Which statements about unbiased tests are true?

A UMP level-α test is unbiased

For N(μ, σ²) with σ known, the equal-tailed two-sided Z-test is UMPU for H₀: μ = μ₀ against μ ≠ μ₀

An unbiased level-α test has power at least α at every point of the alternative

A UMP level-α test is unbiased

For N(μ, σ²) with σ known, the equal-tailed two-sided Z-test is UMPU for H₀: μ = μ₀ against μ ≠ μ₀

An unbiased level-α test has power at least α at every point of the alternative

A UMPU test exists for every composite hypothesis in every family

X₁, …, Xₙ are i.i.d. N(μ, σ²) with μ unknown. The likelihood ratio test of H₀: σ² = σ₀² against σ² ≠ σ₀² rejects H₀ when:

Σ(Xᵢ − X̄)²/σ₀² is too small or too large

Σ(Xᵢ − X̄)²/σ₀² is too large only

Which statements about the Wald, score and likelihood ratio tests of H₀: θ = θ₀ are true (under the usual regularity conditions)?

All three statistics are asymptotically χ² with the same degrees of freedom under H₀

The Wald test is built from the unrestricted MLE

For large n, −2 ln Λ and the Wald statistic are approximately equal under H₀

All three statistics are asymptotically χ² with the same degrees of freedom under H₀

The score test requires the unrestricted MLE

The Wald test is built from the unrestricted MLE

For large n, −2 ln Λ and the Wald statistic are approximately equal under H₀

For a one-parameter exponential family with natural statistic T, the UMPU level-α test of H₀: θ = θ₀ against θ ≠ θ₀ rejects when T c₂. The constants are determined by:

E_{θ₀}φ(T) = α and E_{θ₀}[Tφ(T)] = α E_{θ₀}T

P_{θ₀}(T c₂) = α/2 in every case

c₁ = E_{θ₀}T − z_{α/2} and c₂ = E_{θ₀}T + z_{α/2}

By Wilks’ theorem, the degrees of freedom of the limiting χ² distribution of −2 ln Λ equal:

the number of parameters fixed by H₀ (dim Θ − dim Θ₀)

the number of parameters in the full model

Testing of Hypotheses II: Unbiased and UMPU Tests, Exponential Families, Likelihood Ratio Tests and Large-Sample Tests

The second chapter for Section 9 of the GATE Statistics paper covers what the syllabus names after UMP tests: uniformly most powerful unbiased tests, UMPU tests for exponential families (the two-sided and nuisance-parameter cases), likelihood ratio tests with Wilks’ approximation, and large-sample tests — the Wald, score and likelihood ratio statistics and the z-tests for means, proportions and Poisson rates. The t, chi-square and F tests of normal theory appear here as what they are: likelihood ratio and UMPU tests.

1. Unbiased tests and UMPU tests

A level-α test is unbiased if its power is at least α at every θ in H₁ (and at most α on H₀): it is never likelier to reject when H₀ is true than when it is false. A UMP test is automatically unbiased (compare with φ ≡ α). A test that is UMP among unbiased level-α tests is UMPU. For N(μ, σ²) with σ known, the two-sided test rejecting when |√n(X̄ − μ₀)/σ| > z_α/2 is UMPU for H₀: μ = μ₀ against μ ≠ μ₀, although no UMP test exists.

One-parameter exponential family, f = h(x)c(θ)e^θT(x), H₀: θ = θ₀ against θ ≠ θ₀: the UMPU test rejects when T < c₁ or T > c₂, with c₁, c₂ (and randomisation) fixed by E_θ₀φ = α and E_θ₀[Tφ] = α E_θ₀T — the second condition says the power function has zero derivative at θ₀, so it has a minimum there. For a symmetric null distribution of T this reduces to equal tails; otherwise it does not. For H₀: θ₁ ≤ θ ≤ θ₂ against θ outside, the UMPU test has the same two-sided form with size α at both θ₁ and θ₂.

Nuisance parameters: in multiparameter exponential families, conditioning on the sufficient statistic for the nuisance parameter gives UMPU tests. This is how the classical normal tests are optimal: the one- and two-sample t-tests for means with σ unknown, the chi-square test for σ² with μ unknown (based on (n − 1)S²/σ₀², with unequal tail probabilities in the exact UMPU version), and the F-test for the ratio of two variances.

2. Likelihood ratio tests

The likelihood ratio Λ(x) = sup_θ∈Θ₀ L(θ)/sup_θ∈Θ L(θ) ∈ [0, 1] compares the best fit under H₀ with the best fit overall; the LRT rejects for small Λ. It reproduces the standard tests: for N(μ, σ²) with σ known, −2 ln Λ = n(X̄ − μ₀)²/σ², so it is the two-sided Z-test (n = 25, X̄ = 10.6, μ₀ = 10, σ = 2 gives 2.25); with σ unknown, Λ is a decreasing function of |t|, the t-test; for the variance, it is a test based on Σ(Xᵢ − X̄)²/σ₀² rejecting in both tails.

Wilks’ theorem: under regularity and H₀, −2 ln Λ → χ²_r, where r = dim Θ − dim Θ₀, the number of parameters H₀ fixes. For Poisson data with n = 10, ΣX = 30 and H₀: λ = 2, −2 ln Λ = 2[ΣX ln(X̄/λ₀) − n(X̄ − λ₀)] = 2[30 ln 1.5 − 10] = 4.33, compared with χ²₁ (5% point 3.84): reject.

3. Wald, score and likelihood ratio: the large-sample trio

Three asymptotically equivalent tests of H₀: θ = θ₀
Test	Statistic	Needs
Likelihood ratio	−2 ln Λ = 2[ln L(θ̂) − ln L(θ₀)]	both fits
Wald	(θ̂ − θ₀)² nI(θ̂)	unrestricted MLE θ̂ only
Score (Rao)	U(θ₀)²/(nI(θ₀)), U = ∂ ln L/∂θ	only θ₀ — no MLE

All three are asymptotically χ² with the same degrees of freedom under H₀ and agree to first order for large n; they differ in finite samples and in what must be computed. For a proportion, the score test uses the null standard error √(p₀(1 − p₀)/n) and the Wald test the estimated one √(p̂(1 − p̂)/n).

4. Large-sample z-tests

Large-sample tests (compare z with N(0, 1))
Hypothesis	Statistic
One proportion, p = p₀	(p̂ − p₀)/√(p₀(1 − p₀)/n)
Two proportions, p₁ = p₂	(p̂₁ − p̂₂)/√(p̄(1 − p̄)(1/n₁ + 1/n₂)), p̄ pooled
Poisson mean, λ = λ₀	(X̄ − λ₀)/√(λ₀/n)
Two means	(X̄₁ − X̄₂)/√(S₁²/n₁ + S₂²/n₂)

Worked: 220 successes in 400 against p₀ = 0.5 give z = 0.05/0.025 = 2. Two groups of 200 with 120 and 100 successes pool to p̄ = 0.55, standard error √(0.2475 × 0.01) = 0.0497 and z = 0.1/0.0497 = 2.01. A Poisson sample of 50 with X̄ = 2.5 against λ₀ = 2 gives z = 0.5/0.2 = 2.5. For small normal samples the exact tests replace these: t = (53 − 50)/(8/4) = 1.5 for n = 16, and the variance statistic (n − 1)S²/σ₀² = 10 × 20/10 = 20 for n = 11, S² = 20, σ₀² = 10, to be compared with χ²₁₀.

⚠️ Which variance goes in the denominator

In a test of p = p₀ the standard error is computed under H₀, from p₀ — the null hypothesis tells you the variance. In a confidence interval there is no hypothesised value, so p̂ is used. For two proportions under H₀: p₁ = p₂, pool the samples; the unpooled standard error belongs to an interval for p₁ − p₂.

Key takeaways

Unbiased: power ≥ α on H₁. UMP ⇒ unbiased; the two-sided Z-test is UMPU though not UMP.
Exponential-family UMPU for θ = θ₀: reject outside [c₁, c₂] with E φ = α and E[Tφ] = αE T; conditioning removes nuisance parameters (t, χ², F tests).
LRT: Λ = sup_H₀L/sup L; −2 ln Λ → χ²_r with r the number of constraints (Wilks).
Wald uses the unrestricted MLE, score uses only θ₀, LRT uses both; all three are asymptotically χ² with the same df.
z for a proportion uses p₀ in the standard error; two proportions pool under H₀; Poisson uses √(λ₀/n).

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which statements about unbiased tests are true?
1. A UMP level-α test is unbiased
2. For N(μ, σ²) with σ known, the equal-tailed two-sided Z-test is UMPU for H₀: μ = μ₀ against μ ≠ μ₀
3. An unbiased level-α test has power at least α at every point of the alternative
4. A UMPU test exists for every composite hypothesis in every family
Show answer
Answer: A — A UMP level-α test is unbiased; B — For N(μ, σ²) with σ known, the equal-tailed two-sided Z-test is UMPU for H₀: μ = μ₀ against μ ≠ μ₀; C — An unbiased level-α test has power at least α at every point of the alternative
(A) Its power is at least that of φ ≡ α, i.e. at least α. (B) The normal is an exponential family with symmetric null distribution, so the UMPU conditions give equal tails. (C) Definition. (D) False: the existence theorems cover exponential families (with the two side conditions); outside them a UMPU test need not exist, and nothing guarantees one for an arbitrary family and hypothesis.
For a chi-square test of independence in a 3 × 4 contingency table, the number of degrees of freedom is ____.
Numerical answer — type the value.
Show answer
Answer: 6
(r − 1)(c − 1) = 2 × 3 = 6: the full model has 11 free cell probabilities, independence has (3 − 1) + (4 − 1) = 5, and Wilks’ count is 11 − 5 = 6. Using rc − 1 = 11 forgets the parameters estimated under H₀.
X₁, …, X₁₀ are i.i.d. Poisson(λ) with ΣXᵢ = 30. For H₀: λ = 2 against λ ≠ 2, the likelihood ratio statistic −2 ln Λ, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.33
λ̂ = 3. ln L(λ) = ΣX ln λ − nλ + const, so −2 ln Λ = 2[30 ln(3/2) − 10(3 − 2)] = 2[30 × 0.40547 − 10] = 2 × 2.1640 = 4.33. It exceeds χ²₁,₀.₀₅ = 3.84. Dropping the −n(λ̂ − λ₀) term gives 24.3.
In 400 trials there are 220 successes. For H₀: p = 0.5, the large-sample test statistic z = (p̂ − p₀)/√(p₀(1 − p₀)/n) is ____.
Numerical answer — type the value.
Show answer
Answer: 2
p̂ = 0.55, standard error √(0.25/400) = 0.025, z = 0.05/0.025 = 2. Using p̂ in the standard error (the Wald version) gives √(0.2475/400) = 0.0249 and z = 2.01 — close, but the question specifies the null standard error.
Two independent samples of 200 have 120 and 100 successes. Using the pooled proportion under H₀: p₁ = p₂, the z statistic, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.01
p̄ = 220/400 = 0.55; SE = √(0.55 × 0.45 × (1/200 + 1/200)) = √0.002475 = 0.04975; z = (0.60 − 0.50)/0.04975 = 2.01. Using the unpooled SE √(0.24/200 + 0.25/200) = 0.0495 gives 2.02.
A sample of 16 from a normal population has X̄ = 53 and S = 8. The t statistic for H₀: μ = 50, correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
t = (X̄ − μ₀)/(S/√n) = 3/(8/4) = 1.5, with 15 degrees of freedom. Dividing by S instead of S/√n gives 0.375.
X₁, …, Xₙ are i.i.d. N(μ, σ²) with μ unknown. The likelihood ratio test of H₀: σ² = σ₀² against σ² ≠ σ₀² rejects H₀ when:
1. Σ(Xᵢ − X̄)²/σ₀² is too small or too large
2. Σ(Xᵢ − X̄)²/σ₀² is too large only
3. |X̄ − μ₀| is large
4. ΣXᵢ²/σ₀² is too large
Show answer
Answer: A — Σ(Xᵢ − X̄)²/σ₀² is too small or too large
With V = Σ(Xᵢ − X̄)²/σ₀², Λ = (V/n)^n/2 e^{−(V − n)/2}, which is small when V/n is far from 1 in either direction — a two-tailed test in the χ²_n−1 statistic. The upper-tail-only test is for the one-sided alternative σ² > σ₀²; ΣXᵢ² would be right only if μ were known to be 0.
A sample of 11 from a normal population has S² = 20. For H₀: σ² = 10, the chi-square statistic (n − 1)S²/σ₀² is ____.
Numerical answer — type the value.
Show answer
Answer: 20
(11 − 1) × 20/10 = 20, referred to χ²₁₀; it exceeds the upper 5% point 18.307, so H₀ is rejected against σ² > 10 at 5%. Using n = 11 in place of n − 1 gives 22.
Which statements about the Wald, score and likelihood ratio tests of H₀: θ = θ₀ are true (under the usual regularity conditions)?
1. All three statistics are asymptotically χ² with the same degrees of freedom under H₀
2. The score test requires the unrestricted MLE
3. The Wald test is built from the unrestricted MLE
4. For large n, −2 ln Λ and the Wald statistic are approximately equal under H₀
Show answer
Answer: A — All three statistics are asymptotically χ² with the same degrees of freedom under H₀; C — The Wald test is built from the unrestricted MLE; D — For large n, −2 ln Λ and the Wald statistic are approximately equal under H₀
(A) The standard asymptotic result. (B) False: the score test evaluates U(θ₀) and I(θ₀) at the null value only — its practical advantage. (C) It measures (θ̂ − θ₀) in units of its estimated standard error. (D) A second-order Taylor expansion of ln L about θ̂ gives −2 ln Λ = n I(θ̂)(θ̂ − θ₀)² + o_p(1).
Independent normal samples have S₁² = 12 (n₁ = 11) and S₂² = 4 (n₂ = 16). The F statistic for H₀: σ₁² = σ₂², with the larger variance in the numerator, is ____.
Numerical answer — type the value.
Show answer
Answer: 3
F = S₁²/S₂² = 12/4 = 3, with (n₁ − 1, n₂ − 1) = (10, 15) degrees of freedom. Using the standard deviations gives √3 = 1.73; the F statistic is a ratio of variances.
For a one-parameter exponential family with natural statistic T, the UMPU level-α test of H₀: θ = θ₀ against θ ≠ θ₀ rejects when T < c₁ or T > c₂. The constants are determined by:
1. E_θ₀φ(T) = α and E_θ₀[Tφ(T)] = α E_θ₀T
2. P_θ₀(T < c₁) = P_θ₀(T > c₂) = α/2 in every case
3. E_θ₀φ(T) = α alone
4. c₁ = E_θ₀T − z_α/2 and c₂ = E_θ₀T + z_α/2
Show answer
Answer: A — E_{θ₀}φ(T) = α and E_{θ₀}[Tφ(T)] = α E_{θ₀}T
Size α gives one equation; unbiasedness forces the power function to have a minimum at θ₀, i.e. zero derivative there, which for an exponential family is E_θ₀[Tφ] = αE_θ₀T. Equal tails satisfy the second condition only when T’s null distribution is symmetric; size alone leaves one constant free.
A sample of 50 from a Poisson(λ) population has X̄ = 2.5. For H₀: λ = 2, the large-sample statistic z = (X̄ − λ₀)/√(λ₀/n), correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.5
Under H₀ the variance of X̄ is λ₀/n = 2/50 = 0.04, SE = 0.2, so z = 0.5/0.2 = 2.5. Using √λ₀ without dividing by n gives 0.35.
X₁, …, X₂₅ are i.i.d. N(μ, 4) with X̄ = 10.6. For H₀: μ = 10 against μ ≠ 10, the likelihood ratio statistic −2 ln Λ, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.25
−2 ln Λ = n(X̄ − μ₀)²/σ² = 25 × 0.36/4 = 2.25 — the square of z = 0.6/0.4 = 1.5, which is why the LRT here is exactly the two-sided Z-test. Using σ = 2 in place of σ² = 4 gives 4.5.
By Wilks’ theorem, the degrees of freedom of the limiting χ² distribution of −2 ln Λ equal:
1. the number of parameters fixed by H₀ (dim Θ − dim Θ₀)
2. the sample size minus one
3. the number of parameters in the full model
4. always 1
Show answer
Answer: A — the number of parameters fixed by H₀ (dim Θ − dim Θ₀)
The limit counts the constraints H₀ imposes: testing μ₁ = μ₂ = μ₃ among three free means fixes two contrasts, giving χ²₂. The sample size affects the accuracy of the approximation, not its degrees of freedom.