Statistics Graduate Level Quiz: Bootstrap Resampling

10 questions · exam conditions

0:00

Bootstrap ResamplingQuestion 1 of 10

A linear regression is fit at fixed design points. The errors have conditional mean zero, but diagnostic and scientific considerations indicate that their variances increase substantially with the fitted mean. The investigator wants a bootstrap standard error for a regression coefficient while treating the design points as fixed.

Which resampling procedure is most defensible under these conditions?

Resample centered ordinary residuals identically across all design points and add them to the fitted responses

Resample entire predictor-response pairs, thereby treating the observed design points as a random sample

Multiply appropriately centered residuals by independent mean-zero, unit-variance random weights while keeping the design fixed

Resample the observed responses without their predictors and refit them at the original design points

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Bootstrap Resampling

Practice Bootstrap Resampling 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 Bootstrap Resampling, 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 resampling procedure is most defensible under these conditions?

Resample centered ordinary residuals identically across all design points and add them to the fitted responses
Resample entire predictor-response pairs, thereby treating the observed design points as a random sample
Multiply appropriately centered residuals by independent mean-zero, unit-variance random weights while keeping the design fixed (correct answer)
Resample the observed responses without their predictors and refit them at the original design points

Explanation: When bootstrapping regression with fixed design points, your resampling strategy must respect two constraints simultaneously: keep the predictors fixed (since they're not random) and account for heteroscedasticity (since variance differs across observations). This question tests whether you can identify which procedure handles both. The defensible choice is C — the wild bootstrap. By multiplying each centered residual by an independent mean-zero, unit-variance random weight (e.g., Rademacher or standard normal weights), you resample in place at each design point. This preserves the heteroscedastic structure because the variance of the resampled residual at point

i

scales with

\hat{e}_i^2

, automatically adapting to local variance without assuming homogeneity across points. A fails because pooling residuals across all design points and redistributing them identically assumes homoscedasticity — it treats all residuals as drawn from the same distribution. When variance increases with the fitted mean, transplanting a small-

x

residual onto a large-

x

point distorts the error structure entirely. B violates the fixed-design premise outright. Resampling predictor-response pairs treats the

x

-values as random draws from some population, which contradicts the study design. With a small number of fixed design points, this approach can badly misrepresent the sampling variability of the coefficient. D is incoherent: detaching responses from their predictors destroys the regression relationship. Refitting scrambled responses at original design points produces nonsensical coefficient estimates with no connection to the original model. Study tip: On bootstrap questions, always ask two things first — are predictors fixed or random, and is homoscedasticity plausible? Fixed + heteroscedastic points directly to the wild bootstrap every time.

Question 2

In a cluster-randomized trial, clinics are independently assigned to treatment or control. Outcomes are measured for all enrolled patients within each clinic, and clinic sizes differ. The analysis estimates a treatment effect using patient-level data with clinic indicators.

Which bootstrap scheme best reflects the randomization and dependence structure when estimating the treatment-effect standard error?

Resample individual patients within each treatment arm and retain each patient's original clinic indicator
Resample clinics within each treatment arm and retain all observed patients from every selected clinic (correct answer)
Resample all clinics from a common pool and independently redraw a treatment label for each selected clinic
Resample residuals across all patients without regard to clinic membership or treatment assignment

Explanation: When designing a bootstrap for inference, your guiding principle should be: mimic the original randomization mechanism. In a cluster-randomized trial, the unit of randomization is the clinic, not the individual patient. Clinics are assigned to treatment or control, and patients within a clinic share correlated outcomes because of shared clinic-level effects. Any valid resampling scheme must respect both of these facts. Option B is correct because it resamples clinics within each treatment arm, then retains all patients from each selected clinic. This preserves the within-clinic correlation structure (patients travel with their clinic), honors the arm-specific randomization (you resample treatment clinics separately from control clinics), and naturally accommodates unequal clinic sizes. The resulting bootstrap distribution of the treatment-effect estimator correctly captures variability that arises from clinic-level randomization. Option A goes wrong at the fundamental level: resampling individual patients breaks the intra-clinic correlation. Even though it retains clinic indicators, shuffling patients independently treats the clinic as a fixed label rather than a source of shared variance, underestimating the true standard error. Option C pools clinics and randomly reassigns treatment labels. This ignores the fixed, stratified nature of the original randomization and could produce bootstrap samples where arm sizes differ wildly from the original design, distorting inference. Option D—resampling patient-level residuals without regard to clinic or treatment—is the most naive approach. It destroys both the clustering structure and the treatment assignment mechanism simultaneously, producing a standard error that is almost certainly too small. Study tip: On questions about bootstrap validity, always ask "what was the randomization unit?" — your bootstrap must resample that unit, stratified by the factor that was held fixed in the design.

Question 3

A stratified random sample contains $n_h$ observations from stratum $h$ . The population stratum proportions $W_h$ are known and differ from the sample proportions. The estimator of the population mean is $\hat{\mu}=\sum_h W_h\bar{Y}_h$ .

Which nonparametric bootstrap procedure appropriately estimates the standard error of $\hat{\mu}$ under the stratified sampling design?

Pool all sampled observations, draw a bootstrap sample of the total original size, and compute its unweighted mean
Resample $n_h$ observations within each stratum and recompute $\sum_h W_h\bar{Y}_h^*$ using the known population weights (correct answer)
Resample $n_h$ observations within each stratum and weight the bootstrap stratum means by the realized sample proportions
Draw strata with replacement using probabilities $W_h$ and retain every sampled observation whenever its stratum is selected

Explanation: Whenever you see a bootstrap question involving complex sampling designs, your first instinct should be: does the resampling procedure respect the original sampling structure? Stratified sampling treats each stratum as a separate, independent sampling unit, so the bootstrap must mirror that. In stratified sampling, the population mean estimator

\hat{\mu} = \sum_h W_h \bar{Y}_h

uses known population weights

W_h

— not the sample proportions — to combine stratum-level means. To validly estimate its standard error via bootstrap, you must preserve two things: (1) the within-stratum resampling that reflects each stratum's variability, and (2) the fixed population weights

W_h

used to aggregate. Answer B does exactly this — resampling

n_h

observations with replacement within each stratum and then computing

\sum_h W_h \bar{Y}_h^*

with the known weights. Each bootstrap replicate mimics the original estimator's construction, so the variance across replicates consistently estimates

\text{SE}(\hat{\mu})

. Answer A destroys the stratified structure entirely by pooling observations before resampling. This ignores the design and conflates between-stratum and within-stratum variation, producing an invalid standard error. Answer C correctly resamples within strata but then weights by the realized sample proportions

n_h/n

rather than the known population weights

W_h

. This changes the estimand — you're no longer bootstrapping

\hat{\mu}

as defined. Answer D resamples entire strata rather than observations within strata, which is not a valid procedure for this design and can catastrophically distort variability estimation. Study tip: On bootstrap questions with complex designs, always ask "does the bootstrap estimator match the original estimator formula exactly?" If any piece changes — the weights, the resampling unit, the aggregation — the procedure is invalid.

Question 4

A sample of size $16$ is modeled as independent Poisson observations with unknown mean $\lambda$ . The observed sample mean is $2.25$ . A parametric bootstrap generates each replicate from a Poisson distribution with parameter equal to the observed sample mean and recomputes the sample mean.

As the number of bootstrap replicates becomes large, to what value does the parametric bootstrap standard-error estimate converge, conditional on the observed sample?

$0.375$ , the square root of the fitted Poisson mean divided by the sample size, i.e., $\sqrt{\hat{\lambda}/n}$ (correct answer)
$1.500$ , the square root of the fitted Poisson variance for a single observation
$0.141$ , the fitted conditional variance of the bootstrap sample mean rather than its standard deviation
$0.387$ , obtained by incorrectly replacing the sample size $n=16$ with the degrees of freedom $n-1=15$ in the denominator

Explanation: When you see a question about parametric bootstrap standard errors, anchor yourself to one key principle: the parametric bootstrap estimates uncertainty by re-sampling from a fitted model, so its large-sample behavior is determined entirely by that fitted model's properties — not by resampling variability or degrees-of-freedom corrections. Here, you observe

n = 16

Poisson observations with sample mean

\bar{X} = 2.25

. The parametric bootstrap plugs in

\hat{\lambda} = 2.25

and repeatedly draws

n = 16

i.i.d.

\text{Poisson}(2.25)

observations, computing the sample mean each time. As the number of replicates grows, the bootstrap distribution of that sample mean converges to its true distribution under

\hat{\lambda}

. Since the variance of a single

\text{Poisson}(\hat{\lambda})

observation is

\hat{\lambda}

, the variance of the sample mean of

n = 16

such observations is

\hat{\lambda}/n = 2.25/16 = 0.140625

, giving a standard error of

\sqrt{0.140625} = 0.375

. That confirms A is correct. Choice B mistakes the standard error of the mean for the standard deviation of a single observation —

\sqrt{2.25} = 1.5

ignores the

1/n

factor entirely. Choice C confuses standard error with variance:

0.140625

\hat{\lambda}/n

itself, the variance, not the standard deviation. Choice D replaces

n = 16

with

n - 1 = 15

in the denominator, a degrees-of-freedom correction appropriate for estimating population variance from a sample — it has no role in the bootstrap standard error formula. As a study tip: always distinguish variance of the mean (

\sigma^2/n

) from variance of one observation (

\sigma^2

), and remember that parametric bootstrap SEs reflect the fitted model, not sample-based corrections like

n-1

Question 5

An investigator uses $B=801$ independent bootstrap replicates. The estimated bootstrap standard error of a statistic is $0.80$ , and the conditional distribution of the bootstrap statistic is approximately normal.

Using the large-sample approximation for the sampling variability of a sample standard deviation, what is the approximate Monte Carlo standard error of the reported bootstrap standard error?

$0.800$ , because additional replicates do not reduce conditional bootstrap variability
$0.028$ , using relative Monte Carlo error approximately $1/\sqrt{B-1}$
$0.001$ , using absolute Monte Carlo error approximately equal to $0.80/B$
$0.020$ , using relative Monte Carlo error approximately $1/\sqrt{2(B-1)}$ (correct answer)

Explanation: When bootstrapping, you're using simulation to estimate a statistic, which means your reported value carries its own sampling uncertainty — called Monte Carlo standard error (MCSE). The key insight here is that the bootstrap standard error is itself a sample standard deviation computed from

B

replicates, so you can apply the classical large-sample formula for the standard error of a sample standard deviation. For a normal population with standard deviation

\sigma

, the standard error of the sample standard deviation

s

based on

n

observations is approximately

\sigma/\sqrt{2(n-1)}

. Here,

\sigma

is the estimated bootstrap standard error

0.80

, and

n = B = 801

replicates. Plugging in:

\text{MCSE} \approx \frac{0.80}{\sqrt{2(801-1)}} = \frac{0.80}{\sqrt{1600}} = \frac{0.80}{40} = 0.020

This confirms D is correct. Choice A is conceptually backwards — more replicates absolutely do reduce Monte Carlo error. The conditional bootstrap distribution is fixed given your data, but your estimate of its spread improves with more draws. Choice B uses the formula

1/\sqrt{B-1}

, which omits the factor of

\sqrt{2}

in the denominator. This formula would apply to a standard error of the mean, not of a standard deviation. The factor of 2 under the radical is essential here. Choice C uses

0.80/B

directly, which has no grounding in the standard error formula — it simply divides by

B

rather than

\sqrt{2(B-1)}

, producing a severe underestimate. Study tip: Memorize that the MCSE of a bootstrap standard error scales as

\hat{\sigma}/\sqrt{2(B-1)}

, not

\hat{\sigma}/\sqrt{B-1}

— the factor of 2 distinguishes standard deviation variability from mean variability.

Question 6

The parameter of interest is $\phi=\log(\theta)$ . The observed estimate is $\hat{\theta}=2$ , and three illustrative bootstrap replicates of $\hat{\theta}$ are $1,2,$ and $4$ .

Using the usual sample-standard-deviation definition across the transformed bootstrap replicates, what is the bootstrap standard error of $\hat{\phi}=\log(\hat{\theta})$ ?

$(\log 2)/\sqrt{3}\approx0.400$ , the standard error of the mean of the transformed replicates
$\sqrt{7/3}/2\approx0.764$ , a delta-method transformation evaluated at the observed estimate
$\sqrt{7/3}/(7/3)\approx0.655$ , a delta-method transformation evaluated at the bootstrap mean
$\log 2\approx0.693$ , the standard deviation of $0,\log 2,$ and $2\log 2$ (correct answer)

Explanation: When bootstrapping a transformed parameter, the key principle is simple: transform first, then compute standard deviation directly on the transformed replicates. Don't apply the delta method — the bootstrap already handles nonlinearity automatically. Here, the three bootstrap replicates of

\hat{\theta}

are

1, 2, 4

. Transforming each gives

\hat{\phi}^* \in \{0, \log 2, 2\log 2\}

. The bootstrap standard error is just the sample standard deviation of these three values. Their mean is

\log 2

, and the squared deviations are

(\log 2)^2,\ 0,\ (\log 2)^2

, summing to

2(\log 2)^2

. Dividing by

n-1 = 2

and taking the square root gives

\sqrt{(\log 2)^2} = \log 2 \approx 0.693

, confirming D is correct. Choice A divides

\log 2

\sqrt{3}

, which would be the standard error of the mean of the replicates — a conceptual mix-up. The bootstrap SE of

\hat{\phi}

is the SD of the replicates, not the SE of their mean. Choice B applies the delta method formula

\text{SE}(\hat{\phi}) \approx \frac{1}{\hat{\theta}}\,\text{SE}(\hat{\theta})

, evaluated at

\hat{\theta} = 2

. This approximates what the bootstrap computes exactly — it's the wrong tool here. Choice C also uses the delta method but evaluates it at the bootstrap mean of

\hat{\theta}

(which is

7/3

), compounding the error. Study tip: Whenever a bootstrap SE question involves a transformation, immediately transform the replicates and compute their sample SD directly. Reserve the delta method for closed-form approximations when replicates aren't available.

Question 7

Let $X_1,\ldots,X_n$ be independent observations from a uniform distribution on $[0,\theta]$ , and let $M_n$ be the sample maximum. An investigator uses the ordinary nonparametric bootstrap, drawing samples of size $n$ from the empirical distribution, to estimate the sampling uncertainty of $M_n$ .

Which statement best identifies the fundamental difficulty with this bootstrap procedure?

Conditional on the data, the bootstrap maximum equals $M_n$ with probability approaching $1-e^{-1}$ , creating an atom absent from the target limiting distribution (correct answer)
Because $M_n$ is not asymptotically normal at rate $\sqrt{n}$ , no resampling method can estimate its sampling uncertainty consistently
Conditional on the data, the bootstrap maximum is unbiased for $\theta$ , so its bootstrap variance systematically exceeds the true variance
The procedure fails only because it draws exactly $n$ observations; drawing $n+1$ observations removes the endpoint discrepancy asymptotically

Explanation: When the ordinary bootstrap resamples from the empirical distribution

\hat{F}_n

, which places mass

1/n

on each observed value, the bootstrap maximum

M_n^*

can never exceed

M_n

. This is the core of the problem: the bootstrap is blind to any part of the parameter space beyond the observed data. To see why A is correct, compute the probability that

M_n^* = M_n

. Each bootstrap draw equals

M_n

with probability

1/n

, so the chance of not drawing

M_n

in a single draw is

(1 - 1/n)

. Over

n

draws,

P(M_n^* < M_n) = (1-1/n)^n \to e^{-1}

, meaning

M_n^* = M_n

with probability approaching

1 - e^{-1} \approx 0.632

. This creates a point mass at

M_n

in the bootstrap distribution — a structural atom that has no counterpart in the true sampling distribution of

M_n

, which is continuous. The bootstrap distribution is therefore fundamentally inconsistent as an estimator of the true sampling uncertainty. B is wrong because the issue is not that non-normal rates preclude all resampling methods — subsampling and the

m

-out-of-

n

bootstrap can both work for this problem with appropriate tuning. C is wrong because the bootstrap maximum is biased downward for

\theta

(it cannot exceed

M_n

), not unbiased — and the argument about variance doesn't follow logically in any case. D is wrong because drawing

n+1

observations from

\hat{F}_n

does not resolve the fundamental ceiling at

M_n

; the atom problem persists regardless of resample size. A useful heuristic: whenever a statistic lives at the boundary of the support of

\hat{F}_n

, be immediately suspicious of the standard bootstrap — it cannot simulate draws beyond observed values.

Question 8

A crossover study records each participant's blood pressure after treatment and after placebo. The parameter of interest is the population mean of the within-participant treatment-minus-placebo difference. Dependence between the two measurements from the same participant is substantial.

Which bootstrap procedure most appropriately estimates the standard error of the sample mean difference?

Resample treatment measurements and placebo measurements independently, then subtract the two resampled means
Resample participants with replacement, retaining both measurements for every selected participant, then average the paired differences (correct answer)
Pool all measurements, resample twice from the pooled values, and subtract the resulting sample means
Compute the observed paired differences, randomly permute their signs, and use the standard deviation of the resulting means

Explanation: Whenever you see a bootstrap question involving paired or repeated-measures data, the central issue is preserving the dependence structure. The estimand here is the population mean of within-participant differences, so the sampling unit is the participant, not the individual measurement. The correct approach, B, resamples participants with replacement and keeps both measurements together for each selected participant. This preserves the within-person correlation — the very feature that makes paired designs powerful. After resampling, you compute each participant's treatment-minus-placebo difference and average them. The variability of this mean across bootstrap samples gives you the standard error of the observed mean difference. Because the unit of analysis is the pair, the bootstrap correctly reflects how the estimator would vary across repeated samples from the population. A breaks the pairing entirely by resampling treatment and placebo measurements from separate pools. This discards the within-participant dependence, inflating the estimated standard error — essentially treating the design like two independent groups when it isn't. C compounds this error further by pooling all measurements together, destroying any distinction between treatment and placebo and producing a completely meaningless resampled difference. D is a sign-flip permutation test, which addresses hypothesis testing (specifically testing whether the mean difference is zero) rather than standard error estimation — it answers a different question entirely. A useful rule of thumb: the bootstrap resampling unit must match the unit of randomization or observation. In paired designs, that unit is the participant-pair. Any procedure that breaks up the pair introduces bias into the standard error estimate.

Question 9

A stationary time series has substantial short-range serial correlation. The target is the process mean, estimated by the mean of $n$ consecutive observations. An independent-observation bootstrap is suspected of understating the estimator's standard error.

Which bootstrap construction is asymptotically appropriate under standard weak-dependence conditions?

Resample individual observations independently, because stationarity makes their marginal distributions identical
Resample one block containing the entire observed series, with its starting point chosen uniformly at random
Resample overlapping blocks of length $\ell$ and concatenate them, with $\ell\to\infty$ and $\ell/n\to0$ (correct answer)
Randomly permute all observations before resampling them, so each bootstrap series has negligible serial correlation

Explanation: When you encounter questions about bootstrap methods for dependent data, the key issue is whether your resampling scheme preserves the dependence structure present in the original series. An i.i.d. bootstrap treats observations as exchangeable, which destroys serial correlation and causes the bootstrap variance to underestimate the true variance of the sample mean — exactly the problem described here. The fix is the Moving Block Bootstrap (MBB): divide the series into overlapping blocks of length

\ell

, then randomly draw

\lceil n/\ell \rceil

blocks with replacement and concatenate them to form a bootstrap series of length

n

. Within each block, the serial dependence is intact. The asymptotic validity requires

\ell \to \infty

(so blocks grow long enough to capture the dependence) and

\ell/n \to 0

(so you're drawing many blocks, not just one or two), which is exactly what C states. Under standard mixing (weak-dependence) conditions, this procedure consistently estimates the variance of the sample mean. Choice A fails precisely because stationarity guarantees identical marginal distributions but says nothing about joint dependence. Resampling independently discards the covariance structure, understating the standard error. Choice B resamples a single block — essentially the whole series shifted by a random start — giving you only one possible bootstrap draw, which yields zero bootstrap variance. This is statistically useless. Choice D permutes observations, which actively eliminates serial correlation rather than preserving it, making the bootstrap series even less representative of the true process than the i.i.d. bootstrap. Study tip: On dependent-data bootstrap questions, always ask: "Does this scheme preserve within-block dependence while allowing between-block independence?" If yes, you likely have a valid block bootstrap.

Question 10

A statistic is computed from an observed sample, and four nonparametric bootstrap samples produce the estimates $2, 4, 5,$ and $9$ . Using the usual finite- $B$ bootstrap estimator, what is the estimated bootstrap standard error?

$\sqrt{26/3} \approx 2.94$ , obtained by centering at the mean of the bootstrap estimates (correct answer)
$\sqrt{26/4} \approx 2.55$ , obtained by using the number of bootstrap samples as the divisor
$\sqrt{30/3} \approx 3.16$ , obtained by centering the bootstrap estimates at the observed estimate $4$
$\sqrt{26/3}/\sqrt{4} \approx 1.47$ , obtained by estimating the standard error of their average

Explanation: Whenever you encounter bootstrap standard error questions, your anchor formula should be the sample standard deviation of the bootstrap estimates, treating those

B

replicates as your data. The standard bootstrap estimator centers at their own mean and uses

B - 1

in the denominator — exactly like a sample standard deviation. Here, the four bootstrap estimates are

2, 4, 5, 9

with mean

\bar{\theta}^* = 20/4 = 5

. The estimated bootstrap standard error is:

\widehat{se}_B = \sqrt{\frac{1}{B-1}\sum_{b=1}^{B}(\hat{\theta}^{*b} - \bar{\theta}^*)^2} = \sqrt{\frac{(2-5)^2+(4-5)^2+(5-5)^2+(9-5)^2}{3}} = \sqrt{\frac{9+1+0+16}{3}} = \sqrt{\frac{26}{3}} \approx 2.94

This confirms A is correct. B is wrong because it divides by

B = 4

instead of

B - 1 = 3

. That would be the population variance formula, not the sample standard deviation — a classic off-by-one error. C is wrong because it centers the deviations at

4

, the observed sample estimate, rather than the mean of the bootstrap replicates

\bar{\theta}^* = 5

. The bootstrap SE formula always centers at the bootstrap mean, not the original statistic. D is wrong because dividing by

\sqrt{B}

converts a standard deviation into a standard error of the mean — it answers a completely different question about how precisely the bootstrap estimates average together, not how variable the statistic itself is. Study tip: Memorize the bootstrap SE formula as "sample standard deviation of the

B

replicates." Any question that tempts you to substitute the original estimate or divide by

B

instead of

B-1

is testing whether you know this distinction.