Scatterplots & Line of Fit — 700-800

Scatterplots & Line of Fit: The 700-800 Patterns

Part 1 of 3 — The Archetypes Hard-Tier Items Are Built From

At this tier the arithmetic is trivial. What is being tested is whether you know exactly what a line of best fit claims — and, more often, what it does not claim.

Archetype 1: The Residual, With Its Sign Intact

$\text{residual} = \text{actual} - \text{predicted}$

A positive residual means the point sits above the line; a negative residual means below. Hard items give you the residual and the model and ask for the actual value, which means running the definition backwards:

$\text{actual} = \text{predicted} + \text{residual}$

The trap is subtracting when you should add, and it is available as an option every single time.

Archetype 2: Slope and Intercept as Sentences

For $C = 46m + 890$: the slope $46$ is "each additional machine is associated with $46 more per month"; the intercept $890$ is "the predicted cost when $m = 0$." Note associated with, not "causes" — a line of best fit never establishes causation, and an option that says "causes" is wrong on that word alone.

Archetype 3: Extrapolation Beyond the Data

A model built from cars 1 to 8 years old says nothing reliable about a 15-year-old car. Items plant an input far outside the stated range, and the correct answer is always that the prediction is unreliable because the input is outside the range the model was built from. Frequently the model also predicts something impossible — a negative price, a negative mass — which is the confirming clue.

Archetype 4: Two Competing Models

Two lines, two contexts (two cities, two brands, two groups). The questions are always one of three:

• When are they equal? Set the expressions equal and solve.
• Which changes faster? Compare $|\text{slope}|$.
• Which starts higher? Compare intercepts.

Solving for the input and then reporting the output — or vice versa — is the built-in trap.

Archetype 5: Which Model Fits Better

Given two sets of residuals for the same data, the better model is the one whose residuals are smaller in absolute value. Residuals of a least-squares line always sum to about zero, so "these sum to zero" is evidence of nothing at all — and that is precisely the distractor offered.

Archetype 6: What One New Point Does

• A point far above the others at a typical $x$ raises the intercept and barely moves the slope.
• A point at an extreme $x$ (high leverage) can swing the slope substantially.
• Any point far off the pattern weakens the correlation.

Scatterplots & Line of Fit: Traps & Speed

Part 2 of 3 — Distractor Autopsy

Distractor Species 1: The Sign of the Residual

Every residual item offers both $+r$ and $-r$. Anchor on the picture, not the formula: above the line is positive, below is negative. If the actual value is smaller than the predicted value, the residual is negative — no exceptions.

Distractor Species 2: Input Reported as Output

"After how many hours are the charges equal?" and "what is the charge when they are equal?" have different answers, and both are in the options. Circle the unit in the final clause before you solve.

Distractor Species 3: The Sum-of-Residuals Decoy

The residuals of a least-squares line always sum to approximately zero — for any least-squares fit, good or bad. So "Model 1's residuals sum to zero, therefore Model 1 fits better" is a statement with no content. Compare magnitudes, never the signed sum.

Distractor Species 4: "Causes"

The correct interpretation of a slope is always associated with. An option that says the predictor causes the response, or that changing a change in , is wrong regardless of how well the arithmetic matches.

Scatterplots & Line of Fit: Timed Drill

Part 3 of 3 — Four Items at Test Pace

About 90 seconds each. These cover the four highest-frequency hard-tier skeletons: the two-model crossing, the signed residual, the slope-as-a-sentence interpretation, and the two-points-at-the-same-$x$ residual chain.

Before each answer:

1. What unit does the final clause ask for? Time or value? Actual or predicted?
2. Did I subtract slopes, not add them?
3. Is the residual's sign consistent with above/below the line?