| Term | Estimate |
|---|---|
| Intercept | -105.01 |
| Height (hgt) | 1.02 |
b. Evaluating the Linear Model
Math 115
In 17a we fit a line and learned how to read it. Three things we didn’t settle:
How do we pick the “best” line? Infinitely many lines pass through a cloud of points — some rule has to choose one. That rule is least squares.
How well does the line fit? Strong and weak relationships both produce a line; we need numbers that tell them apart. Those are r and R^2.
Is a line the right tool at all? Some patterns aren’t straight, and a line summarizes them badly. Residual plots are how we check.
In 17a we fit a line to the body-measurements data (n = 507):
\widehat{wgt} = -105 + 1.02 \times hgt
Slope: each extra cm of height raises predicted weight by about 1.02 kg. Intercept: sets the line’s vertical position in the cloud — not a meaningful prediction at height 0.
Try it — the dashed green lines also pass through the cloud and slope upward. What would you say to defend the blue one as best?
Eyeballing narrows it down, but it can’t settle it — two people can prefer different lines and neither can prove the other wrong. “Best” needs a criterion: a number every candidate line can be scored on, so the winner is decided by arithmetic instead of opinion.
For each data point, the residual is the vertical gap between observed and predicted:
e_i = y_i - \hat{y}_i = \text{observed} - \text{predicted}
Try it — our line is \widehat{wgt} = -105 + 1.02 \times hgt. Someone 175 cm tall weighs 90 kg. Find their predicted weight, then their residual. Which side of the line are they on?
Predicted: -105 + 1.02 \times 175 = 73.5 kg. Residual: e = 90 - 73.5 = 16.5 kg. Positive → the point sits above the line, so the model underpredicted this person. A negative residual is the mirror image: below the line, model overpredicted.
The least squares regression line is the one that minimizes the sum of squared residuals:
\sum_{i=1}^{n} e_i^2 = \sum_{i=1}^{n} (y_i - \hat{y}_i)^2
Try it — why square them? Why not just add up the residuals, \sum e_i?
Positive and negative residuals cancel, so the plain sum cannot distinguish a good line from a bad one — even a badly tilted line can sum to zero. Squaring makes every residual positive, so residuals in both directions add together. It also gives more weight to large residuals: a residual of 4 contributes 16, while a residual of 2 contributes 4.
You won’t compute least squares by hand — Jamovi does it for you. The output:
| Term | Estimate |
|---|---|
| Intercept | -105.01 |
| Height (hgt) | 1.02 |
These are the same b_0 = -105 and b_1 = 1.02 from 17a — the line that minimizes \sum e_i^2.
The least-squares line always exists — even when the linear pattern is barely there. We need a separate number to summarize strength (and direction) of the linear relationship.
The correlation coefficient r does that:
-1 \leq r \leq 1
| Value of r | Meaning |
|---|---|
| r close to +1 | strong positive linear relationship |
| r close to -1 | strong negative linear relationship |
| r close to 0 | weak (or no) linear relationship |
Key properties: r is unit-free; swapping x and y doesn’t change r; r only captures linear patterns.
Scatter plots with different correlations. From OpenIntro IMS2, Figure 7.10.
The sign matches the slope direction; the magnitude reflects how tightly the points hug a straight line.
Try it — from the scatter alone, which value is r closest to?
(a) -0.95
(d) 0.3
(b) -0.7
(e) 0.7
(c) -0.3
(f) 0.95
(e) r = 0.717. The cloud slopes upward, so r must be positive: (a), (b), and (c) are out before you judge strength at all. Of the rest, 0.95 would hug the line far more tightly, and 0.3 would be much more scattered, with only a slight upward trend.
The coefficient of determination is just r squared:
R^2 = r^2 \quad \text{(for simple linear regression).}
Interpretation: R^2 is the proportion of variability in y that is explained by x through the linear model.
For our model: R^2 = (0.717)^2 \approx 0.514.
Height explains about 51\% of the variability in weight. The remaining 49\% is due to other factors (build, muscle mass, lifestyle, …).
A random sample of 50 students from one entering class at Elmhurst College. Both variables are in thousands of dollars.
\widehat{\text{Aid}} = 24.32 - 0.0431 \times \text{Income}, \qquad r = -0.499
Interpret the slope in context — then convert it to dollars. Each additional $1,000 of family income is associated with about 0.0431 thousand dollars less gift aid, on average — roughly $43 less.
Predict gift aid for a family income of $100,000. 24.32 - 0.0431 \times 100 = 20.01 thousand — about $20,000.
Compute R^2 and interpret. R^2 = (-0.499)^2 \approx 0.249. About 25% of the variability in gift aid is explained by family income; the other 75% comes from merit awards, athletics, program of study, and need beyond raw income.
A residual plot puts the residual e_i = y_i - \hat{y}_i on the vertical axis and the predicted value \hat{y}_i (or x_i) on the horizontal axis.
What we want to see: points scattered randomly around zero, with no obvious pattern.
Try it: does this plot show random scatter, or a pattern? Is a line the right tool here?
Random scatter around zero — no curve, no fan, roughly constant spread. The linear model is appropriate for these data.
Each column is one dataset: the data with its fitted line on top, the residual plot for those same points directly below.
For each plot: is a line an appropriate summary? If not, what is the problem?
A — random scatter around zero, spread roughly constant. A line is a reasonable summary. B — the fan pattern: residuals sit tightly around zero at the low end, then spread about twice as far once predicted distance gets large. The line still captures the trend, but the model is noticeably more precise for short stopping distances than long ones.
An outlier is a point that stands out from the rest of the data. All three above are outliers — what separates them is how much they affect the line:
Practical check: if you suspect a point is influential, refit the model with it removed and compare the slopes.
Blue line = all points. Dashed line = flagged point (red) removed. Which flagged point actually moved the line — and why?
Only C is influential.
A — an outlier at a typical x, so low leverage. The line barely moves.
B — an outlier with high leverage, but its y agrees with the trend, so the fit is nearly the same with or without it.
C — high leverage and off the pattern. That combination is what makes it influential: the slope falls from 1.12 to 0.77.
By the end of 17b, the regression toolkit describes the relationship in a sample:
| Component | Question it answers |
|---|---|
| b_0, b_1 | What’s the equation of the line? |
| \hat{y} from b_0 + b_1 x | What’s the predicted y? |
| r | How strong is the linear pattern, and in which direction? |
| R^2 | What fraction of variability does x explain? |
| Residual plot | Is a line even the right tool? |
| Refit without a point | Is one point driving the slope? |
Still missing: is the slope real in the population, or could it be sampling noise from a flat true relationship? That’s inference for the slope — topic 18.
bdims in openintro.)elmhurst in openintro.)cars in base R — speeds and stopping distances recorded in the 1920s.)