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Canvas Example MCQ — Q89

• A) approximately 30.9%
• B) approximately 50.0%
• C) approximately 69.1%
• D) approximately 84.1%
• E) None of A-D is close; write the correct answer.

z = -4.60 + .40(12) - .20(2) + .10(7) = 0.50. P = Φ(0.50) ≈ 0.6915 = 69.15%.

Rule: Compute the index first; then apply the standard normal CDF.

Canvas Example MCQ — Q90

• A) approximately -20.0 percentage points
• B) approximately -15.2 percentage points
• C) approximately -7.0 percentage points
• D) approximately +7.0 percentage points
• E) None of A-D is close; write the correct answer.

ME = -0.20 x 0.3521 = -0.07042, i.e. about -7.04 percentage points for a one-unit local increase in MARGIN.

Rule: Probit ME = coefficient x normal PDF at the current index.

Canvas Example MCQ — Q91

• A) approximately -7.0 percentage points
• B) approximately -15.2 percentage points
• C) approximately -20.0 percentage points
• D) approximately +15.2 percentage points
• E) None of A-D is close; write the correct answer.

At MARGIN=2, z=.50 -> P=.6915. At MARGIN=4, z=.10 -> P=.5398. Change = .5398-.6915 = -.1517 = -15.17 pp.

Rule: For a discrete nonlinear change, calculate both probabilities and subtract.

Old Exam — Oct 2023, Q15 · adapted

Given: Probit Model 1 output: NETWORTH coefficient 0.0864, t ≈ 2.44, p-value ≈ 0.017.

• A) No
• B) Yes
• C) Yes because the p-value is higher than 2 %.
• D) No because the p-value lower than 2 %.

p for NETWORTH is about 0.017. Since 0.017<0.02, significant at 2% -> B.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2023, Q16 · adapted

Given: Probit Model 1: z = −7.0166 + 0.5301·FIXRATE − 0.2675·MARGIN + 0.0864·NETWORTH; P(ARM) = Φ(z); marginal effect of X = b_X·φ(z).

• A) ‒0.0030
• B) ‒0.0092
• C) ‒0.0343
• D) ‒0.2675
• E) None of the above (A, B, C, D) is close to be correct.

Model 1: z=-7.0166+0.5301(10)-0.2675(3)+0.0864(3.5). Then ME_MARGIN=-0.2675 phi(z)≈-0.0092 -> B.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2022, Q21

Given: Probit Model 1 from course: NETWORTH coefficient≈0.0864, SE≈0.0354, so t≈2.44 and p≈0.017.

• A) No
• B) Yes
• C) Yes because the p-value is higher than 1%
• D) No because the p-value lower than 1 %

NETWORTH t≈0.0864/0.0354≈2.44, giving p≈0.017. Since 0.017>0.01, it is not significant at 1% -> A.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2022, Q22

Given: Probit Model 1: z=-7.0166+0.5301 FIXRATE-0.2675 MARGIN+0.0864 NETWORTH.

• A) 0.1762
• B) 0.3707
• C) 0.6837 This is when FIXRATE = 15, which is not what the question is about. Correct answer is 0.478 with FIXRATE = 14, therefore D is the correct answer.
• D) 0.4780
• E) None of the above (A, B, C, D) is close to be correct.

z=-7.0166+0.5301(14)-0.2675(3)+0.0864(4)≈-0.055. Phi(z)≈0.478 -> D.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2022, Q23

Given: Probit Model 2: z=-9.8200+0.7535 FIXRATE-0.1945 MARGIN; ME_MARGIN=b_MARGIN x phi(z).

• A) ‒0.0202
• B) ‒1.641
• C) 0.1038
• D) 0.039
• E) None of the above (A, B, C, D) is close to be correct.

Model 2: z=-9.82+0.7535(11.5)-0.1945(2.5)≈-1.64. phi(z)≈0.1038. ME_MARGIN=-0.1945x0.1038≈-0.0202 -> A.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2022, Q24 · adapted

Given: Probit Model 1: z = −7.0166 + 0.5301·FIXRATE − 0.2675·MARGIN + 0.0864·NETWORTH; P(ARM) = Φ(z); marginal effect of X = b_X·φ(z).

• A) ‒0.092
• B) ‒0.0441
• C) 0.0142
• D) 0.039
• E) None of the above (A, B, C, D) is close to be correct.

Model 1: compute z at FIXRATE=11.5, MARGIN=2.5, NETWORTH=3; then ME_NETWORTH=0.0864 phi(z)≈0.0142 -> C.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Dec 2021 re-exam, Q30

Given: Average marginal effect = calculate each observation's marginal effect, then average them.

• A) Taking the marginal effect at the mean of the sample.
• B) Mean marginal effects are always 0.5 in a binary choice model.
• C) Calculate the marginal effects for each observation; then take the mean.
• D) Take the mean of the marginal effects at the 5th and 95th percentile.
• E) None of the above (A, B, C, D) is correct.

AME = (1/n) sum_i ME_i: compute the marginal effect for every observation, then average -> C.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2021, Q25

Given: Probit: z=-0.0788+0.400 DTIME, with DTIME=1.5.

• A) 0.699
• B) 0.627
• C) 0.473
• D) 0.593
• E) None of the above (A, B, C, D) is close to be correct.

z=-0.0788+0.400(1.5)=0.5212. Phi(0.5212)≈0.699 -> A.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Old Exam — Oct 2021, Q26

Given: Probit marginal effect: beta x phi(z) at DTIME=1.5.

• A) 0.348
• B) 0.139
• C) 0.307
• D) 0.123
• E) None of the above (A, B, C, D) is close to be correct.

Probit ME=beta phi(z)=0.400 phi(0.5212). phi(0.5212)≈0.348, so ME≈0.139 -> B.

Rule: Phi(.) gives probability; phi(.) enters Probit marginal effects. Raw Probit coefficients are not percentage-point effects.

Computer Lab 6 — Exercise 3, Model 1 signs

• A) All three: FIXRATE, MARGIN, and NETWORTH
• B) Only NETWORTH
• C) None are significant
• D) Only FIXRATE and NETWORTH

All three absolute t-statistics (2.09, 2.05, 2.44) exceed the critical value of 1.993, so all three coefficients are statistically significant at the 5% level.

Rule: Compare |t| to the critical value for each coefficient individually to assess significance.

Computer Lab 6 — Exercise 3, Model 1 interpretation · adapted

• A) A higher ARM margin makes choosing an adjustable-rate mortgage less attractive (negative sign); the raw coefficient is NOT a direct percentage-point probability effect
• B) A higher ARM margin makes an ARM more attractive, and -0.2675 means exactly a 26.75 percentage-point decrease in probability
• C) The coefficient has no interpretable meaning in a Probit model
• D) -0.2675 is the exact percentage-point change in probability for a one-unit increase in MARGIN

The negative sign indicates direction (higher MARGIN lowers the attractiveness/probability of choosing ARM), but Probit coefficients are index-scale, not probability-scale; the actual probability effect requires multiplying by φ(Xβ), the marginal effect formula.

Rule: Probit coefficient signs indicate direction only; the probability-scale marginal effect requires β_j×φ(Xβ).

Computer Lab 6 — Exercise 3, Task c

• A) ≈17.59%
• B) ≈82.41%
• C) ≈37.21%
• D) ≈93.12%

P = Φ(-0.9312) ≈ 0.1759 = 17.59%, reading the standard normal CDF at z = -0.9312.

Rule: Compute z=Xβ first, then look up Φ(z) for the predicted probability.

Computer Lab 6 — Exercise 3, Task c comparison

• A) +19.62 percentage points
• B) +211.6% relative change only, not expressible in percentage points
• C) -19.62 percentage points
• D) +1.962 percentage points

Change = 37.21% − 17.59% = 19.62 percentage points, consistent with the positive NETWORTH coefficient.

Rule: For discrete changes in nonlinear probability models, compute new probability minus old probability directly.

Computer Lab 6 — Exercise 3, Tasks e/f

• A) ≈-3.93 percentage points per unit (-0.0393)
• B) ≈-19.45 percentage points per unit
• C) ≈-1.167 percentage points per unit
• D) ≈+3.93 percentage points per unit

ME_MARGIN = (-0.1945)(0.2019) ≈ -0.0393, i.e., about a 3.93 percentage-point decrease in probability per one-unit increase in MARGIN, locally at this X.

Rule: Probit marginal effect for continuous X: ME = β_j × φ(Xβ), evaluated at specific covariate values.

Computer Lab 6 — Exercise 3, Task f discrete

• A) -6.18 percentage points; Probit is nonlinear so the marginal effect changes as X changes, making it invalid to simply double the local marginal effect for a large discrete move
• B) -7.86 percentage points (exactly 2×-3.93), because Probit effects are linear
• C) +6.18 percentage points, since probability must rise as MARGIN rises
• D) 0 percentage points, since probability is bounded between 0 and 1

Change = 5.99% − 12.16% = -6.18 percentage points. Because Probit is nonlinear, φ(Xβ) differs at MARGIN=2 versus MARGIN=4, so simply doubling the local marginal effect (which assumes linearity) would be inaccurate; always compute P_new − P_old for discrete changes.

Rule: For discrete changes in nonlinear probability models, always use P(new X) − P(old X), not linear-effect×Δx.

Computer Lab 6 — Exercise 4, LPM weaknesses

• A) Predicted probabilities can fall outside [0,1], and errors are heteroskedastic by construction, so robust standard errors are generally preferred
• B) LPM cannot estimate coefficients at all for binary outcomes
• C) LPM coefficients cannot be interpreted in percentage-point terms
• D) LPM always has a lower R² than Probit or Logit

Because LPM is linear, predicted probabilities are not constrained to [0,1], and the binary nature of Y produces heteroskedastic errors, making robust standard errors advisable, even though the coefficients remain directly interpretable in probability terms.

Rule: Key LPM limitations: possible out-of-range probabilities and heteroskedastic errors; use robust SEs.

Computer Lab 6 — Exercise 4, Probit calc

• A) z = -2.26+2.74(0.35)+0.71(1) = -0.591
• B) z = -2.26+2.74(0.35) = -1.301
• C) z = -2.26+0.71 = -1.55
• D) z = 2.74(0.35)+0.71 = 1.669

z = -2.26 + 2.74×0.35 + 0.71×1 = -2.26 + 0.959 + 0.71 = -0.591.

Rule: Compute the full linear index z=Xβ by summing the intercept and all included terms with their given covariate values.

Computer Lab 6 — Exercise 4, Probit vs LPM nonlinearity

• A) Because Probit is nonlinear: the marginal/discrete effect of PIRAT depends on the predicted probability level itself, which differs by BLACK status even without an explicit interaction term
• B) Because the Probit model actually includes an interaction term between PIRAT and BLACK
• C) Because Probit coefficients differ by group automatically
• D) This is a calculation error; Probit should give identical effects just like LPM

Even without an interaction term, Probit's nonlinear Φ(·) link function makes the implied probability effect of any regressor depend on the overall linear index (and thus on BLACK, which shifts the index), unlike LPM where effects are mechanically constant across groups.

Rule: Nonlinear models (Probit/Logit) naturally produce covariate-dependent marginal effects even without explicit interaction terms, unlike the linear LPM.

Computer Lab 6 — Exercise 4, Logit vs Probit marginal effect

• A) Logit's marginal effect uses the logistic density P(1−P), not the normal PDF φ(·), which belongs to Probit
• B) Logit and Probit always use the exact same marginal effect formula
• C) Logit does not have a marginal effect concept
• D) P(1-P) and φ(Xβ) always give numerically identical results

Logit's link function is the logistic CDF Λ(z), whose derivative is the logistic density P(1-P); the normal PDF φ(·) is specific to Probit's standard normal CDF link, so using φ for Logit's marginal effect is a template/model-mismatch error.

Rule: Match the marginal-effect density to the model: Probit uses φ(Xβ); Logit uses P(1−P).

Computer Lab 6 — Exercise 4, Probit vs Logit comparison

• A) Compare predicted probabilities or marginal effects across models rather than comparing raw coefficients directly, since the two models' coefficient scales differ
• B) Since the coefficients differ so much, one of the two models must be wrong
• C) The Logit model is always exactly twice the Probit coefficient
• D) Raw coefficients from Probit and Logit can be directly compared and interpreted the same way

Probit and Logit use different link functions (normal vs. logistic CDF) with different inherent scales, so their raw coefficients are not directly comparable; however, their implied predicted probabilities and marginal effects typically tell a similar story, as seen here.

Rule: Compare nonlinear-model results via predicted probabilities or marginal effects, not raw coefficient magnitudes, across different link functions.

Formula Battle — method recognition (built from the Formula Book)

• A) φ(Xβ) — the PDF
• B) Φ(Xβ) — the CDF
• C) β·φ(Xβ)
• D) 1/(1+e^(−Xβ))

Probability is a cumulative area. Correct method: Φ(Xβ) — the CDF.

Rule: P = Φ(z)

Formula Battle — method recognition (built from the Formula Book)

• A) β·Φ(z)
• B) β·φ(z)
• C) β
• D) Φ(z)·(1 − Φ(z))

The slope of the CDF is the PDF. Correct method: β·φ(z).

Rule: ME = β·φ(z)