Linear Regression Models

Unit 2 · Exploring Two-Variable Data

What AP Stats asks here

The least-squares regression line (LSRL) fits the line that minimizes the sum of squared residuals. Examiners look for two things: a numerically correct fit and a context-rich interpretation that includes the word 'predicted', the correct units, and a warning about extrapolation when the prediction sits outside the observed range.

LSRL

Equation

\hat{y} = a + bx

Slope

b = r \cdot \dfrac{s_y}{s_x}

Intercept

a = \bar{y} - b\bar{x}

Mean point

(\bar{x}, \bar{y}) \text{ lies on the LSRL}

Goodness of fit

Proportion of variation explained

R^2 = r^2

Standard deviation of residuals

s = \sqrt{\dfrac{1}{n - 2}\sum (y_i - \hat{y}_i)^2}

AP Tip: Every slope interpretation must include the word 'predicted'. 'For each +1 lb of weight, predicted mpg decreases by 0.00708' — not 'mpg decreases by 0.00708' (which would imply a deterministic claim).

Caution: Predictions outside the observed x-range are extrapolations and are not trustworthy. The linear pattern may not extend beyond the data — flag and avoid.

Type 1

Compute the LSRL

Use $b = r s_y / s_x$ for the slope. The intercept comes from $a = \bar{y} - b\bar{x}$ , which is just the algebra that guarantees the line passes through $(\bar{x}, \bar{y})$ .

Example 1

A study of car weight (lbs) and fuel efficiency (mpg) gives

\bar{x} = 3000

s_x = 600

\bar{y} = 28

s_y = 5

r = -0.85

. Which is the LSRL? (A)

\hat{y} = 49.25 - 0.00708x

(B)

\hat{y} = 28 + 0.00708x

(C)

\hat{y} = 0.00708 - 49.25x

(D)

\hat{y} = 49.25 + 0.00708x

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Type 2

Interpret slope and intercept

Use 'predicted', name the units, name the direction, and tie to context. The intercept is meaningful only if $x = 0$ is in or near the observed range and physically interpretable.

Example 2

For

\widehat{\text{mpg}} = 49.25 - 0.00708 \cdot \text{weight}

(weight in lbs, observed range 2000–4500 lbs), which interpretation pair is correct? (A) Slope: mpg decreases by 0.00708 per lb. Intercept: a 0-lb car gets 49.25 mpg. (B) Slope: for each additional pound, predicted mpg decreases by 0.00708. Intercept: not meaningful in context — 0 lb is far outside the observed range. (C) Slope: weight decreases by 0.00708 per mpg. Intercept: 49.25 is the average mpg in the dataset. (D) Slope and intercept cannot be interpreted without the residual plot.

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Type 3

Predict and check for extrapolation

Plug $x$ into the LSRL, but check whether $x$ sits inside the observed range. Outside that range, the model is not validated.

Example 3

An LSRL was fit on ages 18–35:

\widehat{\text{income}} = -10000 + 2500 \cdot \text{age}

. Three predictions are made: age 25, age 50, age 10. Which choice gives the right trust assessment? (A) All three predictions are trustworthy because the formula is fixed. (B) Age 25 is trustworthy (within range); ages 50 and 10 are extrapolations and should not be trusted. (C) Only age 50 is trustworthy because larger ages give richer data. (D) None of the predictions are trustworthy without raw data.

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