Construct and interpret scatterplots, calculate the correlation coefficient and least-squares regression line, and use the line to make predictions.

How to read scatterplots, measure linear association with Pearson's r and the coefficient of determination, fit the least-squares line, and predict while judging interpolation versus extrapolation.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Describing a scatterplot
Correlation coefficient
Least-squares regression line
Interpolation and extrapolation
Correlation is not causation

What this dot point is asking

You must identify the explanatory and response variables, describe the scatterplot, find and interpret $r$ and $r^2$ , write the least-squares equation, and make and judge predictions.

Describing a scatterplot

Plot the explanatory variable $x$ on the horizontal axis and the response variable $y$ on the vertical axis. Then describe:

Form: linear or non-linear.
Direction: positive (rises) or negative (falls).
Strength: how closely points cluster about a line.
Outliers: any points well away from the pattern.

Correlation coefficient

Pearson's correlation coefficient $r$ measures the strength and direction of a linear association. It satisfies $-1 \le r \le 1$ : $r = +1$ is a perfect positive line, $r = -1$ a perfect negative line, and $r = 0$ no linear association.

Least-squares regression line

The least-squares line minimises the sum of the squared vertical distances from the points to the line. Write it as $y = a + bx$ , where $b$ is the slope and $a$ is the $y$ -intercept.

The slope $b$ is the predicted change in $y$ for each one-unit increase in $x$ . The intercept $a$ is the predicted $y$ when $x = 0$ (meaningful only if $x = 0$ is sensible in context).

Worked example

Finding and using the least-squares line

A study of hours studied ( $x$ ) and test score ( $y$ ) gives $\bar{x} = 5$ , $\bar{y} = 62$ , $s_x = 2$ , $s_y = 10$ and $r = 0.8$ .

Slope: $b = r\dfrac{s_y}{s_x} = 0.8 \times \dfrac{10}{2} = 0.8 \times 5 = 4$ .

Intercept: $a = \bar{y} - b\bar{x} = 62 - 4 \times 5 = 62 - 20 = 42$ .

So $\text{score} = 42 + 4 \times \text{hours}$ . Each extra hour of study predicts $4$ more marks. For $6$ hours: $\text{score} = 42 + 4 \times 6 = 66$ . Because $6$ lies inside the data range this is interpolation, so the prediction is reasonably reliable.

Interpolation and extrapolation

Interpolation predicts within the range of the data and is usually reliable. Extrapolation predicts outside the range and is risky, because the linear pattern may not continue.

Correlation is not causation

A strong $r$ shows the variables move together; it does not prove that one causes the other. A lurking third variable, or coincidence, can produce correlation without any causal link.