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CSIR-UGC NET Statistical Inference II: Testing of Hypotheses & Sequential Analysis: practice questions

CSIR-UGC NET - Mathematical Sciences (statistics option) · India · National Testing Agency for CSIR

Original practice questions written to the published syllabus · not past papers
6worked here, free
27written to this exam in this subject
9syllabus units

What this subject covers

The 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)

Where the 27 questions fall

How our questions for CSIR-UGC NET spread across the units of Statistical Inference II: Testing of Hypotheses & Sequential Analysis. Difficulty: 2 easy, 10 medium, 15 hard.

Unbiased tests and UMPU tests for exponential families6
Wald and score (Rao) tests5
Monotone likelihood ratio and uniformly most powerful tests4
Relation between tests and confidence intervals4
Likelihood ratio tests and Wilks' theorem2
Multiple testing: family-wise error rate and false discovery rate2
Sequential probability ratio test: OC and ASN functions2
Hypotheses, the two types of error, size and power1
Neyman-Pearson lemma and most powerful tests1

6 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Neyman-Pearson lemma and most powerful tests · hard

For H₀: p = 0.5 against H₁: p = 0.8 with 10 Bernoulli trials, the MP test rejects for large X. P(X ≥ 9 | H₀) = 11/1024 and P(X = 8 | H₀) = 45/1024. For an exact size of 0.05, with what probability should H₀ be rejected when X = 8?

  1. 1.138
  2. 0.893
  3. 0.050
  4. 0.107
Answer and working

B. 0.893

Size = P(X ≥ 9) + γP(X = 8) = 0.05 gives γ = (0.05 − 11/1024)/(45/1024) = 0.893. Dividing 0.05 by P(X = 8) alone ignores the rejections already made at X ≥ 9.

Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics

2 · Monotone likelihood ratio and uniformly most powerful tests · hard

Why is there no UMP test of a one-sided hypothesis about the location θ of a Cauchy distribution based on one observation?

  1. The Cauchy family has a parameter-dependent support (unlike the normal), which rules out all MP tests
  2. The Cauchy distribution has no mean, so no test of its location can be constructed
  3. The likelihood ratio f(x; θ₁)/f(x; θ₀) is not monotone in x, so the MP region depends on θ₁
  4. Its Fisher information is zero, so every test has power equal to its size
Answer and working

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

3 · Unbiased tests and UMPU tests for exponential families · hard

In what sense is Fisher's exact test for a 2 × 2 table optimal?

  1. It has exactly size α for every table, with no randomisation needed
  2. Its randomised version is UMPU for the odds ratio, by conditioning on the margins
  3. It is the likelihood-ratio test, so Wilks' theorem gives its exact size
  4. It is UMP among all tests of independence, conditionally and unconditionally, for any table size
Answer and working

B. Its randomised version is UMPU for the odds ratio, by conditioning on the margins

The margins are sufficient for the nuisance parameters; conditioning on them gives a hypergeometric null distribution, and the randomised conditional test is UMPU. The non-randomised version is conservative.

Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics

4 · Wald and score (Rao) tests · hard

Which drawback is specific to the Wald test among the three asymptotic likelihood tests?

  1. It needs the model to be refitted under H₀ every time
  2. It has a different asymptotic null distribution from the others
  3. It cannot be used for more than one parameter at a time
  4. It is not invariant to reparameterisation of θ
Answer and working

D. It is not invariant to reparameterisation of θ

Testing θ = 1 and log θ = 0 by Wald can give different statistics, because the standard error is computed on the chosen scale. The LRT is invariant; all three are asymptotically χ² under H₀ and all extend to several parameters.

Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics

5 · Wald and score (Rao) tests · hard

Ten Poisson counts total 30. What are the score and Wald χ² statistics for H₀: λ = 2?

  1. Score 3.333, Wald 5.000
  2. Score 5.000, Wald 5.000
  3. Score 5.000, Wald 3.333
  4. Score 5.000, Wald 1.826
Answer and working

C. Score 5.000, Wald 3.333

Score: U(2) = 30/2 − 10 = 5, I(2) = n/λ₀ = 5, S = 25/5 = 5.000. Wald: (λ̂ − λ₀)² × n/λ̂ = 1 × 10/3 = 3.333. The LRT (4.328) lies between them; all three refer to χ²₁.

Further reading: Lehmann and Romano, Testing Statistical Hypotheses; Casella and Berger, Statistical Inference; Rohatgi and Saleh, An Introduction to Probability and Statistics

6 · Sequential probability ratio test: OC and ASN functions · hard

For the SPRT of μ₀ = 0 against μ₁ = 1 with N(μ, 1) data, α = 0.05 and β = 0.10, E₀(Z) = −1/2. Using Wald's approximation, what is the average sample number under H₀?

  1. 3.99
  2. 9
  3. 1.99
  4. 4.50
Answer and working

A. 3.99

E₀(N) ≈ [(1 − α)log B + α log A]/E₀(Z) = [0.95(−2.251) + 0.05(2.890)]/(−0.5) = 3.99. The fixed-sample one-sided test with the same α and β needs (1.645 + 1.282)² = 8.56, i.e. 9 observations; 4.50 ignores the chance of rejecting.

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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Statistical Inference II: Testing of Hypotheses & Sequential AnalysisThe full study manual, free CSIR-UGC NETFormat, sections and how it has changed

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