Units of Measurement and Functional Form: 10 Exercises with Full Solutions

This workbook accompanies the Units of Measurement and Functional Form article and is built for practice with pen and paper. The ten exercises follow the lesson’s two themes: the mechanical rules for how rescaling a variable changes a regression’s coefficients, and the interpretation of the level-level, log-level, level-log, and log-log functional forms. Each one isolates a single idea so you can see exactly where your understanding needs work.

Attempt every exercise before reading the solutions, and write out the reasoning rather than just the answer, because in this topic the interpretation is the whole point. Remember the two anchors throughout: rescaling never changes the underlying relationship or the R-squared, and logarithms turn slopes into percentages and elasticities.

Part One: The Exercises

Exercise 1 (Rescaling the regressor). A fitted regression is y-hat = 10 + 3x. The variable x is then re-expressed by dividing it by 1,000 (for example, converting grams to kilograms). State the new intercept and the new slope.

Exercise 2 (Rescaling the dependent variable). For the same fitted regression y-hat = 10 + 3x, suppose y is now multiplied by 100 (for example, converting pounds to pence). State the new intercept, the new slope, and what happens to the R-squared.

Exercise 3 (Rescaling both). A fitted regression is y-hat = 50 + 2x. The dependent variable is multiplied by 1,000 and the regressor is multiplied by 10. Find the new slope and the new intercept.

Exercise 4 (Why R-squared does not change). Explain in words why rescaling either variable leaves the R-squared unchanged, even though the coefficients change.

Exercise 5 (Level-level interpretation). A wage regression gives wage-hat = -0.90 + 0.54 educ, where wage is in dollars per hour and educ is years of schooling. Interpret the slope, then predict the change in wage from four additional years of education.

Exercise 6 (Log-level interpretation). A regression gives log(wage)-hat = 0.584 + 0.083 educ. Interpret the slope as a percentage, then give the approximate predicted percentage change in wage from three additional years of education.

Exercise 7 (Log-log elasticity). A regression gives log(salary)-hat = 4.822 + 0.257 log(sales). Interpret the slope, then predict the approximate percentage change in salary if sales rise by 10%.

Exercise 8 (Level-log interpretation). A regression gives y-hat = 200 + 45 log(x). Interpret the slope, and state the approximate change in y for a 1% increase in x.

Exercise 9 (Unit invariance of the log slope). The log-log salary regression in Exercise 7 was estimated with salary measured in thousands of dollars. Suppose salary is re-expressed in dollars by multiplying it by 1,000. What happens to the slope, and what happens to the intercept?

Exercise 10 (Choosing the functional form). For each scenario, name the appropriate functional form and justify it. (a) You believe each extra year of schooling raises wage by roughly the same percentage. (b) You want the elasticity of house price with respect to floor area. (c) You believe each additional bedroom adds a fixed number of dollars to a house’s price.

Part Two: Worked Solutions

Solution 1. Dividing the regressor by a constant leaves the intercept unchanged and multiplies the slope by that constant. With the constant equal to 1,000:

The new line is y-hat = 10 + 3000 times the rescaled x. A one-unit change in the rescaled variable equals a 1,000-unit change in the original, so the coefficient must be 1,000 times larger to predict the same change in y.

Solution 2. Multiplying the dependent variable by a constant multiplies both the intercept and the slope by that constant, and leaves the R-squared unchanged.

The new line is y-hat = 1000 + 300x, and the R-squared is identical to the original, because stretching y proportionally does not change how well the line fits.

Solution 3. With the dependent variable multiplied by c-one equal to 1,000 and the regressor multiplied by c-two equal to 10, the slope is multiplied by the ratio of the two constants and the intercept only by c-one.

So the new line is y-hat = 50000 + 200x. The intercept responds only to rescaling y, while the slope responds to both.

Solution 4. The R-squared is the fraction of the total variation in y that the regression explains, and it depends only on the relative fit of the line to the points, not on the units. Rescaling stretches or shrinks the axes proportionally, which moves every point and the fitted line by the same factor, so the proportion of explained variation is preserved exactly. In short, units change the scale of the relationship but not the quality of the fit, and R-squared measures only the quality of the fit.

Solution 5. In a level-level model the slope is a unit-for-unit effect, so each additional year of education raises predicted wage by 0.54 dollars per hour, a fixed dollar amount regardless of the starting wage. For four additional years:

The predicted wage rises by 2.16 dollars per hour.

Solution 6. In a log-level model, one hundred times the slope is approximately the percentage change in y per one-unit change in x, so each year of education raises predicted wage by about 8.3%.

For three additional years, the approximate percentage change is one hundred times the slope times the change in education.

This is an approximation, accurate for small changes; the exact figure would use the exponential form, but about 25% is the intended reading.

Solution 7. In a log-log model the slope is a constant elasticity, the percentage change in y for a one-percent change in x, so a 1% rise in sales is associated with about a 0.257% rise in salary. For a 10% rise in sales:

Predicted salary rises by approximately 2.57%.

Solution 8. In a level-log model the slope divided by one hundred gives the change in y for a one-percent change in x, so here a 1% increase in x raises y by about 0.45 units.

The interpretation is a level change in y for a percentage change in x, the mirror image of the log-level case.

Solution 9. Because the dependent variable enters as a logarithm, multiplying salary by a constant adds the log of that constant to the intercept and leaves the slope completely unchanged. The slope therefore stays at 0.257, since the added constant cancels out of the deviations-from-the-mean that the slope is built from. The intercept rises by the log of 1,000.

This is exactly why an elasticity is unit-free: the percentage interpretation does not depend on how the variables are measured.

Solution 10. In part (a), a constant percentage effect of a unit change in x calls for the log-level model, with the log of wage regressed on years of schooling, because there the slope reads directly as a percentage change per year. In part (b), an elasticity calls for the log-log model, with the log of price regressed on the log of floor area, because there the slope is the percentage change in price for a one-percent change in area. In part (c), a fixed dollar effect of an extra unit calls for the level-level model, with price regressed on the number of bedrooms, because there the slope is a constant change in price in its own units.

How to Get the Most From This Workbook

Two habits make this material automatic. First, for any rescaling question, separate the two questions “what happened to the slope?” and “what happened to the intercept?” and apply the simple rules: dividing x by a constant scales the slope by that constant and leaves the intercept alone, multiplying y by a constant scales both, and a log on the dependent side protects the slope entirely while shifting only the intercept. Second, for any interpretation question, read the slope through the lens of the functional form before doing any arithmetic, asking whether it is a level change, a percentage change, or an elasticity. Get those two reflexes in place and you can read any regression output correctly no matter how the variables happen to be measured.

See you soon.