| R-squared | F | df (model) | df (residual) | p-value |
|---|---|---|---|---|
| 0.9999 | 784932.9 | 1 | 116 | 0 |
Paper Calibration Curve: Predicting Leaf Area from Mass
The regression from Activity 11, written up as a report
Introduction
To estimate leaf surface area without a leaf-area meter, we built a calibration curve. Paper from one stock has a constant area per gram, so the mass of a paper square is a near-perfect linear predictor of its area. Once we know that relationship, we can trace a leaf onto the same paper, weigh the tracing, and convert its mass to an area.
- Explanatory variable (X): paper mass (g) — easy to measure on a balance
- Response variable (Y): paper area (cm²) — what we actually want
Methods
We weighed 118 paper squares of known area (1–567 cm²) and fit a simple linear regression, area_cm2 ~ mass_g, with lm(). We checked the residuals with base R’s diagnostic plots for the fitted model. All work was done in R with the tidyverse; this report was written in Quarto, so every number and figure below is computed from the data rather than typed in by hand.
Results
Paper mass was an extremely strong predictor of paper area. The overall fit is in Table 1 and the fitted equation is in Table 2.
| Term | Estimate | Std. error | t | p-value |
|---|---|---|---|---|
| (Intercept) | 0.279 | 0.181 | 1.5 | 0.1261386 |
| mass_g | 130.523 | 0.147 | 886.0 | 0.0000000 |
Read the equation off Table 2: predicted area equals the (Intercept) estimate plus the mass_g estimate times the mass. In words, each additional gram of this paper stock adds about the mass_g estimate in cm² of area. The slope’s p-value tests whether mass predicts area at all; the intercept is not distinguishable from zero, which is what physics says it should be — paper with no mass has no area.
The calibration curve

Residual checks

Panel 1 of Figure 2 shows a funnel rather than a flat cloud: the residuals get larger as the squares get larger. Table 3 puts numbers on it.
| Known area (cm²) | Largest residual (cm²) |
|---|---|
| 1 | 0.35 |
| 4 | 0.59 |
| 16 | 0.65 |
| 36 | 0.96 |
| 64 | 1.30 |
| 100 | 3.09 |
| 225 | 2.00 |
| 400 | 7.28 |
| 500 | 7.39 |
| 567 | 8.76 |
The equal-variance assumption is therefore violated at the large-square end of the range, and confidence intervals there should be treated with caution. The leaf tracings we apply the curve to below weigh 0.1–0.4 g, at the tight end of the line, where a typical miss is a fraction of a cm².
Predicting leaf area
| Tracing | Mass (g) | Predicted area (cm²) |
|---|---|---|
| sunny-side | 0.092 | 12.3 |
| shady-side | 0.138 | 18.3 |
The same predict() call converts a whole column at once. Table 5 applies the calibration to every leaf in our class data set, then summarizes the predicted areas by side of the tree.
| Side | n | Mean area (cm²) | SD (cm²) |
|---|---|---|---|
| shady | 21 | 36.7 | 5.8 |
| sunny | 19 | 36.0 | 11.5 |
Table 5 reports the predicted areas in the units we wanted. The two sides differ by about a square centimetre — the same story the Welch’s t-test told us about leaf mass.
Discussion
Write one short paragraph here. Did the calibration fit well — what do R² in Table 1 and Figure 1 say? What did the residual checks in Figure 2 and Table 3 show, and what does that mean for how far you would trust the curve? Why is working in area (cm²) more useful to a reader than raw tracing mass (g) when comparing sunny and shady leaves? This is where you paste the results paragraph you wrote in the activity.
References
- Whitlock, M. & Schluter, D. The Analysis of Biological Data (2nd ed.), Ch. 17.
- Wickham, H. R for Data Science (2nd ed.) — https://r4ds.hadley.nz/