We model the average response as a straight line:

\[ Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i \]

• \(\beta_0\): intercept, the predicted response when \(X=0\).
• \(\beta_1\): slope, the change in predicted response for a one-unit increase in \(X\).
• \(\varepsilon_i\): unexplained variation (error).

We choose the line that minimizes the sum of squared residuals:

\[ \min_{b_0,b_1}\sum_{i=1}^{n}\left(y_i-b_0-b_1x_i\right)^2 \]

For these data, the fitted line is:

\[ \widehat{\mathrm{mpg}} = 37.29 - 5.34\,\mathrm{wt} \]

Interpretation: Each additional 1,000 lb is associated with about 5.34 fewer MPG, on average.

The coefficient of determination, \(R^2\), describes the fraction of variation in the response explained by the model.

\[ R^2 = 1 - \frac{\sum_i (y_i-\hat y_i)^2}{\sum_i(y_i-\bar y)^2} \]

For our example, \(R^2 = 0.753\) — approximately 75.3% of the variation in MPG is explained by a linear relationship with weight.

library(ggplot2)

fit <- lm(mpg ~ wt, data = mtcars)
summary(fit)

ggplot(mtcars, aes(x = wt, y = mpg)) +
  geom_point(color = "#8C1D40", size = 3) +
  geom_smooth(method = "lm", se = TRUE,
              color = "#147D86") +
  labs(x = "Weight (1,000 lb)", y = "MPG") +
  theme_minimal()

Suppose a car weighs 3,000 lb, so its wt value is 3.0.

\[ \widehat{\mathrm{mpg}} = 37.29 - (5.34)(3.0) \approx 21.25 \]

The model predicts approximately 21.3 MPG for a car weighing 3,000 lb.