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WAMathematics ApplicationsSyllabus dot point

How can we measure and use the relationship between two numerical variables?

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 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
The bivariate workflow
Correlation and determination
Predictions and their reliability
Residual plots as the final check

What this dot point is asking

You must construct and read scatterplots, calculate $r$ and the regression line, interpret them, and predict.

The bivariate workflow

Every regression question follows the same order.

Scatterplot: plot the explanatory variable on the horizontal axis and describe direction, form, strength and outliers.
Correlation: find $r$ to quantify the linear association.
Determination: square it for $r^2$ , the proportion of variation explained.
Line: fit $y = a + bx$ with technology.
Predict: substitute, and label the prediction interpolation or extrapolation.

Correlation and determination

Read all four numbers from the calculator's regression output after entering the paired data. The standard interpretation sentence for $r^2$ is " $r^2 \times 100\%$ of the variation in [response] is explained by the variation in [explanatory]".

Worked example

A full regression analysis

A gym records members' weekly training hours ( $x$ ) and their fitness-test score ( $y$ ) for ten members, with hours ranging $2$ to $9$ . Technology returns $r = 0.82$ and $y = 38 + 5.5x$ .

Describe and quantify

The scatterplot is positive, roughly linear and strong. With $r = 0.82$ , $r^2 = 0.82^2 = 0.6724$ , so about $67\%$ of the variation in fitness score is explained by the linear relationship with training hours.

Interpret the slope and intercept

Slope $5.5$ : each extra hour of training per week is associated with a $5.5$ -point increase in predicted fitness score. Intercept $38$ : a member who trains $0$ hours is predicted to score $38$ (a meaningful baseline here, since $0$ hours is realistic).

Predict at 6 hours (interpolation)

$x = 6$ is inside the data range, so interpolation: $y = 38 + 5.5(6) = 38 + 33 = 71$ .

Comment on 15 hours

$x = 15$ is well beyond the data, so predicting there is extrapolation and unreliable; the linear gain may not continue at very high training loads.

Predictions and their reliability

A prediction is only as good as the model and the data range. Interpolation (inside the data) is reliable when $r^2$ is high; extrapolation (outside the data) is always flagged as risky. The reliability also depends on $r^2$ : a model explaining $90\%$ of variation gives tighter predictions than one explaining $40\%$ . Finally, remember that a strong association never proves causation; an observed correlation may be driven by a confounder.

Residual plots as the final check

After fitting the line, a residual plot confirms whether the straight line was appropriate. If the residuals scatter randomly about zero, the linear model holds and predictions are trustworthy within the data range. If they form a curve or a fan, the relationship is non-linear and the data should be transformed (squared, reciprocal or log) before re-fitting. A high $r$ alone does not justify a line, because curved data can still produce a large $r$ ; the residual plot is the more reliable diagnostic and ties the whole regression analysis together.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20218 marksData on

12

houses gives floor area (

x

, square metres) and price (

y

\

thousands). Technology returns

r = 0.88

and the least-squares line

y = 95 + 2.4x

, with areas from

to

260

square metres. (a) Interpret

and

r^2

. (b) Interpret the slope. (c) Predict the price of a

180

square metre house. (d) Comment on predicting the price of a

400$ square metre house.

Show worked answer →

Work through correlation, determination, slope, then predictions.

(a) $r = 0.88$ is a strong positive linear association: larger houses tend to cost more. $r^2 = 0.88^2 = 0.7744$ , so about $77\%$ of the variation in price is explained by the linear relationship with floor area. (3 marks)

(b) Slope $2.4$ : each extra square metre is associated with a $\$2400$ increase in predicted price (since price is in $\$$ thousands). (2 marks)

(c) $x = 180$ is within the data range, so interpolation: $y = 95 + 2.4(180) = 95 + 432 = 527$ , that is $\$527\,000$ . (2 marks)

(d) $x = 400$ is far beyond the data ( $80$ to $260$ ), so this is extrapolation and unreliable; the linear trend may not hold for very large houses. (1 mark)

Markers reward correct $r$ / $r^2$ interpretation, slope in dollars, a labelled interpolation, and identifying extrapolation.

WACE 20235 marksA least-squares line of test score

y

on hours of sleep

x

y = 30 + 6x

with

r = 0.65

. (a) Calculate and interpret the coefficient of determination. (b) A student claims more sleep causes higher scores. Evaluate this claim.

Show worked answer →

Square $r$ , then address causation.

(a) $r^2 = 0.65^2 = 0.4225$ , so about $42\%$ of the variation in test score is explained by the linear relationship with hours of sleep, leaving most of the variation to other factors. (3 marks)

(b) The data shows association, not causation. The claim is not justified from observational data: a confounder such as overall study habits or wellbeing could drive both sleep and scores. (2 marks)

Markers reward the $r^2$ interpretation and rejecting the causal claim with a plausible confounder.