Statistics Graduate Level Quiz: Bootstrap Confidence Intervals

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Bootstrap Confidence IntervalsQuestion 1 of 10

A positive parameter $θ$ is estimated by $\hat{θ}$ . A percentile bootstrap interval for $θ$ is $[2,8]$ . The same bootstrap replicates are transformed to estimate $φ=\log θ$ .

Which statement correctly compares percentile and basic bootstrap intervals under this transformation?

The percentile interval for

φ

[\log 2,\log 8]

, but a basic interval need not transform endpoint by endpoint.

The basic interval for

φ

[\log 2,\log 8]

, but a percentile interval need not transform endpoint by endpoint.

Both intervals for

φ

must equal

[\log 2,\log 8]

because the logarithm is monotone.

Neither interval for

φ

can be obtained from the original endpoints because the logarithm changes bootstrap ranks.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Bootstrap Confidence Intervals

Practice Bootstrap Confidence Intervals 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 Confidence Intervals, 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

A positive parameter $θ$ is estimated by $\hat{θ}$ . A percentile bootstrap interval for $θ$ is $[2,8]$ . The same bootstrap replicates are transformed to estimate $φ=\log θ$ .

Which statement correctly compares percentile and basic bootstrap intervals under this transformation?

The percentile interval for $φ$ is $[\log 2,\log 8]$ , but a basic interval need not transform endpoint by endpoint. (correct answer)
The basic interval for $φ$ is $[\log 2,\log 8]$ , but a percentile interval need not transform endpoint by endpoint.
Both intervals for $φ$ must equal $[\log 2,\log 8]$ because the logarithm is monotone.
Neither interval for $φ$ can be obtained from the original endpoints because the logarithm changes bootstrap ranks.

Explanation: When you see a question about bootstrap intervals and monotone transformations, the key distinction to internalize is that the percentile method and the basic (pivotal) method behave very differently under reparameterization. The percentile interval works directly with quantiles of the bootstrap distribution of

\hat{\theta}^*

. If your percentile interval for

\theta

[2, 8]

, those are simply the 2.5th and 97.5th percentiles of the bootstrap replicates. When you apply a monotone transformation like

\log

, the ranks of the replicates are preserved — the same replicates that defined the endpoints now map to

\log(2)

and

\log(8)

. So the percentile interval for

\phi = \log\theta

is exactly

[\log 2, \log 8]

. This is the transformation equivariance property of the percentile method. The basic (pivotal) interval, by contrast, is constructed from the pivot

\hat{\theta} - \theta

and uses the formula

[2\hat{\theta} - \hat{\theta}^*_{(1-\alpha/2)},\ 2\hat{\theta} - \hat{\theta}^*_{(\alpha/2)}]

. After transforming to

\phi

, the pivot changes form — it becomes

\hat{\phi} - \phi

— and the reflection formula must be recomputed on the log scale. Simply log-transforming the original endpoints does not reproduce the correct basic interval. Answer A correctly captures this asymmetry, making it correct. Answer B reverses the roles — it's the percentile interval, not the basic interval, that transforms equivariantly. Answer C incorrectly extends the equivariance property to the basic method. Answer D is wrong because monotone transformations preserve ranks, so bootstrap quantiles do transform endpoint by endpoint (for the percentile method). Study tip: Remember "percentile = plug-in quantiles = transformation equivariant." Any time you see a monotone transformation question, the percentile interval always maps cleanly; the basic interval requires recomputation.

Question 2

A random-design linear regression is fitted to independent observations. Diagnostic analysis and subject-matter knowledge indicate that the error variance increases substantially with the predictor. The investigator wants a bootstrap percentile interval for the slope without specifying a parametric variance function.

Which resampling plan most directly preserves the relevant heteroskedastic sampling structure?

Resample centered ordinary residuals and add them to fitted values at the original predictor values.
Resample observation pairs consisting of each response and its associated predictor, then refit the model. (correct answer)
Resample predictor values alone while holding each observed response in its original position.
Resample fitted values and residuals independently, then combine them after random permutation.

Explanation: When choosing a bootstrap strategy for regression, the central question is: what sampling unit captures the full probabilistic structure of the data? Under heteroskedasticity, the variance of the error term depends on the predictor value — so the pair

(X_i, Y_i)

is the natural unit of randomness, not the residual alone. Resampling observation pairs (B) is correct because it treats each

(X_i, Y_i)

as a single draw from the joint distribution

F(X, Y)

. When you resample these pairs with replacement and refit the model on each bootstrap sample, the varying spread of

Y

across different

X

values is automatically inherited — no assumptions about the variance function are needed. This is the pairs bootstrap (also called the xy-bootstrap), and it is specifically recommended when heteroskedasticity is present but unmodeled. Choice A describes the residual bootstrap: you resample residuals and re-attach them to fitted values at fixed predictor locations. This implicitly assumes errors are exchangeable across all

X

values — a homoskedasticity assumption. Under heteroskedasticity, swapping a large-

X

residual onto a small-

X

fitted value distorts the variance structure. Choice C resamples only predictor values while keeping responses fixed, which breaks the

(X_i, Y_i)

linkage entirely and produces nonsensical design matrices with no valid inference. Choice D independently resamples fitted values and residuals before combining them. This destroys the mean-variance relationship and introduces arbitrary mixing that has no grounding in the data-generating process. The key study tip: whenever heteroskedasticity is present and no parametric form is assumed, default to the pairs bootstrap — it respects the joint distribution without requiring you to model the variance structure.

Question 3

For a scalar estimator, $70\%$ of the bootstrap replicates are below the observed estimate, and jackknife calculations indicate substantial nonzero acceleration. The ordinary percentile interval is noticeably asymmetric.

Which interpretation most accurately describes what a bias-corrected and accelerated, or BCa, interval would do?

It replaces the empirical bootstrap distribution by a symmetric normal distribution with a jackknife variance estimate.
It subtracts the bootstrap mean bias from both percentile endpoints while leaving their tail probabilities unchanged.
It adjusts the percentile probabilities for both median bias and the rate at which standard error changes with the parameter. (correct answer)
It widens both percentile tails by the same amount whenever more than half the replicates lie below the estimate.

Explanation: When you encounter a question about bootstrap confidence intervals, focus on what specific corrections each method applies and why those corrections are needed. The BCa interval addresses two distinct sources of distortion. First, median bias: if 70% of bootstrap replicates fall below the observed estimate (as stated here), the bootstrap distribution is shifted, meaning the estimate doesn't sit at the median of its bootstrap distribution. BCa captures this with a bias-correction constant

\hat{z}_0 = \Phi^{-1}(0.70) \approx 0.524

. Second, acceleration: the jackknife-estimated acceleration constant

\hat{a}

accounts for the rate at which the estimator's standard error changes as the parameter itself changes — a form of skewness in the sampling distribution. BCa incorporates both constants to shift the tail probabilities used when reading off percentile endpoints, producing an interval that adapts to both effects. This is exactly what C describes. A is wrong because BCa never replaces the bootstrap distribution with a normal approximation — that would be a different method entirely (like the bootstrap-t or a simple normal interval). BCa stays fully nonparametric. B is wrong because BCa does not arithmetically subtract a bias quantity from the endpoints. It adjusts the probability levels (quantiles) at which you read the bootstrap distribution — the correction is in the tail probabilities, not a shift of the values themselves. D is wrong because BCa does not widen both tails equally. The adjustment is asymmetric and depends on both

\hat{z}_0

and

\hat{a}

, meaning the two tails are shifted by different amounts. Remember: BCa = two corrections to quantile levels, not to the values. If a question mentions jackknife acceleration alongside skewed replicates, BCa is the only standard method addressing both simultaneously.

Question 4

Independent observations come from a distribution known only to be uniform on $[0,θ]$ , and $θ$ is estimated by the sample maximum $M$ . A researcher forms a nonparametric percentile bootstrap interval by resampling the observations and recalculating the maximum.

What is the most important problem with this proposed interval?

The bootstrap maximum has infinite variance conditionally on the data, so empirical percentile endpoints do not stabilize.
Every bootstrap maximum is at least $M$ , so the lower percentile endpoint cannot fall below the unknown value $θ$ .
The bootstrap maxima are symmetric around $M$ , so the percentile interval becomes unnecessarily conservative near the boundary.
Every bootstrap maximum is at most $M$ , so the upper percentile endpoint cannot exceed $M$ even though $θ$ generally does. (correct answer)

Explanation: When evaluating bootstrap methods, always ask: what values can the bootstrap statistic actually take? The answer reveals whether the resampling scheme can meaningfully approximate the sampling distribution. Here, your data consists of observations

x_1, \ldots, x_n

all drawn from

[0, \theta]

, with sample maximum

M = \max_i x_i

. When you resample with replacement from these observed values, every bootstrap draw is some

x_i^* \in \{x_1, \ldots, x_n\}

, meaning every bootstrap observation is

\leq M

. Therefore every bootstrap maximum

M^* \leq M

. The bootstrap distribution of

M^*

is entirely confined to

(-\infty, M]

, yet the true parameter

\theta \geq M

almost surely. The upper percentile endpoint of your interval is capped at

M

, guaranteeing the interval systematically undershoots

\theta

. This is answer D, and it exposes a fundamental failure: the bootstrap distribution cannot even reach the true parameter. A is wrong because bootstrap maxima are bounded and well-defined — there is no infinite-variance issue here. The problem is bias, not instability. B describes the opposite of what happens. Bootstrap maxima are at most

M

, never exceeding it, so this statement is factually backwards. C is wrong because bootstrap maxima are not symmetric around

M

— they are entirely left-skewed, bounded above by

M

. Symmetry is not the issue. As a strategy: whenever an estimator sits at the boundary of its parameter space (like a maximum or minimum), ask whether bootstrap resamples can ever exceed that boundary. If not, the percentile interval is structurally broken from the start.

Question 5

A researcher estimates a population parameter by $\hat{θ}=12$ . From a large number of nonparametric bootstrap samples, the estimated $0.025$ and $0.975$ quantiles of $\hat{θ}^{*}$ are $9$ and $16$ , respectively.

Which pair gives, in order, the $95\%$ percentile bootstrap interval and the $95\%$ basic bootstrap interval?

$[9,16]$ and $[8,15]$ (correct answer)
$[8,15]$ and $[9,16]$
$[9,15]$ and $[8,16]$
$[8,16]$ and $[9,15]$

Explanation: When you see a bootstrap confidence interval question, the key is keeping the two methods' formulas straight — they look similar but pivot differently around

\hat{\theta}

. The percentile interval is the simplest: just read off the empirical quantiles of the bootstrap distribution directly. Here, the 0.025 and 0.975 quantiles of

\hat{\theta}^*

are 9 and 16, so the 95% percentile interval is simply

[9, 16]

. The basic (pivot) interval corrects for bias by reflecting the bootstrap quantiles around the original estimate

\hat{\theta} = 12

. The formula is

[2\hat{\theta} - q_{0.975}^*, \ 2\hat{\theta} - q_{0.025}^*]

. Plugging in:

[2(12) - 16, \ 2(12) - 9] = [24 - 16, \ 24 - 9] = [8, 15]

. So the basic interval is

[8, 15]

. That makes A correct: percentile

[9,16]

, basic

[8,15]

. Choice B reverses the two intervals — a direct swap that reflects confusing which method uses the quantiles directly versus which reflects them. Choice C mixes endpoints from both methods (taking 9 and 15 together, or 8 and 16 together), which has no valid derivation. Choice D similarly scrambles the endpoints without following either formula. A reliable memory trick: the basic interval pivots, so you subtract the bootstrap quantiles from

2\hat{\theta}

; the percentile interval just reads the quantiles. Notice that because

\hat{\theta} = 12

sits closer to 9 than to 16, the basic interval shifts left relative to the percentile interval — a quick sanity check on your arithmetic.

Question 6

Using the same observed data, an analyst computes a $95\%$ percentile bootstrap interval with $2{,}000$ replicates. Repeating the calculation with different random seeds produces visibly different upper endpoints. The analyst proposes increasing the number of replicates to $20{,}000$ .

Which statement best describes the likely consequence of this change?

It makes the bootstrap distribution closer to normal by roughly a factor of $10$ but leaves quantile simulation error unchanged.
It reduces the sampling standard error of the original estimator by roughly a factor of $\sqrt{10}$ and improves coverage automatically.
It reduces Monte Carlo variability of the estimated quantiles by roughly a factor of $\sqrt{10}$ but does not repair systematic coverage error. (correct answer)
It eliminates both bootstrap approximation error and sampling uncertainty because the empirical quantiles become effectively exact.

Explanation: Whenever you see a question about bootstrap intervals, separate two distinct sources of error in your mind: Monte Carlo simulation error (from using a finite number of replicates) and statistical error (from the original sample size). These are independent, and fixing one does not fix the other. The instability across random seeds is a telltale sign of Monte Carlo variability — the estimated quantiles are bouncing around because 2,000 replicates isn't enough to pin them down precisely. The standard error of an estimated quantile from

B

bootstrap replicates shrinks at rate

1/\sqrt{B}

. Going from 2,000 to 20,000 replicates multiplies

B

by 10, so the simulation noise drops by a factor of

\sqrt{10} \approx 3.16

. That's exactly what C describes. Critically, however, percentile bootstrap intervals can have systematic undercoverage in small samples regardless of how many replicates you use — more replicates just compute the wrong interval more precisely. Choice A is wrong because increasing replicates doesn't push the bootstrap distribution toward normality; that's governed by the central limit theorem applied to the original data, not the simulation count. Choice B conflates Monte Carlo error with the sampling distribution of the estimator — the original sample size hasn't changed, so the standard error of your estimator is completely unaffected. Coverage improvement is not automatic even with Monte Carlo precision. Choice D overstates the case dramatically: exact empirical quantiles still reflect the limitations of the original sample and can carry bias. Your study tip: on bootstrap questions, always ask which kind of error is being discussed. Monte Carlo error scales with

1/\sqrt{B}

; statistical error scales with

1/\sqrt{n}

. They live in completely different places.

Question 7

In a crossover study, each of $40$ independently sampled patients receives treatments A and B in randomized order. The estimand is the population mean within-patient difference in outcomes. Outcomes from the same patient are strongly positively correlated.

Which nonparametric bootstrap procedure is appropriate for constructing a confidence interval for this estimand?

Hold the A outcomes fixed, resample only the B outcomes, and recompute the paired mean difference.
Resample all A outcomes and all B outcomes independently, then subtract the two resampled means.
Resample the eighty individual outcomes jointly, then randomly reassign half to each treatment condition.
Resample patients, retaining both outcomes for each selected patient, and recompute the mean paired difference. (correct answer)

Explanation: When you see a bootstrap question involving paired or clustered data, the key principle is: the bootstrap must resample the same unit that was originally sampled. Here, the study sampled 40 patients — not 80 outcomes — so the bootstrap must treat each patient (with both their A and B outcomes attached) as the indivisible sampling unit. In a crossover design, the estimand is the mean within-patient difference

\bar{D} = \frac{1}{n}\sum_{i=1}^n (Y_{iA} - Y_{iB})

. The strong positive correlation between a patient's two outcomes is precisely what makes the paired design efficient — and it's exactly what the bootstrap must preserve. The correct procedure, D, resamples patients with replacement (giving a bootstrap sample of 40 patients), keeps both outcomes together for each selected patient, and recomputes the mean difference. This respects the dependence structure and correctly reflects the variability in the original sampling process. A is wrong because holding A outcomes fixed and resampling only B outcomes breaks the pairing entirely and introduces artificial asymmetry — you'd be underestimating the true variability of the estimator. B independently resamples the two treatment arms, which destroys the within-patient correlation. The resulting bootstrap variance will be too large (it ignores the correlation benefit), yielding an interval that is anti-conservative in the wrong direction relative to the paired design's efficiency. C resamples all 80 outcomes jointly and randomly reassigns them, which is closer to a permutation test for a null hypothesis — not a confidence interval procedure. It also destroys the pairing and treats outcomes as exchangeable across patients. Study tip: On bootstrap questions, always identify what was originally sampled and resample that unit intact. For clustered or paired data, the cluster is the unit — never split it.

Question 8

For an observed estimate $\hat{θ}=10$ with estimated standard error $\widehat{SE}=2$ , bootstrap samples produce the studentized statistic $T^{*}=(\hat{θ}^{*}-\hat{θ})/\widehat{SE}^{*}$ . Its estimated $0.025$ and $0.975$ quantiles are $-1.5$ and $2.0$ .

What is the corresponding equal-tailed bootstrap- $t$ confidence interval for $θ$ ?

$[6,13]$ , obtained by inverting the studentized quantiles (correct answer)
$[7,14]$ , obtained by retaining the quantile order
$[6,14]$ , obtained from the larger absolute quantile
$[7,13]$ , obtained from the smaller absolute quantile

Explanation: When constructing a bootstrap-

t

interval, the key insight is that you must invert the studentized pivot — and inversion flips the quantile order. The pivot is

T^* = (\hat{\theta}^* - \hat{\theta})/\widehat{SE}^*

, and the confidence interval comes from solving the inequality

q_{0.025} \leq T \leq q_{0.975}

for

\theta

, substituting

T = (\hat{\theta} - \theta)/\widehat{SE}

. Rearranging gives

\hat{\theta} - q_{0.975} \cdot \widehat{SE} \leq \theta \leq \hat{\theta} - q_{0.025} \cdot \widehat{SE}

. Notice the quantiles swap sides during inversion. Plugging in

\hat{\theta} = 10

\widehat{SE} = 2

q_{0.025} = -1.5

, and

q_{0.975} = 2.0

Lower bound: $10 - (2.0)(2) = 10 - 4 = 6$
Upper bound: $10 - (-1.5)(2) = 10 + 3 = 13$

This gives

[6, 13]

, confirming A is correct. Choice B arrives at

[7, 14]

by naively plugging quantiles in their original order without inverting — a fundamental procedural error that ignores the algebra of solving for

\theta

. Choice C uses only the larger absolute quantile (2.0) symmetrically, producing

[6, 14]

; this confuses the bootstrap-

t

with a normal-theory interval that assumes symmetry. Choice D uses the smaller absolute quantile (1.5) symmetrically, giving

[7, 13]

, compounding both errors simultaneously. A reliable memory aid: the bootstrap-

t

interval is

[\hat{\theta} - q_{1-\alpha/2}\cdot\widehat{SE},\ \hat{\theta} - q_{\alpha/2}\cdot\widehat{SE}]

. The upper quantile goes on the left, and the lower quantile goes on the right — always invert.

Question 9

There are $999$ ordered bootstrap estimates. Software defines the empirical quantile at probability $p$ by linear interpolation at rank $h=1+(B-1)p$ . The estimates at ranks $25$ and $26$ are $-1.8$ and $-1.7$ ; those at ranks $974$ and $975$ are $2.5$ and $2.6$ .

Using this software convention, what is the equal-tailed $95\%$ percentile bootstrap confidence interval?

$[-1.800,2.600]$ , using the nearest integer ranks
$[-1.705,2.505]$ , using linear interpolation (correct answer)
$[-1.795,2.595]$ , using reversed interpolation weights
$[-1.700,2.500]$ , using the inner adjacent ranks

Explanation: When software computes empirical quantiles via linear interpolation, it maps probability

p

to a fractional rank

h = 1 + (B-1)p

, then interpolates between the floor and ceiling ranks. For a 95% equal-tailed interval with

B = 999

, you need the 2.5th and 97.5th percentiles. For the lower bound,

p = 0.025

h = 1 + 998(0.025) = 1 + 24.95 = 25.95

. This lies between ranks 25 and 26, with fractional part 0.95. Interpolating:

-1.8 + 0.95(-1.7 - (-1.8)) = -1.8 + 0.95(0.1) = -1.8 + 0.095 = -1.705

. For the upper bound,

p = 0.975

h = 1 + 998(0.975) = 1 + 973.05 = 974.05

. This lies between ranks 974 and 975, with fractional part 0.05. Interpolating:

2.5 + 0.05(2.6 - 2.5) = 2.5 + 0.005 = 2.505

. This confirms B,

[-1.705, 2.505]

, is correct. A is wrong because simply rounding to the nearest integer rank ignores the fractional position entirely — ranks 26 and 975 would give

[-1.700, 2.600]

, which mismatches the stated values anyway. C reverses the interpolation weights (using 0.05 at the lower end and 0.95 at the upper), which gets the fractional arithmetic backwards. D uses the inner ranks 26 and 974 directly without any interpolation, discarding the software convention altogether. The key strategy: always compute

h = 1 + (B-1)p

first, isolate the fractional part, and apply it as the weight on the upper neighboring rank. The fractional part tells you how far above the lower rank you are — a common source of sign errors on exam questions.

Question 10

An estimator has observed value $\hat{θ}=5.0$ . Its bootstrap replicates have mean $5.8$ and standard deviation $1.2$ . A researcher uses the bias estimate $\widehat{b}=\overline{\hat{θ}^{*}}-\hat{θ}$ and constructs a bias-corrected normal interval centered at $\hat{θ}-\widehat{b}$ with critical value $1.96$ .

Which interval does the researcher obtain?

$[1.85,6.55]$ , after shifting the center downward (correct answer)
$[2.65,7.35]$ , without correcting the estimated bias
$[3.45,8.15]$ , after shifting the center upward
$[2.83,5.57]$ , after dividing by the bootstrap spread

Explanation: When you see bootstrap inference questions, focus on three quantities: the observed estimator

\hat{\theta}

, the bootstrap mean

\overline{\hat{\theta}^*}

, and the bootstrap standard deviation. The bias-corrected estimator subtracts the estimated bias from the original estimate, then builds a normal interval around that corrected center. Here, the estimated bias is

\widehat{b} = \overline{\hat{\theta}^*} - \hat{\theta} = 5.8 - 5.0 = 0.8

. Since the bootstrap replicates average above the observed estimate, the estimator is positively biased, so you correct downward. The bias-corrected center becomes

\hat{\theta} - \widehat{b} = 5.0 - 0.8 = 4.2

. Using the bootstrap standard deviation as the standard error, the margin of error is

1.96 \times 1.2 = 2.352

. The interval is

[4.2 - 2.352,\ 4.2 + 2.352] = [1.848, 6.552] \approx [1.85, 6.55]

. This confirms answer A — the center shifted downward because the bias was positive. Answer B uses the uncorrected center

\hat{\theta} = 5.0

, yielding

[5.0 - 2.352, 5.0 + 2.352] = [2.65, 7.35]

— a common mistake of forgetting to apply the bias correction entirely. Answer C shifts the center to

5.0 + 0.8 = 5.8

, producing

[3.45, 8.15]

— this adds the bias rather than subtracting it, reversing the correction direction. Answer D applies some nonstandard rescaling unrelated to any standard bootstrap procedure. A reliable study tip: always check the direction of bias correction. If the bootstrap mean exceeds

\hat{\theta}

, the bias is positive and you shift the center down, not up.