Does the error variance change with the regressors?
Regress $\hat\varepsilon_i^2$ on the regressors, then compare $N \times R^2_{\text{aux}}$ to a $\chi^2$ critical value.
Spring 2026 Midterm (Boswell), Question 17
2
Goldfeld-Quandt test
Is the error variance larger at one end of the data?
Sort, split into two groups, then compare the two error variances with an $F$ test.
Spring 2026 Final (Boswell), Question 29
3
OLS vs WLS
What do similar point estimates and different standard errors tell us?
Both estimators are unbiased. Only the variances differ.
Spring 2026 Midterm (Boswell), Question 19
Problem 1 · Breusch-Pagan test
The Question: Fuel Consumption and Truck Weight
A logistics company analyzes fuel consumption $Y_i$ based on truck weight $X_i$:
$Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i$
The Breusch-Pagan auxiliary regression is:
Call: lm(formula = I(eh^2) ~ weight)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.77682 0.05336 14.56 < 2e-16 ***
weight 0.22286 0.05422 4.11 4.63e-05 ***
Residual standard error: 1.193 on 498 degrees of freedom
Multiple R-squared: 0.03281, Adjusted R-squared: 0.03086
F-statistic: 16.89 on 1 and 498 DF, p-value: 4.627e-05
If $N = 500$ and the 5% critical value is $\chi^2_{(1)} = 3.84$, what is your conclusion?
aReject $H_0$; LM $= 16.4$, so heteroskedasticity is present.
bFail to reject $H_0$; LM $= 0.035$, not enough evidence to show homoskedasticity is present with respect to weight.
cReject $H_0$; LM $= 3.84$, so heteroskedasticity is present.
dFail to reject $H_0$; LM $= 16.4$, not enough evidence to show homoskedasticity is present with respect to weight.
eCannot conclude; need the F-statistic instead.
Source: Spring 2026 Midterm (Boswell), Question 17
Breusch-Pagan · The idea
Does the Spread Around the Line Increase with Weight?
The dataFuel consumption $Y_i$ against truck weight $X_i$, with the OLS line $\hat Y_i = \hat\beta_0 + \hat\beta_1 X_i$.
Spread around the lineEach bar runs two standard deviations above and below the line, so it covers about 95% of the trucks at that weight. The bars get longer as weight rises.
GoalDecide whether the spread increases with weight. We can see it in the picture, but we want a formal rule that answers yes or no. The Breusch-Pagan test is that rule.
Illustration: 200 trucks whose error variance rises with weight.
Breusch-Pagan · Step 1
Regress the Squared Residuals on Weight
ResidualsThe residual $\hat\varepsilon_i = Y_i - \hat Y_i$ is the vertical distance from truck $i$ to the OLS line.
Plot the residualsPlot $\hat\varepsilon_i$ against weight. The OLS line becomes the zero line; the spread is unchanged.
Square them$E(\varepsilon_i^2) = \sigma_i^2$, so $\hat\varepsilon_i^2$ tracks the error variance of truck $i$.
Auxiliary regressionRegress $\hat\varepsilon_i^2$ on weight: $\hat\varepsilon_i^2 = 0.777 + 0.223\,\text{weight}_i$
$H_0$: $\alpha_2 = 0$ $\iff$ flat line $\iff$ homoskedastic$H_1$: $\alpha_2 \neq 0$ $\iff$ the variance changes with weight
Breusch-Pagan · Step 2
Compute the Test Statistic
Call: lm(formula = I(eh^2)~ weight)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.77682 0.05336 14.56 < 2e-16 ***
weight 0.22286 0.05422 4.11 4.63e-05 ***
Residual standard error: 1.193 on 498 degrees of freedom
Multiple R-squared: 0.03281, Adjusted R-squared: 0.03086
F-statistic: 16.89 on 1 and 498 DF, p-value: 4.627e-05
lm( )linear model: estimate by OLSeh“e hat”: the residuals $\hat\varepsilon_i$ saved from the original regressionI(eh^2)square the residuals: the dependent variable is $\hat\varepsilon_i^2$~ weightregressed on weight $\implies$ this is the auxiliary regression
On the exam: output like this may appear without any explanation. Recognize I(eh^2) ~ weight as the Breusch-Pagan auxiliary regression.
Use the Multiple $R^2$, not the adjusted $R^2$.
Formula
$LM = N \times R^2_{\text{aux}} \sim \chi^2_{(S-1)}$
Plug in
$LM = 500 \times 0.03281 = 16.4$
Degrees of freedom$S - 1$ counts the regressors in the auxiliary regression. The intercept is not counted. One regressor (weight) $\implies$ $\chi^2_{(1)}$.
Breusch-Pagan · Step 3
Compare $LM$ to the $\chi^2_{(1)}$ Distribution
Under $H_0$$LM$ follows a $\chi^2_{(1)}$ distribution. Most draws are small.
Rejection regionOnly 5% of the distribution lies above $3.84$.
Our statistic$LM = 16.4$ lies far inside the rejection region ($p \approx 0.00005$).
REJECT $H_0$$16.4 > 3.84$ $\implies$ the error variance changes with truck weight.
Breusch-Pagan · Interpretation
Checking Each Answer Choice
bFail to reject $H_0$; LM $= 0.035$, not enough evidence to show homoskedasticity is present with respect to weight.$0.035$ is about $R^2$ itself. $LM$ multiplies it by $N = 500$.
cReject $H_0$; LM $= 3.84$, so heteroskedasticity is present.$3.84$ is the critical value, not the test statistic.
dFail to reject $H_0$; LM $= 16.4$, not enough evidence to show homoskedasticity is present with respect to weight.Right statistic, wrong decision: $16.4 > 3.84$.
eCannot conclude; need the F-statistic instead.The $F$-statistic is printed and agrees: $F = t^2 = 4.11^2 \approx 16.9$, $p = 4.6 \times 10^{-5}$.
aReject $H_0$; LM $= 16.4$, so heteroskedasticity is present.Answer: (a)
WordingFailing to reject would mean no evidence of heteroskedasticity, not “evidence of homoskedasticity.”
InterpretationThe slope on weight is positive, so the error variance rises with truck weight. The usual OLS standard errors are invalid $\implies$ use robust standard errors, or model the variance with WLS.
Problem 2 · Goldfeld-Quandt test
The Question: R&D Spending and Firm Size
The original regression: $\text{r\&d}_i = \alpha + \beta\, \text{size}_i + \gamma\, X_i + \varepsilon_i$
A researcher estimates $\widehat{\text{r\&d}}_i = \hat\alpha + \hat\beta\, \text{size}_i + \hat\gamma\, X_i$ to find the determinants of R&D spending, using a cross-section of 40 companies. Observations are sorted by size, the middle 4 observations dropped, and OLS is estimated separately on the large size group ($RSS_1 = 13{,}984.61$) and the small size group ($RSS_2 = 5{,}708.51$).
At the 5% level, $F_{(18,18),\,0.95} = 2.217$. If the GQ test is carried out, the correct conclusion is:
aReject $H_0$; the large size group has significantly larger variance.
bFail to reject $H_0$; there is insufficient evidence of heteroskedasticity of this form.
cReject $H_0$; the model is misspecified because the slope differs across groups.
dFail to reject $H_0$; the error term is homoskedastic with certainty.
eReject $H_0$; the small size group has significantly larger variance.
Source: Spring 2026 Final (Boswell), Question 29
Goldfeld-Quandt · The idea
Is the Spread Around the Line Larger for Large Firms?
The dataR&D spending against firm size for 40 companies, with the OLS line.
Spread at each endEach bar runs two standard deviations above and below the line. The bar for the smallest firms is short; the bar for the largest firms is long.
GoalDecide whether the spread is larger among large firms. The Goldfeld-Quandt test answers yes or no by comparing the error variance of the smallest firms with the error variance of the largest firms.
Illustration: 40 firms whose error variance rises with size.
Goldfeld-Quandt · Step 1
Sort by Size, Drop the Middle, Fit Each Group
Start40 firms, in the order they appear in the data.
SortOrder the firms by size, the variable we suspect drives the variance.
Drop the middleRemove the middle 4 firms. This leaves 18 small and 18 large firms.
Fit each groupRun the regression separately on each group.
Residual sums of squaresAdd up the squared residuals in each group:
$RSS_2 = 5{,}708.51$ (small),
$RSS_1 = 13{,}984.61$ (large).
Illustration: 40 firms whose error variance rises with size, plotted against size only.
Goldfeld-Quandt · Step 2
Compute the Test Statistic
RSSresidual sum of squares: $RSS = \sum \hat\varepsilon_i^2$, added up within one groupRSS₁the 18 largest firms: $13{,}984.61$RSS₂the 18 smallest firms: $5{,}708.51$
On the exam: the same quantity may be called the sum of squared errors (SSE) or the sum of squared residuals. It is always $\sum \hat\varepsilon_i^2$.
Hypotheses$H_0$: $\sigma^2_{\text{large}} = \sigma^2_{\text{small}}$ vs $H_1$: $\sigma^2_{\text{large}} > \sigma^2_{\text{small}}$
Variance in each groupDivide by the degrees of freedom: $\hat\sigma^2_j = RSS_j / \text{df}_j$.
StatisticPut the group suspected of the larger variance in the numerator. Both groups have the same degrees of freedom, so they cancel:
Each group has $18 - 3 = 15$ degrees of freedom (18 firms, 3 coefficients) $\implies$ $GQ \sim F_{(15,15)}$ under $H_0$. The exam printed $F_{(18,18),\,0.95} = 2.217$, which uses the wrong degrees of freedom. The decision is the same with either critical value.
Goldfeld-Quandt · Step 3
Compare $GQ$ to the $F_{(15,15)}$ Distribution
Under $H_0$The two variances are equal, so the ratio is close to 1.
Rejection regionOnly 5% of the distribution lies above $2.40$.
Our statistic$GQ = 2.450$: the large-firm variance is about 2.5 times the small-firm variance.
REJECT $H_0$$2.450 > 2.40$ $\implies$ the error variance rises with firm size.
Goldfeld-Quandt · Interpretation
Checking Each Answer Choice
bFail to reject $H_0$; there is insufficient evidence of heteroskedasticity of this form.The original answer key printed (b), but $2.450$ exceeds both $2.40$ and $2.217$.
cReject $H_0$; the model is misspecified because the slope differs across groups.GQ compares error variances, not slopes.
dFail to reject $H_0$; the error term is homoskedastic with certainty.A test never establishes $H_0$ “with certainty.”
eReject $H_0$; the small size group has significantly larger variance.The large group is in the numerator, and the ratio is above 1.
aReject $H_0$; the large size group has significantly larger variance.Answer: (a)
InterpretationR&D spending is more variable among large firms than among small firms. The usual OLS standard errors are invalid $\implies$ use robust standard errors, or WLS with a variance that rises with size.
Problem 3 · OLS vs WLS
The Question: Similar Estimates, Different Standard Errors
The original regression: $\text{wage}_i = \beta_0 + \beta_1\, \text{educ}_i + \varepsilon_i$
An economist estimates a cross-sectional wage model and suspects heteroskedasticity. Comparing OLS and WLS (weighted by firm size), the slope on education is $\hat\beta_{OLS} = 1.42$ with standard error $0.31$, while $\hat\beta_{WLS} = 1.39$ with standard error $0.180$. A colleague argues that because the two point estimates are nearly identical, heteroskedasticity is not a concern and OLS should be preferred for simplicity. Which best identifies the flaw in this reasoning?
aSimilarity in point estimates does not imply OLS standard errors are valid; heteroskedasticity leaves OLS coefficients unbiased but renders standard errors inefficient, and the narrower WLS standard error reflects a genuine efficiency gain.
bThe colleague is correct; when slope estimates are similar across OLS and WLS, heteroskedasticity is confirmed absent and OLS is BLUE.
cWLS always produces smaller standard errors than OLS by construction, so the difference in standard errors carries no informational content.
dWLS estimates are always biased when the variance function is misspecified, making OLS preferable under any uncertainty about the form of heteroskedasticity.
eOLS is preferred here because WLS is only appropriate when the sample size exceeds 1,000 observations.
Source: Spring 2026 Midterm (Boswell), Question 19
OLS vs WLS · An aside
Whose Advice Do You Trust?
The questionHow many drinks should you have tonight? You ask 8 friends: 4 responsible friends and 4 party animals.
Treat everyone equallyAverage all 8 answers: drinks. The party animals pull the average up, and you overdo it.
Reweight the adviceThe party animals' answers swing far from sensible, so give them less weight (smaller dots). The weighted average is drinks.
Ask 100 friend groupsEvery group gives different advice. The equal averages pile up around 6.5 drinks and spread out; the weighted averages pile up near 3 and stay close together.
OLS vs WLS · The idea
WLS Gives More Weight to Observations with Small Error Variance
The dataWage against education, with the OLS line. OLS gives every worker the same weight.
Firm sizePoint size shows firm size. Workers at large firms (big points) stay close to the line; workers at small firms scatter far from it $\implies$ heteroskedasticity. WLS gives the large-firm workers more weight.
WLS lineThe weighted line is close to the OLS line: both estimate the same $\beta_1$.
GoalThe two point estimates barely differ, but the standard errors do. Decide what that comparison says about heteroskedasticity.
Illustration: 100 workers; the error variance is smaller at larger firms.
OLS vs WLS · Step 1
Two Questions About Each Estimator
The same 100 workers; point size shows firm size. Dashed gray: OLS line (slope 1.42). Blue: WLS line (slope 1.39).
sestandard error: how much the estimate would move from one sample to the next
Right on average?(unbiased)
Moves from sample to sample?(standard error)
OLS
Yes 1.42
se $= 0.31$
WLS
Yes 1.39
se $= 0.18$
Left columnThe two lines nearly coincide: both estimators are right on average.
The colleague's mistakeHeteroskedasticity changes only the right column. The colleague looked only at the left column.
OLS vs WLS · Step 2
Repeat the Study 200 Times
One sample100 workers; point size shows firm size. Big points stay near the line, small points scatter. OLS line: dashed gray. WLS line: blue.
New samplesEach new sample gives new lines. The OLS lines swing more than the WLS lines.
True lineBoth sets of lines center on the true line (black) $\implies$ both estimators are unbiased.
SlopesEach line's slope becomes one dot in its row.
SpreadThe width of each pile is the standard error: 0.32 for OLS, 0.17 for WLS (exam: 0.31 and 0.18).
The exam's 1.42 and 1.39 are one sample: the first one shown here. Both piles land near $\beta_1$, so similar estimates are expected.
OLS vs WLS · Step 3
With Constant Error Variance, WLS Cannot Be More Precise Than OLS
One sampleSame firms and weights, but now every worker has the same error variance: big and small points scatter equally.
New samplesNow the WLS lines swing as much as the OLS lines, or more.
SlopesOLS is BLUE here $\implies$ the WLS pile cannot be narrower.
SpreadThese weights do not match the error variance here, so the WLS pile is wider: 0.47 against 0.27 for OLS.
CONCLUSION
If the error variance were constant, the WLS standard error could not be smaller than the OLS standard error.
On the exam it is much smaller (0.18 against 0.31) $\implies$ evidence of heteroskedasticity.
OLS vs WLS · Interpretation
Checking Each Answer Choice
bThe colleague is correct; when slope estimates are similar across OLS and WLS, heteroskedasticity is confirmed absent and OLS is BLUE.Both estimators are unbiased under heteroskedasticity, so similar slopes are expected either way.
cWLS always produces smaller standard errors than OLS by construction, so the difference in standard errors carries no informational content.With constant error variance, the WLS pile was not narrower (Step 3).
dWLS estimates are always biased when the variance function is misspecified, making OLS preferable under any uncertainty about the form of heteroskedasticity.WLS is unbiased with any weights, as long as $E(\varepsilon_i \mid x_i) = 0$. Wrong weights can give a larger variance than OLS.
eOLS is preferred here because WLS is only appropriate when the sample size exceeds 1,000 observations.No such rule exists. (The original answer key printed (e), a typo later corrected to (a).)
aSimilarity in point estimates does not imply OLS standard errors are valid; heteroskedasticity leaves OLS coefficients unbiased but renders standard errors inefficient, and the narrower WLS standard error reflects a genuine efficiency gain.Answer: (a)
In plain wordsThe estimates agree, but WLS is more precise, and that difference is evidence of heteroskedasticity.
Careful with the wording of (a)Standard errors are not “inefficient.” Under heteroskedasticity the OLS estimator is inefficient, and its usual standard errors are invalid.