How to solve SAT scatterplot and line of best fit questions

Two-variable data questions show how one quantity relates to another, usually through a scatterplot with a line or curve of best fit, or through a table of paired values. On the Digital SAT you interpret the slope and intercept of a model, make predictions, compare actual values with predicted ones, and decide whether a linear, exponential, or quadratic model fits best. Problem-Solving and Data Analysis is about 15% of the Math section, or 5 to 7 of the 44 questions. Those questions are split across seven skills, so expect about one question on this skill, sometimes two.

What the questions look like

Interpret a line of best fit. You get a scatterplot and the equation of its line of best fit, and you choose what the slope or $y$ -intercept means, or use the line to predict a value.

Actual versus predicted. You compare a data point with the value the model predicts, often by finding the residual, which is the difference between them.

Choose a model. From a scatterplot or a table, you decide whether the relationship is linear, exponential, or quadratic, or choose the equation that best fits the data shown.

Step-by-step method

Read both axes first. Note the variable, the units, and the scale on each axis. Axes on the SAT do not always start at 0, and gridlines may count by 2, 5, or 10.
Interpret the slope as a predicted rate. In $y = mx + b$ , $m$ is the predicted change in $y$ for each increase of 1 in $x$ . Use "predicted" or "estimated," because the line describes a trend, not every point.
Interpret the intercept as the predicted value at $x = 0$ . That value may not make sense in context, and the SAT sometimes asks you to recognize that.
Predict by substitution. Plug the $x$ value into the model. The result is an estimate.
Find residuals as actual minus predicted. A positive residual means the point lies above the line. A negative residual means it lies below.
Pick the model type from the pattern.
- Equal differences in $y$ for equal steps in $x$ mean linear.
- Equal ratios in $y$ for equal steps in $x$ mean exponential.
- A curve that rises and then falls, or falls and then rises, suggests quadratic.
Match equations to graphs. The sign of the slope or growth factor and the approximate $y$ -intercept usually eliminate all but one choice.

Worked example

Example 1. For a group of students, the line of best fit relating hours studied, $h$ , to test score, $s$ , is $s = 4.2h + 61$ . One student studied for 6 hours and scored 83.

(a) What is the best interpretation of 4.2? (b) What is the residual for this student?

(a) The slope 4.2 means that for each additional hour of study, the model predicts the test score will be 4.2 points higher. It does not mean each student gained exactly 4.2 points per hour.

(b) The predicted score at $h = 6$ is

$s = 4.2(6) + 61 = 25.2 + 61 = 86.2$

The residual is actual minus predicted: $83 - 86.2 = -3.2$ . The student scored 3.2 points below what the model predicts, so the point lies below the line.

Example 2. The table shows a quantity $y$ at several values of $x$ . Which type of model fits best?

$x$	0	1	2	3
$y$	200	260	338	439.4

Check differences first: $60$ , $78$ , and $101.4$ . They are not constant, so the relationship is not linear.

Check ratios: $\frac{260}{200} = 1.3$ , $\frac{338}{260} = 1.3$ , and $\frac{439.4}{338} = 1.3$ . The ratios are constant, so the model is exponential:

$y = 200(1.3)^x$

The quantity starts at 200 and increases by 30% for each increase of 1 in $x$ .

Common traps

Reversing the residual. Residual is actual minus predicted. In Example 1, $86.2 - 83 = 3.2$ has the wrong sign and puts the point above the line.
Claiming cause and effect. A line of best fit shows an association. Choices that say studying "causes" higher scores go beyond what a scatterplot shows. Look for "is associated with" or "predicts."
Treating predictions as exact. The model gives estimated values. An answer choice that says a specific student "will score" a value is too strong.
Extrapolating far outside the data. If the data cover 0 to 10 hours of study, the line's prediction at 40 hours is unreliable, and it may predict an impossible score.
Misreading the scale. If the vertical axis starts at 50 instead of 0, slopes look steeper than they are. Compute the slope from labeled values, not from how the graph looks.

Desmos tip

Enter the data in a table with columns $x_1$ and $y_1$ . Type y1 ~ mx1 + b for a linear fit or y1 ~ ab^x1 for an exponential fit. Desmos reports the parameters and a fit statistic ( $r$ or $R^2$ ), and for the table above the exponential fit returns $a = 200$ and $b = 1.3$ . Desmos also stores the residuals in a list called $e_1$ , which you can plot to see which points fall above or below the model.