix (4) is the optimality test
• Optimal assignment 1-2, 2-1, 3-3, 4-4
• Minimum total cost = Rs 13

(a) Explain how the operating characteristic (OC) curve of a single sampling plan with n = 5 and c = 2 is constructed when the lot size is large. The answer must state the sampling distribution to use (binomial, since N is large), give the acceptance-probability function, tabulate P_a for a range of incoming quality levels p, describe the plot, and interpret the curve (AQL, LTPD, producer's and consumer's risks, effect of n and c). (b) Three demands: (i) choose and construct the suitable control chart for 24 daily samples of constant size n = 50 with the number of defectives recorded - this is an np chart (equivalently a p chart); (ii) test whether the process is in statistical control; (iii) recommend revised control limits for subsequent use after eliminating any assignable cause. (c(i)) Distinguish slack from surplus variables and discuss the role each plays in formulating and solving a linear programming problem (conversion to standard form, starting basis, need for artificial variables, economic interpretation). (c(ii)) Solve the given minimisation LPP by the DUAL SIMPLEX method specifically - convert all constraints to the 'less than or equal to' form, set up a dual-feasible but primal-infeasible starting tableau, and iterate using the dual-simplex leaving-then-entering rule until primal feasibility is attained. State the optimal x1, x2, x3 and Z.

• Type-B OC curve; binomial model justified by large lot size
• P_a(p) = (1-p)^5 + 5p(1-p)^4 + 10p^2(1-p)^3
• Table of P_a: 0.9988 at p = 0.05; 0.9914 at 0.10; 0.9421 at 0.20; 0.8369 at 0.30; 0.5000 at 0.50
• AQL, LTPD, producer's risk alpha, consumer's risk beta
• Ideal OC curve is a vertical step; finite n makes the curve flat
• np chart (or p chart) with constant n = 50
• Sum of defectives = 88, p-bar = 0.0733, n p-bar = 3.667
• Trial UCL = 9.20, LCL = 0
• Sample 11 (12 defectives) is out of control
• Revised n p-bar = 3.304, UCL = 8.57, LCL = 0 (p-chart: p-bar = 0.0661, UCL = 0.1715)
• 3-sigma limits based on the binomial standard deviation
• Slack added to 'less than or equal to'; surplus subtracted from 'greater than or equal to'
• Both >= 0, zero coefficient in the objective function
• Slack columns form the identity matrix / initial basic feasible solution
• Artificial variable, Big-M (penalty) method, Two-Phase method
• Complementary slackness / shadow price interpretation
• Dual simplex: start dual feasible (c_j - z_j >= 0) and primal infeasible (negative RHS)
• Multiply 'greater than or equal to' rows by -1; slacks s1, s2, s3; no artificial variables needed
• Leaving variable = most negative basic value (s1 = -4)
• Entering variable = min |(c_j - z_j)/a_rj| over negative a_rj only; x1 enters with ratio 2
• Optimal x1 = 4, x2 = 0, x3 = 0
• Z_min = 8

(a(i)) Two linked demands: (i) set out the theoretical basis of Wald's sequential probability ratio test as applied to acceptance sampling by attributes, and (ii) explain precisely how the acceptance and rejection numbers (the two boundary lines) are determined so that the producer's risk alpha and consumer's risk beta are met. (a(ii)) Obtain the maximum likelihood estimate of the exponential failure rate lambda from a type-I (time-truncated) censored life test: 10 units placed on test for a fixed 200 hours, 4 failures observed at 50, 80, 120 and 150 hours, the remaining 6 units still running at truncation. (b(i)) Two parts: state what a transportation problem is (its structure and purpose), and write down its general mathematical model - objective function, supply constraints, demand constraints, non-negativity and the balance condition. (b(ii)) Solve the given 3 x 4 balanced transportation problem using Vogel's Approximation Method to obtain the initial basic feasible solution, and complete the solution by testing optimality (MODI / u-v method) and improving if necessary; state the final allocation and minimum cost. (c(i)) Give a formal definition of a Poisson process (its postulates or equivalent characterisations) and state at least three genuine applications. (c(ii)) A single-server M/M/1 queue with Poisson arrivals at 10 per hour and exponential service averaging 4 minutes, FCFS and infinite capacity. Compute four quantities: (I) the probability that there is no queue, (II) the average queue size, (III) the expected waiting time in the queue, and (IV) the probability that a customer spends less than 12 minutes in the queue.

• Wald's sequential probability ratio test (SPRT), 1943-47
• H0: p = p1 (AQL), H1: p = p2 (LTPD)
• A = (1 - beta)/alpha, B = beta/(1 - alpha)
• Acceptance line d = s n - h1, rejection line d = s n + h2
• s = ln[(1-p1)/(1-p2)] / ln[p2(1-p1)/(p1(1-p2))]
• h1 = ln[(1-alpha)/beta] / ln[p2(1-p1)/(p1(1-p2))], h2 = ln[(1-beta)/alpha] / ln[...]
• Wald-Wolfowitz optimality; ASN; truncation
• Type-I (time-truncated) censoring; censored contribution e^{-lambda T0}
• L(lambda) = lambda^r e^{-lambda T}
• lambda-hat = r/T = number of failures / total time on test
• T = 400 + 6(200) = 1600 unit-hours
• lambda-hat = 0.0025 per hour; MTTF = 400 hours
• m origins with supplies a_i, n destinations with demands b_j, unit costs c_ij, decision variables x_ij
• Minimise Z = sum_i sum_j c_ij x_ij
• sum_j x_ij = a_i; sum_i x_ij = b_j; x_ij >= 0
• Balance condition sum a_i = sum b_j; dummy origin/destination with zero cost
• m + n - 1 basic cells; a special LPP solved by NWCR/LCM/VAM then MODI
• Balanced problem: total supply = total demand = 17
• Vogel's Approximation Method penalties = difference between the two lowest costs in a row/column
• VAM allocations: (P2, D4) = 1, (P1, D2) = 5, (P1, D1) = 1, (P3, D1) = 6, (P3, D3) = 3, (P3, D4) = 1
• VAM initial cost = Rs 102
• MODI / u-v optimality test, opportunity cost d_ij = c_ij - u_i - v_j
• Optimal cost = Rs 100 with x12 = 5, x13 = 1, x23 = 1, x31 = 7, x33 = 1, x34 = 2
• Independent and stationary increments; orderliness (no simultaneous events)
• P[N(t) = n] = e^{-lambda t}(lambda t)^n/n!
• E[N(t)] = Var[N(t)] = lambda t
• Exponential inter-arrival times with mean 1/lambda
• Three applications: queueing arrivals, radioactive decay, reliability/failures (or insurance claims, defects per unit)
• M/M/1 model, FCFS, lambda = 10/hour, mu = 15/hour, rho = 2/3
• P(no queue): P₀ = 1 − ρ = 1/3 (server idle), or P(N ≤ 1) = 1 − ρ² = 5/9 if read as 'nobody waiting' — interpretation stated
• Lq = rho^2/(1 - rho) = 1.33 customers (Ls = 2)
• Wq = 2/15 hour = 8 minutes (Ws = 12 minutes)
• P(Wq > t) = rho e^{-(mu - lambda)t}; P(Wq < 12 min) = 0.755
• P(W_system < 12 min) = 1 - e^{-1} = 0.632

(a) Explain the two-stage least squares estimation procedure for the given two-equation Keynesian system - a stochastic consumption function C_t = alpha + beta Y_t + u_t and the identity Y_t = C_t + Z_t. The explanation must identify the endogenous and predetermined variables, show why OLS fails, derive the reduced form, and set out both stages explicitly. (b) Two demands: (i) survey the methods available for detecting heteroscedasticity, and (ii) explain the Goldfeld-Quandt test in detail, including the assumptions that must hold for it to be applicable. (c) For the normal linear model Y = X beta + u with u ~ N(0, sigma^2 I_n), X of order n x k with rank k < n: (i) obtain the maximum likelihood estimator beta-hat of beta, and (ii) with e = Y - X beta-hat and M = I - X(X'X)^{-1}X', prove that e'e/sigma^2 ~ chi-square with (n - k) degrees of freedom. (d) Three linked demands: (i) name the usual sources of data on vital events; (ii) describe the types of age error that commonly occur in census data; and (iii) explain how such errors are measured and adjusted. (e) Two parts: (i) explain why raw scores are converted to standard scores, and (ii) for the given frequency distribution of test scores of 62 children, compute the mean and standard deviation and transform every raw score to a standard score having mean 50 and standard deviation 10 (i.e. T = 50 + 10 z).

• Classify variables (C, Y endogenous; Z exogenous; identity) and show Y is correlated with u, so OLS is inconsistent (Cov(Y,u) = σᵤ²/(1−β))
• Identification: exactly identified by order and rank conditions
• Stage 1: regress Y on Z; Stage 2: regress C on Ŷ
• Result: β̂ = Σ(Z−Z̄)(C−C̄)/Σ(Z−Z̄)(Y−Ȳ), equal to IV and ILS; consistent but biased in small samples
• At least four detection methods named with how each works (graphical, Park, Glejser, Spearman, BPG, White)
• GQ procedure: order by X, omit c central observations, two OLS fits, F = RSS₂/RSS₁
• GQ df = (n−c−2k)/2 each; reject when F exceeds critical value
• GQ assumptions: normality, no autocorrelation, correct specification, variance related to one variable
• Likelihood and normal equations giving β̂ = (X′X)⁻¹X′Y (and σ̂² = e′e/n)
• e = Mu since MX = 0; e′e = u′Mu with M symmetric idempotent
• rank M = tr M = n − k
• Orthogonal diagonalisation to a sum of n−k squared independent N(0,1), so χ²(n−k)
• Sources: civil registration, SRS, census, surveys, hospital records
• Errors: age heaping, misstatement, under-enumeration of children, differential coverage
• Detection: Whipple's and Myers' indices, UN Joint Score
• Adjustment: grouping, smoothing formulae (Carrier-Farag, Karup-King, Sprague, Beers), stable-population and PES methods
• At least two reasons (comparability, relative position, additivity, common scale)
• n = 62, mean 5.726 and σ = 1.724 correctly computed
• T = 50 + 10(x − 5.726)/1.724
• All eight standard scores: 74.8, 69.0, 63.2, 57.4, 51.6, 45.8, 40.0, 34.2

(a) Three demands: define the autocorrelation of order k and the correlogram, and then DERIVE the correlogram (the autocorrelation function) of an infinite series generated as a simple equally-weighted moving average of extent m of independent random components - the Slutzky-Yule set-up. (b) Four demands: explain the uses and limitations of index numbers; compute Laspeyres', Paasche's and Fisher's price indices for the three-commodity data; and examine whether the PAASCHE and FISHER indices satisfy (i) the Time Reversal Test and (ii) the Factor Reversal Test. (c(i)) Define morbidity and discuss the different measures used to quantify it - incidence, prevalence, attack rates, case fatality, duration and disability measures - with their formulae and the relationship between them. (c(ii)) Two parts: explain what a stable population is (its defining conditions and properties), and then PROVE that under stability the age composition C(x, t) - the proportion of the population aged x at time t - does not depend on t, i.e. C(x, t) = C(x).

• ρ_k = γ_k/γ₀ and the correlogram as the plot of ρ_k against k
• γ₀ = σ²/m and γ_k = (m−k)σ²/m² from the m − k common components
• ρ_k = (m − k)/m for k ≤ m and 0 for k ≥ m
• Correlogram drawn: linear fall from 1 at k = 0 to 0 at k = m, zero beyond
• At least three uses and three limitations
• Laspeyres 82.76, Paasche 77.78 and Fisher 80.23 with the four sums 145, 120, 135, 105
• Time reversal: Fisher satisfies (P₀₁P₁₀ = 1); Paasche fails (0.9398)
• Factor reversal: Fisher satisfies (P·Q = 0.7241); Paasche fails (0.6806)
• Definition of morbidity as departure from well-being
• Incidence and prevalence with formulae and their relation (P ≈ I × D)
• Attack rate, case-fatality rate or duration of illness
• Disability-based or composite measures (DALY, disability days) and data sources
• Definition: fixed fertility and mortality for a long time, closed to migration, constant growth rate r
• B(t) = B(0)e^{rt} from Lotka's equation
• N(x, t) = B(t−x)l(x) = B(0)e^{r(t−x)}l(x)
• C(x, t) = e^{-rx}l(x)/∫e^{-ra}l(a)da, independent of t

(a(i)) Define a life table - what it is, its columns and the assumptions behind it - and list its principal uses. (a(ii)) Given d_75 = 476 and complete expectations of life e_75 = 3.92 and e_76 = 3.66 years, find l_75 and l_76 by using the life-table recurrence connecting T_x, L_x and e_x. (b(i)) From the tabulated numbers of women (in thousands) and total births in each five-year age group of the reproductive span, compute (i) the Age-Specific Fertility Rate for each group, (ii) the Total Fertility Rate, and (iii) the Gross Reproduction Rate, given that 46.2 per cent of births are female. All three must be computed. (b(ii)) Define the reproduction rates (Gross and Net) and then critically assess how far they can legitimately be treated as indices of population growth, i.e. state their limitations as growth measures. (c) For the general linear model Y = X beta + u with E(u) = 0, E(uu') = sigma^2 I_n and X a fixed matrix of rank k (k < n), three things are required: obtain the OLS estimator beta-hat, obtain its dispersion (variance-covariance) matrix, and PROVE the Gauss-Markov theorem that beta-hat is the best linear unbiased estimator of beta.

• Definition: cohort surviving under fixed age-specific mortality, radix l₀
• Columns l_x, d_x, q_x, p_x, L_x, T_x, e_x
• Assumptions (closed cohort, constant rates, uniform deaths)
• At least three uses (life expectancy, insurance and pensions, projections, stable-population and NRR)
• Relation e_x = ½ + p_x(e_{x+1} + ½)
• p₇₅ = 3.42/4.16 = 0.8221 and q₇₅ = 0.1779
• l₇₅ = 476/0.1779 ≈ 2,676
• l₇₆ = 2,676 − 476 = 2,200
• ASFR = births/women × 1,000 for each of the seven groups: 16.25, 136.83, 119.87, 86.84, 61.89, 18.67, 10.00
• ΣASFR = 450.35 per 1,000
• TFR = 5 × ΣASFR = 2.252 per woman
• GRR = TFR × 0.462 = 1.040
• GRR and NRR defined (daughters, mortality allowance)
• NRR = 1 replacement; >1 growth; <1 decline
• Limitations: eventual rather than current growth, age-structure momentum, constant rates, migration, tempo effects
• Balanced conclusion
• β̂ = (X′X)⁻¹X′Y from the normal equations, with rank condition
• Unbiasedness E(β̂) = β
• D(β̂) = σ²(X′X)⁻¹
• Gauss-Markov: β* = CY with C = (X′X)⁻¹X′ + D, DX = 0 from unbiasedness
• D(β*) = σ²[(X′X)⁻¹ + DD′], DD′ positive semi-definite so β̂ has minimum variance

(a) Apply BOTH the order condition and the rank condition of identification to each of the four structural equations of the given system (endogenous Y1-Y4, predetermined X1-X3), and classify every equation as under-identified, exactly (just) identified or over-identified. Both conditions must be applied to all four equations. (b) From l_91 = 871 and the d_x column for ages 91 to 101, first build the l_x column, then compute five life-table probabilities: (i) 3q_93, (ii) 4p_92, (iii) the probability that three independent lives aged 92, 93 and 94 all survive 4 years, (iv) the probability that exactly two of three lives aged 93, 94 and 95 survive 4 years, and (v) the probability that all three of those lives die within 4 years. All five sub-parts must be answered. (c(i)) Explain what parallel tests are in classical test theory and the purposes they serve in psychological studies - chiefly the estimation of reliability, but also equating, test security and the lengthening of tests. (c(ii)) Explain the method of combining the rankings given by several judges to a number of subjects into a single composite ranking - the rank-sum procedure - together with the test of agreement among judges (Kendall's coefficient of concordance), the treatment of ties, and alternatives when judges differ in competence.

• Counting: M = 4, K = 3 and each equation's G_i, k_i
• Order condition K − k_i ≥ G_i − 1 satisfied with equality for all four
• Rank condition applied with correct coefficient matrices: equations 1, 2 and 3 have rank 2 < 3 and are not identified
• Equation 4 has a non-zero determinant, rank 3, exactly identified; conclusion that order is necessary, not sufficient
• Construction of l_x: 871, 575, 366, 222, 129, 71, 37, 19, 9, 4, 1
• (i) 295/366 = 0.806 and (ii) 71/575 = 0.1235
• (iii) 0.00107 using independent survival probabilities
• (iv) exactly two of three = 0.0199 and (v) all die = 0.7646
• Definition: same true score, equal error variance, X = T + E
• Reliability as correlation between parallel forms
• At least two uses (equivalent forms, avoiding practice effect, split-half with Spearman-Brown, SEM)
• Rank sums R_i and Kendall's W = 12S/[m²(n³−n)] with S = Σ(R_i − R̄)²
• Range of W and its meaning; tie correction
• Significance test χ² = m(n−1)W
• Combined ranking by ordering rank sums, provided W is significant