Linear Models (lm)

Reviewed & published by Brayan K

A linear model fits the best straight-line relationship between a response and one or more predictors — R's lm() function makes regression a single, readable line of code.

Part of the free R course at LearnCodingFast — hands-on lessons with worked examples and the output they print, plus practice exercises and a quick quiz.

By the end of this lesson you'll fit a model with lm(), read its coefficients and R-squared, make predictions on new data, and build models with multiple predictors.

What You'll Learn in This Lesson

• How lm() fits a regression with response ~ predictor
• How to read the intercept and slope
• How summary() reports p-values and R-squared
• What R-squared measures
• How to predict() on new data
• How to model with multiple predictors

1️⃣ Fitting a Model

lm(response ~ predictor, data = df) fits the model. coef() pulls out the intercept and slope, which define the best-fit line.

# lm() fits a linear model. The formula is: response ~ predictor.

df <- data.frame(
  hours = c(1, 2, 3, 4, 5),
  score = c(52, 61, 68, 80, 89)
)

model <- lm(score ~ hours, data = df)
coef(model)          # intercept and slope
# (Intercept)       hours
#        42.1         9.3

# Read it as: score ≈ 42.1 + 9.3 * hours
# (each extra hour adds about 9.3 points)

2️⃣ Reading the Model with summary()

summary(model) reports the full results — coefficient estimates, their p-values, and R-squared, the share of variation the model explains.

df <- data.frame(
  hours = c(1, 2, 3, 4, 5),
  score = c(52, 61, 68, 80, 89)
)
model <- lm(score ~ hours, data = df)

# summary() gives the full picture: coefficients, p-values, and R-squared.
summary(model)$r.squared      # how much variation is explained (0-1)
# -> [1] 0.9941...

# Extract pieces:
summary(model)$coefficients[, "Estimate"]   # the fitted coefficients

3️⃣ Predicting and Multiple Predictors

predict(model, newdata) applies the fitted rule to new cases. Add predictors with + in the formula — mpg ~ wt + hp uses both weight and horsepower.

df <- data.frame(
  hours = c(1, 2, 3, 4, 5),
  score = c(52, 61, 68, 80, 89)
)
model <- lm(score ~ hours, data = df)

# predict() applies the model to NEW data.
new_data <- data.frame(hours = c(6, 7))
predict(model, new_data)
#    1     2
# 97.9 107.2

# Multiple predictors use + in the formula:
mtcars_model <- lm(mpg ~ wt + hp, data = mtcars)
round(coef(mtcars_model), 3)
# (Intercept)          wt          hp
#      37.227      -3.878      -0.032

Your turn. Fill in the ___ blank, run it, and compare with the expected output.

# 🎯 YOUR TURN — replace each ___ then run it.

ads <- data.frame(
  spend = c(10, 20, 30, 40, 50),
  sales = c(100, 140, 190, 230, 280)
)

# 1) Fit a model predicting sales from spend
model <- lm(___, data = ads)   # 👉 replace ___ with sales ~ spend

# 2) Print the slope (the "spend" coefficient), rounded to 2 decimals
cat("Slope:", round(coef(model)["spend"], 2), "\n")

# 3) Predict sales for a spend of 60
cat("Predicted at 60:", round(predict(model, data.frame(spend = 60)), 1), "\n")

# ✅ Expected output:
# Slope: 4.5
# Predicted at 60: 323

Mini-Challenge: Model and Report

Write it from the outline, run it, and check it against the example output. Fitting, reporting R-squared, and predicting is the full regression workflow in miniature.

# 🎯 MINI-CHALLENGE: model and report.
#
# Use this data:
#   temp <- c(15, 18, 20, 25, 30, 35)
#   sales <- c(20, 24, 27, 36, 45, 52)   (ice cream sales)
#
# 1. Fit lm(sales ~ temp).
# 2. Print the intercept and slope, each rounded to 2 decimals.
# 3. Print the R-squared from summary(model)$r.squared, rounded to 3.
# 4. Predict sales when temp = 28 with predict().
#
# ✅ Example output (approximate):
#    Intercept: -7.5   Slope: 1.71
#    R-squared: 0.996
#    At 28 degrees: 40.4

# your code here

Common Errors (and the fix)

• Formula reversed — lm(y ~ x) predicts y FROM x. Swapping them models the wrong relationship.
• predict() column mismatch — newdata must use the same predictor names as the formula. data.frame(hours = 6), not x = 6.
• Extrapolating too far — predicting well outside the training range is unreliable; the line may not hold there.
• Forgetting data = — pass the data frame so R finds the columns: lm(y ~ x, data = df).

Pro Tips

• 💡 Plot the fit: in ggplot2, geom_smooth(method = "lm") draws the same regression line visually.
• 💡 Check residuals: plot(model) produces diagnostic plots to test the model's assumptions.
• 💡 Categorical predictors work: include a factor in the formula and R creates the dummy variables for you.

📋 Quick Reference — Linear Models

🎉 Lesson Complete!

• ✅ lm(y ~ x, data = df) fits a regression
• ✅ The intercept and slope define the best-fit line
• ✅ summary() gives p-values and R-squared
• ✅ R-squared is the share of variation explained
• ✅ predict() applies the model to new data

Practice quiz

Continue this course

• Previous: Checkpoint: Data Analysis
• Next: Writing Functions & Packages — Package reusable code, document it, and share it with others
• Quick reference: R cheat sheet › Stats & Models

Frequently asked questions

What does lm() do and how do I read its formula?

lm() fits a linear regression — it finds the straight-line relationship that best predicts a response variable from one or more predictors. The formula syntax is response ~ predictor, read as 'response explained by predictor'. So lm(score ~ hours) models score as a function of hours. The result is an object holding the fitted coefficients (an intercept and a slope per predictor) plus diagnostics you can inspect with summary().

How do I interpret the intercept and slope?

The intercept is the model's predicted value when all predictors are zero — the line's starting height. Each slope (coefficient) tells you how much the response changes for a one-unit increase in that predictor, holding others fixed. If the 'hours' slope is 9.1, then each extra hour of study is associated with about 9.1 more points. Signs matter: a negative slope means the response falls as the predictor rises.

What is R-squared and what counts as 'good'?

R-squared, in summary(model)$r.squared, is the fraction of the response's variation that the model explains, from 0 to 1. An R-squared of 0.85 means the model accounts for 85% of the variability. Higher is generally better, but 'good' depends entirely on the field — physics experiments may expect 0.99, while human-behaviour data might be impressive at 0.3. Don't chase R-squared blindly; a model can fit the sample well yet generalise poorly.

How do I make predictions with a fitted model?

Use predict(model, newdata), where newdata is a data frame with the same predictor column names the model was trained on. For lm(score ~ hours), you'd pass data.frame(hours = c(6, 7)) to predict scores for 6 and 7 hours. Be cautious about predicting far outside the range of your training data (extrapolation) — the linear relationship that held within your data may not continue beyond it.