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 $C = 62v + 410$, where $v$ is the number of vans in a fleet: the slope $62$ is "each additional van is associated with $62 more per month"; the intercept $410$ is "the predicted cost when $v = 0$." 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 children 2 to 10 years old says nothing reliable about a 30-year-old adult. 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. Adding the residuals with their signs proves nothing: a miss of $+5$ and a miss of $-5$ cancel to $0$, so a line with big misses in both directions can still have residuals that "sum to zero." That signed sum 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 association.

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

A line of best fit runs through the middle of the data, with points above it and points below it, so positive and negative residuals cancel and the signed sum lands near zero for a good fit and a bad fit alike. 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 $x$ will produce a change in $y$, is wrong regardless of how well the arithmetic matches.

Distractor Species 5: The Plausible Irrelevant Truth

In extrapolation and interpretation items, one distractor is a true statement that answers a different question — "the slope is negative, so value decreases." True, and beside the point. Ask whether the statement addresses the specific concern the question raised.

Distractor Species 6: Anchor-Point Confusion

When a line is defined by two points, the intercept is not one of the given $y$-values. Using $52$ from $(4, 52)$ as the $y$-intercept produces a clean, wrong prediction.

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Speed Techniques

Residual, one line. actual $=$ predicted $+$ residual. Write it down before computing anything.

Slope from two points, then anchor. Compute $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, then use point-slope from either given point — never assume a given point is the intercept.

Two models, one subtraction. For $y = a_{1} + b_{1}t$ and $y = a_{2} + b_{2}t$, the crossing is at $t = \frac{a_{2} - a_{1}}{b_{1} - b_{2}}$ Dividing by the sum of the slopes is a planted option, so check that you subtracted.

Better fit, at a glance. Scan the two residual lists for the largest absolute value. The model whose worst miss is smaller almost always wins, and you rarely need to compute anything.

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?