Original public practice · MS-1 · October 2026

Mathematical statistics: cumulative practice

Derive estimators and tests, inspect calibration, fit small datasets and defend model decisions. This form contains 12 multipart questions and 120 practice points.

This is calculus-based undergraduate practice. Review sums, derivatives, integrals and conditional probability; statistics also uses vectors and matrices. Every lesson remains directly accessible. These original problems sample the published material; they are not an official university paper, a calibrated grade prediction or an examination of every course objective.

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State the sampling model, parameter domain, units and theorem hypotheses. Distinguish exact finite-sample calculations, asymptotic approximations and empirical evidence. You may cite results taught in the linked lessons unless a part asks for their proof. Supply intermediate reasoning and the requested calculations; a theorem name or numerical match alone is not a proof. A calculator or local numerical code may help check arithmetic; no paid tool is required. Supplied theorem statements may be used only under their stated assumptions. Alternative valid methods can earn the same points.

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1. Models and estimation

40 practice points.

Question 1: Identification and a moving support

10 points

Consider two separate experiments. In experiment I, X₁,…,Xₙ are iid Normal(θ²,1), θ∈R. In experiment II, Y₁,…,Yₙ are iid Uniform(0,b), b>0, and M=max Yᵢ. In experiment II use the density b⁻¹1{0≤y≤b}, and assume M>0.

(a) · 5 points

In experiment I determine all MLEs of θ in terms of X̄. Which function of θ is identifiable? Explain why a likelihood maximum cannot resolve the sign.

(b) · 5 points

In experiment II derive the MLE, the law of M/b, and an unbiased estimator of b. Explain why solving a zero-score equation and quoting regular Fisher-information theory fails here.

Review question 1: worked argument and scoring criteria

(a)

Ignoring constants, the log likelihood is −Σ(Xᵢ−θ²)²/2. Decomposition around X̄ leaves n(X̄−θ²)². Set μ=θ²≥0; its minimizing value is max(X̄,0). If X̄>0 the maximizers are ±√X̄; otherwise the unique maximizer is 0. The mean μ=θ² is identifiable, but θ and −θ induce the same joint distribution. More data in this experiment cannot distinguish those two signs.

2 points: Reduces the optimization to a constrained mean.
2 points: Gives the complete set of maximizers in all cases.
1 point: Explains identification separately from optimization.

(b)

L(b)=b^(−n)1{b≥M}, so it decreases on the feasible set and is maximized at b=M. For 0≤u≤1, P(M/b≤u)=uⁿ, giving E(M/b)=∫₀¹u·nu^(n−1)du=n/(n+1). Thus (n+1)M/n is unbiased for b. The feasible support moves with b and the optimum is on its boundary. The interior score −n/b never vanishes; differentiating while ignoring the indicator misses the moving support and violates the usual regularity arguments.

2 points: Derives the boundary maximum and ratio CDF.
2 points: Derives the expectation and unbiased correction.
1 point: Identifies the moving-support regularity failure.

Targeted lesson returns:

Question 2: Improvement without losing unbiasedness

10 points

X₁,…,Xₙ are iid Bernoulli(p), 0<p<1, n≥2. Let S=ΣXᵢ. Estimate p under squared-error loss, starting from the estimator X₁.

(a) · 5 points

Prove S is sufficient and compute E(X₁|S). Compare the exact squared-error risks of X₁ and the conditional estimator.

(b) · 5 points

Prove completeness of S using the polynomial identity obtained from E_p[g(S)]=0 for every p∈(0,1). Deduce that S/n has minimum variance among all unbiased estimators based on the sample.

Review question 2: worked argument and scoring criteria

(a)

The joint mass p^S(1−p)^(n−S) depends on the data only through S; equivalently, given S=s, every binary vector with s ones is equally likely independently of p. Symmetry gives E(X₁|S=s)=s/n. Both estimators are unbiased. Their risks are their variances, p(1−p) and p(1−p)/n, so conditioning reduces risk by p(1−p)(1−1/n).

2 points: Proves sufficiency from the mass or conditional law.
1 point: Derives the conditional estimator.
2 points: Computes both exact risks and their difference.

(b)

The assumption is Σₛ₌₀ⁿg(s)C(n,s)pˢ(1−p)^(n−s)=0. Divide by (1−p)ⁿ and set t=p/(1−p)>0: the polynomial Σg(s)C(n,s)tˢ vanishes on an interval, so every coefficient is zero and g(s)=0. For any unbiased estimator U, its conditional mean h(S)=E(U|S) is parameter-independent by sufficiency and unbiased. Completeness applied to h(S)−S/n forces h(S)=S/n. Total variance then gives Var(U)≥Var(S/n). All moments here are finite for real-valued estimators on the finite sample space.

3 points: Establishes the polynomial identity and coefficient conclusion.
1 point: Uses sufficiency and completeness to identify the conditional estimator.
1 point: Uses total variance to prove the minimum-variance conclusion.

Targeted lesson returns:

Question 3: Information and a vanishing derivative

10 points

X₁,…,Xₙ are iid Normal(μ,1), μ∈R. The target is ψ=μ². You may use that √n(X̄−μ) is exactly standard normal, and the continuous-mapping theorem. Write χ²₁ for the distribution of Z² when Z is standard normal.

(a) · 5 points

Derive the sample score and Fisher information for μ. Compute the exact bias and variance of X̄² as an estimator of ψ. You may use E(Z⁴)=3 and the zero odd moments of a standard normal.

(b) · 5 points

For μ≠0 find the limiting distribution of √n(X̄²−μ²). At μ=0 find the correct nondegenerate scale and distribution. Repeat the μ=0 result for the unbiased estimator X̄²−1/n; explain the danger of substituting μ=0 into a first-order standard error.

Review question 3: worked argument and scoring criteria

(a)

The score is Σ(Xᵢ−μ), with mean zero and variance n, so information is n. Write X̄=μ+Z/√n. Then E(X̄²)=μ²+1/n, so bias is 1/n. Squaring the centered expansion 2μZ/√n+(Z²−1)/n and using the supplied moments gives variance 4μ²/n+2/n²; the cross term vanishes.

2 points: Derives score and information.
1 point: Obtains the bias.
2 points: Expands and derives the exact variance.

(b)

√n(X̄²−μ²)=2μZ+Z²/√n converges to Normal(0,4μ²) for μ≠0. At μ=0 that scale converges to zero, but nX̄²=Z² has exactly χ²₁ distribution. For the corrected estimator n(X̄²−1/n)=Z²−1, exactly χ²₁−1. The derivative of μ² is zero at zero; a vanishing first-order variance does not mean no estimation uncertainty. A first-order normal interval at this point is unjustified.

2 points: Derives the regular nonzero-μ limit.
2 points: Gives both exact zero-μ scaled laws.
1 point: Explains the failed first-order approximation.

Targeted lesson returns:

Question 4: A posterior is not a decision rule

10 points

A Bernoulli success probability p has Uniform(0,1) prior. Three conditionally independent trials yield two successes. For actions labeled release or hold, release costs 4 if p<1/2 and zero otherwise; hold costs 1 if p≥1/2 and zero otherwise. For a separate point estimate of p, use squared-error loss. You may use ∫₀¹pᵃ(1−p)ᵇdp=a!b!/(a+b+1)! for nonnegative integers a,b.

(a) · 5 points

Normalize the posterior density, find its mean, and derive the posterior probability p<1/2. Find the squared-error Bayes estimate.

(b) · 5 points

Choose release or hold by posterior expected loss. Derive the probability threshold for release. Explain why the posterior mean exceeding 1/2 is insufficient and why this posterior does not by itself check the Bernoulli design.

Review question 4: worked argument and scoring criteria

(a)

The likelihood is proportional to p²(1−p), whose integral is 2!1!/4!=1/12. Thus the posterior density is 12p²(1−p). Its mean is 12·3!1!/5!=3/5. The posterior lower-half probability is [4p³−3p⁴]₀^(1/2)=5/16. The expected squared loss decomposes as Var(p|data)+(a−E[p|data])², so the optimal point estimate is 3/5.

2 points: Normalizes the density.
1 point: Computes the mean.
1 point: Computes 5/16.
1 point: Justifies the squared-error optimum.

(b)

Let q=P(p<1/2|data)=5/16. Release has expected loss 4q=5/4; hold has loss 1−q=11/16, so hold is preferred. Release is optimal when 4q≤1−q, or q≤1/5 (ties allowed). The action depends on a tail probability and asymmetric losses, not the posterior mean. The posterior is conditional on the prior and iid Bernoulli design; dependence, selection or a changing success rate require design investigation or predictive checks, not confidence from normalization alone.

2 points: Computes both losses and chooses hold.
1 point: Derives the 1/5 threshold.
1 point: Separates point estimation from the action.
1 point: States a concrete model/design limitation.

Targeted lesson returns:

2. Inference and calibration

40 practice points.

Question 5: Which uncertainty interval?

10 points

Five independent Normal(μ,σ²) observations have sample mean 10 and unbiased sample variance s²=4. σ² is unknown. A future observation is independent from the same distribution. You may use that (X̄−μ)/(s/√n) and (Xnew−X̄)/(s√(1+1/n)) each have Student t distribution with n−1 degrees of freedom. The 0.975 quantile of t₄ is 2.776.

(a) · 5 points

Construct a two-sided 95% interval for μ and a 95% prediction interval for a future observation. Show the two standard errors and interpret the difference.

(b) · 5 points

Explain the repeated-sampling meaning of the mean interval and use its dual two-sided test to assess H₀:μ=13 at level 0.05. Would the same exact t calibration hold for five successive readings with strong positive serial correlation?

Review question 5: worked argument and scoring criteria

(a)

The mean standard error is 2/√5≈0.8944, giving 10±2.776·2/√5≈[7.517,12.483]. The prediction standard error is 2√(6/5)≈2.1909, giving 10±6.082≈[3.918,16.082]. The second includes new individual variability as well as estimation uncertainty; it does not estimate a different population mean.

2 points: Computes the mean standard error and interval.
2 points: Computes the prediction standard error and interval.
1 point: Explains the extra prediction variability.

(b)

Under repeated independent normal samples, the interval procedure covers the fixed μ with probability 0.95; the realized interval does not assign a frequentist probability 0.95 to a random μ. The test statistic (10−13)/(2/√5)=−3√5/2≈−3.354 has magnitude greater than 2.776, so reject H₀. Equivalently 13 is outside the interval. Serial correlation invalidates the stated iid pivot: the variance of the mean includes covariance terms and the usual variance independence/chi-square argument need not hold. Exact t₄ calibration is not justified.

2 points: Gives a correct coverage interpretation.
2 points: Computes the test and interval-duality decision.
1 point: Identifies why serial dependence breaks the pivot.

Targeted lesson returns:

Question 6: A test with a discrete boundary

10 points

S is Binomial(2,p). Test the simple null p=1/2 against p=3/4 with size exactly 0.30. Randomized rejection means using an auxiliary uniform random variable independent of S. You may use the Neyman–Pearson lemma: rejecting larger likelihood ratios first, with boundary randomization if necessary, maximizes power among tests of the given size.

(a) · 5 points

List both masses and all likelihood ratios. Construct the most powerful size-0.30 test and compute its power.

(b) · 5 points

What is the largest power achievable by a nonrandomized test with size at most 0.30? State the benefit of randomization. Explain why this single simple-null test is not itself a 70% confidence interval procedure for all p.

Review question 6: worked argument and scoring criteria

(a)

For S=0,1,2 the null masses are (1/4,1/2,1/4); alternative masses are (1/16,3/8,9/16). Ratios are (1/4,3/4,9/4). Reject always at 2, then reject with probability r at 1. Size 1/4+r/2=0.30 gives r=0.10. Never reject at 0. Power is 9/16+(1/10)(3/8)=3/5.

2 points: Lists masses and likelihood-ratio order correctly.
2 points: Derives the randomization and exact size.
1 point: Computes power 3/5.

(b)

A deterministic rejection subset of {0,1,2} can have null probability at most 0.30 only if it is empty, {0} or {2}. The most powerful is {2}, of size 1/4 and power 9/16. Randomization uses the otherwise unused size budget and increases power by 3/80 to 3/5. A confidence-set procedure requires a specified test for every candidate p₀, with rejection probability at most 0.30 under each p₀, and consistent treatment of auxiliary randomness. One test at p₀=1/2 only specifies membership at that candidate; it does not construct the entire confidence set.

2 points: Enumerates feasible deterministic tests and identifies the best.
1 point: Computes the power improvement.
2 points: Explains the family of tests needed for inversion.

Targeted lesson returns:

Question 7: A discrepancy and its reference law

10 points

A predeclared null says four exact independent measurements are Uniform(0,1). The observed values are 0.1,0.3,0.6,0.8. Define D=sup_x|F₄(x)−F₀(x)|, where F₄ is the empirical CDF and F₀ the null CDF. A different analyst chooses a Uniform(a,b) model only after examining the data.

Ordered measurements for the fixed-null discrepancy
i	x(i)
1	0.1
2	0.3
3	0.6
4	0.8

(a) · 5 points

Compute D by evaluating both sides of every empirical jump. Is D alone a p-value? Explain how a valid fixed-null simulation reference would be produced.

(b) · 5 points

For the analyst fitting a and b, explain why fixed-null critical values cannot simply be reused. Outline a fitted-model simulation check, identifying both the refitting step and the limitation of claiming exact calibration for a data-chosen model.

Review question 7: worked argument and scoring criteria

(a)

At the ordered values, right gaps i/4−x(i) are 0.15,0.20,0.15,0.20; left gaps x(i)−(i−1)/4 are 0.10,0.05,0.10,0.05. The supremum is D=0.20; between jumps the differences are linear and beyond the sample range their extrema are included in these endpoints. D is a discrepancy, not a tail probability. Simulate independent samples of four from the predeclared Uniform(0,1), recompute the identical D for each, and compare the observed D with that null distribution, reporting Monte Carlo uncertainty.

2 points: Computes both sets of one-sided jump gaps.
1 point: Obtains D and justifies checking endpoints.
2 points: Distinguishes discrepancy from calibrated simulation evidence.

(b)

Fitting a and b uses the same data and changes the distribution of the discrepancy. A parametric simulation check generates samples of size four from the fitted model, refits a and b using exactly the original algorithm on each sample, and recomputes the discrepancy against each refitted CDF. Any model-selection step must also be replicated if the target is the whole selected procedure. This is not automatically an exact finite-sample test: estimated support is a nonregular problem, and a parametric bootstrap needs validity analysis or a justified invariant reference. Omitting refitting answers a different question; even a precisely simulated reference cannot repair a false sampling model.

2 points: Explains the effect of data-dependent fitting.
2 points: Specifies simulation, refitting and repeated selection where applicable.
1 point: States the nonregular/finite-sample validity limitation.

Targeted lesson returns:

Question 8: A bootstrap atom and a selected winner

10 points

X₁,…,Xₙ are independent observations from a continuous distribution and have a unique sample maximum M, n≥2. A nonparametric bootstrap draws n times with replacement from these n observed values and has maximum M*. Separately, a team compares 20 candidate models on one validation set and reports only the smallest validation error.

(a) · 5 points

Derive P*(M*=M) conditional on the observed sample and its limit as n→∞. Compare this with a new independent sample maximum equaling this fixed observed M. Explain the warning for bootstrap approximations near a support endpoint.

(b) · 5 points

Explain why the reported minimum validation error is optimistic for the selected model. Give a defensible protocol for estimating its future error and a separate protocol for assessing the entire selection procedure when fresh data are limited.

Review question 8: worked argument and scoring criteria

(a)

M* equals M exactly when the unique maximum is drawn at least once. The conditional probability is 1−(1−1/n)ⁿ→1−e^(−1). A fresh sample from a continuous distribution equals the fixed observed M with probability zero. The bootstrap maximum also cannot exceed M. Its persistent atom and truncated support warn that smooth-estimator bootstrap arguments do not automatically apply to extremes. This does not prove every possible bootstrap method fails for every tail functional; it identifies a concrete mismatch in this procedure.

2 points: Derives the finite-n probability.
1 point: Obtains the limiting atom.
1 point: Gives the continuous fresh-sample comparison.
1 point: States the precise limitation for extremes.

(b)

The selected index depends on random validation fluctuations; taking a minimum preferentially retains favorable fluctuations. Refit/freeze the chosen model using only the declared development procedure, then evaluate once on a new independent test set from the target population, preserving any known exposure and reporting sampling uncertainty. If data are limited, use outer held-out folds with the complete candidate fitting and inner selection repeated using only each outer training fold; assess the selected model on the untouched outer fold. Preprocessing, tuning and feature selection must stay inside that boundary. Dependence or distribution shift requires a design-appropriate split; random folds are not universally valid.

2 points: Explains selection-induced optimism.
1 point: Specifies a genuinely fresh frozen-model evaluation.
2 points: Specifies nested selection with preprocessing and design boundaries.

Targeted lesson returns:

3. Fitted models and decisions

40 practice points.

Question 9: A fitted line and two kinds of uncertainty

10 points

Fit y=β₀+β₁x+ε to the four synthetic rows below. For conditional inference assume fixed x, independent Normal(0,σ²) errors and a correct linear mean. Use unbiased residual variance SSE/(n−2). For a design row v=(1,x₀), mean variance is σ²v(XᵀX)⁻¹vᵀ and future-observation variance adds σ². The 0.975 quantile of t₂ is 4.303.

Synthetic line-fitting data
x	y
−1	0
0	1
1	1
2	4

(a) · 5 points

Compute XᵀX, fitted coefficients, all residuals, SSE, residual variance and the estimated coefficient covariance. Verify both normal-equation residual conditions.

(b) · 5 points

At x₀=1 compute the fitted mean and 95% mean-response and prediction intervals. Explain why zero residual sum neither verifies linearity nor establishes causal effects, and what changes under heteroscedastic errors.

Review question 9: worked argument and scoring criteria

(a)

XᵀX=[[4,2],[2,6]] with inverse [[0.3,−0.1],[−0.1,0.2]]. Xᵀy=(6,9), so β̂=(0.9,1.2). Fitted responses are (−0.3,0.9,2.1,3.3), residuals (0.3,0.1,−1.1,0.7), whose sum and x-weighted sum are zero. SSE=0.09+0.01+1.21+0.49=1.8, s²=0.9, and estimated coefficient covariance is [[0.27,−0.09],[−0.09,0.18]].

1 point: Computes the design cross-product and inverse.
1 point: Computes coefficients and residuals.
1 point: Checks both residual orthogonality conditions.
1 point: Computes SSE and residual variance.
1 point: Computes the full covariance matrix.

(b)

The fitted mean is 2.1. v(XᵀX)⁻¹vᵀ=0.3, so estimated mean variance is 0.27 and future-response variance is 1.17. The intervals are 2.1±4.303√0.27≈[−0.136,4.336] and 2.1±4.303√1.17≈[−2.554,6.754]. Residual orthogonality follows mechanically from least squares with an intercept and cannot verify the mean specification; observational association also does not establish intervention effects. Under unequal variances the covariance is (XᵀX)⁻¹XᵀΩX(XᵀX)⁻¹, not σ²(XᵀX)⁻¹, and the stated exact t reference is unjustified.

2 points: Computes the two variances and intervals.
1 point: Explains mechanical residual orthogonality.
1 point: Distinguishes association from causation.
1 point: Identifies the heteroscedastic covariance and calibration change.

Targeted lesson returns:

Question 10: Coordinates are part of the analysis

10 points

Four synthetic two-variable observations are (−2,0),(0,−1),(0,1),(2,0). Use centered sample covariance with divisor n−1. Principal directions have unit Euclidean length. The sign of a direction is arbitrary but must agree with the scores and reconstruction.

(a) · 5 points

Compute the mean, covariance, eigenvalues and a leading loading. Give all first-component scores, rank-one reconstructed rows, explained-variance fraction and total squared reconstruction error.

(b) · 5 points

Multiply the second measured coordinate by 3 and redo the leading-direction and explained-variance calculation. What happens if each coordinate is standardized by its sample standard deviation first? Explain what neither analysis proves about a population.

Review question 10: worked argument and scoring criteria

(a)

The mean is (0,0). The covariance is diag(8/3,2/3). Eigenvalues are 8/3 and 2/3, with leading loading (1,0) (or its negative). Scores are (−2,0,0,2), reconstructing (−2,0),(0,0),(0,0),(2,0). Explained variance is (8/3)/(10/3)=4/5. The squared reconstruction error is 0+1+1+0=2=(n−1) times the discarded eigenvalue.

1 point: Computes centering and covariance with the correct divisor.
1 point: Obtains eigenvalues and a unit loading.
1 point: Gives consistent scores and reconstructions.
1 point: Computes 4/5 explained variance.
1 point: Computes and checks the reconstruction error.

(b)

After rescaling, covariance is diag(8/3,6), so the leading direction becomes (0,1), with explained fraction 6/(6+8/3)=9/13. Standardizing either version yields the identity correlation matrix, whose eigenvalues are both 1. Every unit direction then maximizes sample projected variance; there is no unique first component, and any one component explains half the standardized variance. Unit changes affect covariance PCA. None of these sample calculations proves a population latent factor, a causal mechanism or stable loadings on future data; external validation and uncertainty assumptions are separate.

2 points: Computes the rescaled leading direction and 9/13 fraction.
2 points: Explains the standardized tie and nonuniqueness.
1 point: States a concrete statistical interpretation limit.

Targeted lesson returns:

Construct scores and reconstruction from data

Question 11: A fitting step and a nonfinite optimum

10 points

For grouped binomial observations, yᵢ successes out of mᵢ trials are conditionally independent with logit pᵢ=xᵢᵀβ, where logit p=ln[p/(1−p)]. Ignoring binomial constants, ℓ(β)=Σ[yᵢxᵢᵀβ−mᵢln(1+exp(xᵢᵀβ))]. First consider an intercept-only model with two rows (m,y)=(4,1),(6,3).

(a) · 5 points

Derive the vector score and negative Hessian from the supplied likelihood. For the intercept-only data compute one Newton step from β=0 and the finite MLE. State the conditional success-probability estimate.

(b) · 5 points

Now use two Bernoulli rows with intercept and slope: (x,y)=(−1,0),(1,1). Prove there is no finite maximum along intercept 0 and increasing slope, despite full design rank. Contrast this with an intercept-only Poisson log-link model having independent counts 0,1,2 and unit exposures; derive its MLE and estimated inverse information.

Review question 11: worked argument and scoring criteria

(a)

Differentiation gives U=Xᵀ(y−mp) and I=XᵀWX with Wᵢᵢ=mᵢpᵢ(1−pᵢ). The Newton ascent step is βnew=β+I⁻¹U when I is invertible. At β=0 all pᵢ=1/2, U=4−10/2=−1 and I=10/4=2.5, so the first step is −0.4. The intercept score vanishes at p=4/10=2/5, hence β̂=logit(2/5)=ln(2/3)≈−0.405465. The first iterate is close to, but not equal to, the maximum. With successes and failures both present, the strictly concave scalar likelihood has this finite maximum.

2 points: Derives the vector score and curvature.
1 point: States the correctly signed Newton update.
1 point: Computes the step and fitted probability.
1 point: Justifies existence and uniqueness in this case.

(b)

At intercept 0 and slope b, both observed outcomes have fitted probability sigmoid(b), so ℓ(b)=2ln sigmoid(b) increases toward 0 as b→∞, never attaining it at finite b. Each Bernoulli log probability is at most 0, so this is the global supremum and no finite maximizer exists. The two design rows (1,−1),(1,1) have full rank: rank does not preclude separation. For the Poisson model ℓ(β)=3β−3e^β plus constants; score 3−3e^β gives β̂=0 and fitted mean 1. Negative Hessian at the optimum is 3, so inverse information is 1/3. That last quantity is an asymptotic variance estimate, not a justified exact normal interval for three counts.

2 points: Proves separation and nonattainment of the likelihood supremum.
1 point: Separates rank from finite-MLE existence.
1 point: Derives the Poisson fit and inverse information.
1 point: Labels the uncertainty approximation accurately.

Targeted lesson returns:

Question 12: A frozen comparison with a changed target

10 points

Two models were fitted and frozen before observing the four synthetic validation responses below. The intended loss is mean squared error in the original response units. The validation cases are drawn independently from the intended population for this exercise. No extra test set has yet been collected.

Public synthetic validation evidence; predictions were frozen
Case	Observed y	Model A	Model B
1	1	0	1
2	2	2	3
3	4	3	5
4	5	7	5

(a) · 5 points

Compute both mean squared errors and the four paired differences (squared error of A minus that of B). Compute their sample mean and sample standard deviation. What does pairing preserve, and what population superiority claim is justified without further assumptions?

(b) · 5 points

The team now selects B and revises its predictions after inspecting these errors. Explain what must change before claiming fresh performance evidence. If the target changes from y to 1/y for positive y, prove why reciprocating a conditional-mean predictor generally fails to produce the new conditional mean, using Y=1 or 3 with equal probabilities as a counterexample.

Review question 12: worked argument and scoring criteria

(a)

A has squared errors (1,0,1,4), MSE 6/4=1.5. B has (0,1,1,0), MSE 2/4=0.5. Paired differences are (1,−1,0,4), mean 1. Their squared deviations from 1 are (0,4,1,9), so sample variance is 14/3 and sample standard deviation √(14/3)≈2.160. Pairing compares both models on the same cases and retains shared case difficulty. B has lower observed loss by 1; four cases and these calculations alone do not prove population superiority. An exact t claim would require a suitable normal difference model, not merely independence of cases.

2 points: Computes both error vectors and MSEs.
1 point: Computes paired differences and their mean.
1 point: Computes the sample spread with divisor 3.
1 point: Explains pairing and the inferential limit.

(b)

The four responses have become development evidence for selection and revision; relabeling or resetting them cannot make them fresh. Freeze the entire revised pipeline and evaluate on new independent target-population cases, recording prior exposure, the selected loss and uncertainty. For the reciprocal target, E(1/Y|x) is generally not 1/E(Y|x). In the equally likely {1,3} example, E(Y)=2 so 1/E(Y)=1/2, whereas E(1/Y)=(1+1/3)/2=2/3. Squared-error optimality for each target requires its own conditional expectation; changing the target can therefore change the optimal model and assessment.

2 points: Preserves exposure and specifies a fresh frozen-pipeline evaluation.
2 points: Derives the reciprocal counterexample.
1 point: Connects target changes to conditional-mean optimality and loss.

Targeted lesson returns:

Turn the review into a new argument

Award points for reasoning actually present. Allow valid later reasoning based on a carried-forward value unless its criterion requires the correct value; do not award correct-result points for an incorrect result. Record missing hypotheses, invalid conditioning, unsupported calibration and arithmetic errors separately.

The total is out of 120, with no university-grade conversion. Preserve your first attempt, consult the targeted lesson, explain the repair and attempt a fresh case. A teacher or peer may review the complete argument; if nobody did, label it self-review. Public worked solutions cannot certify independent proficiency.

Return to Mathematical statistics: cumulative practice for focused practice. The sustained statistics investigation supplies a separate report and changed-case task. Independent assessment remains a separate requirement.

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