Statistics Graduate Level Quiz: Forecasting And Prediction Intervals

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Forecasting And Prediction IntervalsQuestion 1 of 10

A simple linear regression is fitted to $n=25$ observations. The residual standard error is $s=2$ , the predictor mean is $\bar{x}=10$ , and $S_{xx}=\sum_{i=1}^{n}(x_i-\bar{x})^2=100$ . At $x_0=14$ , the fitted response is $\hat{y}_0=18$ . Assume the usual Gaussian linear-model conditions, and use $t_{0.975,23}=2.064$ .

Which interval is the approximate $95\%$ prediction interval for a single new response at $x_0=14$ ?

[13.48,\ 22.52]

, using both new-response variation and leverage.

[16.15,\ 19.85]

, using only uncertainty in the fitted mean response.

[13.79,\ 22.21]

, using new-response variation but omitting leverage.

[13.71,\ 22.29]

, using a normal rather than a Student critical value.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Forecasting And Prediction Intervals

Practice Forecasting And Prediction 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 Forecasting And Prediction 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

Which interval is the approximate $95\%$ prediction interval for a single new response at $x_0=14$ ?

$[13.48,\ 22.52]$ , using both new-response variation and leverage. (correct answer)
$[16.15,\ 19.85]$ , using only uncertainty in the fitted mean response.
$[13.79,\ 22.21]$ , using new-response variation but omitting leverage.
$[13.71,\ 22.29]$ , using a normal rather than a Student critical value.

Explanation: Whenever you see a prediction interval question, you must distinguish it from a confidence interval for the mean response — these two intervals serve fundamentally different purposes and have different formulas. A prediction interval for a single new observation accounts for two sources of uncertainty: (1) uncertainty in estimating the true mean

\hat{y}_0

, and (2) the irreducible variance of a new individual observation around that mean. The correct formula is:

\hat{y}_0 \pm t_{\alpha/2,\, n-2} \cdot s\sqrt{1 + \frac{1}{n} + \frac{(x_0 - \bar{x})^2}{S_{xx}}}

Plugging in:

s=2

n=25

x_0-\bar{x}=4

S_{xx}=100

\text{SE} = 2\sqrt{1 + \frac{1}{25} + \frac{16}{100}} = 2\sqrt{1.2} \approx 2\sqrt{1.2} = 2(1.095) = 2.19

Margin:

2.064 \times 2.19 \approx 4.52

, giving interval

[18-4.52,\ 18+4.52] = [13.48,\ 22.52]

. That confirms A is correct. B is wrong because it omits the critical "

+1

" under the radical — this gives only the confidence interval for the mean response (width ≈ ±1.85), not a prediction interval for a new observation. C is wrong in a subtle way: it includes new-response variation (the

+1

) but omits the leverage term

\frac{(x_0-\bar{x})^2}{S_{xx}}

, producing a slightly too-narrow interval of

[13.79, 22.21]

. D is wrong because it substitutes the normal critical value

z_{0.975}=1.96

instead of the correct

t_{23,0.975}=2.064

, shrinking the margin inappropriately for small samples. Study tip: Memorize that prediction intervals always include that leading "1" inside the square root. If your interval looks as narrow as a confidence interval for the mean, you've almost certainly forgotten it.

Question 2

An analyst selects an ARIMA model by minimizing an information criterion and then constructs prediction intervals by fixing the selected model and its estimated parameters. Future innovations are simulated, but model-selection and parameter-estimation uncertainty are ignored. Repeated-sample coverage is below the nominal level.

Which modification most directly addresses the stated source of undercoverage?

Simulate future innovations with larger variance while keeping the selected specification and coefficients fixed.
Resample or simulate complete series, reselect and refit the model, and then generate forecasts in each replicate. (correct answer)
Replace the Student critical value by a normal critical value after fitting the selected model once.
Condition on the observed residuals and vary only the last innovation used in each forecast replicate.

Explanation: Whenever you see a question about prediction interval undercoverage in time series, ask yourself: what sources of uncertainty is the procedure ignoring? Valid frequentist coverage requires that the interval-construction procedure accounts for every random component that varies across repeated samples — including which model gets selected and what parameter estimates result. The passage explicitly identifies the culprit: model-selection uncertainty and parameter-estimation uncertainty are both ignored. The analyst fixes one model and one set of coefficients, then only simulates future innovations. This treats the model as if it were known with certainty, which it isn't. The correct fix, answer B, is a full bootstrap or simulation loop where you (1) generate a complete artificial series from the data-generating process, (2) rerun the entire model-selection procedure on that series, (3) refit the chosen model, and (4) produce a forecast. Repeating this across many replicates embeds all three sources of uncertainty — innovations, estimation, and selection — into the resulting interval, restoring honest repeated-sample coverage. A inflates innovation variance while keeping the model and coefficients fixed, which is an ad hoc patch that doesn't target the stated problem; it can over- or under-correct without ever touching estimation or selection variability. C swapping a Student-

t

critical value for a normal one actually narrows the interval (the normal quantile is smaller), making undercoverage worse, not better. D conditioning on observed residuals and varying only the last innovation is even more restrictive than the original procedure — it suppresses nearly all uncertainty rather than incorporating more of it. The key study takeaway: bootstrap validity requires bootstrapping the entire inferential pipeline, not just the part that feels most "random."

Question 3

A prediction method is evaluated on two observable regimes. The low-variance regime occurs with probability $0.75$ and has conditional coverage $0.95$ . The high-variance regime occurs with probability $0.25$ and has conditional coverage $0.75$ .

Which conclusion about the method's coverage is correct?

Its marginal coverage is $0.95$ because the most frequently observed regime determines long-run calibration.
Its marginal coverage is $0.85$ because the two conditional coverage probabilities should be averaged equally.
Its conditional coverage is $0.90$ because marginal calibration forces calibration within every observable regime.
Its marginal coverage is $0.90$ , even though it does not have $0.90$ conditional coverage in either regime. (correct answer)

Explanation: Whenever you see a question mixing conditional probabilities across subgroups, your instinct should be the law of total probability: marginal probability equals a weighted average of conditional probabilities, where the weights are the subgroup probabilities. Here, the marginal coverage is computed as:

P(\text{covered}) = P(\text{low}) \cdot P(\text{covered} \mid \text{low}) + P(\text{high}) \cdot P(\text{covered} \mid \text{high})

= (0.75)(0.95) + (0.25)(0.75) = 0.7125 + 0.1875 = 0.90

So the marginal coverage is exactly

0.90

, even though neither regime individually achieves

0.90

coverage — the low-variance regime overshoots at

0.95

and the high-variance regime undershoots at

0.75

. This confirms D is correct. A is wrong because long-run calibration is not determined by the most frequent regime alone — all regimes contribute proportionally to their probability. Ignoring the rare regime would only be valid if its probability were zero. B applies a simple arithmetic mean:

(0.95 + 0.75)/2 = 0.85

. This is the classic equal-weighting trap. You must weight by regime probabilities

(0.75, 0.25)

, not by

(0.5, 0.5)

. C confuses the direction of implication. Marginal calibration does not force conditional calibration within every subgroup — that's a strictly stronger requirement. A method can be marginally well-calibrated while being miscalibrated in specific subpopulations, which is precisely the failure mode D illustrates. Your study tip: marginal calibration ≠ conditional calibration. On exam questions involving subgroups, always check whether coverage guarantees are marginal or conditional — they are not interchangeable, and conflating them is the most common trap in conformal prediction and calibration problems.

Question 4

A forecaster will report intervals for the next four horizons. Each ordinary pointwise interval has marginal coverage $0.95$ , but the client requires the probability that all four future values are simultaneously covered to be at least $0.95$ . No dependence structure among forecast errors is assumed.

Which adjustment provides the requested simultaneous-coverage guarantee?

Construct each interval at level ${0.95}^{1/4}$ , which gives exact joint coverage only under independence, not arbitrary dependence.
Retain four marginal intervals at level $0.95$ because positive horizon dependence ensures joint coverage.
Construct each marginal interval at level $1-0.05/4=0.9875$ by a Bonferroni adjustment. (correct answer)
Retain the first three intervals at level $0.95$ and widen only the fourth to level $0.9875$ .

Explanation: Whenever you see a question about simultaneous coverage across multiple intervals, think about the difference between marginal coverage (each interval individually) and joint coverage (all intervals together). These require different tools. The core challenge here: you need

P(\text{all four future values covered}) \geq 0.95

. The Bonferroni inequality gives you a clean, assumption-free path. It states that for any events

A_1, \ldots, A_k

P\!\left(\bigcap_{i=1}^k A_i\right) \geq 1 - \sum_{i=1}^k P(A_i^c)

If each marginal non-coverage probability is

\alpha/k = 0.05/4 = 0.0125

, then the joint non-coverage probability is bounded above by

4 \times 0.0125 = 0.05

, guaranteeing joint coverage of at least

0.95

. This means constructing each interval at level

1 - 0.05/4 = 0.9875

, exactly what C prescribes. Crucially, Bonferroni makes no assumptions about dependence structure — it works in the worst case. A is tempting but subtly wrong: setting each level to

0.95^{1/4}

achieves exact joint coverage of

0.95

only under independence. Since the problem explicitly says no dependence structure is assumed, this provides no guarantee under arbitrary dependence. B is wrong because positive dependence can either help or hurt joint coverage depending on the configuration — it does not reliably ensure the target is met without precise knowledge of the dependence structure. D is wrong because widening only the fourth interval does nothing to control the failure probabilities of the first three, leaving joint coverage unguaranteed. Study tip: On exam questions about simultaneous inference, Bonferroni is your go-to when dependence is unknown — it's conservative but always valid. Recognize that "exact under independence" is a trap phrase signaling an assumption-dependent method.

Question 5

A stationary process follows the known model $X_t=0.8X_{t-1}+\varepsilon_t$ , where the innovations are independent $N(0,9)$ variables. The current observation is $X_t=10$ , and model parameters are treated as known.

Which interval is the approximate $95\%$ conditional prediction interval for $X_{t+3}$ ?

$[-3.30,\ 13.54]$ , based on accumulated innovations through horizon $3$ . (correct answer)
$[-4.68,\ 14.92]$ , based on the unconditional stationary variance.
$[-5.06,\ 15.30]$ , based on three innovations with equal weights.
$[-0.76,\ 11.00]$ , based only on the one-step innovation variance.

Explanation: When forecasting from an AR(1) model, the key concept is multi-step forecast uncertainty, which accumulates differently at each horizon rather than staying fixed or growing uniformly. For

X_t = 0.8X_{t-1} + \varepsilon_t

with

\varepsilon_t \sim N(0,9)

, the three-step-ahead point forecast given

X_t = 10

is:

\hat{X}_{t+3} = (0.8)^3 \cdot 10 = 0.512 \cdot 10 = 5.12

The conditional forecast variance at horizon

h

accumulates through the MA representation:

\text{Var}(X_{t+3} \mid X_t) = \sigma^2\left(1 + \phi^2 + \phi^4\right) = 9\left(1 + 0.64 + 0.4096\right) = 9(2.0496) \approx 18.45

So the standard error is

\sqrt{18.45} \approx 4.29

, giving the 95% interval:

5.12 \pm 1.96(4.29) \approx [-3.30,\ 13.54]

This confirms A is correct — it properly accumulates innovation variance weighted by successive AR coefficients. B uses the unconditional stationary variance

\sigma^2/(1-\phi^2) = 9/0.36 = 25

, which is appropriate only when you have no current observation and forecast from the marginal distribution. Here you're conditioning on

X_t = 10

, so B overstates uncertainty. C treats all three innovation weights as equal (all 1's), ignoring that contributions shrink as

\phi^k

. This inflates variance incorrectly. D uses only the one-step variance

\sigma^2 = 9

, forgetting that uncertainty at horizon 3 must include innovations at steps

t+1

t+2

, and

t+3

. This severely underestimates the interval. Study tip: For multi-step AR forecasts, always build variance by summing

\sigma^2(1 + \phi^2 + \cdots + \phi^{2(h-1)})

— the weights shrink geometrically, and confusing this with stationary or one-step variance is the most common trap on these questions.

Question 6

Forecast errors for two regional demands, $A$ and $B$ , are jointly normal with means zero, variances $4$ and $9$ , and correlation $0.5$ . Their point forecasts are $40$ and $60$ , respectively. The total demand is $T=A+B$ , and parameter uncertainty is ignored.

Which interval is the approximate $95\%$ prediction interval for total demand?

$100\pm1.96\sqrt{16}$ , adding the covariance only once to the marginal variances.
$100\pm1.96\sqrt{13}$ , treating the two regional forecast errors as independent.
$100\pm1.96(5)$ , adding the two regional forecast-error standard deviations.
$100\pm1.96\sqrt{19}$ , incorporating the positive covariance between regional errors. (correct answer)

Explanation: When combining forecast errors from multiple sources, the key concept is variance addition under correlation. The forecast error for total demand

T = A + B

e_T = e_A + e_B

, and you need the variance of this sum, which depends on both marginal variances and the covariance between them. The variance formula is

\text{Var}(e_T) = \text{Var}(e_A) + \text{Var}(e_B) + 2\text{Cov}(e_A, e_B)

. Given

\sigma_A^2 = 4

\sigma_B^2 = 9

, and

\text{Cov}(e_A, e_B) = \rho \cdot \sigma_A \cdot \sigma_B = 0.5 \times 2 \times 3 = 3

, the total variance is

4 + 9 + 2(3) = 19

. The forecast mean is

40 + 60 = 100

, so the 95% prediction interval is

100 \pm 1.96\sqrt{19}

, confirming D is correct. A is tempting because it mentions covariance, but adding covariance only once (without the factor of 2) gives

4 + 9 + 3 = 16

. This forgets that

\text{Cov}(e_A, e_B)

appears twice in the expansion of

\text{Var}(e_A + e_B)

— once for each cross term. B assumes independence, setting covariance to zero and yielding

\sqrt{13}

. This underestimates risk whenever errors are positively correlated, which is common when regional demands share common drivers. C adds standard deviations directly:

\sigma_A + \sigma_B = 2 + 3 = 5

. This is only valid when errors are perfectly correlated (

\rho = 1

), representing the worst-case upper bound — not the general case. A reliable strategy: always write out

\text{Var}(X+Y) = \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y

explicitly, so you never forget the factor of 2 or the covariance term.

Question 7

A split-conformal prediction interval is calibrated using absolute forecast residuals from a stable historical period. The calibration residuals are treated as exchangeable. It is then applied to a future period known to have a larger error variance because of a structural regime change, without reweighting or recalibration.

Which statement best describes the nominal conformal coverage claim in this application?

Finite-sample coverage is still guaranteed because conformal validity is unaffected by changes in the residual distribution.
Coverage is not generally guaranteed because the future score is not exchangeable with the calibration scores. (correct answer)
Conditional coverage is guaranteed within the new regime, although marginal coverage may fall below nominal.
Coverage remains exact if the point forecasting model is unbiased during both the historical and future periods.

Explanation: Whenever you see a question about conformal prediction intervals, the key concept to anchor yourself to is exchangeability — the mathematical condition that makes conformal coverage guarantees valid. Split-conformal prediction sets achieve finite-sample marginal coverage exactly because the calibration scores and the test score are assumed to be exchangeable: any permutation of them is equally likely under the null. The coverage guarantee flows entirely from this symmetry, not from any distributional assumptions about normality or stationarity. Here, a structural regime change increases the error variance in the future period. This means the future absolute residual is drawn from a fundamentally different distribution than the calibration residuals. A score drawn from a heavier-tailed distribution is systematically more likely to exceed any fixed quantile threshold computed from the lighter-tailed calibration set — breaking the exchangeability assumption and, with it, the coverage guarantee. The nominal

1 - \alpha

claim no longer holds in general, which is exactly what B states. A is wrong because it inverts the logic: conformal validity is entirely dependent on exchangeability. A change in the residual distribution is precisely the condition that destroys the guarantee. C is doubly wrong — neither conditional nor marginal coverage is guaranteed when exchangeability fails; in fact, coverage will tend to under-cover in the new regime (not over-cover), because calibration quantiles are too small for the inflated variance. D is wrong because unbiasedness of the point forecast is irrelevant; coverage depends on the score distribution's exchangeability, not on mean-zero errors. As a study habit, always ask yourself: "Are the calibration and test scores truly exchangeable?" Any distributional shift — in variance, autocorrelation, or regime — is a red flag that the formal guarantee has been voided.

Question 8

A forecasting model fitted to a long daily series leaves residuals with substantial short-range serial dependence. An analyst uses an ordinary residual bootstrap that independently resamples individual residuals and obtains prediction intervals that are too narrow.

Which bootstrap procedure is most defensible for improving the prediction intervals while retaining the observed dependence structure?

Resample individual residuals independently, but center and standardize them before recursively generating forecasts.
Resample contiguous residual blocks, recursively generate pseudo-series, and refit the forecasting model in each replicate. (correct answer)
Resample fitted values independently, add the original residual sequence, and hold all estimated coefficients fixed.
Randomly permute the residual sequence, recursively forecast from the permuted order, and refit only the intercept.

Explanation: When residuals exhibit serial dependence, the ordinary bootstrap's core assumption — that observations are exchangeable — is violated. Resampling individual residuals destroys the autocorrelation structure, which is exactly why the analyst's prediction intervals are too narrow: they fail to capture the additional uncertainty propagated by correlated errors over the forecast horizon. The question is testing your knowledge of block bootstrap methods designed specifically to preserve dependence structure. B is correct because the moving or non-overlapping block bootstrap resamples contiguous chunks of residuals, keeping short-range dependence intact within each block. By then recursively generating pseudo-series and refitting the model in each replicate, you capture both parameter uncertainty and the genuine autocorrelation structure — producing prediction intervals that honestly reflect forecast uncertainty. A is wrong because centering and standardizing individual residuals before resampling still resamples them independently. This does nothing to restore serial dependence; it only adjusts moments, leaving the fundamental exchangeability violation untouched. C is wrong on two counts: resampling fitted values independently discards the residual dependence structure, and holding all coefficients fixed ignores parameter uncertainty entirely — both of which contribute to underestimating forecast uncertainty. D is wrong because randomly permuting the residual sequence actively destroys whatever dependence structure exists. A permuted sequence is essentially i.i.d., which is the same mistake as the original failed procedure, only reframed. Study tip: When you see "serial dependence in residuals," immediately think block bootstrap. The guiding principle is: preserve the structure you want to replicate — if dependence matters, resample in blocks and always refit the model.

Question 9

For a future positive response $Y$ , a fitted model gives $\log(Y)\mid\mathcal{D}\sim N(2,0.4^2)$ . Parameter uncertainty is negligible, and the analyst wants an equal-tailed $95\%$ prediction interval on the original scale.

Which construction is appropriate?

$\left[\exp(2-1.96(0.4)),\ \exp(2+1.96(0.4))\right]$ . (correct answer)
$\left[\exp(2+0.4^2/2-1.96(0.4)),\ \exp(2+0.4^2/2+1.96(0.4))\right]$ .
$\left[\exp(2)-1.96\exp(0.4),\ \exp(2)+1.96\exp(0.4)\right]$ .
$\left[\exp(2-1.96(0.4^2)),\ \exp(2+1.96(0.4^2))\right]$ .

Explanation: When working with lognormal prediction intervals, the key insight is that you should construct the interval on the log scale first, then transform. If

\log(Y) \mid \mathcal{D} \sim N(\mu, \sigma^2)

, then a 95% equal-tailed interval for

\log(Y)

is simply

[\mu - 1.96\sigma,\ \mu + 1.96\sigma]

. Because

\exp(\cdot)

is a monotone increasing function, exponentiating the endpoints preserves both the coverage probability and the equal-tailed property — giving you

[\exp(\mu - 1.96\sigma),\ \exp(\mu + 1.96\sigma)]

. With

\mu = 2

and

\sigma = 0.4

, this yields

[\exp(2 - 1.96(0.4)),\ \exp(2 + 1.96(0.4))]

, which is exactly Answer A — the correct choice. Answer B adds the lognormal mean-correction term

\sigma^2/2

to the center before exponentiating. This would be relevant if you were constructing an interval centered at the mean of

Y

(which is

\exp(\mu + \sigma^2/2)

), but a prediction interval should be centered at the median,

\exp(\mu)

. The mean-correction is a distractor that conflates point estimation with interval construction. Answer C incorrectly applies a linear

\pm 1.96\sigma

structure directly on the original scale, as if

Y

itself were normally distributed. This ignores the lognormal nature of

Y

entirely and produces a symmetric (non-lognormal) interval. Answer D uses

\sigma^2 = 0.16

instead of

\sigma = 0.4

in the exponent — a classic variance-vs-standard-deviation mix-up that substantially narrows the interval. Study tip: Always build prediction intervals on the scale where the distributional assumption holds, then apply the monotone inverse transformation. Never transform first and compute intervals after.

Question 10

In a Bayesian normal model, a future observation satisfies $Y_{n+1}\mid\mu\sim N(\mu,4)$ . After observing the data, the posterior distribution is $\mu\mid\mathcal{D}\sim N(10,1)$ .

Which interval is the central $95\%$ posterior predictive interval for $Y_{n+1}$ ?

$10\pm1.96\sqrt{1}$ , using only uncertainty about the unknown mean.
$10\pm1.96\sqrt{4}$ , conditioning on the posterior mean as if it were known.
$10\pm1.96\sqrt{5}$ , combining observation noise and posterior uncertainty. (correct answer)
$10\pm1.96(3)$ , adding the two component standard deviations before scaling.

Explanation: Whenever you see a Bayesian predictive inference question, ask yourself: what sources of uncertainty contribute to the distribution of a future observation? Here, there are two — the randomness in

Y_{n+1}

given

\mu

(observation noise), and your remaining uncertainty about

\mu

itself (posterior uncertainty). To find the posterior predictive distribution, you marginalize over

\mu

Y_{n+1} \mid \mathcal{D} = \int p(Y_{n+1}\mid\mu)\, p(\mu\mid\mathcal{D})\, d\mu

Because both distributions are normal, the result is also normal. The predictive mean equals the posterior mean,

10

, and the predictive variance adds the two variance components:

\text{Var}(Y_{n+1}\mid\mathcal{D}) = \underbrace{4}_{\text{observation noise}} + \underbrace{1}_{\text{posterior uncertainty}} = 5

. The central 95% interval is therefore

10 \pm 1.96\sqrt{5}

, confirming C is correct. Choice A uses only the posterior variance (

\sigma^2 = 1

), ignoring that even if you knew

\mu

exactly, future observations would still scatter around it with variance 4. This would be appropriate for a credible interval on

\mu

, not a predictive interval for

Y_{n+1}

. Choice B uses only the observation variance (

\sigma^2 = 4

), treating the posterior mean as though it were the true

\mu

— effectively pretending there's no remaining uncertainty about

\mu

. This undercovers the true predictive spread. Choice D adds the standard deviations (

\sqrt{4} + \sqrt{1} = 3

) instead of the variances. Variances add for independent components; standard deviations do not. Study tip: Always remember that predictive intervals are wider than credible intervals — you're predicting a new random outcome, not just estimating a parameter.