State statistical services Statistical Inference I: Estimation: practice questions

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 papers

10worked here, free

32written to this exam in this subject

11syllabus units

What this subject covers

How to estimate well and know it: sufficiency and completeness, unbiasedness and minimum variance, the Cramer-Rao bound, the Rao-Blackwell and Lehmann-Scheffe theorems, the methods of moments, maximum likelihood and minimum chi-square, and confidence intervals.

In the exams: AICE-JRF Ag Stat Unit 1 (heavy: each theorem named); GATE ST 'Estimation' (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I (heavy); ISI and CSIR-NET (present)

Point estimation: unbiasedness, consistency and efficiency
Sufficiency, the factorisation theorem and minimal sufficiency
Completeness, ancillarity and Basu's theorem
Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems
Cramer-Rao inequality and Fisher information
Method of moments
Maximum likelihood: properties, invariance and asymptotic normality
Minimum chi-square and other methods
Interval estimation: pivotal quantities and coverage
Large-sample confidence intervals
Iterative maximum likelihood: Newton-Raphson and Fisher scoring

Where the 32 questions fall

How our questions for State statistical services spread across the units of Statistical Inference I: Estimation. Difficulty: 16 easy, 8 medium, 8 hard.

Large-sample confidence intervals	6
Point estimation: unbiasedness, consistency and efficiency	4
Interval estimation: pivotal quantities and coverage	4
Cramer-Rao inequality and Fisher information	3
Maximum likelihood: properties, invariance and asymptotic normality	3
Minimum chi-square and other methods	3
Sufficiency, the factorisation theorem and minimal sufficiency	2
Completeness, ancillarity and Basu's theorem	2
Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems	2
Method of moments	2
Iterative maximum likelihood: Newton-Raphson and Fisher scoring	1

10 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 · Point estimation: unbiasedness, consistency and efficiency · medium

For large normal samples, what is the efficiency of the sample median relative to the sample mean as an estimator of μ?

1.571
1.000
0.500
0.637

Answer and working

D. 0.637

Var(median) ≈ πσ²/(2n) and Var(mean) = σ²/n, so efficiency = (σ²/n)/(πσ²/(2n)) = 2/π ≈ 0.637. 1.571 is the inverse (how much larger the median's variance is).

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

2 · Completeness, ancillarity and Basu's theorem · hard

X₁, ..., Xₙ are iid N(θ, θ²), θ > 0. Which statement about the minimal sufficient statistic T = (ΣXᵢ, ΣXᵢ²) is correct?

It is not sufficient, since the mean and variance are linked through θ
It is complete and ancillary, since its distribution is free of θ
It is minimal sufficient but not complete, since the family is a curved exponential family
It is complete, since every exponential family of any form has a complete sufficient statistic

Answer and working

C. It is minimal sufficient but not complete, since the family is a curved exponential family

E[2(ΣX)² − (n + 1)ΣX²] = 2(nθ² + n²θ²) − (n + 1)(2nθ²) = 0 for all θ, a non-zero function with zero mean, so T is not complete. A two-dimensional statistic for a one-dimensional parameter (curved family) need not be complete.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

3 · Cramer-Rao inequality and Fisher information · medium

A sample of 18 is drawn from N(μ, σ²) with μ known and σ² = 9. What is the Cramér-Rao lower bound for an unbiased estimator of σ²?

9.53
9.0
4.5
1.0

Answer and working

B. 9.0

I(σ²) = 1/(2σ⁴) per observation, bound = 2σ⁴/n = 162/18 = 9.0, attained by Σ(Xᵢ − μ)²/n. 2σ⁴/(n − 1) = 9.53 is the variance of s² when μ is unknown.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

4 · Cramer-Rao inequality and Fisher information · medium

For a normal sample of size 10 with μ unknown, what is the efficiency of s² relative to the Cramér-Rao bound 2σ⁴/n for estimating σ²?

0.9
1.111
1.0
0.8

Answer and working

A. 0.9

Var(s²) = 2σ⁴/(n − 1), so efficiency = (2σ⁴/n)/(2σ⁴/(n − 1)) = (n − 1)/n = 0.9. s² is nevertheless the UMVUE: the bound is not attainable when μ is unknown.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

5 · Maximum likelihood: properties, invariance and asymptotic normality · easy

From 100 seeds, 30 germinate. Using the asymptotic variance of the MLE, 1/[nI(p)], what is the estimated standard error of p̂?

0.0021
0.0548
0.1449
0.0458

Answer and working

D. 0.0458

1/[nI(p̂)] = p̂(1 − p̂)/n = 0.21/100 = 0.0021; SE = √0.0021 = 0.0458. 0.0021 is the variance, not the standard error.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

6 · Interval estimation: pivotal quantities and coverage · easy

Ten plots of a new wheat variety give mean 4.2 t/ha and s = 0.5 t/ha. With t₀.₀₂₅,₉ = 2.262, what is the 95% confidence interval for the mean yield?

4.09 to 4.31 t/ha
3.07 to 5.33 t/ha
3.89 to 4.51 t/ha
3.84 to 4.56 t/ha

Answer and working

D. 3.84 to 4.56 t/ha

x̄ ± t s/√n = 4.2 ± 2.262 × 0.5/√10 = 4.2 ± 0.358. Using z = 1.96 ignores that σ is estimated; omitting √n gives the range for single plots, not for the mean.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

7 · Large-sample confidence intervals · medium

Why is the Wilson score interval preferred to the Wald interval for a binomial proportion?

It is exact, guaranteeing at least nominal coverage for every p and n
It is always narrower than the Wald interval at the same confidence level and sample size
It needs no normal approximation at all, unlike the Wald interval
It inverts the score test, keeps coverage near nominal for small n and stays inside [0, 1]

Answer and working

D. It inverts the score test, keeps coverage near nominal for small n and stays inside [0, 1]

The Wilson interval uses the null standard error √(p(1 − p)/n) and solves for p, which corrects the Wald interval's undercoverage near 0 and 1. It is still an approximation and is not always narrower; exact (Clopper-Pearson) intervals guarantee coverage.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

8 · Large-sample confidence intervals · easy

How many farms must be surveyed to estimate an adoption proportion within ±0.05 with 95% confidence, when nothing is known about p?

385
1537
2000
20

Answer and working

A. 385

n = 1.96² p(1 − p)/d² with the worst case p = 0.5: 3.8416 × 0.25/0.0025 = 384.2, so 385. 2000 is the Chebyshev requirement; 20 forgets to square; 1537 omits the factor p(1 − p) = 0.25.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

9 · Large-sample confidence intervals · hard

A Poisson rate is estimated by X̄ from n counts. Using the delta method on the log scale, which 95% interval for λ results?

exp(X̄ ± 1.96/√n)
X̄ ± 1.96√(X̄/n)
X̄ exp(±1.96/√(nX̄))
X̄ exp(±1.96√(X̄/n))

Answer and working

C. X̄ exp(±1.96/√(nX̄))

SE(log X̄) ≈ SE(X̄)/X̄ = √(X̄/n)/X̄ = 1/√(nX̄). The interval log X̄ ± 1.96/√(nX̄) back-transforms to X̄ exp(±1.96/√(nX̄)), which is always positive. X̄ ± 1.96√(X̄/n) is the Wald interval on the original scale.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

10 · Iterative maximum likelihood: Newton-Raphson and Fisher scoring · hard

θ̃ is a √n-consistent estimator of θ. What can be said of the one-step estimator θ̃ + ℓ′(θ̃)/[nI(θ̃)]?

It is no better asymptotically than θ̃ itself
It is unbiased for θ in samples of every size
It is asymptotically efficient, with the same limiting distribution as the MLE
It is exactly equal to the MLE whenever n is large enough for the central limit theorem to apply

Answer and working

C. It is asymptotically efficient, with the same limiting distribution as the MLE

A single scoring (or Newton) step from a √n-consistent start yields an estimator with √n(θ₁ − θ) → N(0, 1/I(θ)), the MLE's limit. It is not exactly the MLE, and not unbiased in finite samples.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

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