Exercise 2 (20 minutes)

You practice: the log₂ transformation, comparing it to a polynomial, and visualising the fit. When you're done, assist some of the others …

Same ESS data as exercise 1 (Denmark & Bulgaria; run that setup first). In exercise 1 you fitted a polynomial of age. Now you try the log₂ transformation instead — and decide which you prefer.

1. Fit a log₂ model. Replace the age polynomial with log2(agea), keeping the same controls, weighted by pspwght.

Hint

lm_robust(imwbcnt ~ cntry + gndr + eduyrs + log2(agea), data = ESS, weights = pspwght)

R solution — don't peek too early ;) !

ols_log <- lm_robust(imwbcnt ~ cntry + gndr + eduyrs + log2(agea),
                     data = ESS, weights = pspwght)

modelsummary(list("Xenophobia" = ols_log),
             coef_rename = c("log2(agea)" = "Age (log2)"),
             stars = TRUE, gof_map = c("nobs", "r.squared"), output = "kableExtra")

2. The log2(agea) coefficient is . Because the predictor is on a log₂ scale, this reads as: a doubling of age (say from 20 to 40, or 40 to 80) is associated with about scale points more xenophobialess xenophobiano change in xenophobia.

3. Compare the two curves. Put the log₂ model next to the quadratic from exercise 1 and look at the fit statistics.

R solution — don't peek too early ;) !

ols_poly <- lm_robust(imwbcnt ~ cntry + gndr + eduyrs + poly(I(agea / 10), 2, raw = TRUE),
                      data = ESS, weights = pspwght)

modelsummary(list("Quadratic" = ols_poly, "log2" = ols_log),
             gof_map = c("nobs", "r.squared", "adj.r.squared"),
             stars = TRUE, output = "kableExtra")

According to \(R^2\) and adjusted \(R^2\), which model fits better? The quadratic is clearly betterThe log₂ is clearly betterThey are essentially equally good

4. Given that the two fit about equally well, which would you report — and why? The log₂ model — it is simpler (one coefficient) and reads as 'per doubling of age'The quadratic — more coefficients always means a better modelIt does not matter at all which you report

5. Write one sentence as a # comment justifying your choice — then compare:

Model answer

The log₂ and quadratic models fit the age–xenophobia relationship about equally well (nearly identical \(R^2\) and adjusted \(R^2\)). When fit is a tie, prefer the simpler, more interpretable model: the log₂ needs a single coefficient and reads cleanly as "a doubling of age is associated with about 0.4 points more xenophobia", whereas the quadratic needs two coefficients and a turning point to interpret. The one reason to prefer the quadratic would be a genuine rise-then-fall in age — which we did not find here.

6. Visualise the log₂ fit. Make predictions across the age range and plot the implied curve.

R solution

# 1. Fictional data spanning the observed ages, controls held constant
fict <- tibble(
  agea   = seq(15, 90, by = 1),
  cntry  = "Denmark",
  gndr   = "Female",
  eduyrs = mean(ESS$eduyrs)
)

# 2. Predicted xenophobia + 95% interval
fict <- predict(ols_log, newdata = fict, interval = "confidence")$fit %>%
  as_tibble() %>%
  bind_cols(fict, .)

# 3. Plot
ggplot(fict, aes(x = agea, y = fit)) +
  geom_ribbon(aes(ymin = lwr, ymax = upr), alpha = 0.25) +
  geom_line(linewidth = 1, colour = "#901A1E") +
  labs(x = "Age", y = "Predicted xenophobia (0–10)") +
  theme_minimal()

The curve rises quickly among the young and flattens among the old — the signature shape of a log predictor.

7. Bonus, for the fast. The log of a variable is only defined for positive values. Which variables in a typical survey could you not log without a fix — and what is the usual trick?

Talking points

log() is undefined at 0 and for negatives. Income with zeros, counts that include 0, or anything centred around 0 cannot be logged directly. Common fixes: add a small constant (log(x + 1), i.e. log1p), or use a different transformation. Age is safe because everyone is older than zero.

8. Discuss with your neighbour.

   • The deck's health-spending example was a case where log₂ clearly won. Here it only ties the polynomial. What is different about the two relationships?
   • "The coefficient on log2(agea) is 0.4." Why is "per doubling" a more honest way to say this than "per year"?
   • If you logged the outcome instead of the predictor, how would the interpretation change?

Talking points

• Health spending spans a huge range (from a few dollars to over $10,000) and flattens sharply, so the log dramatically straightens it. Age spans a much narrower range (15–90, barely more than two doublings) and bends only gently, so both tools capture the mild curve about equally.
• A log predictor means the effect is not constant per year — it is larger among the young and smaller among the old. "Per doubling" states that honestly; "per year" would falsely imply a constant effect, which is exactly the linearity assumption the log was meant to relax.
• Logging the outcome makes coefficients read as approximate percentage changes in \(Y\) per unit of \(X\) — the convention you already met with log wages in Lecture 10.