Correlation coefficient and coefficient of determination in WACE Mathematics Applications Unit 3

Syllabus dot point

“Calculate and interpret Pearson's correlation coefficient r and the coefficient of determination r squared, and state their limitations.”

WACEMathematics ApplicationsUnit 37 min readUpdated 2 Jun 2026

Quick answer

Pearson's correlation coefficient $r$ measures the strength and direction of a linear association on a scale from $-1$ to $+1$ . The sign gives the direction and the size gives the strength. Squaring it gives the coefficient of determination $r^2$ , the proportion of variation in the response variable explained by the linear relationship with the explanatory variable. Both measure linear association only and both are distorted by outliers.

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What this dot point is asking
Pearson's correlation coefficient
The coefficient of determination
Limitations of r and r squared
Exam-style questions

What this dot point is asking

You must calculate $r$ with technology, interpret its sign and size, convert to $r^2$ and interpret it as explained variation, and state the limitations of each.

Pearson's correlation coefficient

The correlation coefficient $r$ is a single number summarising how closely points cluster about a straight line.

The correlation coefficient r

$r$ lies between $-1$ and $+1$ . The sign matches the direction of the association (positive or negative) and the magnitude measures strength: $|r|$ near $1$ is strong, near $0$ is weak. $r = 0$ means no linear association.

A working scale for describing strength:

$0.75 \le |r| \le 1$ : strong.
$0.5 \le |r| < 0.75$ : moderate.
$0.25 \le |r| < 0.5$ : weak.
$|r| < 0.25$ : little or no linear association.

In SCSA Mathematics Applications you find $r$ with the statistics menu of your calculator after entering the paired data, never by hand. Always describe it with three words: strength, direction and (after checking the scatterplot) form.

The coefficient of determination

Squaring $r$ gives $r^2$ , which has a sharper interpretation than $r$ itself.

Coefficient of determination

$r^2 = (r)^2$ , always between $0$ and $1$ . Interpreted as a percentage, $r^2$ is the proportion of the variation in the response variable that is explained by the linear relationship with the explanatory variable. The remaining $1 - r^2$ is explained by other factors.

If $r = 0.8$ then $r^2 = 0.64$ , so $64\%$ of the variation in the response variable is explained by the linear relationship and $36\%$ is due to other factors. The standard sentence is " $r^2 \times 100\%$ of the variation in [response variable] is explained by the variation in [explanatory variable]".

Worked example

From r to a full interpretation

A nurse records patients' age ( $x$ , years) and recovery time ( $y$ , days). Technology returns $r = 0.71$ .

Describe the correlation

$r = 0.71$ is positive and moderate-to-strong: older patients tend to take longer to recover, in a roughly linear way (after confirming the scatterplot looks linear).

Calculate the coefficient of determination

$r^2 = 0.71^2 = 0.5041$ , that is about $50.4\%$ .

Interpret in context

About $50\%$ of the variation in recovery time is explained by the linear relationship with age. The other $50\%$ is due to other factors such as general health, type of illness and treatment.

Comment on a prediction

Because only half the variation is explained, a least-squares prediction of recovery time from age alone carries real uncertainty. A higher $r^2$ would give more confidence in predictions.

Limitations of r and r squared

Both measures assume a linear pattern and are easily misread.

They measure linear association only. On curved data, $r$ can be high yet a straight line is wrong; always inspect the scatterplot first.
Outliers distort them. One extreme point can pull $r$ up or down sharply.
They do not prove causation. A strong $r$ shows association, not that one variable drives the other.
They ignore the slope. Two data sets with different slopes can share the same $r$ ; $r$ is about scatter about the line, not the steepness of the line.

Describe the scatterplot before trusting r

The correlation coefficient is only meaningful once the form is confirmed linear. A strong $r$ on a curved plot is a warning to transform the data, not evidence of a linear relationship.

Common errors

Forgetting the sign: $r = -0.9$ is a strong negative association; dropping the minus sign reverses the meaning.
Reading $r^2$ as a fraction of points: $r^2$ is the proportion of variation explained, not the share of data points lying on the line.
Quoting $r$ on curved data: Check the form first; a high $r$ on a curve is misleading.
Confusing $r$ and $r^2$: Strength of association is read from $r$ ; explained variation is read from $r^2$ . Use the right one for the question.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style5 marks

For a set of bivariate data the least-squares analysis gives a correlation coefficient

r = -0.92

between rainfall (

x

, mm) and number of bushfires (

y

). (a) Describe the association. (b) Calculate the coefficient of determination and interpret it in context.

Show worked answer →

Read the sign and size of $r$ , then square it.

(a) With $r = -0.92$ the association is strong, negative and (assuming linear form) close to a straight line: as rainfall increases, the number of bushfires tends to decrease. (2 marks)

(b) $r^2 = (-0.92)^2 = 0.8464 = 84.6\%$ . So about $85\%$ of the variation in the number of bushfires can be explained by the linear relationship with rainfall. (3 marks)

Markers reward direction-strength-form for $r$ , and an $r^2$ interpretation phrased as the proportion of variation in the response variable explained.

WACE 2023-style4 marks

A student calculates

r = 0.95

for a clearly curved scatterplot and concludes the variables have a strong linear relationship. Explain two limitations of the correlation coefficient that this conclusion ignores.

Show worked answer →

Pearson's $r$ has assumptions the student has overlooked.

Limitation 1: $r$ only measures linear association. A high $r$ on curved data is misleading, because the points may follow a strong non-linear pattern that a straight line does not capture. Form must be checked first. (2 marks)

Limitation 2: $r$ is sensitive to outliers, so a single extreme point can inflate or deflate $r$ and make a poor linear fit look strong. (2 marks)

Markers reward the linearity assumption and outlier sensitivity, each tied back to the student's error.