Which Topics Appear on the Midterm

Four Kinds of Questions and an Approach for Each

The Ten Problems

The Question: Which Variable Should Be Dropped?

                 Followers Posts Hashtags AvgCaptionLength
Followers             1.00  0.22     0.35             0.88
Posts                 0.22  1.00     0.41             0.20
Hashtags              0.35  0.41     1.00             0.29
AvgCaptionLength      0.88  0.20     0.29             1.00

    Followers Posts Hashtags AvgCaptionLength
VIF       6.1   1.5      1.9              5.8

VIF: The Other Regressors' Fit to $x_j$

Checking Each Answer Choice

1. aNo variables should be removed; all VIFs are under 10. 10 is one cutoff, but 6.1 and 5.8 are above 5, and the 0.88 correlation shows the two variables carry nearly the same information.
2. bRemove Posts and Hashtags because their VIFs are low. A low VIF is good: the variable adds information the other regressors do not have.
3. cKeep all; correlation does not imply multicollinearity. High correlation between regressors is what multicollinearity means. The phrase being borrowed is "correlation does not imply causation."
4. dRemove Hashtags because it is strongly correlated with Followers. Their correlation is 0.35, which is not strong, and the VIF of Hashtags is 1.9.
5. eFollowers and AvgCaptionLength are collinear and one of them should be dropped. Answer: (e)

The Question: The Analyst Drops Both Variables

Multicollinearity vs Omitted Variables

Checking Each Answer Choice

1. aHasChildren becomes unbiased; removing correlated predictors purges confounding. Removing a variable that affects $y$ and is related to HasChildren creates confounding. It does not remove it.
2. bStandard errors decrease, but joint explanatory power is unchanged. Income and Age explained part of enrollment. Without them, $R^2$ and the joint explanatory power fall.
3. cThe model gains efficiency, which corrects the remaining coefficient. HasChildren had VIF 1.2, so its standard error was not inflated much. Any gain in precision comes with bias.
4. eOnly $R^2$ falls; HasChildren is uncorrelated with Income and Age. VIF = 1.2 means $R^2 = 0.17$ with the others, so it is correlated with them.
5. dHasChildren absorbs the excluded variables' effects: biased and inconsistent. The efficiency gain does not offset the omitted variable bias. Answer: (d)

The Question: Training Error vs Test Error

1. aThe gap is negligible; an $R^2$ of 0.87 is sufficiently high that out-of-sample performance is guaranteed to be similar.
2. bThe gap proves the model is misspecified and must be re-estimated using the full sample before any conclusions can be drawn.
3. cA doubling of MAE on the test set is normal and expected; train/test gaps of this magnitude are standard in econometric models and require no further investigation.
4. dThe gap most likely reflects overfitting: the model has been tuned too closely to features of the training data which do not generalise to new observations.
5. eThe gap arises because MAE is not a valid out-of-sample metric; the student should report MSE instead, which would show no gap.

Add Terms: Training Error Falls, Test Error Does Not

Checking Each Answer Choice

1. aThe gap is negligible; $R^2 = 0.87$ guarantees similar out-of-sample performance. $R^2$ is measured on the training data. An overfit model has a high training $R^2$ by construction.
2. bThe gap proves misspecification; re-estimate on the full sample. Using the full sample removes the test set, so the gap can no longer be measured. A gap shows overfitting; it does not prove misspecification.
3. cDoubling of MAE is normal and needs no investigation. Test MAE 2.4 times training MAE is a warning sign. Some gap is normal; this one is large.
4. eMAE is not a valid out-of-sample metric; MSE would show no gap. MAE is a valid out-of-sample measure. MSE squares the same test errors, so it would show a gap too.
5. dOverfitting: the model is tuned too closely to features of the training data that do not generalize. Answer: (d)

The Question: Does the Variance Grow with Followers?

Goldfeld-Quandt: Compare the Variance in Two Groups

Run the Goldfeld-Quandt Test on Simulated Data

Checking Each Answer Choice

1. a$F_{GQ} = 1.46$; reject. The statistic is right, but $1.46 < 1.96$, so we fail to reject.
2. b$F_{GQ} = 0.69$; reject. $460/670 = 0.69$ puts the smaller variance on top. And 0.69 is not above any critical value.
3. d$F_{GQ} = 0.69$; fail to reject. The decision matches, but the ratio is upside down: the high-Followers group's SSE goes in the numerator.
4. eWe need more information. Both groups have the same $n - k = 46$, so the SSEs are enough.
5. c$F_{GQ} = 1.46$; fail to reject: no evidence of heteroskedasticity with respect to followers. Answer: (c)

The Question: What Is the FGLS Weight?

The FGLS Weight: Check the Units First

Checking Each Answer Choice

1. a1.0051; valid. $1.0051 = \hat\sigma^2$, the variance itself, not the weight. And the GQ test gave no reason to use FGLS.
2. b1.0025; invalid. $1.0025 = \hat\sigma$. The weight is $1/\hat\sigma$.
3. c0.0805; invalid. Uses 12,000 instead of 12. Followers is in 1,000s.
4. eNone of the weights are right. Choice (d) has the correct weight.
5. d0.9975; no, this transformation is invalid and worsens precision. Answer: (d)

The Question: Larger Robust SEs and Their Effect on $t$ and $p$

1. athe absolute value of the t-statistics will increase; p-values will decrease
2. bthe absolute value of the t-statistics will increase; p-values will remain the same
3. cBoth the absolute value of the t-statistics and p-values will remain unchanged
4. dthe absolute value of the t-statistics will decrease; p-values will decrease
5. ethe absolute value of the t-statistics will decrease; p-values will increase

Checking Each Answer Choice

The Question: Is the Variance Related to Truck Weight?

Call:
lm(formula = I(eh^2) ~ weight)

Coefficients:
            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

Breusch-Pagan: $N \times R^2$ From the Auxiliary Regression

Checking Each Answer Choice

1. bFail to reject; LM = 0.035. 0.035 is about the size of $R^2$ itself (0.033). The statistic is $N \times R^2 = 16.4$.
2. cReject; LM = 3.84. 3.84 is the critical value, not the statistic.
3. dFail to reject; LM = 16.4. $16.4 > 3.84$, so we reject.
4. eNeed the F-statistic. $N R^2$ is enough. The F-statistic is printed too and gives the same answer.
5. aReject $H_0$; LM = 16.4: heteroskedasticity is present with respect to weight. Answer: (a)

The Question: Is There Heteroskedasticity at the 1% Level?

lm(formula = res2 ~ age + sqft100)
            Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.0115081  0.0153786  -0.748   0.4544
age          0.0035429  0.0003100  11.429   <2e-16 ***
sqft100      0.0009758  0.0005287   1.846   0.0652 .
Residual standard error: 0.1734 on 1077 degrees of freedom
Multiple R-squared: 0.1082, Adjusted R-squared: 0.1066
F-statistic: 65.35 on 2 and 1077 DF,  p-value: < 2.2e-16

Checking Each Answer Choice

The Question: Which Standard Error Goes in the Interval?

Checking Each Answer Choice

The Question: Same Estimates, So Is OLS Fine?

1. 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.
2. bThe colleague is correct; when slope estimates are similar across OLS and WLS, heteroskedasticity is confirmed absent and OLS is BLUE.
3. cWLS always produces smaller standard errors than OLS by construction, so the difference in standard errors carries no informational content.
4. dWLS estimates are always biased when the variance function is misspecified, making OLS preferable under any uncertainty about the form of heteroskedasticity.
5. eOLS is preferred here because WLS is only appropriate when the sample size exceeds 1,000 observations.

Draw Many Samples: Compare the Spread of OLS and WLS

Checking Each Answer Choice

1. bSimilar estimates confirm no heteroskedasticity; OLS is BLUE. Under heteroskedasticity OLS is still unbiased, so similar estimates are expected either way. They say nothing about the variance.
2. cWLS SEs are always smaller by construction. WLS SEs are not smaller by construction: with badly wrong weights WLS can be less precise than OLS. The smaller WLS SE fits weights that are about right, but that SE assumes the weights are correct.
3. dWLS is always biased if the variance function is misspecified. Wrong weights cost efficiency, not unbiasedness, as long as the weights depend only on the regressors.
4. eWLS needs more than 1,000 observations. There is no such rule.
5. aSimilar estimates do not make OLS SEs valid; OLS is unbiased but inefficient, and the narrower WLS SE is a real efficiency gain. Answer: (a)

The Question: What Does the ADF Test Say?

Augmented Dickey-Fuller Test
Dickey-Fuller = -1.6311, Lag order = 2, p-value = 0.7298
alternative hypothesis: stationary

Checking Each Answer Choice

The Question: Which Autocorrelations Are Significant?

Testing One Autocorrelation: $r_k\sqrt{T}$

Checking Each Answer Choice

The Question: Which Model Produced This Series?

Set $\phi_1$ and $\phi_2$: Watch the ACF and PACF

Checking Each Answer Choice

1. aACF cuts off after lag 1; MA(1). The ACF decays over several lags; it is the PACF that cuts off after lag 1.
2. bLags 11–13 cross the bounds; ARIMA(12,0,0). The PACF has one spike, so one AR term is enough. ACF bars far out belong to the gradual decay that an AR(1) produces (or to chance: with a 5% band, about 1 lag in 20 crosses), so they do not justify 12 AR terms.
3. cBoth decay; ARIMA(1,0,1). The PACF does not decay: it has one spike and then stays inside the bounds.
4. dACF and PACF are insufficient; always use AIC and BIC. The plots are the standard first step for choosing the order; AIC and BIC are a check, not a requirement.
5. eOne PACF spike at lag 1 and a gradually decaying ACF: ARIMA(1,0,0), with ARIMA(1,0,1) if residual checks show remaining structure. Answer: (e)

The Question: A 95% Prediction Interval for $\hat Y_{t+1}$

Forecast, Then Add $\pm 1.96\,\sigma$

Checking Each Answer Choice

1. b[2.32, 3.28] Uses the variance 0.25 in place of $\sigma = 0.5$.
2. c[1.32, 3.28] Centered at 2.30, not at the forecast 2.8.
3. d[2.02, 3.98] Centered at 3.0, the current value $Y_t$, instead of the forecast.
4. eNone are correct. Choice (a) is correct.
5. a[1.82, 3.78] $= 2.8 \pm 1.96 \times 0.5$ Answer: (a)

The Question: Effect of Autocorrelated Errors on the Forecast

Breusch-Godfrey test for serial correlation of order up to 2
data:  ar1_model
LM test = 15.152, df = 2, p-value = 0.000514

Breusch-Godfrey Test: A Lagged $Y$ Makes OLS Inconsistent

Checking Each Answer Choice

1. a(ii) and (iv) only Leaves out (i): with a lagged dependent variable, autocorrelation makes OLS inconsistent.
2. b(iii) and (v) only Both are false: the forecast is affected, and $R^2$ does not help.
3. c(i) and (iii) only (i) and (iii) contradict each other.
4. e(i), (ii), (iv), and (v) only (v) is false.
5. d(i), (ii), and (iv) only Answer: (d)

The Question: Which Is Not a Correction for Autocorrelation?

1. aApplying HAC (Newey-West) robust standard errors to account for the serial correlation in inference.
2. bAdding lagged dependent variables to the model to directly capture the remaining autocorrelation structure.
3. cApplying Cochrane-Orcutt feasible GLS, which estimates the autocorrelation parameter in the error term and re-estimates the transformed model.
4. dFirst-differencing the series
5. eApplying Weighted Least Squares (WLS) using the inverse of the squared residuals as weights

Checking Each Answer Choice

The Question: Write the ARDL(1,3) With the Lag Operator

Checking Each Answer Choice

The Question: The Rate of Decline and Stability

Checking Each Answer Choice

The Question: The Total (Long-Run) Multiplier

Move $\lambda$: Add Up the Lag Weights

Checking Each Answer Choice

1. a0.5 0.5 is $\lambda$, the ratio between weights.
2. b1.2 1.2 is the intercept. It does not respond to $X$.
3. d1.8 Not the sum of the lag weights: $0.8 + 0.4 + 0.2 + 0.1 + \cdots = 1.6$.
4. eNone are correct. Choice (c) is correct.
5. c$1.6 = 0.8/(1 - 0.5)$ Answer: (c)

The Question: Which Way Does Granger Causality Run?

1. aBoth advertising expenditure and sales revenue Granger cause each other so we need a VAR model to represent this relationship.
2. bAdvertising expenditure Granger causes sales revenue, but sales revenue does not Granger cause advertising expenditure.
3. cSales revenue Granger causes advertising expenditure, but advertising expenditure does not Granger cause sales revenue.
4. dNeither advertising expenditure nor sales revenue Granger cause each other.
5. eThere is no significant relationship between advertising expenditure and sales revenue.

Checking Each Answer Choice

The Question: What Does the Logit Coefficient 0.6 Mean?

Probability, Odds, and Log-Odds

Checking Each Answer Choice

1. a(i) only 0.6 is the change in log-odds. The odds ratio is $e^{0.6} = 1.82$.
2. b(ii) only A logit coefficient is not a marginal effect on probability.
3. c(iii) only (iv) is also correct: it states the same fact on the log-odds scale.
4. d(iv) only (iii) is also correct: it states the same fact on the odds scale.
5. e(iii) and (iv) only Answer: (e)

The Question: Probability and Marginal Effect at Income = 30

The Marginal Effect Is the Slope of the Curve

Checking Each Answer Choice

1. aPP = 0.289; ME = 0.040 0.289 is $\Lambda(-0.9)$, not the probability at $z = -0.3$; 0.040 is the coefficient.
2. bPP = 0.289; ME = 0.008 Wrong probability, and 0.008 is the LPM's Income coefficient.
3. dPP = 0.426; ME = 0.040 Right probability, but the ME multiplies 0.04 by $\Lambda(1 - \Lambda) = 0.245$.
4. eNone are correct. Choice (c) is correct.
5. cPP = 0.426; ME = 0.010 Answer: (c)

The Question: Is an LPM Prediction of 0.22 Fine?

Checking Each Answer Choice

Three Ways to Report a Marginal Effect

The Question: Why Is the AME Smaller Than the MEM?

1. aThe AME must be computed incorrectly: in any model, the average of the marginal effects equals the marginal effect at the average.
2. bAt the mean DTIME, $z$ is close to 0, where $\phi$ is largest. Many people have DTIME far from 0, where $\phi(z_i)$ is small, so their marginal effects pull the average down.
3. cThe AME uses the logit density, which is flatter than the normal density.
4. dThe MEM is larger because it ignores the sign of the coefficient.
5. eThe two numbers measure different variables.

Checking Each Answer Choice

1. aThe average of MEs equals the ME at the average. True only when the marginal effect is the same for everyone, as in the LPM. Here it depends on $\phi(z_i)$.
2. cThe AME uses the logit density. Both use the probit density $\phi$.
3. dThe MEM ignores the sign. Both are positive because the coefficient 0.3 is positive.
4. eDifferent variables. Both are the marginal effect of DTIME.
5. b$\phi$ is largest near $z = 0$; people far from 0 have small marginal effects and pull the average down. Answer: (b)

The Question: The LPM Prediction for 5 Ads and 20 Views

Checking Each Answer Choice

The Question: The Probit Prediction for the Same User

Checking Each Answer Choice

The Question: The Effect of One More Ad

Two Ways to Get the Effect of One More Ad

Checking Each Answer Choice

1. c0.05 Not either method.
2. d0.06 Not either method.
3. eNone are correct. 0.03 and 0.04 are both correct, by the two methods.
4. a0.03: the difference $\Phi(1.8) - \Phi(1.5)$ Accepted
5. b0.04: the derivative $\phi(1.5) \times 0.3$ Accepted

The Question: The Probability of Failing

1. a$\Phi(0.5)$
2. b$\phi(0.5)$
3. c$\Phi(-0.5)$
4. d$\phi(2)$
5. eNone of the given answer options are correct.

Checking Each Answer Choice

The Question: Which Likelihood Is Maximized?

1. a$L = \sum_{i=1}^{334} [\Phi(\beta_0 + \beta_1 x_i)]^{y_i} + [1 - \Phi(\beta_0 + \beta_1 x_i)]^{1 - y_i}$
2. b$L = \prod_{i=1}^{334} [\Phi(\beta_0 + \beta_1 x_i)]^{y_i}\,[1 - \Phi(\beta_0 + \beta_1 x_i)]^{1 - y_i}$
3. c$L = \sum_{i=1}^{334} y_i \log[\Phi(\beta_0 + \beta_1 x_i)] + (1 - y_i)\log[1 - \Phi(\beta_0 + \beta_1 x_i)]$
4. d$L = \prod_{i=1}^{334} [\Phi(\beta_0 + \beta_1 x_i)]^{y_i}$

Checking Each Answer Choice

The Question: Rank the Models by Accuracy

Checking Each Answer Choice

The Question: Which Model Catches the Most Cases?

Precision, Recall, Specificity, F1

Checking Each Answer Choice

The Question: Which Model Performs Best?

Move the Threshold: One ROC Curve, Many Confusion Matrices

Checking Each Answer Choice

The Question: Match Each Scenario to a Model

1. aScenario 1: Ordered Logit; Scenario 2: Binary Logit
2. bScenario 1: Multinomial Logit; Scenario 2: Poisson Regression
3. cScenario 1: Binary Probit; Scenario 2: Multinomial Probit
4. dScenario 1: Rank-Ordered Logit; Scenario 2: Count Model
5. eScenario 1: Poisson Regression; Scenario 2: Ordered Logit

Match the Dependent Variable to the Model

Checking Each Answer Choice

1. aOrdered logit; binary logit. Visits are a count, and the job offers are not a yes/no outcome.
2. bMultinomial logit; Poisson. Reversed: Poisson is for the count.
3. cBinary probit; multinomial probit. Visits are not 0/1, and multinomial models ignore the ranking.
4. dRank-ordered logit; count model. Reversed.
5. eScenario 1: Poisson regression; Scenario 2: ordered logit Answer: (e)

The Question: A New Plan Changes the Old Choices

1. aThe model violated the assumption of linearity in parameters.
2. bThe new choice introduces perfect multicollinearity.
3. cThe independence of irrelevant alternatives (IIA) assumption may not hold.
4. dThe logistic model overfits the original set of insurance options.
5. eThe multinomial logit model is appropriate to use

Checking Each Answer Choice

The Question: Speedometers Stop at 85 mph

1. aOLS, because top speed is a continuous variable
2. bTobit, because top speed is censored from above at 85
3. cTruncated regression, because cars above 85 mph are missing from the data
4. dPoisson, because speed takes whole-number values
5. eOrdered probit, because faster cars rank above slower cars

Checking Each Answer Choice

The Question: The DiD Estimate from a 2×2 Table

Two Differences: Over Time, Then Between Cities

Checking Each Answer Choice

1. a−5 LA's own change: ignores the common trend.
2. b−1 Humboldt's change: the control group's trend, not the effect.
3. d−6 $75 - 81$: the after-only gap, which includes the 2-point gap that existed before.
4. e+4 Wrong sign: LA fell by more than Humboldt.
5. c−4 percentage points Answer: (c)

The Question: The Predicted Post-Bill Wage in a High-Funded State

Each Coefficient Is a Cell or a Difference

Checking Each Answer Choice

1. a23.5 $\beta_0 + \beta_1$: the low-funding state after the bill.
2. b26.5 $\beta_0 + \beta_1 + \beta_2$: leaves out $\delta$.
3. c27.8 $\beta_0 + \beta_2 + \delta$: leaves out the common trend $\beta_1$.
4. eNone are correct. Choice (d) is correct.
5. d$29.3 = 22.0 + 1.5 + 3.0 + 2.8$ Answer: (d)

The Question: The High-Funded State's Own Change

Checking Each Answer Choice

The Question: High-Funded States Were Already Growing Faster

1. aThis challenges the parallel trends assumption; it can be assessed by plotting pre-treatment wage trends for both groups and, if sufficient pre-periods exist, testing for differential pre-trends.
2. bThis describes omitted variable bias in the cross-section; it can be resolved by adding more control variables to the post-period regression only.
3. cThis is a measurement error problem; wages in high-funded states are systematically mismeasured prior to treatment, biasing our difference-in-difference estimate downward.
4. dThis describes reverse causality: high pre-treatment wages caused states to receive high funding, which is corrected by instrumenting for HighFund using geographic allocation formulas.
5. eThe parallel trends assumption only applies to the post-period, so pre-existing differences in growth rates are irrelevant to the validity of the DiD estimate.

Break Parallel Trends: The DiD Estimate Includes the Extra Growth

Checking Each Answer Choice

1. bOmitted variable bias; add controls to the post period. The problem is a difference in trends before treatment; controls in one period cannot reveal or remove it.
2. cMeasurement error. Nothing suggests wages were measured wrong.
3. dReverse causality; use IV. The critique is about growth rates before the bill, not about how funding was assigned.
4. eParallel trends only applies after treatment. It is about what would have happened without treatment; pre-period trends are the only evidence we can see.
5. aParallel trends is in doubt: plot pre-treatment trends and test for different pre-trends. Answer: (a)

Summary of the Ten Problems