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6 questions · 1 formulas

Quadratic

ME = b1 + 2·b2·X; turning point = −b1/(2b2)

Logs & DummiesD–A levelNormal priority

Suppose that she obtained 1 2 3 4 ˆ 3.8 0.23 1.31 0.032 1.05 y x x x x = + − + + by using data from the past 24 months. How would we interpret the coefficient on 4x ?

→ B) On an average, we would expect an additional $1.05 million in loans during June, July, and August.

Old Exam — Dec 2021 re-exam, Q20

Given: Y is monthly residential lending in $m; X4=1 for Jun-Jul-Aug. Coefficient on X4 is +1.05.

  • A) On an average, we would expect an additional 1.05 loans per month during June, July, and August.
  • B) On an average, we would expect an additional $1.05 million in loans during June, July, and August.
  • C) On an average, we would expect an additional 1.05% increase in loans during June, July, and August.
  • D) There is no difference in the total residential loan amount between the months of June and September.
  • E) None of the above (A, B, C, D) is correct.

X4 is a seasonal dummy and Y is measured in millions of dollars. Holding other regressors fixed, X4=1 adds 1.05 million dollars of expected loans -> B.

Rule: Interactions and quadratics mean the marginal effect is not constant. Log coefficients require percentage interpretation.

InteractionsD–A levelNormal priority

Suppose that she obtained 1 2 3 4 3 4 ˆ 3.8 0.23 1.31 0.032 1.05 0.22 y x x x x x x = + − + + − by using data from the past 24 months. How would we interpret the coefficient of 3 4 x x ?

→ D) Advertising is less effective in June, July, and August.

Old Exam — Dec 2021 re-exam, Q21

Given: Interaction coefficient on X3xX4 is -0.22; it changes the advertising slope in summer.

  • A) On an average, we would expect $220,000 less in loans during the months of June, July, and August.
  • B) Advertising is more effective during June, July, and August.
  • C) On an average, we would expect $220,000 more in loans during the months of June, July, and August.
  • D) Advertising is less effective in June, July, and August.
  • E) None of the above (A, B, C, D) is correct.

Advertising slope outside summer is 0.032. In summer it becomes 0.032-0.22, so the interaction lowers the advertising effect -> D.

Rule: Interactions and quadratics mean the marginal effect is not constant. Log coefficients require percentage interpretation.

QuadraticsD–A levelNormal priority

What can the manager infer about the relationship between the square of the level of output and the average costs?

→ D) As the level of output increases, average costs increase at an increasing rate.

Old Exam — Dec 2021 re-exam, Q22

Given: A manager estimated average costs (AC) as a function of output: AC = b0 + 0.34·output + 0.007·output². Both coefficients are positive and significant; output is always positive.

  • A) As the level of output increases, average costs decrease at an increasing rate.
  • B) As the level of output increases, at first average costs decrease, then will start to increase.
  • C) As the level of output increases, at first average costs increase, then will start to decrease.
  • D) As the level of output increases, average costs increase at an increasing rate.
  • E) None of the above (A, B, C, D) is correct.

Marginal effect = 0.34 + 2(0.007)·output = 0.34 + 0.014·output, which is positive and rising as output rises. So average costs increase at an increasing rate → D.

Rule: Quadratic: marginal effect = b1 + 2·b2·X; turning point = −b1/(2·b2).

⚠ The original answer key highlights B, which conflicts with the displayed equation. D follows mathematically.

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.

QuadraticsE-levelNormal priority

Y = b0 + b1·AGE + b2·AGE². You need the AGE where the effect turns.

→ B) −b1 / (2b2)

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

  • A) −b1 / b2
  • B) −b1 / (2b2)
  • C) b1 / (2b2)
  • D) −b2 / (2b1)

Set the derivative to zero. Correct method: −b1 / (2b2).

Rule: X* = −b1/(2b2)