Econometrics Exam: 5 Worked Problems Explained

Econometrics study guide • Five worked problems

An econometrics exam tests more than your ability to remember formulas. You also need to choose the right estimator, check whether the supplied numbers agree, and explain what each result means. This guide works through five connected problems on ordinary least squares, demand estimation, autocorrelation, simultaneous equations, and ARIMA model selection.

First, we rebuild a small regression from its original observations. Next, we examine a demand equation whose summary statistics require extra care. Finally, we connect time-series diagnostics with identification and forecasting, showing why a plausible numerical answer can still rest on an unsupported assumption.

Data provenance: the numerical exercises come from the six supplied photographs, covering pages 294–299 of an “Econometrics Examination” section. The photographs do not establish the book’s full bibliographic details, so this article does not guess an author or edition. Explanations, calculations, charts, and checks below are newly developed. These are instructional data and reported historical examples, not current economic estimates.

What these econometrics exam problems test

Each problem asks a different question about a model. For example, a coefficient test concerns a specific parameter, whereas an identification check asks whether the available variation can reveal that parameter at all. Therefore, choose the economic and statistical question before choosing the formula.

Throughout this article, n denotes the regression sample size, while K counts all estimated regression coefficients, including the intercept. Thus, a model with three explanatory variables has K = 4. Lowercase x and y denote deviations from sample means when they appear in centered sums.

Problem 1: Estimate an OLS regression from eight observations

The first exercise relates quantity supplied, Y, to price, X. We use the eight observations below and include an intercept. Although the exercise describes other influences as constant, real market data would require additional reasoning before we interpreted this fitted slope as a causal supply response.

Step 1: Center the data and calculate the slope

First, add the observations and calculate both means. The price total equals 64, while the quantity total equals 104. Therefore, X̄ = 8 and Ȳ = 13.

Next, divide the centered cross-product by the centered sum of squares of X. This gives the slope without requiring a matrix calculation.

Holding the model’s other conditions fixed, a one-unit increase in price corresponds to about 0.824 additional units of fitted quantity. However, the intercept describes the fitted value at X = 0, which lies outside the observed price range of 5 to 11. Its mathematical role is clear, but its economic interpretation requires caution.

Step 2: Keep full precision until the final answer

The photographed solution rounds the slope to 0.82 before calculating the intercept. Consequently, it reports 13 − 0.82(8) = 6.44 rather than 6.411765. Both lines look similar, but the second calculation no longer uses the exact OLS slope.

For example, the rounded line gives Ŷ = 14.64 at X = 10. The exact fit gives 14.647059 instead. Therefore, retain at least several decimal places internally and round only your displayed answers.

Step 3: Calculate residual variance and standard errors

For each observation, subtract its fitted quantity from its actual quantity. Next, square those residuals and sum them. Because the regression estimates two parameters, the residual degrees of freedom equal 8 − 2 = 6.

Under the classical constant-variance assumptions, the coefficient standard errors follow from the same estimated error variance. In particular, the intercept’s uncertainty depends on how far the regressor mean lies from zero.

Step 4: Test both coefficients

To test a zero coefficient, divide its estimate by its standard error. With six degrees of freedom, the two-sided 5% critical value is approximately 2.4469. Therefore, both absolute t-statistics exceed the critical value.

These finite-sample t results assume independent, normally distributed errors with constant variance and a correctly specified conditional mean. With only eight observations, that assumption set matters. Moreover, statistical significance alone does not establish that the effect is large, stable, or causal.

Step 5: Explain the coefficient of determination

The line accounts for approximately 64.05% of the sample variation in quantity around its mean. However, this percentage does not measure the fraction of quantity that price “causes.” It describes in-sample fit for this specification.

For the underlying decomposition, see Penn State’s regression model evaluation notes. Their sum-of-squares framework explains why total variation separates into fitted variation and residual variation when OLS includes an intercept.

Step 6: Separate a confidence interval from a prediction interval

At X = 10, the point estimate equals 14.6471. Nevertheless, two different uncertainty questions remain: how precisely do we estimate average quantity at that price, and how widely might one new quantity observation vary?

The prediction interval includes the extra variation of a new disturbance. Consequently, it is wider than the confidence interval for the conditional mean. Penn State’s estimation and prediction lesson provides the corresponding formulas and assumptions.

Problem 2: Interpret demand and audit the model’s fit

The second exercise supplies a demand equation from 24 annual observations. Quantity Y uses kilograms per year, own price X₁ and substitute price X₃ use dollars, and income X₂ uses thousands of dollars.

Step 1: Compare the signs with economic theory

A negative own-price coefficient matches the usual downward-sloping demand relationship. Similarly, positive income sensitivity matches a normal good. However, a substitute’s price would ordinarily have a positive cross-price effect, so the estimated coefficient of −4 runs against that expectation.

Still, the unexpected sign does not prove that the goods are complements. Sampling noise, collinearity, omitted influences, or a mistaken economic classification could contribute. In addition, observational demand estimation can face price endogeneity if market supply and demand jointly determine prices.

Step 2: Test the slopes with the correct denominator

With three regressors and an intercept, K = 4 and the residual degrees of freedom equal 20. Next, compare the absolute t-statistics with approximately 2.086, the two-sided 5% critical value.

Therefore, own price and income reject their zero nulls, whereas substitute price does not. The photographed answer gives the third t-statistic a positive sign, but the estimated coefficient is negative. Although this sign error does not change a two-sided significance decision, it does change the direction of the reported statistic.

Step 3: Find the missing term in the R² calculation

The exercise provides Σy² = 40, Σx₁y = 10, and Σx₂y = 45. Its answer key then calculates (−7 × 10 + 2.4 × 45)/40 = 0.95. However, the displayed regression also contains X₃, so a third cross-product belongs in that calculation.

For example, a hypothetical Σx₃y of 1 would change the arithmetic to R² = 0.85. A hypothetical value of 2 would give 0.75. These are sensitivity illustrations of the missing term, not reconstructed data sets or claims that every remaining printed statistic would stay unchanged.

Alternatively, an analyst could refit a restricted model without X₃. However, that refit would generally change the intercept and the remaining slopes. Simply crossing out one variable while retaining every other full-model estimate does not perform a restricted regression.

Step 4: Recover the answer key’s downstream calculations conditionally

To understand the intended exercise, suppose we accept R² = 0.95 as an additional supplied summary. Under that condition, the adjusted R², joint F-statistic, and residual standard error follow mechanically. Nevertheless, each result inherits the same unresolved input problem.

The joint test concerns the null that all three population slopes equal zero. Therefore, a large F-statistic can coexist with an individually insignificant substitute-price coefficient. Those tests answer different questions.

For background on joint tests, see Penn State’s multiple-regression evaluation lesson. The model-building lesson also explains the degrees-of-freedom adjustment behind adjusted R².

Step 5: Calculate price and income elasticities

A linear slope measures an absolute change, whereas an elasticity measures a proportional response. Consequently, elasticity depends on where we evaluate a linear demand curve. Using the supplied means Ȳ = 32, X̄₁ = 8, and X̄₂ = 16 gives:

Locally, a 1% increase in own price corresponds to approximately 1.75% lower demand, holding the other regressors fixed. Similarly, a 1% income increase corresponds to about 1.20% higher demand. Thus, these plug-in calculations describe price-elastic and income-elastic demand at the stated evaluation point.

However, neither elasticity stays constant across a linear demand surface. By contrast, a log-log specification gives slope coefficients a direct elasticity interpretation, subject to its model assumptions. Always check the functional form before labeling a coefficient an elasticity.

Step 6: Check whether the supplied means fit the regression

An OLS regression with an intercept passes through its sample means. Therefore, the stated means should satisfy the fitted equation when they come from the same estimation sample.

A negative average substitute price conflicts with the ordinary positive-price interpretation of this exercise. Consequently, the reported intercept, coefficients, means, or variable definitions cannot all describe the same standard positive-price OLS data set as written. The elasticity arithmetic remains reproducible, but the supplied summaries need correction before we treat them as a coherent empirical model.

Problem 3: Diagnose and address autocorrelation

Our third exercise reports an investment regression for 1960–1979, a span of 20 annual observations. Its fitted relationship uses GNP and a consumer price index:

Step 1: Recognize the pattern without overstating the test

Serial correlation means that disturbances at different dates relate to one another. For example, an omitted investment cycle may leave several positive residuals in a row. Positive first-order correlation makes adjacent residuals similar, which tends to reduce the Durbin–Watson statistic.

Using this approximation, d = 0.77 suggests ρ̂ ≈ 0.615. However, that number is a diagnostic approximation rather than an exact residual-regression estimate. A formal Durbin–Watson decision needs the appropriate bounds or p-value for the sample and design.

Moreover, the ordinary Durbin–Watson test has limitations when the regression contains a lagged dependent variable. SAS’s autocorrelation testing documentation explicitly discusses this restriction. A suitable alternative, such as a Breusch–Godfrey test, can accommodate broader dynamic specifications under its assumptions.

Step 2: Separate unbiasedness from valid standard errors

Autocorrelation alone does not necessarily bias OLS coefficients. Under strict exogeneity, E(u | X) = 0, the usual conditional unbiasedness argument still applies. Nevertheless, the conventional covariance formula generally fails when the errors correlate across observations.

Consistency also requires suitable exogeneity and dependence conditions. In contrast, if omitted dynamics or an endogenous regressor correlate with the error, OLS can become biased and inconsistent. Therefore, “autocorrelation never biases OLS” is too broad a statement.

Step 3: Estimate an AR(1) error relationship

First, estimate the original regression and collect residuals in chronological order. Next, regress each residual on its lag, typically without an intercept for this estimator. Finally, use the resulting ρ̂ to transform the equation and, when appropriate, iterate the procedure.

Step 4: Transform every column, including the intercept

The transformed equation has intercept contribution β₀(1 − ρ̂). Therefore, transforming only Y and X while ignoring the constant column can produce the wrong implementation. This detail matters especially when we retain the first observation.

Cochrane–Orcutt drops the first observation. By comparison, Prais–Winsten retains it through a stationary AR(1) adjustment:

Stata’s Prais–Winsten and Cochrane–Orcutt manual documents these GLS methods and alternative estimates of ρ. Dedicated software also handles the transformed constant correctly, which reduces avoidable implementation errors.

Step 5: Understand the expanded dynamic equation

Substituting the AR(1) disturbance into the regression gives another useful representation. In particular, the intercept becomes β₀(1 − ρ), and the lag coefficients obey common-factor restrictions.

For example, the coefficient on lagged X₁ equals minus ρ times its contemporaneous coefficient. An unrestricted dynamic regression does not automatically impose that relationship. Consequently, a dynamic specification and a regression with AR(1) errors need not represent the same statistical model.

Step 6: Investigate the remaining dependence

The transformed regression reports d = 0.89, compared with 0.77 originally. Although the statistic moves upward, it remains far below the informal no-first-order-correlation benchmark of about 2. The approximation now gives ρ̂ ≈ 0.555, suggesting that substantial dependence may remain.

Next, examine the residual time plot and ACF. Possible explanations include higher-order dynamics, structural change, omitted variables, an inappropriate functional form, or a poorly fitting AR(1) correction. However, the two Durbin–Watson values alone cannot identify the cause.

If the conditional mean is credible and the goal is inference, an appropriate heteroskedasticity-and-autocorrelation-consistent covariance estimator may help. In contrast, forecasting may require an explicit dynamic error model. Neither approach repairs endogenous regressors or an incorrect economic specification automatically.

Problem 4: Solve a simultaneous-equations model

The fourth problem links an interest rate R with income Y, while treating money supply M as exogenous. Because R and Y appear in each other’s equations, their values emerge jointly.

Step 1: Explain why ordinary OLS generally fails

A shock to the income equation changes Y and, through the first equation, changes R. Consequently, R generally correlates with the disturbance in the income equation. Similarly, the endogenous Y regressor generally correlates with the first equation’s disturbance.

That feedback creates the usual simultaneity problem for separate OLS regressions. However, the diagnosis depends on the structural assumptions; calling a variable endogenous does not substitute for explaining the channel. Here, the two-equation system makes that channel explicit.

Step 2: Derive the reduced forms

First, insert the income equation into the interest-rate equation. Then collect the terms containing R on the left. Provided D = 1 − a₂b₁ is nonzero, division gives:

Next, substitute this reduced form into the income equation. The resulting expression also uses only the exogenous M as its observed explanatory variable.

This denominator condition ensures an algebraic solution for the contemporaneous system. However, it does not by itself establish the time-series stability of a richer dynamic model. Keep algebraic solvability separate from dynamic stability.

Step 3: Check identification equation by equation

The R equation contains M and Y, leaving no excluded exogenous variable to instrument Y. Therefore, under the stated restrictions, it is underidentified. Additional observations cannot fix a missing identification restriction.

By contrast, the Y equation excludes M, making money supply a potential instrument for R. If M is exogenous and has a nonzero reduced-form effect on R, that equation is exactly identified. Thus, the order count needs the corresponding relevance, or rank, condition.

Step 4: Recover the identified coefficients through ILS

Indirect least squares estimates the reduced forms and then uses these algebraic identities. However, the supplied pages give no numerical reduced-form estimates. Therefore, the formulas identify how to recover b₀ and b₁, but they do not support numerical structural estimates.

The same reduced-form coefficients do not separately reveal a₁ and a₂. For example, π₁ = a₁/(1 − a₂b₁) combines both. Consequently, knowing b₁ still leaves multiple possible combinations of a₁ and a₂.

Step 5: Add a lagged variable and reconsider 2SLS

Now suppose the first equation also includes Y_t−1, while the second equation excludes it. Under a valid instrument assumption, the Y equation has two excluded instruments for its one endogenous regressor. Therefore, it becomes overidentified if the rank condition holds.

Meanwhile, the first equation contains both M and lagged Y. It still lacks an excluded instrument for contemporaneous Y, so it remains underidentified under this information set. The identification status belongs to each equation, not to the system as one undifferentiated label.

For the identified income equation, the first 2SLS stage predicts R using an intercept, M, and lagged Y. Next, the structural stage uses the instrument-generated variation in R to estimate its effect on Y. However, dedicated IV software must calculate the standard errors; ordinary OLS standard errors from a manually fitted second stage are generally incorrect.

Stata’s instrumental-variable regression manual documents 2SLS estimation and instrument specification. In practice, also assess first-stage strength and the economic credibility of the exclusion restrictions.

Problem 5: Read the ACF and compare ARIMA models

The final exercise supplies SAS time-series output for 220 observations. Its working-series mean is approximately 0.033797, and its reported standard deviation is 2.122958. However, those descriptive statistics alone do not establish stationarity or select an ARIMA order.

Step 1: Interpret the autocorrelation pattern

The lag-one ACF equals 0.82315, indicating strong positive short-run dependence. Next, the ACF falls, crosses below zero around lag seven, and reaches a pronounced negative region around lags ten and eleven. Consequently, a simple positive AR(1) model does not explain the entire observed pattern well.

The PACF begins with a large lag-one value, but several later values also deserve attention. Therefore, treating the output as a perfectly clean PACF cutoff at lag one would overstate the evidence. Use the correlogram to propose candidates, then estimate and diagnose them.

A rough pointwise white-noise benchmark is ±1.96/√220, or approximately ±0.132. However, the SAS table displays lag-dependent standard errors, so its marked limits need not match that fixed benchmark. Moreover, examining twenty lags introduces a multiple-comparison issue; isolated spikes do not carry the same meaning as a coherent pattern.

For model-identification background, see Hyndman and Athanasopoulos’s ARIMA chapter. Its discussion connects stationarity, ACF/PACF patterns, estimation, and model comparison.

Step 2: Calculate the Box–Pierce statistic

The Box–Pierce test combines squared autocorrelations across the selected lags. Here, the null states that the population autocorrelations from lag 1 through lag 20 all equal zero. Because this calculation concerns the supplied series rather than fitted ARMA residuals, the exercise uses 20 degrees of freedom.

The answer key prints 649.56, which differs slightly from the sum reconstructed from the displayed five-decimal ACF entries. Also, its printed intermediate value of 2.9523 yields 649.506 when multiplied by 220. These small discrepancies do not affect the decision.

At the 5% level, the chi-square critical value for 20 degrees of freedom is approximately 31.4104. Since 649.5139 is much larger, we reject the joint zero-autocorrelation null decisively. Nevertheless, this result does not identify a unique AR or MA order.

Step 3: Compare the Ljung–Box adjustment

The Ljung–Box statistic adjusts the weighting of the sample autocorrelations. Using the same transcribed values, it reaches the same practical conclusion. However, after fitting an ARMA model, the reference degrees of freedom usually account for estimated dynamic parameters rather than automatically remaining at m.

Forecasting: Principles and Practice, chapter 5 explains portmanteau tests and residual diagnostics. A model should leave little usable serial structure in its residuals, even when its original series shows substantial dependence.

Step 4: Distinguish full-order, subset, and factored models

SAS syntax contains a subtle but consequential distinction. For example, p=10 estimates a full AR model with lags one through ten. In contrast, p=(1 10) estimates only lags one and ten, while p=(1)(10) multiplies two AR factors.

Here, the backshift operator B moves a series back one period. Expanding the factored AR specification reveals why its lag structure differs from a two-lag subset model:

Thus, the factored model includes a lag-11 interaction with a coefficient constrained by the two main parameters. Furthermore, an MA coefficient only at lag ten is not the same as ten freely estimated MA coefficients. The SAS ARIMA procedure manual explains these subset and factored specifications.

Step 5: Compare the reported AIC values correctly

For model selection, the supplied answer key compares three candidates and reports the following AIC values. These numbers come from the photographed solution; we cannot refit the models because the original 220 observations are unavailable. Therefore, the table is a comparison of reported estimates, not a newly reproduced SAS analysis.

Among these candidates, the AR-lag-one plus MA-lag-ten model has the smallest reported AIC. However, that ranking assumes comparable data, likelihood conventions, and estimation settings. It does not prove that this candidate is the true process or that its forecasts dominate every alternative.

This relative-support calculation strongly favors candidate two within the stated comparison. Nevertheless, it is not the probability that candidate two is correct. A model set can omit a better specification entirely.

Next, check residual autocorrelation, parameter uncertainty, and the relevant stationarity and invertibility restrictions. If forecasting is the goal, also compare forecasts on a later holdout period or through rolling-origin evaluation. A low AIC provides a useful ranking, while those additional checks address different aspects of model quality.

Step 6: Write clear SAS code for the candidate models

/* Requires the original ordered series in WORK.SERIES, variable Y.
   The supplied ACF table is not a replacement for those observations. */
proc arima data=work.series;
    identify var=y nlag=20;

    /* Candidate 1: factored AR model */
    estimate p=(1)(10) method=ml;

    /* Candidate 2: AR lag 1, subset MA lag 10 */
    estimate p=(1) q=(10) method=ml;

    /* Candidate 3: subset MA lags 1 and 10 */
    estimate q=(1 10) method=ml;
run;
quit;

This template makes the likelihood method explicit so that the comparison uses consistent settings. However, the code does not guarantee the printed historical AIC values, because the original series and all original estimation details are missing. Check residual diagnostics for each candidate before choosing a final forecasting specification.

The answer-key audit: details that change the interpretation

Many study guides stop when their arithmetic matches a printed solution. However, an econometrics exam can reward a more careful habit: checking whether the answer uses the correct information and assumptions. The following audit brings the main findings together.

Reproduce the verifiable calculations in Python

Below, the script reproduces the OLS estimates, both intervals, and the two autocorrelation statistics. It uses the eight raw observations and the twenty displayed ACF values. However, it cannot reconstruct the missing demand cross-product or recreate the original ARIMA likelihoods.

import numpy as np
from scipy.stats import t, chi2

x = np.array([5, 11, 7, 8, 11, 7, 6, 9], dtype=float)
y = np.array([12, 14, 10, 13, 17, 12, 11, 15], dtype=float)
n = len(x)
xc, yc = x - x.mean(), y - y.mean()
sxx, sxy = xc @ xc, xc @ yc
b1 = sxy / sxx
b0 = y.mean() - b1 * x.mean()
e = y - (b0 + b1 * x)
sse = e @ e
s2 = sse / (n - 2)
r2 = 1 - sse / (yc @ yc)
se = np.sqrt([s2 * (1/n + x.mean()**2/sxx), s2/sxx])
tstats = np.array([b0, b1]) / se
pvalues = 2 * t.sf(np.abs(tstats), df=n-2)

x0 = 10.0
yhat = b0 + b1 * x0
h0 = 1/n + (x0 - x.mean())**2/sxx
critical = t.ppf(0.975, df=n-2)
ci = yhat + np.array([-1, 1]) * critical * np.sqrt(s2*h0)
pi = yhat + np.array([-1, 1]) * critical * np.sqrt(s2*(1+h0))
print("Intercept, slope:", b0, b1)
print("SSE, R-squared:", sse, r2)
print("Standard errors, t, p:", se, tstats, pvalues)
print("Prediction, mean CI, observation PI:", yhat, ci, pi)

acf = np.array([
    .82315, .64539, .49820, .36657, .24378,
    .11572, -.02956, -.19275, -.34779, -.48502,
    -.48492, -.44583, -.43919, -.39035, -.31819,
    -.25245, -.19351, -.14885, -.06968, -.00019
])
T = 220
lags = np.arange(1, len(acf) + 1)
q_bp = T * np.sum(acf**2)
q_lb = T * (T+2) * np.sum(acf**2 / (T-lags))
print("Box-Pierce, Ljung-Box:", q_bp, q_lb)
print("Chi-square critical value:", chi2.ppf(.95, df=20))

As a practical check, the code should return an OLS slope near 0.823529, R² near 0.640523, and a Box–Pierce statistic near 649.5139. Small differences from printed textbook answers can arise from rounding. Larger differences should prompt a check of the data transcription, formula, and degrees of freedom.

How to approach an econometrics exam efficiently

First, write down the target: a slope, a mean response, a future observation, or a structural effect. Next, list the available data and the assumptions each calculation needs. This short setup prevents many errors before the arithmetic begins.

1. Count parameters explicitly. Include the intercept when calculating regression residual degrees of freedom.
2. Separate estimates from uncertainty. A coefficient, standard error, t-statistic, and confidence interval each convey different information.
3. Preserve full numerical precision. Round final displays rather than intermediate slopes and residuals.
4. Check identities. With an intercept, residuals sum to zero and the fitted equation passes through the sample means.
5. Label conditional results. If a missing input prevents a unique answer, state the necessary assumption.
6. Interpret before concluding. Explain the units, sign, statistical uncertainty, and limits of each result.
7. Diagnose after correction. Re-estimation does not guarantee that the original model problem has disappeared.

Finally, avoid using a single statistic as an all-purpose quality score. A high R² does not prove causality, a small p-value does not establish economic importance, and the lowest AIC does not guarantee useful forecasts. Good econometrics combines those tools with a coherent model and transparent assumptions.

Frequently asked questions

Can I calculate R² when one regressor’s cross-product is missing?

Not from the listed cross-products alone. However, other sufficient information, such as SSE and total variation, could provide an alternative route. In this demand exercise, the answer key’s shortcut requires an unstated condition or corrected input.

Does an insignificant coefficient equal zero?

No. A test that fails to reject zero does not establish an exactly zero estimate or population effect. Therefore, retain the actual estimate in algebraic identities unless you explicitly refit a restricted model.

Why does a prediction interval remain wider than a confidence interval?

A new observation includes its own disturbance in addition to estimation uncertainty. Consequently, the new-observation formula includes an extra error-variance term. Even a precisely estimated conditional mean does not eliminate individual outcome variation.

Does a high R² rule out autocorrelation?

No. The investment example reports R² = 0.98 alongside a low Durbin–Watson statistic. Thus, strong in-sample fit and serially dependent residuals can occur together.

Is a lagged variable always a good instrument?

No. A valid instrument needs both relevance and an appropriate exogeneity restriction. For example, persistent structural disturbances can correlate with lagged outcomes, undermining instrument validity.

Can an ACF plot uniquely identify an ARIMA model?

Usually not in a finite sample. Instead, it helps narrow a set of plausible candidates. Estimation, residual diagnostics, and forecast evaluation then provide additional evidence.

Is an MA coefficient at lag ten the same as MA(10)?

A subset model with only lag ten uses one MA coefficient. By contrast, an unrestricted MA(10) estimates ten MA coefficients. Therefore, state the exact lag set or software syntax to avoid ambiguity.

Conclusion: Solve the problem and verify its assumptions

A strong econometrics exam answer combines correct computation with a clear statement of what the data support. The OLS example shows how to calculate a complete regression and distinguish two prediction targets. Meanwhile, the demand example demonstrates why checking identities can reveal problems that a polished answer key overlooks.

The remaining exercises connect serial dependence, structural identification, and dynamic model selection. Together, they reinforce a practical habit: write the model, check the information, calculate carefully, and return to the assumptions before interpreting the result. That habit transfers directly from exam preparation to applied economic research.

Method note: Numerical results described as recalculated use the supplied observations or printed summary values. Reported AIC values come from the supplied answer page. Linked university materials and official software documentation support the methods; they do not independently validate the exercise’s numerical inputs. References checked November 9, 2026.