Statistics Graduate Level Quiz: Simulation For Validation

10 questions · exam conditions

0:00

Simulation For ValidationQuestion 1 of 10

A researcher plans to use the large-sample test statistic $T_n=\sqrt{n}(\bar X-1)/S$ and reject the null hypothesis of mean $1$ when $|T_n|>1.96$ . Under the scientific null model, the observations are independent and satisfy $\log X_i\sim N(-1/2,1)$ , so the population mean is exactly $1$ . The sample size is $n=40$ .

Which simulation design most directly assesses the finite-sample validity of the proposed asymptotic rejection rule under the stated null model?

Generate normal observations with mean

1

and variance matching the lognormal variance, then record how often the proposed statistic exceeds

1.96

Resample the observed values without centering them, then record how often the proposed statistic exceeds

1.96

across bootstrap samples.

Generate lognormal observations using parameters estimated without imposing mean

1

, then compare the statistic with a Student

t

critical value.

Generate observations from

\log X_i\sim N(-1/2,1)

, recompute

S

in every sample, and record how often

|T_n|>1.96

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Simulation For Validation

Practice Simulation For Validation in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Simulation For Validation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which simulation design most directly assesses the finite-sample validity of the proposed asymptotic rejection rule under the stated null model?

Generate normal observations with mean $1$ and variance matching the lognormal variance, then record how often the proposed statistic exceeds $1.96$ .
Resample the observed values without centering them, then record how often the proposed statistic exceeds $1.96$ across bootstrap samples.
Generate lognormal observations using parameters estimated without imposing mean $1$ , then compare the statistic with a Student $t$ critical value.
Generate observations from $\log X_i\sim N(-1/2,1)$ , recompute $S$ in every sample, and record how often $|T_n|>1.96$ . (correct answer)

Explanation: When evaluating whether an asymptotic test achieves its nominal size at a specific finite sample, you need a Monte Carlo study that perfectly mirrors the stated null model — same distribution, same parameters, same statistic, same critical value. Any deviation introduces a confound that makes it impossible to isolate the finite-sample behavior of the proposed rule. Option D does exactly this: it draws

X_i

with

\log X_i \sim N(-1/2, 1)

, which guarantees

E[X_i]=1

(the null is true by construction), recomputes

S

fresh in every replicate just as the test requires, and checks whether

|T_n| > 1.96

. The empirical rejection frequency across many replicates estimates the true Type I error at

n=40

, directly answering whether the asymptotic

z

-approximation is valid here. Option A substitutes normal data for lognormal data. Even though the variance can be matched, the shape is completely different — lognormal is right-skewed, which is precisely why asymptotic approximation might struggle. Replacing the distribution defeats the purpose of the study. Option B describes a bootstrap procedure on observed data. This assesses variability of the statistic given a particular dataset, not the null-sampling behavior of the test across hypothetical repetitions of the experiment under the true null. Option C introduces two errors simultaneously: parameters are estimated without imposing the null (so the true mean may not equal 1), and the

t

critical value replaces the proposed 1.96 threshold. You can't assess the validity of a specific rule by changing both the data-generating model and the decision boundary. Strategy tip: For simulation-validity questions, ask yourself whether the design faithfully reproduces every element of the stated problem — null distribution, test statistic formula, and critical value. Any substitution is a red flag.

Question 2

For each sample size, an analyst validates a likelihood-ratio test of a scalar restriction in a model containing a nuisance parameter $\eta$ . The analyst estimates $\eta$ under the null from one observed data set, simulates many samples from that fitted null model, and compares the simulated statistic with a $\chi^2_1$ distribution.

What is the strongest generally justified conclusion from close agreement between the simulated and $\chi^2_1$ distributions?

The chi-square approximation is uniformly accurate over the entire nuisance-parameter space because the simulation uses a consistent estimate of $\eta$ .
The chi-square approximation is exact at every sample size for the fitted value of $\eta$ because the simulations impose the null restriction.
The chi-square approximation appears pointwise accurate near the fitted value of $\eta$ , but poor behavior elsewhere in the nuisance space remains possible. (correct answer)
The chi-square approximation is validated only under alternatives near the fitted value of $\eta$ because the nuisance parameter was estimated under the null.

Explanation: When validating asymptotic approximations through simulation, you must think carefully about what the simulation actually varies versus what it holds fixed. Here, the analyst fixes

\hat{\eta}

— a single estimated value of the nuisance parameter — and generates all simulated samples from that one point in the nuisance space. The simulation is essentially a pointwise check: "Does the

\chi^2_1

approximation work when

\eta = \hat{\eta}

?" This makes C the strongest justified conclusion. Close agreement tells you the approximation is accurate near the fitted value, but because the entire simulation is anchored to one estimate of

\eta

, you have no information about how the approximation behaves at other values of

\eta

. The nuisance-parameter space can be large, and the LRT's finite-sample behavior can vary substantially across it — a phenomenon called non-uniform size distortion. A is wrong because using a consistent estimate of

\eta

does not imply uniformity over the nuisance space. Consistency only guarantees convergence to the true value; it says nothing about approximation quality at other parameter points you never simulated. B is wrong because "exact at every sample size" is far too strong. The

\chi^2

approximation is asymptotic — it improves with

n

but is not exact at finite samples even under perfect conditions. Imposing the null restriction in simulation ensures you're sampling from the null, but that doesn't grant exactness. D is wrong because the question concerns approximation accuracy under the null (Type I error), not under alternatives. The nuisance parameter being estimated under the null is the setup, not a limitation on null-distribution validation. Study tip: Whenever a simulation study fixes nuisance parameters at a single estimated value, immediately recognize it as a pointwise validation — and flag any answer claiming uniform or global validity as overreach.

Question 3

An asymptotic approximation gives a tail probability for a test statistic at an extreme cutoff. Under a proposed null model, the true probability is expected to be around $10^{-5}$ . A direct simulation with $100{,}000$ independent replicates produces one exceedance.

Which next step would most efficiently provide a defensible simulation-based validation of the extreme-tail approximation?

Treat the observed proportion $10^{-5}$ as conclusive because it exactly matches the expected asymptotic order of magnitude.
Use antithetic random numbers under the same null distribution because they guarantee a large reduction in variance for every tail indicator.
Retain only simulated samples that exceed the cutoff and estimate the tail probability from their unweighted relative frequency.
Use an importance-sampling distribution that makes exceedances more common and weight each replicate by the appropriate likelihood ratio. (correct answer)

Explanation: Whenever you see a question about validating an extreme-tail probability via simulation, your central challenge is variance: with a true probability of

p = 10^{-5}

, a naive simulation needs millions of replicates just to observe a handful of exceedances. The question is really asking which method makes rare-event estimation statistically credible and efficient. Importance sampling (D) is the correct tool here. You shift the sampling distribution toward the tail so that exceedances occur far more frequently, then correct each observation by multiplying by the likelihood ratio

w_i = dP_0/dQ(x_i)

, where

P_0

is the null and

Q

is the proposal. This yields an unbiased estimator with dramatically lower variance, giving you a defensible confidence interval around your tail probability estimate with far fewer replicates. A is wrong because a single exceedance out of 100,000 trials produces an estimate with enormous relative uncertainty — the 95% confidence interval spans roughly an order of magnitude. "Matching the asymptotic order of magnitude" with

n=1

event is not statistical validation. B is wrong because antithetic variates reduce variance by exploiting negative correlation between pairs of replicates. This works well for smooth integrands but provides little to no guaranteed variance reduction for a rare binary indicator, since both antithetic samples typically equal zero simultaneously. C is wrong because retaining only exceedances and using their unweighted frequency is just naive Monte Carlo restricted to observed events — it does nothing to increase the rate at which rare events are generated, so it solves nothing. Your study tip: whenever

p \ll 1/n

, naive simulation is unreliable — reach immediately for importance sampling or other variance-reduction techniques designed specifically for rare events.

Question 4

A time-series regression uses a heteroskedasticity-and-autocorrelation-consistent Wald statistic. Under the null, the regression errors follow a stationary $\operatorname{AR}(1)$ process with autocorrelation $0.7$ . The bandwidth for the covariance estimator is selected separately from each data set.

Which design most directly assesses whether the chi-square approximation is accurate for the stated procedure?

Simulate null errors with autocorrelation $0.7$ , rerun the bandwidth selection, and recompute the Wald statistic separately in every replicate. (correct answer)
Simulate independent errors with the same marginal variance, retain the original bandwidth, and recompute the Wald statistic in every replicate.
Randomly permute the observed regression residuals, retain their marginal distribution, and reuse the covariance matrix from the original sample.
Simulate null errors with autocorrelation $0.7$ , but replace the robust covariance estimator by the exact covariance known in the simulation.

Explanation: When assessing whether a chi-square approximation is accurate for a specific inferential procedure, you must replicate every stochastic component of that procedure — not a simplified or modified version of it. The question describes a Wald statistic whose covariance estimator uses a data-driven bandwidth. That bandwidth selection is itself a random variable, so its variability is part of the procedure's finite-sample behavior. Choice A is correct because it faithfully reproduces the full procedure under the null: errors are generated with the specified

\text{AR}(1)

autocorrelation of

0.7

, bandwidth selection is rerun on each replicate, and the Wald statistic is recomputed fresh. The empirical distribution of these statistics can then be compared against the

\chi^2

reference to assess approximation quality. This is the only design that matches what would happen in practice. Choice B fails because simulating independent errors ignores the autocorrelation structure entirely. This tests a different scenario — one where HAC correction may be unnecessary — so any size distortion found (or not found) is irrelevant to the stated procedure. Choice C is flawed in two ways: permuting residuals cannot reproduce the correct temporal dependence structure, and reusing a single covariance matrix eliminates the variability from estimation, artificially understating uncertainty in the test statistic's distribution. Choice D undermines the assessment by replacing the HAC estimator with the exact known covariance. This removes estimation error from the covariance matrix, which is central to whether the chi-square approximation holds for the actual procedure. You'd be evaluating a cleaner, oracle version that no practitioner could use. The key strategy: whenever a procedure involves adaptive or data-driven components (like bandwidth selection), a valid simulation study must re-execute those components in every replicate — otherwise you're evaluating a phantom procedure, not the real one.

Question 5

A logistic-regression analyst evaluates a nominal $95\%$ Wald interval by simulation. Complete or quasi-complete separation occurs in $8\%$ of simulated samples, causing the unpenalized maximum-likelihood estimate and Wald interval to be undefined. Among the remaining samples, the interval covers the true coefficient in $94\%$ of cases.

Which conclusion best reflects a valid simulation assessment of the proposed asymptotic interval procedure?

The procedure has approximately $94\%$ unconditional coverage because failed fits are computational events unrelated to statistical coverage.
The procedure has exactly $0.92(0.94)=0.8648$ coverage because every separated sample must be counted as noncoverage.
The procedure has nominal coverage after conditioning on existence, which is sufficient to validate its unconditional asymptotic approximation.
The reported $94\%$ is conditional coverage only; unconditional performance requires a prespecified rule for failed fits and inclusion of those outcomes. (correct answer)

Explanation: When evaluating any statistical procedure by simulation, you must think carefully about the unconditional coverage probability — the probability that the interval covers the true parameter across all samples the procedure will encounter in practice, not just the "nice" ones where estimation succeeds. The key insight is that unconditional coverage is a weighted average over every possible sample outcome:

P(\text{cover}) = P(\text{cover} \mid \text{no sep.}) \cdot P(\text{no sep.}) + P(\text{cover} \mid \text{sep.}) \cdot P(\text{sep.})

Here,

P(\text{no sep.}) = 0.92

and the conditional coverage among those samples is

0.94

. But what happens in the

8\%

of separated samples? The problem doesn't say — the analyst never specified a fallback rule. Without knowing

P(\text{cover} \mid \text{sep.})

, the unconditional coverage is simply unknown. This is exactly what D captures: the

94\%

figure is conditional coverage only, and completing the assessment requires a prespecified rule (e.g., report no interval, use a penalized estimator) that determines outcomes for failed fits. A is wrong because separation is not merely a "computational nuisance" — those samples are drawn from the same data-generating process and their coverage outcomes count toward the unconditional rate. Ignoring them inflates the apparent coverage. B is tempting because it attempts an unconditional calculation, but it incorrectly assumes separated samples always fail to cover (assigning them coverage probability zero). That assumption is unjustified without a specified rule. C is the most seductive distractor: conditional coverage being near nominal does not validate unconditional performance, because the conditioning event itself (non-separation) is sample-dependent. A useful habit: whenever a simulation reports coverage "among converged samples," immediately ask what happened to the rest — those outcomes are never free to ignore.

Question 6

A likelihood-ratio test concerns a variance component constrained to be nonnegative. Under the null, the variance component equals zero. Theory predicts the limiting null distribution $\tfrac12\chi^2_0+\tfrac12\chi^2_1$ . A researcher instead uses the usual $\chi^2_1$ critical value $3.84$ . Null simulations at a large sample size produce a rejection rate of approximately $0.026$ .

Which conclusion is most consistent with the simulations and the boundary-limit theory?

The simulations validate the usual $\chi^2_1$ approximation because $0.026$ is reasonably close to the nominal level $0.05$ .
The simulations contradict the mixture limit because that limit assigns probability one-half to statistics larger than every positive critical value.
The simulations support the mixture limit because its tail probability above $3.84$ is approximately $\tfrac12(0.05)=0.025$ . (correct answer)
The simulations show that the correct asymptotic law is $\chi^2_0$ because roughly one-half of the limiting distribution is concentrated at zero.

Explanation: Whenever a likelihood-ratio test involves a parameter on the boundary of its parameter space (here, a variance component constrained to be ≥ 0), the standard

\chi^2_1

null distribution no longer applies. Instead, the limiting null distribution is the mixture

\tfrac{1}{2}\chi^2_0 + \tfrac{1}{2}\chi^2_1

, where

\chi^2_0

places a point mass of 1 at zero. Intuitively, about half the time the MLE hits the boundary (zero), producing a test statistic of exactly 0, and half the time it moves interior, producing a

\chi^2_1

statistic. The key calculation: the tail probability of this mixture above the

\chi^2_1

critical value

3.84

\tfrac{1}{2} \cdot P(\chi^2_0 > 3.84) + \tfrac{1}{2} \cdot P(\chi^2_1 > 3.84) = \tfrac{1}{2}(0) + \tfrac{1}{2}(0.05) = 0.025

. The simulated rejection rate of ≈ 0.026 matches this almost exactly, confirming C is correct. A is wrong because calling 0.026 "reasonably close" to 0.05 misses the point — the discrepancy is systematic and theoretically predicted, not random noise. The test is anti-conservative by exactly the factor the mixture theory explains. B is wrong because it misreads the mixture.

\chi^2_0

places all its mass at zero, so it contributes nothing to the upper tail — the mixture does not assign probability one-half above every positive value. D is wrong because the correct limit isn't purely

\chi^2_0

; it's the half-half mixture. Observing that half the mass concentrates at zero is only part of the picture. Study tip: On boundary-parameter problems, always check whether a mixture distribution applies — the rejection rate under the standard critical value will be exactly half the nominal level, a telltale signature you should recognize immediately.

Question 7

A heteroskedastic linear regression uses a sandwich-covariance Wald statistic for testing $H_0:R\beta=r$ . The intended asymptotic reference distribution is chi-square. The observed design matrix is treated as fixed, and the conditional error variance is believed to depend strongly on the fitted mean.

Which simulation scheme best validates the chi-square approximation for the procedure conditional on the observed design?

Impose $R\beta=r$ , generate heteroskedastic errors at the fixed design points, refit the model, and recompute the sandwich covariance in every replicate. (correct answer)
Keep the fitted coefficients unrestricted, generate homoskedastic normal errors at the fixed design points, and reuse the original sandwich covariance in every replicate.
Impose $R\beta=r$ , randomly resample rows of the design matrix, and compare coefficient estimates using the covariance from the original data.
Keep the responses fixed, generate new design matrices from their empirical distribution, and recompute only the model-based homoskedastic covariance matrix.

Explanation: When validating an asymptotic reference distribution via simulation, the golden rule is: your simulation must mimic exactly the conditions under which the statistic is derived. For a sandwich-covariance Wald statistic testing

H_0: R\beta = r

, this means three things must hold simultaneously — the null must be true, the error structure must match reality, and the covariance estimator must be the one actually used. Option A satisfies all three. By imposing

R\beta = r

, you ensure the simulated data genuinely come from the null distribution, so any rejection is a false positive — exactly what you're calibrating. By generating heteroskedastic errors at the fixed design points, you respect both the conditional-on-design assumption and the real error structure the sandwich estimator was built to handle. Recomputing the sandwich covariance in every replicate ensures the statistic is constructed identically to how it would be in practice. The resulting empirical distribution of the Wald statistic should then track

\chi^2

if the approximation is valid. Option B fails because homoskedastic errors violate the very heteroskedasticity the test is designed for; you'd be validating the wrong scenario and potentially finding chi-square agreement that disappears in practice. Option C resamples the design matrix, breaking the conditional-on-design framework — the simulation no longer reflects the fixed-

X

setting the asymptotic theory assumes. Option D keeps responses fixed (eliminating all Monte Carlo variation in the outcome) and uses a misspecified homoskedastic covariance, making it doubly invalid. Your study tip: whenever a simulation question involves hypothesis testing, ask yourself — "Is the null imposed? Is the data-generating process realistic? Is the estimator recomputed fresh?" All three must say yes.

Question 8

A statistician proposes an asymptotic normal confidence interval $\widehat\theta\pm1.96\widehat{\operatorname{se}}(\widehat\theta)$ for a nonlinear parameter. The standard error is computed by a complex data-dependent procedure, and the statistician wants to estimate the interval's finite-sample coverage under a known data-generating model with parameter value $\theta_0$ .

Which simulation procedure estimates the coverage of the interval actually being proposed?

Generate one data set, repeatedly resample it, and calculate the proportion of bootstrap estimates lying within $1.96$ original standard errors of the original estimate.
Generate many data sets at $\theta_0$ , recompute both the estimate and its standard error, and record whether each resulting interval contains $\theta_0$ . (correct answer)
Generate many data sets at $\theta_0$ , recompute the estimate, and assess whether each estimate lies within $1.96$ true standard deviations of its Monte Carlo mean.
Generate many data sets at the fitted parameter, retain the standard error from the observed data, and record whether intervals contain the fitted estimate.

Explanation: Whenever you see a question about coverage estimation, anchor yourself to this definition: the coverage probability of a confidence interval procedure is the long-run frequency with which the procedure — applied fresh to new data — captures the true parameter. Every ingredient of the procedure must be recomputed from scratch on each simulated dataset. B is correct because it faithfully mirrors the actual proposal. You fix

\theta_0

, generate many independent datasets from that model, and on each one you apply the full procedure: recompute

\widehat{\theta}

, recompute

\widehat{\operatorname{se}}(\widehat{\theta})

using the same complex data-dependent method, form the interval

\widehat{\theta} \pm 1.96\,\widehat{\operatorname{se}}

, and check containment of

\theta_0

. The proportion of successes estimates finite-sample coverage — exactly what was asked. A describes a bootstrap resampling diagnostic applied to a single observed dataset. It approximates the sampling distribution of

\widehat{\theta}

via resampling but never generates new data from

\theta_0

, so it cannot estimate coverage in the frequentist sense. C uses the true standard deviation and the Monte Carlo mean — neither of which is available in practice or part of the proposed interval. You would be evaluating an idealized procedure, not the one actually proposed. D fixes the standard error at its observed-data value instead of recomputing it on each new dataset. This ignores variability in

\widehat{\operatorname{se}}

, which is a key source of finite-sample coverage distortion — precisely what the statistician wants to study. Strategy tip: When evaluating a simulation procedure, mentally check a checklist — correct distribution, correct parameter value, and all procedure components recomputed independently. Any shortcut (fixed SE, single dataset, oracle quantities) means you're evaluating a different procedure than the one proposed.

Question 9

To assess a nominal level- $0.05$ Wald test, an analyst performs $20{,}000$ independent null simulations and observes rejection in proportion $0.0540$ of them. Assume simulation replicates are independent and generated correctly.

Which interpretation best accounts for Monte Carlo uncertainty?

The approximation is adequately calibrated because any rejection estimate within one percentage point of $0.05$ is indistinguishable from nominal size.
There is evidence of modest size inflation because the Monte Carlo standard error is about $0.0016$ , placing $0.05$ just below a two-sided $95\%$ interval. (correct answer)
There is no evidence of size inflation because the relevant standard error is based on the original data sample size rather than the number of simulations.
The approximation is severely invalid because $0.0540/0.05=1.08$ implies an accurately estimated relative size distortion of exactly $8\%$ .

Explanation: When you simulate a null hypothesis to assess test size, your observed rejection rate is itself a random variable with its own sampling distribution. The key framework here is treating the proportion of rejections across

B

simulations as a binomial estimate, and applying standard error logic to determine whether any deviation from nominal size is statistically meaningful. Under the null, each simulation replicate is a Bernoulli trial with probability

p_0 = 0.05

. The Monte Carlo standard error of the estimated rejection rate is

\sqrt{p_0(1-p_0)/B} = \sqrt{(0.05)(0.95)/20{,}000} \approx 0.00154

, roughly

0.0016

. A two-sided 95% confidence interval around the observed rate of

0.0540

is approximately

0.0540 \pm 1.96(0.0016)

, or about

(0.0509,\ 0.0571)

. Since

0.05

falls just below this interval, there is modest but detectable evidence of size inflation — confirming B as the best answer. A is wrong because declaring any estimate within one percentage point "indistinguishable" ignores the actual precision of the simulation. With 20,000 replicates, you have enough resolution to detect deviations far smaller than one percentage point. C is wrong because the relevant standard error for evaluating the simulated rejection rate depends on the number of simulation replicates, not the original data sample size. These are completely separate sources of variability. D is wrong because computing

0.054/0.05 = 1.08

without accounting for Monte Carlo uncertainty overstates precision. The ratio is an estimate, not an exact measurement, and calling the distortion "exactly 8%" ignores simulation noise. Study tip: Whenever a question gives you a simulated rejection rate and a number of replicates, immediately compute

\sqrt{\hat{p}(1-\hat{p})/B}

— the Monte Carlo SE is the lens through which all conclusions must pass.

Question 10

For a nominal level- $0.05$ test, simulations give estimated null rejection probabilities $0.080$ at sample size $100$ and $0.060$ at sample size $400$ . The analyst claims that the size distortion is of order $n^{-1/2}$ because the excess rejection probability fell from $0.030$ to $0.010$ .

Which assessment of this claim is most appropriate?

The claim is established because quadrupling the sample size must reduce every first-order asymptotic error by exactly one-half, and the observed decline is consistent with that law.
The claim is disproved because an $n^{-1/2}$ error would require the rejection probability itself, rather than its excess over $0.05$ , to halve when the sample size quadruples.
The claim is not established; assessing a convergence rate requires examining several sample sizes and accounting for Monte Carlo uncertainty in each estimated rejection probability. (correct answer)
The claim is established provided both simulations use the same number of replicates, since equal replication ensures the two estimated distortions are directly and reliably comparable.

Explanation: When evaluating claims about asymptotic convergence rates in simulation studies, you need to think carefully about what constitutes statistical evidence versus a plausible-looking pattern. A single pair of data points is almost never sufficient to establish a rate of convergence. Here's why C is correct: the analyst observes the excess rejection probability dropping from

0.030

0.010

when

n

goes from

100

400

. An

n^{-1/2}

rate would predict the excess to scale by

\sqrt{100}/\sqrt{400} = 1/2

, giving

0.015

, not

0.010

. The observed drop is actually larger than what

n^{-1/2}

predicts — more consistent with

n^{-1}

. But even setting that aside, two simulation estimates cannot establish a rate. Each estimated rejection probability carries Monte Carlo standard error (roughly

\sqrt{\hat{p}(1-\hat{p})/R}

), so the difference of two estimates inherits compounded uncertainty. You need multiple sample sizes to fit a trend and confidence intervals around each estimate to make a credible rate claim. A is wrong because quadrupling

n

halves an

n^{-1/2}

error only in the limit, and this "law" doesn't apply exactly at finite

n

— nor does two points establish adherence to any asymptotic law. B is wrong because it misidentifies what should halve. Under

n^{-1/2}

distortion, the excess over the nominal level scales as

n^{-1/2}

, not the rejection probability itself, so B's criterion is simply incorrect. D is wrong because equal replication controls one source of variance but doesn't resolve the fundamental problem: two data points are insufficient to establish a convergence rate regardless of replication balance. Study tip: When a question asks you to assess a convergence rate claim, always ask: how many data points were used, and was Monte Carlo uncertainty quantified? One pair of simulations is never enough.