State statistical services: Statistical Officer / Assistant Statistical Officer (State PSCs and selection boards) · India · State Public Service Commissions and Subordinate Service Selection Boards
Original practice questions written to the published syllabus · not past papersThe logic and optimality of tests: errors and power, the Neyman-Pearson lemma, uniformly most powerful and unbiased tests, likelihood-ratio tests and their large-sample forms, and Wald's sequential probability ratio test.
In the exams: AICE-JRF Ag Stat Units 1-2 incl. SPRT (heavy); GATE ST 'Testing of Hypotheses' (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I-II (heavy)
ICAR AIEEA-PG Group 07, CUET-PG Statistics, SSC JSO and state statistical services
Two types of error, level of significance, and the z, t, chi-square and F tests applied to agricultural data.
How our questions for State statistical services spread across the units of Statistical Inference II: Testing of Hypotheses & Sequential Analysis. Difficulty: 22 easy, 18 medium, 6 hard.
| Tests based on Z, t, chi-square and F distributions | 12 | |
| Power and sample-size calculation for common tests | 8 | |
| Hypotheses, the two types of error, size and power | 7 | |
| Likelihood ratio tests and Wilks' theorem | 4 | |
| Neyman-Pearson lemma and most powerful tests | 3 | |
| Monotone likelihood ratio and uniformly most powerful tests | 3 | |
| Multiple testing: family-wise error rate and false discovery rate | 3 | |
| Unbiased tests and UMPU tests for exponential families | 2 | |
| Wald and score (Rao) tests | 2 | |
| Relation between tests and confidence intervals | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
What is the power of a test at a particular alternative θ₁?
B. The probability of rejecting H₀ when θ₁ is the true value
Power = 1 − β(θ₁) = P_θ₁(reject H₀). Accepting H₀ under θ₁ is the Type II error probability β; rejecting a true H₀ has probability α. The posterior probability of H₁ is a Bayesian quantity, not power.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
Plot yields are N(μ, 16). From 16 plots, H₀: μ = 50 is rejected when x̄ > 51. What is the size of the test?
B. 0.1587
Under H₀, X̄ ~ N(50, 1), so α = P(X̄ > 51) = P(Z > 1) = 0.1587. Using σ = 4 instead of SE = 1 gives z = 0.25; doubling gives a two-sided figure that does not apply to this one-sided rule.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
X₁, ..., Xₙ are iid N(μ, σ²) with μ known. For H₀: σ² = σ₀² against H₁: σ² = σ₁² with σ₁² > σ₀², the most powerful test rejects when:
A. Σ(Xᵢ − μ)² is large
log Λ = −(n/2)log(σ₁²/σ₀²) + (1/2)(1/σ₀² − 1/σ₁²)Σ(Xᵢ − μ)², increasing in Σ(Xᵢ − μ)² when σ₁² > σ₀². The test uses χ²ₙ critical values since μ is known.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
Why is there no UMP test of a one-sided hypothesis about the location θ of a Cauchy distribution based on one observation?
C. The likelihood ratio f(x; θ₁)/f(x; θ₀) is not monotone in x, so the MP region depends on θ₁
f(x; θ₁)/f(x; θ₀) rises and then falls back to 1 as x → ±∞, so the MP region against θ₁ is a bounded interval whose location depends on θ₁; no single region is best for all θ₁ > θ₀. Tests do exist; the support is the whole line; information is positive.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
A 2 × 2 table of adoption (yes/no) by training (yes/no) has cells a = 30, b = 20, c = 10, d = 40. What is Pearson's χ² without continuity correction?
A. 16.67
χ² = n(ad − bc)²/[(a + b)(c + d)(a + c)(b + d)] = 100 × 1000²/(50 × 50 × 40 × 60) = 16.67, with 1 df. With Yates' correction it is 15.04; omitting n gives 0.17.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
Germination is 45 of 100 seeds for treated and 30 of 100 for untreated seed. What is the pooled two-proportion z-statistic?
A. 2.191
Pooled p̄ = 75/200 = 0.375; SE = √(0.375 × 0.625 × (1/100 + 1/100)) = 0.0685; z = 0.15/0.0685 = 2.191. The unpooled SE gives 2.218; forgetting the two sample sizes gives 3.098.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
In a 2 × 2 table one expected frequency is 3. What is the recommended analysis?
A. Fisher's exact test, because the χ² approximation fails with small expected counts
With expected counts below 5, the χ² approximation can be poor; Fisher's exact test (conditioning on the margins) is the standard alternative. Yates' correction is a cruder fix. The z-test for two proportions is equivalent to the uncorrected χ² and has the same problem.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
A 95% two-sided confidence interval for a mean difference is 0.1 to 0.5 t/ha. What does the corresponding 5% two-sided test conclude?
C. It rejects a difference of 0 and does not reject a difference of 0.3
A two-sided level-α test rejects θ₀ exactly when θ₀ lies outside the 1 − α interval: 0 is outside, 0.3 inside.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
In a variety trial, eight new entries are each to be compared only with one check variety, with control of the family-wise error rate. Which procedure is designed for this?
C. Dunnett's test
Dunnett's procedure handles many-to-one comparisons with a control and is more powerful there than Tukey's all-pairs method or Scheffé's all-contrasts method.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
For a one-sided z-test of H₀: μ = μ₀ at level α with known σ and n observations, what is the power at μ₀ + δ (δ > 0)?
C. Φ(δ√n/σ − z_α)
Power = P(X̄ > μ₀ + z_ασ/√n | μ₀ + δ) = P(Z > z_α − δ√n/σ) = Φ(δ√n/σ − z_α). 1 − Φ(...) is β; omitting √n ignores the sample size.
Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics
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