Lecture 11 — Linear Regression

Using one measurement to predict another (Whitlock & Schluter Ch. 17)

Fitting a linear regression with lm(), reading the summary() output, checking the residuals, and using the slope to turn a tracing mass into a leaf area.

Where we left off (Lectures 09–10)

• Five steps — hypotheses → check normality → check variance → run the test → interpret and report
• Welch’s t-test compares the means of two groups
• Pseudoreplication — average up to the thing you actually sampled before you test
• Our leaves — mean leaf mass did not differ between the shady and sunny sides
• Our sugar maples — shady-side leaves were significantly larger in area than sunny-side leaves on the same tree

Note

✅ Where today comes in

• A t-test compares two groups.
• Today we ask a different question:
  • can one number predict another number?
  • we weighed our leaf tracings, but what we actually want is leaf area in cm²

Goals for Today

• Understand what a linear regression does: fit a straight line, \(\hat{Y} = a + bX\)
• See what a residual is, in a real plot
• Fit a model with lm()
• Read and interpret the summary() output — slope, p-value, R²
• Check the residuals the way R shows them to you
• Use the slope to turn a tracing mass into a leaf area
• Report the result in a standard sentence

• Tools today:
  • lm() — fit the line
  • summary() — read the result
  • plot(model) — check the residuals
  • reading 7.660e-03 as 0.007660
  • geom_smooth(method = "lm") — draw the line
• Reading:
  • 📖 Whitlock & Schluter, Ch. 17 — Regression
    • §17.1 Linear regression
    • §17.3 Testing the slope
    • §17.4 R²
    • §17.5 Assumptions
• Today’s naming:
  • models → _model
  • plots → _plot

🧩 Chunk 1 of 3 · What Regression Does

• We will cover:

• the question regression answers,

• the equation \(\hat{Y} = a + bX\),

• what a residual is,

• what R² tells you,

• and the assumptions.

• No code in this chunk — just the idea.

What Is Linear Regression?

Linear regression fits the best straight line through a cloud of points so you can predict one number (Y) from another number (X).

Role	Variable	On the plot
Explanatory (X)	what you can measure easily	x-axis
Response (Y)	what you actually want to know	y-axis

X is usually the easy measurement. Y is the expensive one. Regression gives you an equation that converts X into Y.

📖 W&S §17.1 p. 540–541

Regression vs. the t-test:

Test	The question
t-test	Do two groups have different means?
Regression	Does X predict Y?

Use regression when:

• both variables are numeric
• X could reasonably predict Y
• you want an equation, not just a yes/no

A Real Example from the Textbook

W&S Example 17.1 — “the lion’s nose.” 32 male lions of known age.

Aging a wild lion is hard — you have to have watched it since it was a cub.

But a male lion’s nose darkens as it ages, and you can photograph a nose from a truck.

• X = proportion of the nose that is black — easy
• Y = age in years — what you want

Fit the line once on lions of known age, and from then on you can estimate a lion’s age from a photograph.

📖 W&S §17.1, Example 17.1

Our Version of the Same Problem

The problem: we want leaf area in cm², but we don’t have a leaf-area meter. We have a balance.

The fix — a calibration curve:

1. Cut paper squares of known area (1, 4, 16 … 567 cm²)
2. Weigh each square
3. Fit the line: area ~ mass
4. Trace a leaf onto the same paper and weigh the tracing
5. Put the tracing mass into the equation → predicted leaf area

Why it works: paper from one stock has a constant area per gram, so mass and area move together almost perfectly.

Known area (cm²)	Squares weighed
1	25
4	32
16	24
36	3
64	7
100	14
225–567	13

118 paper squares in all.

Remember paper_mass_g in our leaf data? That is the tracing mass. By the end of today it becomes an area.

The Regression Equation

\[\hat{Y} = a + bX\]

Symbol	Name	What it means
\(\hat{Y}\)	predicted Y	what the line says Y should be
a	intercept	Y when X = 0 — where the line crosses
b	slope	how much Y changes per 1 unit of X

Ours will read:

\[\widehat{\text{area}} = a + b \times \text{mass}\]

📖 W&S §17.1 pp. 543–545

The slope is the useful one.

b = the extra cm² you get for each extra gram of paper

That is just the area density of the paper stock — and it is the number that does the converting.

The intercept is a bookkeeping number.

a = the predicted area of a piece of paper weighing nothing

Physically that should be 0. A small value near zero is fine; it is measurement noise, not biology.

What Is a Residual?

Red = the residuals for the lion data.

A residual is one point’s miss:

\[\text{residual} = \text{observed} - \text{predicted}\]

• A point above the line → positive residual (that lion was older than its nose suggested)
• A point below the line → negative residual
• A point on the line → residual of 0

The line R gives you is the one that makes those red segments as small as possible overall. That is all “least squares” means — and R does the work.

📖 W&S §17.1 p. 542

R² — How Tight Is the Fit?

R² runs from 0 to 1. It is the fraction of the variation in Y that the line accounts for.

R²	Reads as
1.00	points sit on the line
0.62	X explains 62% of the variation
0.00	the line tells you nothing

Both of these are good regressions. Nose blackness really does predict a lion’s age — just with scatter, because lions vary. Paper is manufactured to be uniform, so it has almost none.

In biology, R² of 0.3–0.7 is normal and useful. Don’t expect 0.99.

📖 W&S §17.4 p. 555

R² and the p-value Answer Different Questions

	The question it answers
p-value	Is there really a relationship, or could this slope be chance?
R²	How tightly do the points follow the line?

You can have a significant slope with a low R² — a real relationship with a lot of scatter. That is most of ecology.

Important

The slope is the hypothesis test.

H₀: the slope is 0 — X does not predict Y

Hₐ: the slope is not 0 — X does predict Y

summary() gives you the p-value for exactly that test, and we will find it on the next few slides.

📖 W&S §17.3 pp. 551–553

Assumptions of Regression

Four things need to be roughly true for the p-value and the intervals to mean anything (W&S §17.5):

#	Assumption	How you check it
1	Linear — Y really does follow a straight line	the scatter plot, then plot(model) panel 1
2	Independent — one observation doesn’t depend on another	your study design
3	Equal variance — the scatter is about the same all the way along	plot(model) panels 1 and 3
4	Normal residuals — the misses are bell-shaped	plot(model) panel 2

Important

Check the residuals, not the raw data.

Normality here is about the misses — the red segments — not about Y itself. Y is supposed to go up with X; of course it isn’t bell-shaped.

The good news: once you have a fitted model, plot() draws all of these for you. No extra code.

Same habit as the t-test: check → decide → report.

📖 W&S §17.5 pp. 557–560

🧩 Chunk 2 of 3 · Fit It and Read It

We will cover: loading the paper data, the scatter plot, fitting with lm(), and reading the summary() output line by line.

🛑 After this chunk: Activity Parts 1–3 (the study, load, plot, fit, read the summary).

Load Libraries and Data

# Load packages at the top ------------------------------
library(readxl) # read Excel files
library(tidyverse) # dplyr + ggplot2
library(janitor) # clean_names()

# Load the paper calibration data ----------------------
paper_df <- read_excel("data/paper_area_weights.xlsx")

Two columns, 118 rows: area_cm2 is the known area of each square, mass_g is what it weighed.

Always Plot It First

# Scatter plot of mass against known area --------------
scatter_plot <- paper_df %>%
  ggplot(aes(x = area_cm2, y = mass_g)) +
  geom_point() +
  geom_smooth(method = "lm")
scatter_plot

Three plain lines. aes(x = , y = ) puts the variable you set on x and the one you measured on y, and geom_point() draws the dots.

What you are looking for:

• Straight? → a line is the right tool
• Scatter the same all the way along?
• Any point way off on its own?

Here: straight, tight, nothing strange. Fit the line.

Note

Never fit a regression you haven’t looked at. A curve, a funnel, or one wild point all change what you should do next.

Fit the Model with lm()

# Fit the regression line ------------------------------
paper_model <- lm(mass_g ~ area_cm2, data = paper_df)

Nothing prints. lm() hands back a model object that is holding the answer — summary() is what shows it to you.

The formula is the whole trick:

lm(Y ~ X, data = df)

• Y — the response — goes on the left of ~
• X — the explanatory variable — goes on the right
• ~ reads as “is predicted by”

So mass_g ~ area_cm2 is “mass is predicted by area.”

Warning

⚠️ Which way round? We cut the areas and measured the masses, so mass is the response. lm(area_cm2 ~ mass_g) runs fine too — it just asks R to treat ten areas we chose on purpose as if they were the noisy measurement.

summary() — the Whole Output

# Read the model ---------------------------------------
summary(paper_model)

Call:
lm(formula = mass_g ~ area_cm2, data = paper_df)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.066548 -0.001774  0.000907  0.002267  0.056127 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -2.058e-03  1.389e-03  -1.481    0.141    
area_cm2     7.660e-03  8.646e-06 885.964   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.01351 on 116 degrees of freedom
Multiple R-squared:  0.9999,    Adjusted R-squared:  0.9999 
F-statistic: 7.849e+05 on 1 and 116 DF,  p-value: < 2.2e-16

There are only four numbers in here that you need, and we will take them one at a time:

1. the slope (area_cm2 → Estimate)
2. the intercept ((Intercept) → Estimate)
3. the slope’s p-value (area_cm2 → Pr(>|t|))
4. the R² (Multiple R-squared)

Reading It — 1. The Coefficients Table

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.058e-03  1.389e-03  -1.481    0.141
area_cm2     7.660e-03  8.646e-06 885.964   <2e-16

The Estimate column is the equation.

• (Intercept) = -0.002058 → that is a
• area_cm2 = 0.007660 → that is b, the slope

7.660e-03 means 7.660 × 10⁻³ — move the decimal point three places left.

Tip

Prefer plain decimals? options(scipen = 999) turns scientific notation off and options(scipen = 0) turns it back on. We leave it on — with it off, p prints as <0.0000000000000002.

So the fitted line is:

\[\widehat{\text{mass}} = -0.0021 + 0.00766 \times \text{area}\]

Say the slope out loud, with units:

Every extra 1 cm² of this paper weighs another 0.00766 g.

Note

×10,000 → 76.6 g/m² — the paper’s grammage, the number on the side of the ream. Copy paper is sold at 75–80 g/m², so the slope is a property you can go and check.

The other three columns, in one line each:

Column	What it is
Std. Error	how uncertain that estimate is
t value	Estimate ÷ Std. Error
Pr(>|t|)	the p-value for that row

📖 W&S §17.3 p. 551

Reading It — 2. The Two p-Values

(Intercept) -2.058e-03  ...   0.141
area_cm2     7.660e-03  ...  <2e-16

The slope row is the one that matters. Pr(>|t|) for area_cm2 is <2e-16 — R’s way of writing “smaller than I can print.”

• H₀: the slope is 0 → area tells you nothing about mass
• p < 0.001 → reject H₀. Area predicts mass.

The intercept row: p = 0.141, so the intercept is not distinguishable from 0 — exactly what physics says it should be. A piece of paper with no area has no mass.

Tip

Which p-value do I report?

The slope’s. That is the biological question. Nobody reports the intercept’s p-value unless zero means something special.

Note

Never write p = 0. When R prints <2e-16, you write p < 0.001.

Reading It — 3. R² and the Bottom Line

Residual standard error: 0.01351 on 116 degrees of freedom
Multiple R-squared:  0.9999, Adjusted R-squared:  0.9999
F-statistic: 7.849e+05 on 1 and 116 DF,  p-value: < 2.2e-16

• Multiple R-squared = 0.9999 → area explains 99.99% of the variation in mass. That is manufactured paper, not biology.
• F-statistic with 1 and 116 DF, p < 2.2e-16 → the overall test of the model. With one X it tells you the same thing as the slope’s p-value.
• Residual standard error = 0.0135 g → a typical miss is about 13 mg.

Where the df come from:

• 1 = one explanatory variable
• 116 = 118 squares − 2 numbers we estimated (a and b)

Use Adjusted R-squared once you have more than one X; with a single X, report Multiple R-squared.

Important

R² = 0.9999 does not mean “everything is fine.” It means the points are close to the line. Whether the assumptions hold is a separate question — and we check it two slides from now.

→ ACTIVITY 11 Parts 1–3 now

🛑 Do Activity Parts 1–3 now

Load the paper data, make the scatter plot, fit lm(mass_g ~ area_cm2), and read the slope, the p-value and R² off the summary output.

🧩 Chunk 3 of 3 · Plot It, Check It, Use It

We will cover: drawing the fitted line, reading plot(model), converting the whole column of tracings, averaging away the pseudoreplication, and the paired t-tests.

🛑 After this chunk: Activity Parts 4–7 (check the residuals, convert, average, paired t-tests).

Draw the Line

# Scatter plot with the fitted line --------------------
calibration_plot <- paper_df %>%
  ggplot(aes(x = area_cm2, y = mass_g)) +
  geom_point() +
  geom_smooth(method = "lm")
calibration_plot

One extra line. geom_smooth(method = "lm") fits the same regression again, just for drawing, and adds:

• the fitted line
• a grey 95% confidence band around it

The band is narrowest in the middle of the data and flares at the ends — the line is pinned down best where you have the most data. Here it is so narrow you can barely see it.

Note

geom_smooth() defaults to a wiggly curve. method = "lm" is what makes it a straight regression line.

The Same Plot, Dressed Up

# The publication version — for reference ---------------
calibration_plot <- paper_df %>%
  ggplot(aes(x = area_cm2, y = mass_g)) +
  geom_point() +
  geom_smooth(method = "lm") +
  labs(x = "Paper area (cm²)",
       y = "Paper mass (g)") +
  theme_minimal()
calibration_plot

This is the version you’d put in a report — and it is only two additions to the plot you already made:

• labs() — real axis labels with units
• theme_minimal() — drop the grey background

Tip

Don’t type this today. Get the plain geom_point() + geom_smooth(method = "lm") working first. Labels and themes are the last thing you add, once you know the plot is right.

Check the Residuals — plot(model)

# All four residual plots in one call ------------------
par(mfrow = c(2, 2))
plot(paper_model)

par(mfrow = c(2, 2)) just says “put the next four plots in a 2×2 grid.”

Panel	Name	Checks
1	Residuals vs Fitted	linear? equal scatter?
2	Normal Q-Q	residuals bell-shaped?
3	Scale-Location	equal scatter again
4	Residuals vs Leverage	any one point running the show

Panel 1 is the one to look at first.

• ✅ Good: a flat, shapeless cloud around 0
• ❌ Curved: the relationship isn’t straight
• ❌ Funnel: the scatter grows as Y grows

What Our Residual Plot Is Telling Us

Look at panel 1 again. The points fan out — a 1 cm² square misses by hundredths of a cm², a 567 cm² square misses by several cm².

That is a funnel: assumption 3 (equal variance) does not hold.

# How big are the misses at each square size? ---------
paper_df %>%
  mutate(residual = residuals(paper_model)) %>%
  group_by(area_cm2) %>%
  summarize(biggest_miss = round(max(abs(residual)), 2))

# A tibble: 10 × 2
   area_cm2 biggest_miss
      <dbl>        <dbl>
 1        1         0   
 2        4         0   
 3       16         0   
 4       36         0.01
 5       64         0.01
 6      100         0.02
 7      225         0.02
 8      400         0.06
 9      500         0.06
10      567         0.07

Important

A great R² and a bad residual plot can live together. R² says the points are near the line. The residual plot says the size of the miss depends on how big the square is — which it does, because a balance has percentage error, not fixed error.

So what do we do?

• For predicting a leaf tracing (0.1–0.4 g) we are down in the tight end of the line, where a typical miss is a fraction of a cm². The calibration is fine for the job.
• The confidence intervals at the big-square end are the part you shouldn’t trust.
• The proper fix — fitting on a log scale, or weighting the points — is a graduate topic. Knowing to look, and saying what you saw, is the undergraduate job.

From Tracing Mass to Leaf Area

The model says

\[\text{mass} = a + b \times \text{area}\]

We have the mass and want the area, so rearrange:

\[\text{area} = \frac{\text{mass} - a}{b}\]

# Typed off the summary(), out of sci notation --------
intercept <- -0.002058   # from -2.058e-03
slope     <-  0.007660   # from  7.660e-03

# 1 m2 = 10000 cm2
slope * 10000

[1] 76.6

• Type the two numbers in. They are right there in the Coefficients table, and the arithmetic stays in plain sight
• Rounding to six decimals changes nothing — same t, df and p at the end
• Or let R hand them to you: coef(paper_model), then [1] and [2] — and it prints plain decimals

Note

×10,000 → 76.6 g/m² — the paper’s grammage, the number on the side of the ream. Copy paper is sold at 75–80 g/m², so the slope is a property you can check against the label.

A tracing weighing 0.138 g is

\[\frac{0.138 - (-0.002058)}{0.007660} = 18.3 \text{ cm}^2\]

Convert a Whole Column of Tracings

# Load our leaf data ----------------------------------
leaf_df <-
  read_excel("data/2026_09_03_data_sci_leaf_area.xlsx") %>%
  clean_names()

# Every tracing mass becomes an area ------------------
leaf_df <- leaf_df %>%
  mutate(area_cm2 = (paper_mass_g - intercept) / slope)

leaf_df %>%
  select(teams, shade, mass_g, paper_mass_g, area_cm2)

# A tibble: 53 × 5
   teams shade mass_g paper_mass_g area_cm2
   <chr> <chr>  <dbl>        <dbl>    <dbl>
 1 12345 sunny   0.4          0.21     27.7
 2 12345 sunny   0.48         0.21     27.7
 3 12345 sunny   0.34         0.21     27.7
 4 12345 sunny   0.65         0.24     31.6
 5 12345 sunny   0.27         0.15     19.9
 6 12345 sunny   0.43         0.34     44.7
 7 12345 shady   0.39         0.29     38.1
 8 12345 shady   0.45         0.25     32.9
 9 12345 shady   0.6          0.23     30.3
10 12345 shady   0.57         0.28     36.8
# ℹ 43 more rows

This is the payoff. paper_mass_g was the mass of each leaf’s tracing — now every leaf has an area_cm2.

• One line of arithmetic, all 53 rows at once
• No new function to learn

Note

The NAs are honest. One team never weighed their tracings, so those leaves have no area. R keeps the gap instead of inventing a number.

One Value per Team per Side

# Average away the pseudoreplication ------------------
team_df <- leaf_df %>%
  group_by(teams, shade) %>%
  summarize(area_cm2 = mean(area_cm2, na.rm = TRUE),
            mass_g   = mean(mass_g,   na.rm = TRUE),
            .groups  = "drop")

team_df

# A tibble: 10 × 4
   teams              shade area_cm2 mass_g
   <chr>              <chr>    <dbl>  <dbl>
 1 12345              shady     34.0  0.448
 2 12345              sunny     29.9  0.428
 3 Oscar_Karl         shady    NaN    0.400
 4 Oscar_Karl         sunny    NaN    0.339
 5 fighting_mongooses shady     35.7  0.653
 6 fighting_mongooses sunny     32.2  0.544
 7 leaf_oglers        shady     36.3  0.588
 8 leaf_oglers        sunny     37.4  0.599
 9 still_loading      shady     42.5  0.526
10 still_loading      sunny     53.7  0.789

• Six leaves off one branch are not six independent measurements
• Using all 53 is pseudoreplication — it makes p look better than the study earned
• Averaging gives each team one sunny and one shady value

Note

You met this in T-Tests II. Same tree, same pairing — now in cm² instead of grams.

Paired t-Tests on Area and Mass

# One row per team ------------------------------------
wide_df <- team_df %>%
  pivot_wider(names_from = shade,
              values_from = c(area_cm2, mass_g))

t.test(wide_df$area_cm2_shady,
       wide_df$area_cm2_sunny, paired = TRUE)

t.test(wide_df$mass_g_shady,
       wide_df$mass_g_sunny, paired = TRUE)

Each team measured both sides, so the team is the pair.

	mean diff	t	df	p
area	−1.2 cm²	−0.34	3	0.76
mass	−0.016 g	−0.25	4	0.81

• Neither is significant — with 5 teams there is not much power
• The area test has df = 3: one team never traced their leaves

Note

A real answer of “we could not detect a difference” is still an answer.

How to Report a Regression

The template — fill in from your summary() output:

[What predicted what] (linear regression: F(df₁, df₂) = ___, p ___, R² = ___). The fitted equation was Y = ___ + ___ × X.

Ours:

Paper area was a very strong predictor of paper mass (linear regression: F(1, 116) = 784,900, p < 0.001, R² = 0.9999). The calibration equation was mass = −0.0021 + 0.00766 × area, equivalent to 76.6 g/m². Residual scatter increased with square size, so the largest squares are the least well described.

Where every number comes from:

In the sentence	In summary()
F and both df	the F-statistic: line
p	that same line
R²	Multiple R-squared
intercept	(Intercept) → Estimate
slope	area_cm2 → Estimate

Always include: the test name, F with both df, p, R², and the equation. Add a line about the assumptions you checked.

📖 W&S §17.3 p. 553

→ ACTIVITY 11 Parts 4–7 now

🛑 Do Activity Parts 4–7 now

• Run plot(model) and say whether the assumptions are met
• Convert the whole paper_mass_g column into areas
• Average to one value per team per side, then run both paired t-tests
• Your four answers in the script are what you hand in

What We Learned Today

The ideas:

• Regression fits a line so you can predict Y from X
• \(\hat{Y} = a + bX\) — the slope converts X into Y
• A residual is one point’s miss: observed − predicted
• R² = how tight the fit is; p = whether the slope is real
• Check the residuals — and say what you saw

The code:

• lm(Y ~ X, data = df) — fit it
• summary(model) — slope, intercept, p, R²
• geom_smooth(method = "lm") — draw it
• plot(model) — check it
• 7.660e-03 = 0.007660 — reading scientific notation

• Hand in: your finished scripts/11_regression.R — the code is given; your answers are graded

📖 W&S Ch. 17:

• §17.1 Linear regression · §17.2 Prediction
• §17.3 Testing the slope · §17.4 R²
• §17.5 Assumptions

Up next — Lecture 12:

• Pulling real climate data into R with GSODR
• Summarizing daily weather into yearly values
• The same lm() — temperature predicted by date
  • the slope becomes a rate of change per year
  • and you write that regression yourself