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22 questions · 2 formulas

Log interpretations

level-level: ΔY = b·ΔX log-level: %ΔY ≈ 100b·ΔX level-log: ΔY ≈ (b/100)·%ΔX log-log: b = elasticity

Dummy in log-Y model

approx = 100b%; exact = 100(e^b − 1)%

Normal DistributionD–A level★ High priority — official Canvas example

In a regression with ln(price) as dependent variable, pool dummy coefficient = 0.0774. Best interpretation?

→ A) A house with a pool is on average about 7.74% more expensive than a house without a pool, holding the other regressors fixed.

Canvas Example MCQ — Q53

  • A) A house with a pool is on average about 7.74% more expensive than a house without a pool, holding the other regressors fixed.
  • B) A house with a pool is on average $7,740 more expensive than a house without a pool.
  • C) A house with a pool is on average 7.74% less expensive than a house without a pool.
  • D) A house with a pool is on average $77,740 more expensive than a house without a pool.
  • E) None of A-D is close; write the correct answer.

For a dummy in log-Y, the standard approximation is 100b% = 7.74%. Exact effect = 100(e^0.0774-1) ≈ 8.05%. The source MCQ is testing the approximation.

Rule: Know both approximate and exact dummy-in-log effects.

Normal DistributionD–A level★ High priority — official Canvas example

In ln(price) on ln(sqft), coefficient on ln(sqft) = 0.7244. Interpret.

→ A) A 1% increase in house size is associated with about a 0.7244% increase in sales price.

Canvas Example MCQ — Q54

  • A) A 1% increase in house size is associated with about a 0.7244% increase in sales price.
  • B) A 1,000 sqft increase in size increases sales price by about $7,244.
  • C) A 1% increase in size increases sales price by about $72.44.
  • D) A 10% increase in size increases sales price by about 0.8925%.
  • E) None of A-D is close; write the correct answer.

This is a log-log specification. The slope is elasticity directly.

Rule: ln(Y) on ln(X): coefficient = elasticity.

Normal DistributionD–A level★ High priority — official Canvas example

Interpret b1 in ln(Y)=b0+b1X+u.

→ B) A one-unit change in X is associated with approximately a 100*b1% change in Y.

Canvas Example MCQ — Q57

  • A) A 1% change in X is associated with a b1% change in Y.
  • B) A one-unit change in X is associated with approximately a 100*b1% change in Y.
  • C) A 1% change in X is associated with a change in Y of 0.01*b1.
  • D) A one-unit change in X is associated with a b1-unit change in Y.

This is a log-level model. For small b, multiply the coefficient by 100 to get the approximate percent change in Y for one unit of X.

Rule: ln(Y) on level X -> 100b% per unit X.

Normal DistributionD–A level★ High priority — official Canvas example

Interpret b1 in ln(Y)=b0+b1 ln(X)+u.

→ A) A 1% change in X is associated with a b1% change in Y.

Canvas Example MCQ — Q58

  • A) A 1% change in X is associated with a b1% change in Y.
  • B) A one-unit change in X is associated with a b1-unit change in Y.
  • C) A one-unit change in X is associated with a 100*b1% change in Y.
  • D) A 1% change in X is associated with a change in Y of 0.01*b1.

This is log-log. b1 is the elasticity.

Rule: ln-ln coefficient = elasticity.

Model Fit & RMSED–A level★ High priority — official Canvas example

Training model: ln(Y-hat)=0.50+0.80X. Hold-out observations (X, lnY): (1,1.10), (2,2.30), (3,2.80), (4,4.00). Find RMSE.

→ C) 0.212

Canvas Example MCQ — Q88

  • A) 0.045
  • B) 0.150
  • C) 0.212
  • D) 0.450
  • E) None of the above.

Predictions: 1.30, 2.10, 2.90, 3.70. Errors actual-predicted: -0.20, +0.20, -0.10, +0.30. Squared errors: .04,.04,.01,.09; mean = .18/4 = .045; RMSE = √.045 = .2121.

Rule: Do not stop at MSE=.045. RMSE is the square root, giving .212.

Normal DistributionD–A levelVery high priority

The 99% confidence interval for β-coefficient for the lsqrft variable is closest to__________.

→ A) (0.5027, 0.9026)

Old Exam — Dec 2022, Q21

Given: For ln(sqft): coefficient=0.7026, SE=0.0759. Use a 99% critical value about 2.63.

  • A) (0.5027, 0.9026)
  • B) (0.5517, 0.8535)
  • C) (0.5887, 0.8174)
  • D) Not possible to compute given the information in the summary output.
  • E) None of the above (A, B, C, D) is close to be correct.

99% CI ≈ b ± 2.63 SE = 0.7026 ± 2.63(0.0759) ≈ (0.503,0.902), matching A.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

Which of the following interpretations of the coefficient of TRADITIONAL is most correct?

→ A) A traditional house in on average about 15.5 % less expensive than non-traditional houses.

Old Exam — Oct 2022, Q16

Given: TRADITIONAL coefficient=-0.1547 in a log-price model.

  • A) A traditional house in on average about 15.5 % less expensive than non-traditional houses.
  • B) A waterfront house in on average about 1.55 % less expensive than non-traditional houses.
  • C) A waterfront house in on average about 1.55 times less expensive than non-traditional houses.
  • D) A waterfront house in on average about 0.15% less expensive than non-traditional houses.
  • E) None of the above (A, B, C, D) is close to be correct.

TRADITIONAL is a dummy in ln(PRICE), b=-0.1547. Approximate effect=-15.47%. Exact effect=100(e^-0.1547-1)≈-14.3%. Among the choices, A is the intended approximation.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

You have estimated a model of two variables related such that: ln(y) = 17.3 - 0.04x. If x decreases by 2 units, what is the expected change in y?

→ B) y increases by 8 percent.

Old Exam — Dec 2021 re-exam, Q24

Given: ln(y)=17.3-0.04x. A 1-unit fall in x changes ln(y) by +0.04; 2-unit fall -> about +8% in y.

  • A) y decreases by 0.08 units.
  • B) y increases by 8 percent.
  • C) y increases by 4 units.
  • D) y decreases by 8 percent.
  • E) None of the above (A, B, C, D) is correct.

Delta x=-2. Delta ln(y)=(-0.04)(-2)=+0.08, so y rises by approximately 8% -> B.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

What is the income elasticity of your firm’s product?

→ A) 1.212

Old Exam — Dec 2021 re-exam, Q25

Given: ln(Q)=3.418+1.212 ln(I): the ln-ln slope is elasticity.

  • A) 1.212
  • B) 2.206
  • C) 3.418
  • D) 4.630
  • E) None of the above (A, B, C, D) is correct.

In a log-log model, the slope is elasticity. Income elasticity=1.212 -> A.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

You want to test the null hypothesis that parameter (coefficient) of the ln(I) variable is equal to 1. Then the value of the test statistic (t- statistic) is

→ D) 0.54

Old Exam — Dec 2021 re-exam, Q26

Given: Test H0: beta=1. t=(1.212-1)/0.392≈0.54.

  • A) 3.09
  • B) 0.392
  • C) 3.418
  • D) 0.54
  • E) None of the above (A, B, C, D) is correct.

t=(1.212-1)/0.392=0.212/0.392≈0.54 -> D.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

The t-statistic for the ln(SQFT) (see ln_sqft) is closest to_________.

→ C) 56.99

Old Exam — Oct 2021, Q19

Given: ln(SQFT) coefficient=.892479, SE=.015659.

  • A) 5.69
  • B) 0.00
  • C) 56.99
  • D) Not possible to compute given the information in the summary output.

t=b/SE=0.892479/0.015659≈56.99 -> C.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Normal DistributionD–A levelVery high priority

Which of the following interpretations of the coefficient of WATERFRONT is most correct?

→ B) A waterfront house in on average 13.5 % more expensive than not waterfront houses.

Old Exam — Oct 2021, Q21

Given: WATERFRONT is a dummy in a log-price model with coefficient≈0.135.

  • A) A waterfront house in on average 1.35 times more expensive than not waterfront houses.
  • B) A waterfront house in on average 13.5 % more expensive than not waterfront houses.
  • C) A waterfront house in on average 13.5 % more expensive than traditional houses.
  • D) A waterfront house in on average 135 % more expensive than not waterfront houses.
  • E) None of the above (A, B, C, D) is close to be correct.

Dummy in ln(price): b≈0.135. Approx effect≈13.5% higher; exact=100(e^0.135-1)≈14.5%. Intended answer B.

Rule: Identify model units and form before calculating. Never confuse coefficient size, significance, R2 and prediction accuracy.

Model Fit & RMSED–A levelVery high priority

Using the above model, predict the prices of a house with 2500 square feet that is not close to the university. Use the natural predictor (not the corrected predictor). The predicted price is closest to

→ B) $212 021

Old Exam — Dec 2022, Q23

Given: Historical software-dependent prediction question. Original model: ln(PRICE)=b0+b1 ln(SQFT)+d1 CLOSE+d2[CLOSE x ln(SQFT)].

  • A) $207 479
  • B) $212 021
  • C) $218 756
  • D) $225 335
  • E) None of the above (A, B, C, D) is close to be correct.

Current exam rule: prediction uses the stated fitted equation; lower RMSE/AIC/SC is preferred when the criterion is being minimized; hold-out data must remain outside estimation/selection. Source answer: B) $212 021.

Rule: The old exam supplied a spreadsheet/software. For the 2026 paper exam, expect the necessary values/table/output to be printed. Use the method, not the historical data file.

Logs & DummiesE-levelNormal priority

In ln(PRICE) = 4.8918 + 0.9284·ln(SQFT) + ..., how should the coefficient on ln(SQFT) be interpreted?

→ A) A 1% increase in SQFT is associated with about a 0.93% increase in PRICE, holding other variables constant

Computer Lab 5 — Exercise 4, Log-log

  • A) A 1% increase in SQFT is associated with about a 0.93% increase in PRICE, holding other variables constant
  • B) A 1-unit increase in SQFT increases PRICE by 0.9284 dollars
  • C) A 1% increase in SQFT increases PRICE by 92.84 dollars
  • D) SQFT explains 92.84% of variation in PRICE

In a log-log model, the coefficient on ln(X) is an elasticity: a 1% change in X is associated with approximately a b% change in Y, here about 0.93%.

Rule: ln(Y) = ... + b·ln(X) ⇒ coefficient b is an elasticity (percent change in Y per 1% change in X).

QuadraticsD–A levelNormal priority

With ln(PRICE) including age (b_age=-0.0134253) and age² (b_age2=0.000121292), what is the turning point (in years) where the marginal effect of age on ln(PRICE) switches from negative to positive, using AGE* = -b_age/(2·b_age2)?

→ A) ≈55.34 years

Computer Lab 5 — Exercise 4, Quadratic

  • A) ≈55.34 years
  • B) ≈27.67 years
  • C) ≈110.68 years
  • D) ≈-55.34 years

AGE* = -(-0.0134253)/(2×0.000121292) = 0.0134253/0.000242584 ≈ 55.34 years.

Rule: Turning point of a quadratic term: X* = -b1/(2b2).

QuadraticsD–A levelNormal priority

Given the age and age² coefficients, for a house that is 30 years old, is the marginal effect of an additional year of age on ln(PRICE) positive or negative (turning point ≈55.34 years)?

→ A) Negative, because 30 is below the turning point of about 55.34 years

Computer Lab 5 — Exercise 4, Quadratic

Given: Model: ln(PRICE) = … + b_age·AGE + b_age2·AGE² + …, with b_age < 0 and b_age2 > 0 (U-shape). Turning point = −b_age/(2·b_age2) ≈ 55.34 years.

  • A) Negative, because 30 is below the turning point of about 55.34 years
  • B) Positive, because age always has a positive quadratic coefficient
  • C) Zero, since AGE² exactly offsets AGE at any age
  • D) Cannot be determined without knowing SQFT

Since 30 years is less than the turning point of 55.34 years, the marginal effect b_age + 2·b_age2·AGE is still negative at AGE=30 (the U-shape has not yet turned positive).

Rule: Before the turning point, the marginal effect of a U-shaped quadratic term is negative; after it, positive.

Logs & DummiesE-levelNormal priority

The WATERFRONT dummy coefficient in a ln(PRICE) regression is 0.2036. What is the exact percentage price difference for waterfront vs non-waterfront houses, using 100×[exp(b)−1]?

→ A) ≈22.58%

Computer Lab 5 — Exercise 4, Dummy in log model

  • A) ≈22.58%
  • B) ≈20.36%
  • C) ≈2.036%
  • D) ≈45.16%

Exact effect = 100×[exp(0.2036)−1] = 100×[1.2258−1] ≈ 22.58%.

Rule: For a dummy in a log(Y) model, use the quick approximation 100b% or the exact 100[exp(b)−1]%; the exact value is always somewhat larger in magnitude for positive b.

Regression BasicsE-levelVery high priority

Switching from Excel's standard OLS output to Python's HC1 robust standard errors for the same ln(PRICE) regression, which outputs should remain identical, and which can change?

→ A) Coefficients and fitted values stay the same; standard errors, t-statistics and p-values can differ

Computer Lab 5 — Exercise 4, Robust SE

  • A) Coefficients and fitted values stay the same; standard errors, t-statistics and p-values can differ
  • B) Everything, including coefficients, changes
  • C) Only the R² changes
  • D) Coefficients change but standard errors stay identical

OLS point estimates and fitted values depend only on the OLS fitting procedure, which is unaffected by the choice of variance estimator; only the estimated standard errors (and resulting t-stats/p-values) can change with HC1.

Rule: HC1 robust SEs change only the measure of estimation uncertainty, not the OLS coefficients themselves.

Model Fit & RMSEE-levelVery high priority

What does RMSE = sqrt[(1/N)Σ(y−ŷ)²] measure, and in what units is it expressed if Y = ln(PRICE)?

→ A) Typical magnitude of prediction errors, expressed in log-price units (not dollars)

Computer Lab 6 — Exercise 1, Task 2

  • A) Typical magnitude of prediction errors, expressed in log-price units (not dollars)
  • B) The percentage of correctly classified observations, expressed as a percent
  • C) Typical prediction error, always expressed in dollars regardless of how Y is defined
  • D) The correlation between predicted and actual prices

RMSE summarizes typical prediction error magnitude, with squaring penalizing large misses more heavily; its units match the dependent variable, so if Y=ln(PRICE), RMSE is in log-price units, not dollars.

Rule: RMSE units always match the dependent variable's units/scale.

Returns & CompoundingE-levelVery high priority

An index goes from 100 to 160 over 10 years. You want the average annual growth. Which formula?

→ B) (1.6)^(1/10) − 1

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

  • A) (160 − 100)/100/10
  • B) (1.6)^(1/10) − 1
  • C) 1.6 / 10
  • D) ln(1.6) × 10

Multi-period growth compounds. Correct method: (1.6)^(1/10) − 1.

Rule: CAGR = (End/Start)^(1/n) − 1

Logs & DummiesE-levelNormal priority

ln(Price) regressed on a dummy with coefficient 0.25. You need the EXACT percentage effect.

→ B) 100(e^0.25 − 1)%

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

  • A) 25%
  • B) 100(e^0.25 − 1)%
  • C) 0.25 SEK
  • D) e^0.25

Exponentiate the coefficient. Correct method: 100(e^0.25 − 1)%.

Rule: Exact = 100(e^b − 1)%

Logs & DummiesE-levelNormal priority

ln(Y) regressed on ln(X). How do you read the coefficient?

→ B) Elasticity: % change in Y per 1% change in X

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

  • A) Change in Y per unit X
  • B) Elasticity: % change in Y per 1% change in X
  • C) % change in Y per unit X
  • D) Change in Y per 1% change in X

Both sides in logs. Correct method: Elasticity: % change in Y per 1% change in X.

Rule: log-log → elasticity