Statistics Graduate Level Quiz: Regression Diagnostics

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Regression DiagnosticsQuestion 1 of 10

An ordinary least squares regression has $p=4$ fitted coefficients and residual mean square $s^2=2$ . Observation $i$ has residual $e_i=2$ and leverage $h_{ii}=0.50$ . Observation $j$ has residual $e_j=4$ and leverage $h_{jj}=0.10$ .

Based on Cook's distance, which comparison of the two observations is correct?

Observation

i

is more influential because its Cook's distance is approximately

1.00

, versus

0.25

for observation

j

Observation

j

is more influential because its Cook's distance is approximately

2.00

, versus

0.50

for observation

i

The observations are equally influential because the larger residual for

j

exactly offsets the larger leverage for

i

Observation

j

is more influential because Cook's distance depends on residual magnitude but not on leverage after fitting.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Regression Diagnostics

Practice Regression Diagnostics 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 Regression Diagnostics, 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

Based on Cook's distance, which comparison of the two observations is correct?

Observation $i$ is more influential because its Cook's distance is approximately $1.00$ , versus $0.25$ for observation $j$ . (correct answer)
Observation $j$ is more influential because its Cook's distance is approximately $2.00$ , versus $0.50$ for observation $i$ .
The observations are equally influential because the larger residual for $j$ exactly offsets the larger leverage for $i$ .
Observation $j$ is more influential because Cook's distance depends on residual magnitude but not on leverage after fitting.

Explanation: Cook's distance measures how much all fitted values change when observation

i

is deleted, capturing both leverage (how unusual the predictor values are) and residual magnitude (how poorly the model fits). The formula is:

D_i = \frac{e_i^2}{p \cdot s^2} \cdot \frac{h_{ii}}{(1 - h_{ii})^2}

where

p

is the number of fitted coefficients,

s^2

is the residual mean square,

e_i

is the residual, and

h_{ii}

is the leverage. For observation

i

D_i = \frac{(2)^2}{4 \cdot 2} \cdot \frac{0.50}{(0.50)^2} = \frac{4}{8} \cdot \frac{0.50}{0.25} = 0.5 \cdot 2 = 1.00

For observation

j

D_j = \frac{(4)^2}{4 \cdot 2} \cdot \frac{0.10}{(0.90)^2} = \frac{16}{8} \cdot \frac{0.10}{0.81} \approx 2 \cdot 0.123 = 0.247

So observation

i

has the larger Cook's distance (~1.00 vs ~0.25), confirming A is correct. The high leverage of

h_{ii} = 0.50

creates an enormous multiplier through the

\frac{h}{(1-h)^2}

term, more than compensating for observation

j

's larger raw residual. Choice B incorrectly computes Cook's distance, likely omitting the leverage ratio or miscalculating the

(1-h)^2

denominator. Choice C is tempting but wrong — the two quantities don't simply "offset"; leverage enters nonlinearly through

(1-h_{ii})^2

, which amplifies influence dramatically near

h = 0.5

. Choice D is fundamentally incorrect; Cook's distance explicitly depends on leverage, not just residuals. The key study tip: always remember that leverage enters Cook's distance as

\frac{h}{(1-h)^2}

, which grows explosively as

h \to 1

. A moderate residual with high leverage can easily dominate a large residual with low leverage.

Question 2

In a Bernoulli logistic regression, the final iteratively reweighted least squares leverage can be written as $h_i=w_i x_i^{\mathsf T}(X^{\mathsf T}WX)^{-1}x_i$ , where $w_i=\hat p_i(1-\hat p_i)$ . For observation $A$ , $w_A=0.001$ and $x_A^{\mathsf T}(X^{\mathsf T}WX)^{-1}x_A=40$ . For observation $B$ , the corresponding values are $0.25$ and $0.8$ . Both observations have outcome $1$ , with $\hat p_A=0.001$ and $\hat p_B=0.50$ .

Which diagnostic comparison is most accurate?

Observation $A$ has higher leverage and a smaller Pearson residual, so it must be less influential than observation $B$
Observation $A$ has lower leverage but a far larger Pearson residual, so it may still be more influential than observation $B$ (correct answer)
Observation $B$ has lower leverage and a larger Pearson residual, so it necessarily dominates every influence diagnostic
The two observations have equal leverage because the weight and predictor-space terms cancel in both cases

Explanation: When diagnosing influential observations in logistic regression, you need to track two separate quantities — leverage and residual size — and understand that influence typically depends on both. Neither alone tells the whole story. Start by computing the leverages. For observation

A

h_A = 0.001 \times 40 = 0.04

. For observation

B

h_B = 0.25 \times 0.8 = 0.20

. So

B

has higher leverage. Now compute the Pearson residuals, defined as

r_i = (y_i - \hat{p}_i)/\sqrt{\hat{p}_i(1-\hat{p}_i)}

. For

A

(outcome 1,

\hat{p}_A = 0.001

r_A = (1 - 0.001)/\sqrt{0.001 \times 0.999} \approx 0.999/0.0316 \approx 31.6

. For

B

r_B = (1 - 0.50)/\sqrt{0.25} = 0.5/0.5 = 1.0

. Observation

A

has dramatically larger residual despite lower leverage, meaning it could still exert substantial influence — making B the most accurate statement. Choice A is wrong because it concludes that lower leverage necessarily implies less influence, ignoring that a massive residual can compensate and drive influence metrics like Cook's distance upward. Choice C incorrectly claims

B

"dominates every influence diagnostic" — with its residual near 1 and moderate leverage,

B

is actually fairly unremarkable compared to

A

's enormous residual. Choice D is simply false; the products

0.04

and

0.20

are not equal. The key study tip: influence = leverage × residual (conceptually). An observation in an unusual predictor region and poorly fit by the model is the dangerous combination — never dismiss a point just because one component looks small.

Question 3

In an ordinary least squares regression, an observation has leverage $h_{ii}=0.80$ but residual $e_i=0$ . The design matrix remains full rank if this observation is removed.

Which statement best describes the observation's current and potential influence?

Its case-deletion coefficient change and Cook's distance are zero, but a small response perturbation could make it influential because leverage is high (correct answer)
Its case-deletion coefficient change must be large because leverage alone determines influence regardless of the residual
Its Cook's distance is infinite because the factor $1/(1-h_{ii})^2$ diverges for any high-leverage observation
Its zero residual implies low leverage after refitting, so perturbing its response cannot materially affect the fitted model

Explanation: Influence in regression depends on two separate ingredients: leverage (how unusual an observation's predictors are) and the residual (how far its response deviates from the fitted value). High leverage alone is not sufficient for influence — the residual must also be nonzero. This distinction is precisely what this question tests. The standard formulas make the logic clear. Cook's distance is proportional to

h_{ii} \cdot e_i^2

, and the case-deletion coefficient change (DFBETAS) also scales with the residual. When

e_i = 0

, both quantities equal zero regardless of how large

h_{ii}

is. So right now, this observation — despite sitting at

h_{ii} = 0.80

— exerts no actual influence on the estimated coefficients. Answer A captures this perfectly: current influence is zero, but the high leverage means the observation sits in a powerful position. If its response were nudged even slightly, that nonzero residual would be amplified by the large leverage factor

h_{ii}/(1-h_{ii})

, potentially producing a dramatic shift in the fitted model. A is correct. Answer B is wrong because leverage alone does not determine influence — the residual is an equally essential factor, not an afterthought. Answer C misreads the Cook's distance formula: the

1/(1-h_{ii})^2

term appears in the denominator of the variance, but it multiplies

e_i^2

, which is zero here, so Cook's distance is exactly zero, not infinite. Answer D inverts the logic entirely — zero residual does not imply low leverage, and high leverage means the observation can be influential if perturbed. Study tip: Memorize that influence = leverage × residual. Neither alone is sufficient. A high-leverage, zero-residual point is a "loaded gun" — harmless now, dangerous if the response shifts.

Question 4

A binomial generalized linear model has residual deviance $252$ on $90$ residual degrees of freedom. Residual checks do not reveal a specific omitted predictor or link-function defect, and the analyst elects to retain the working mean model while allowing for overdispersion.

Under a quasi-binomial analysis using the Pearson or deviance ratio as a dispersion estimate, which adjustment is most appropriate?

Retain the coefficient estimates and divide their model-based standard errors by approximately $\sqrt{2.8}\approx1.67$
Retain the coefficient estimates and multiply their model-based standard errors by approximately $\sqrt{2.8}\approx1.67$ (correct answer)
Multiply every coefficient estimate and standard error by $2.8$ to account for excess residual variation
Refit with ordinary least squares because a deviance ratio above $1$ invalidates all binomial mean estimates

Explanation: Whenever you encounter overdispersion in a GLM, your first instinct should be to separate two questions: what happens to the coefficients? and what happens to the standard errors? These are answered independently under quasi-likelihood theory. Here, the estimated dispersion parameter is

\hat{\phi} = \frac{\text{Residual Deviance}}{\text{Residual df}} = \frac{252}{90} = 2.8

. Under a quasi-binomial model, the mean structure — and therefore the coefficient estimates themselves — remains exactly as fitted. What changes is the variance: quasi-likelihood inflates the variance of each coefficient by a factor of

\hat{\phi}

, so standard errors are multiplied by

\sqrt{\hat{\phi}} = \sqrt{2.8} \approx 1.67

. This is choice B, and it is the correct adjustment. Wider standard errors reflect that the data are more variable than a pure binomial model assumes, making inference more conservative. Choice A gets the direction exactly backwards — dividing standard errors would narrow them, falsely increasing precision when you actually have excess variation. This is a trap for students who confuse "correcting downward" with "correcting for overdispersion." Choice C is doubly wrong: it incorrectly adjusts the coefficient estimates (which are unaffected under quasi-likelihood) and applies a linear rather than square-root scaling to the standard errors. Choice D confuses a diagnostic signal with model invalidity. A dispersion ratio above 1 does not invalidate the mean model — it simply signals that variance needs rescaling, which quasi-likelihood handles without abandoning the binomial mean structure entirely. Study tip: Remember the quasi-likelihood mantra — coefficients stay, standard errors scale by

\sqrt{\hat{\phi}}

. Overdispersion is a variance problem, not a bias problem.

Question 5

A linear regression of a response on a single continuous predictor produces residuals that are predominantly positive for both low and high predictor values and predominantly negative for intermediate predictor values. The residual spread is roughly constant across predictor values, and no individual observation has unusually high Cook's distance.

Which modification is most directly supported by this residual structure?

Delete the observations at both predictor extremes because their residual signs establish undue influence
Use inverse-variance weights because the residual pattern primarily indicates increasing conditional variance
Add a positive quadratic term in the predictor to represent curvature in the conditional mean (correct answer)
Retain the linear mean and use heteroskedasticity-consistent standard errors to remove the curvature

Explanation: Whenever you see a residual plot described in terms of where residuals are positive or negative, your first instinct should be to ask: "What pattern in the mean function does this reveal?" The key diagnostic here is the sign structure — positive residuals at both extremes and negative residuals in the middle — not the spread or any individual point's influence. This sign pattern is the classic fingerprint of a concave-down (∩-shaped) relationship between the response and predictor. When you fit a straight line through a curve that bends downward, the line overshoots in the middle (hence negative residuals there) and undershoots at both tails (hence positive residuals at the extremes). The fix is to let the mean function itself bend — specifically, by adding a term

\beta_2 x^2

with an expected negative coefficient. This is exactly what C proposes with a positive quadratic term (positive in the sense of adding the squared predictor; the coefficient's sign is estimated from data). C directly addresses the misspecified conditional mean. A is wrong because residual sign does not indicate influence or leverage. Cook's distance — which the passage explicitly tells you is unremarkable — measures influence. Deleting extreme observations based solely on residual sign is a methodological error. B is wrong because heteroskedasticity means non-constant spread. The passage explicitly states that residual spread is roughly constant, so inverse-variance weighting addresses a problem that doesn't exist here. D is wrong for a subtle but important reason: heteroskedasticity-consistent (HC) standard errors correct inference under non-constant variance, but they do nothing to fix a misspecified mean function. You'd still have biased estimates of the mean. A useful rule of thumb: mean misspecification (curved residuals) calls for a better model structure; variance misspecification (fan-shaped spread) calls for weighted least squares or HC standard errors. Don't confuse the two.

Question 6

An ordinary least squares model with an intercept has $n=30$ observations and $p=3$ fitted coefficients. For observation $i$ , the raw residual is $e_i=2$ , the leverage is $h_{ii}=0.40$ , and the residual mean square is $s^2=4$ .

Using the full-sample residual mean square, which pair gives the internally studentized residual and the deleted prediction residual for observation $i$ , respectively?

$1.29$ and $3.33$ , respectively (correct answer)
$1.67$ and $2.58$ , respectively
$0.77$ and $3.33$ , respectively
$1.29$ and $1.20$ , respectively

Explanation: When you see a question involving residual diagnostics in OLS regression, keep two formulas straight: the internally studentized residual and the deleted prediction residual (also called the PRESS residual). The internally studentized residual is

r_i = \frac{e_i}{\sqrt{s^2(1 - h_{ii})}}

. Plugging in:

r_i = \frac{2}{\sqrt{4(1 - 0.40)}} = \frac{2}{\sqrt{4 \times 0.60}} = \frac{2}{\sqrt{2.4}} = \frac{2}{1.549} \approx 1.29

. The deleted prediction residual is

d_i = \frac{e_i}{1 - h_{ii}}

. Plugging in:

d_i = \frac{2}{1 - 0.40} = \frac{2}{0.60} \approx 3.33

. This confirms answer A is correct:

1.29

and

3.33

, respectively. Now for the distractors. Answer B (1.67 and 2.58) results from omitting

1 - h_{ii}

inside the square root — computing

2/\sqrt{4} = 1.0

is wrong, but 1.67 suggests dividing by

s

alone without the leverage correction; 2.58 is

2/\sqrt{0.6}

rather than

2/0.6

, confusing the deleted residual formula with another variance adjustment. Answer C (0.77 and 3.33) gets the deleted residual right but inverts the studentized formula — using

\sqrt{s^2/(1-h_{ii})}

in the denominator instead of

\sqrt{s^2(1-h_{ii})}

. Answer D (1.29 and 1.20) correctly computes the studentized residual but then applies an extra square root to the deleted residual:

\sqrt{3.33} \approx 1.83

, or confuses the formula entirely. A reliable study habit: memorize that the deleted residual inflates

e_i

by dividing by

(1-h_{ii})

, while the studentized residual deflates it by multiplying

s^2

by the same factor before taking the square root — opposite operations, easy to mix up under pressure.

Question 7

In a regression with a correctly specified linear conditional mean, diagnostic residuals show that variability increases in proportion to the magnitude of the fitted mean. Scientific knowledge supports the variance model $\operatorname{Var}(Y_i\mid X_i)=\sigma^2\mu_i^2$ , where $\mu_i=E(Y_i\mid X_i)$ . The researcher wants efficient coefficient estimates while retaining the original mean model.

Which response is most appropriate if the stated variance relationship is credible?

Fit feasible weighted least squares using weights proportional to $1/\hat\mu_i^2$ and reassess the weighted residuals (correct answer)
Fit weighted least squares using weights proportional to $\hat\mu_i^2$ so high-variance observations receive greater emphasis
Keep ordinary least squares and use robust standard errors because this necessarily restores coefficient efficiency
Log-transform the response because this preserves the same conditional-mean parameters while stabilizing the variance

Explanation: Whenever you see heteroskedasticity with a known variance structure, your goal is to exploit that structure to recover efficiency — not just protect against it. The key principle: weighted least squares (WLS) downweights observations with high variance, assigning weight

w_i \propto 1/\operatorname{Var}(Y_i \mid X_i)

. Here, since

\operatorname{Var}(Y_i \mid X_i) = \sigma^2 \mu_i^2

, the optimal weights are

w_i \propto 1/\mu_i^2

. Because

\mu_i

is unknown, you substitute

\hat{\mu}_i

from an initial OLS fit, yielding feasible WLS. After fitting, you must check weighted residuals to verify the variance model was appropriate — if the model is right, those residuals should appear homoskedastic. This is exactly what A prescribes, making it correct. B inverts the logic fatally. Weighting by

\hat{\mu}_i^2

gives more influence to high-variance observations, which is precisely backwards — it amplifies the problem rather than correcting it. C is a common conceptual trap. Robust (sandwich) standard errors correct inference under heteroskedasticity but do not restore estimator efficiency. OLS with robust SEs remains inefficient when the variance structure is known and exploitable; only WLS achieves the Gauss-Markov efficiency bound in this setting. D is tempting but wrong. A log-transform changes the mean model itself —

E[\log Y_i \mid X_i] \neq \log E[Y_i \mid X_i]

by Jensen's inequality — so the original conditional-mean parameters are not preserved. Study tip: On problems involving known heteroskedasticity, always set up weights as

w_i = 1/\operatorname{Var}(Y_i)

and distinguish between correcting inference (robust SEs) versus correcting efficiency (WLS). Only WLS actually improves estimation.

Question 8

In an ordinary least squares regression, a researcher considers the contrast $c^{\mathsf T}\beta$ . The full-data estimate is $c^{\mathsf T}\hat\beta=1.70$ . For observation $i$ , the quantities needed for case deletion are $c^{\mathsf T}(X^{\mathsf T}X)^{-1}x_i=0.12$ , $e_i=3$ , and $h_{ii}=0.25$ .

What is the value of the contrast after deleting observation $i$ ?

$2.06$ , because deletion reverses the sign but does not apply the leverage denominator
$2.18$ , because deletion adds the case's leverage-adjusted contribution to the full-data estimate
$1.34$ , because deletion uses the unadjusted contribution based only on the raw residual
$1.22$ , because deletion subtracts the case's leverage-adjusted contribution from the full-data estimate (correct answer)

Explanation: When working with case-deletion diagnostics in OLS, the key formula to internalize is how removing a single observation shifts a contrast estimate. The exact update formula states:

c^\mathsf{T}\hat\beta_{(i)} = c^\mathsf{T}\hat\beta - \frac{c^\mathsf{T}(X^\mathsf{T}X)^{-1}x_i \cdot e_i}{1 - h_{ii}}

This tells you that deleting observation

i

adjusts the full-data estimate by subtracting a leverage-weighted residual contribution. Plugging in the given values —

c^\mathsf{T}\hat\beta = 1.70

c^\mathsf{T}(X^\mathsf{T}X)^{-1}x_i = 0.12

e_i = 3

, and

h_{ii} = 0.25

— you get:

c^\mathsf{T}\hat\beta_{(i)} = 1.70 - \frac{(0.12)(3)}{1 - 0.25} = 1.70 - \frac{0.36}{0.75} = 1.70 - 0.48 = 1.22

This confirms D is correct: deletion subtracts the leverage-adjusted contribution. Choice A is wrong because it incorrectly reverses the sign of the adjustment and ignores the

(1 - h_{ii})

denominator entirely, treating deletion as an addition. Choice B arrives at 2.18 by adding the adjustment rather than subtracting it — a sign error that would describe what happens if you included an extra influential observation, not removed one. Choice C omits the leverage correction, computing

1.70 - (0.12)(3) = 1.34

; this forgets that high-leverage points have their contributions amplified by the

(1 - h_{ii})

denominator. A useful memory anchor: case deletion always divides by

(1 - h_{ii})

, which inflates the correction for high-leverage points — the closer

h_{ii}

is to 1, the more sensitive the estimate is to that observation's removal.

Question 9

Two observations have nearly identical, highly unusual predictor values and responses that lie close to the fitted regression surface. Each observation has a small ordinary residual and a modest Cook's distance when assessed individually. When both observations are removed, however, one estimated slope changes substantially.

Which diagnostic strategy most directly addresses the phenomenon described?

Inspect standardized residual normality because a nearly normal residual distribution rules out masked influence
Compute only externally studentized residuals because masking can occur only through an underestimated error variance
Replace Cook's distance with leverage because any high-leverage observation is necessarily influential after deletion
Compute joint or subset-deletion diagnostics for the two observations because individual deletion may be affected by masking (correct answer)

Explanation: When multiple observations share unusual predictor values, they can mask each other's influence — each looks harmless in isolation, but together they distort the fitted model. This is the core phenomenon being tested: masking in influence diagnostics. The passage describes a classic masking scenario. Two high-leverage points with small individual residuals each produce modest Cook's distances when deleted alone, yet jointly removing them shifts a slope substantially. This happens because when one influential point remains in the data while the other is deleted, it anchors the fit, concealing how much both observations together are driving the estimates. The diagnostic that directly addresses this is joint or subset-deletion, where you remove both observations simultaneously and measure the aggregate impact — making D the correct answer. Choice A is a red herring. Residual normality plots assess distributional assumptions, not influence. Masked influential points can produce perfectly well-behaved residuals precisely because they're pulling the surface toward themselves, so normality provides no protection against masking. Choice B misidentifies the mechanism. Masking is not solely a function of underestimated error variance — it arises from the geometry of leverage. Externally studentized residuals correct for one observation's effect on variance, but they still evaluate observations individually and cannot detect joint influence. Choice C contains a subtle but important error: high leverage is necessary but not sufficient for influence. A high-leverage point that happens to lie near the true regression surface may not be influential. Leverage alone does not replace deletion diagnostics. Study tip: On regression diagnostics questions, whenever you see multiple outliers or clustered unusual observations, immediately think about masking — individual diagnostics will systematically understate influence, and only subset-deletion methods reveal the true joint effect.

Question 10

For six consecutively ordered observations from a regression containing an intercept, the ordinary residuals are $1,2,1,-1,-2,-1$ . Treating this ordering as potentially temporal, an analyst computes the Durbin–Watson statistic $D=\sum_{t=2}^{6}(e_t-e_{t-1})^2/\sum_{t=1}^{6}e_t^2$ .

Which value and interpretation are correct?

$D\approx2.67$ , providing evidence of negative serial correlation in adjacent residuals
$D\approx1.33$ , providing evidence of positive serial correlation in adjacent residuals
$D\approx0.67$ , providing evidence of positive serial correlation in adjacent residuals (correct answer)
$D\approx0.67$ , providing evidence of negative serial correlation in adjacent residuals

Explanation: The Durbin–Watson statistic tests for serial correlation in regression residuals, and its value always falls between 0 and 4: values near 0 signal strong positive autocorrelation, near 2 signal no autocorrelation, and near 4 signal strong negative autocorrelation. When you see residuals that follow a smooth wave pattern — rising then falling, or vice versa — your instinct should immediately be "positive autocorrelation." Start by computing the numerator: the sum of squared successive differences. The consecutive differences

(e_t - e_{t-1})

are

(2-1), (1-2), (-1-1), (-2-(-1)), (-1-(-2)) = 1, -1, -2, -1, 1

. Squaring and summing gives

1+1+4+1+1 = 8

. The denominator is

\sum e_t^2 = 1+4+1+1+4+1 = 12

. Therefore

D = 8/12 \approx 0.67

. Since

0.67 \ll 2

, this indicates positive serial correlation — adjacent residuals move together rather than alternating. The residual pattern

1, 2, 1, -1, -2, -1

visually confirms this: values drift smoothly before reversing, a hallmark of positive autocorrelation. Choice A is doubly wrong: the computed value

2.67

is incorrect (that would be

32/12

), and even if it were right,

D > 2

would suggest negative, not general, correlation. Choice B gets the direction of interpretation right (positive autocorrelation) but miscalculates

D

1.33

— likely from dividing 8 by 6 instead of 12. Choice D correctly computes

D \approx 0.67

but misidentifies the direction, confusing "small D" with negative correlation. A reliable memory anchor: think of

D \approx 2

as your neutral baseline. Anything below 2 means residuals are too similar (positive autocorrelation); anything above 2 means they alternate too sharply (negative autocorrelation).