Bivariate data analysis and correlation (QCE General Mathematics Unit 3)

Syllabus dot point

“Identify the explanatory and response variables, construct and use a scatterplot to describe an association between two numerical variables in terms of direction, form and strength, calculate Pearson's correlation coefficient r and the coefficient of determination R^2 from raw data using technology and interpret them, and use r to determine R^2 and vice versa”

14 min readUpdated 23 Sep 2026

Quick answer

Put the explanatory variable on the $x$ -axis and the response variable on the $y$ -axis, then describe the scatterplot by direction (positive or negative), form (linear or non-linear) and strength (strong, moderate or weak), noting outliers. Pearson's $r$ , calculated with technology, measures the direction and strength of a linear association from $-1$ to $1$ . The coefficient of determination $R^2 = r^2$ is the proportion of the variation in the response variable explained by the explanatory variable; to go back, $r = \pm\sqrt{R^2}$ with the sign from the direction.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

Bivariate data are pairs of measurements taken on the same individuals, such as the temperature and the number of pies sold on each day. The QCAA General Mathematics 2025 syllabus (Unit 3, Topic 1, Bivariate data analysis 1, "Identifying and describing associations between two numerical variables") asks you to

identify the explanatory variable and the response variable,
construct and use a scatterplot to identify the association,
describe the association in terms of direction (positive or negative), form (linear or non-linear) and strength (strong, moderate or weak),
calculate Pearson's correlation coefficient, $r$ , from raw data using technology, and interpret it to quantify the strength of a linear association,
calculate the coefficient of determination, $R^2$ , from raw data using technology, and interpret it in terms of explained variation, and
use $r$ to determine $R^2$ , and vice versa.

The same topic covers associations between two categorical variables (two-way tables), which has its own page. Fitting a line and making predictions comes next, in Topic 2.

The answer

Explanatory and response variables

The explanatory variable is the one that may explain, influence or be used to predict the other. The response variable is the one that responds. Ask: "which one would I use to predict the other?"

Context	Explanatory ( $x$ )	Response ( $y$ )
Hours of study and exam mark	hours of study	exam mark
Age of a car and its price	age	price
Temperature and ice-creams sold	temperature	ice-creams sold

The explanatory variable goes on the horizontal axis of a scatterplot and into the calculator's $x$ list.

Constructing a scatterplot

Each pair becomes one point. Put the explanatory variable on the horizontal axis, choose an even scale on each axis that covers the data (an axis need not start at 0), label each axis with the variable and its units, and plot every point accurately.

Describing an association

Describe three features, always in context:

Direction. Positive if $y$ tends to increase as $x$ increases (points rise from left to right); negative if $y$ tends to decrease as $x$ increases.
Form. Linear if the points are scattered about a straight line; non-linear if they follow a curve.
Strength. How closely the points follow the pattern: strong, moderate or weak. If there is no pattern at all, say there is no association.

Also mention any outliers: points well away from the rest.

A full description sounds like: "There is a strong, negative, linear association between evening screen time and hours of sleep: students who spend more time on screens after 6 pm tend to sleep less."

Pearson's correlation coefficient, $r$

$r$ measures the direction and strength of a linear association on a scale from $-1$ to $+1$ . You calculate it from the raw data with your calculator's statistics (two-variable or linear regression) mode.

The sign gives the direction: $r > 0$ positive, $r < 0$ negative.
The size, $|r|$ , gives the strength: $|r| = 1$ is a perfect linear association (all points on a line) and $r = 0$ means no linear association.

A commonly used guide for describing strength:

$\lvert r \rvert$	Strength of linear association
0.75 to 1	strong
0.5 to 0.75	moderate
0.25 to 0.5	weak
below 0.25	none (or very weak)

Two important limits:

$r$ describes linear association only. Always look at the scatterplot first. Points on a perfect U-shaped curve (such as $y = x^2$ for $x$ from $-3$ to $3$ ) have $r = 0$ , even though the relationship is exact.
$r$ is sensitive to outliers. One unusual point can greatly strengthen or weaken it.

The coefficient of determination, $R^2$

$R^2$ is the square of $r$ (for a linear model), and it lies between 0 and 1. It tells you the proportion of the variation in the response variable that is explained by the variation in the explanatory variable. Quote it as a percentage and use the variable names:

" $R^2 = 0.923$ , so about 92.3% of the variation in students' hours of sleep can be explained by the variation in their evening screen time."

The remaining $1 - R^2$ is due to other factors and random variation.

Converting between $r$ and $R^2$

R^2 = r^2, \qquad r = \pm\sqrt{R^2}.

Squaring loses the sign, so when you go from $R^2$ back to $r$ , choose the sign from the direction of the association (the scatterplot or the slope of the line). For example, $R^2 = 0.64$ with a negative association gives $r = -0.8$ .

Key fact

Explanatory variable on the $x$ -axis, response on the $y$ -axis. Describe an association by direction, form and strength (and outliers). Pearson's $r$ , from technology, measures the direction and strength of a linear association ( $-1 \le r \le 1$ ). $R^2 = r^2$ ( $0 \le R^2 \le 1$ ) is the proportion of the variation in the response variable explained by the explanatory variable. $r = \pm\sqrt{R^2}$ , with the sign from the direction.

Worked examples: describing and measuring association

1. Using the calculator

Use the screen time and sleep data (screen time 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 5.0, 5.5, 6.0 hours; sleep 8.9, 8.6, 8.8, 8.1, 7.9, 8.2, 7.4, 7.3, 6.8, 6.9 hours) to find $r$ and $R^2$ .

Variables: Screen time is explanatory ( $x$ ); sleep is the response ( $y$ ).
Calculator: Enter the lists in statistics mode and choose linear regression: $r = -0.96078$ .
$R^2$: $(-0.96078)^2 \approx 0.923$ .
Answer: $r \approx -0.961$ , $R^2 \approx 0.923$ .

Marker's note: write $r$ to three decimal places and square the unrounded value.

2. Describing the association in context

Describe the association in Example 1.

Direction: $r < 0$ and the points fall from left to right: negative.
Form: The points are scattered about a straight line: linear.
Strength: $|r| = 0.961 \ge 0.75$ : strong.
Answer: There is a strong, negative, linear association between evening screen time and hours of sleep, with no obvious outliers. Students who spend more time on screens after 6 pm tend to sleep less.

3. Interpreting $R^2$

Interpret $R^2 \approx 0.923$ in Example 1.

Answer. About 92.3% of the variation in the students' hours of sleep can be explained by the variation in their evening screen time. The other 7.7% is due to other factors (such as homework, sport or caffeine) and random variation.

Marker's note: "92.3% of the sleep is caused by screen time" is wrong on two counts. $R^2$ is about variation, and it does not show causation.

4. From $R^2$ to $r$

For a group of houses, a scatterplot of distance from the city centre (km) against price shows prices falling as distance increases, with $R^2 = 0.5476$ . Find $r$ .

Magnitude: $\sqrt{0.5476} = 0.74$ .
Sign: Prices fall as distance increases: negative.
Answer: $r = -0.74$ , a moderate negative linear association (just under the 0.75 guide for "strong").

5. The effect of an outlier

An eleventh student in Example 1 reports 2 hours of screen time and only 5 hours of sleep (they were unwell). Recalculate $r$ .

Calculator. With the extra point, $r \approx -0.401$ .

Interpretation. One unusual point has changed the description from strong to weak. Investigate outliers before trusting $r$ , and report the analysis with and without the point if it is unrepresentative.

6. When $r$ misleads

The points $(-3, 9)$ , $(-2, 4)$ , $(-1, 1)$ , $(0, 0)$ , $(1, 1)$ , $(2, 4)$ , $(3, 9)$ lie exactly on $y = x^2$ . Find $r$ and comment.

Calculator. $r = 0$ .

Comment. There is no linear association, but there is a perfect non-linear one. This is why you always check the form on a scatterplot before using $r$ .

Common traps

Swapping the variables: Decide which variable explains the other before plotting or entering data.
Describing with numbers only: " $r = -0.961$ " is not a description. Use the words direction, form and strength, in context.
Giving $r$ as the percentage explained: The percentage of variation explained comes from $R^2$ , not $r$ . $r = 0.9$ means 81% explained, not 90%.
Losing the sign: $R^2$ is never negative. When converting $R^2$ to $r$ , take the sign from the direction of the association.
Using $r$ for a curved pattern: $r$ only measures linear association. Check the scatterplot first.
Ignoring outliers: A single outlier can change $r$ a lot.
Claiming causation: A strong association does not show that one variable causes the other; see the association and causation page.

Exam technique

For "describe the association" questions, write one sentence with the three words (direction, form, strength) and the variable names, then add a second sentence in plain language. For $R^2$ , use the sentence frame "About ...% of the variation in [response] can be explained by the variation in [explanatory]". When given $R^2$ and asked for $r$ , write $r = \pm\sqrt{R^2}$ and justify the sign from the scatterplot or context. If the question gives two possible explanatory variables, calculate $r$ (or $R^2$ ) for each and compare them explicitly.

Note

Suppose you write down, for each student in your class, how long they spent on their phone last night and how long they slept. If you plot these as dots, you might see that the dots slope downwards: more phone, less sleep. That downward trend is a "negative association". The correlation number $r$ tells you how close the dots are to a straight line, from $-1$ (perfect downhill line) to $+1$ (perfect uphill line). Squaring it gives $R^2$ , which tells you what share of the differences in sleep can be linked to differences in phone time.

Exam-style questions

Questions in the style of QCAA 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.

Practice5 marks

The maximum temperature and the number of pies sold each day at a bakery are provided in the table. Maximum temperature (degrees C): 29, 20, 31, 27, 23, 25, 22, 33. Number of pies sold: 32, 39, 25, 33, 37, 35, 37, 30. a) Construct a scatterplot to display the data on the grid provided. [3 marks] b) Describe the association between the maximum temperature and the number of pies sold in terms of direction and strength. [2 marks]

Show worked answer →

a) Scatterplot (3 marks). Maximum temperature is the explanatory variable, so it goes on the horizontal axis; pies sold goes on the vertical axis (1 mark). Scale both axes evenly to cover the data (for example 20 to 34 °C and 24 to 40 pies) and label them with units (1 mark). Plot all eight points accurately, such as $(29, 32)$ , $(20, 39)$ and $(31, 25)$ (1 mark).

b) Description (2 marks). As the maximum temperature increases, the number of pies sold tends to decrease, so the association is negative (1 mark). The points lie close to a straight line ( $r \approx -0.905$ ), so it is strong (1 mark).

A full description is "a strong, negative, linear association between maximum temperature and pies sold", but the two marks here are for direction and strength.

Practice4 marks

Hiroki believes more fish are caught on warmer days. Jiro believes the number of fish caught is more dependent on the number of people fishing. Temperature, t (degrees C): 32, 26, 20, 27, 23, 29. Number of fish caught, f: 530, 400, 320, 220, 180, 120. Number of people fishing, p: 46, 58, 38, 34, 30, 28. Calculate the correlation coefficient for each dataset and use the results to identify the explanatory variable for the stronger linear association. Use the least-squares line equation for the stronger association to predict the number of fish caught on a 25 degrees C day when 50 people are fishing.

Show worked answer →

Step 1: two correlations (1 mark): From the calculator, $r \approx 0.309$ for temperature and fish caught, and $r \approx 0.807$ for people fishing and fish caught.
Step 2: choose the explanatory variable (1 mark): $0.807$ is much closer to 1, so the number of people fishing has the stronger linear association: Jiro is correct, and $p$ is the explanatory variable.
Step 3: least-squares line (1 mark): $f = 10.89p - 129.8$ .
Step 4: predict (1 mark): The temperature is a distractor. $f = 10.89 \times 50 - 129.8 \approx 414.7$ , so about 415 fish.

$R^2 \approx 0.65$ for people fishing, so about 65% of the variation in fish caught is explained by the number of people fishing, compared with under 10% by temperature.

Original4 marks

The age (years) and price ($'000) of eight used cars of the same model are: age 1, 2, 3, 4, 5, 6, 7, 8; price 31.5, 27.0, 27.4, 22.1, 22.8, 17.2, 18.9, 13.6. (a) Identify the explanatory and response variables. (b) Use your calculator to determine

r

and

R^2

. (c) Describe the association. (d) Interpret

R^2

in context.

Show worked answer →

(a) Age is the explanatory variable; price is the response variable (age helps explain price, not the reverse). (1 mark)

(b) $r \approx -0.963$ and $R^2 \approx 0.928$ . (1 mark)

(c) A strong, negative, linear association: older cars of this model tend to sell for less. (1 mark)

(d) About 92.8% of the variation in the price of these cars can be explained by the variation in their age. (1 mark)

Original3 marks

A scatterplot of hours of weekly exercise against resting heart rate shows a negative linear association. The coefficient of determination is

R^2 = 0.64

. (a) Determine the value of

r

. (b) Describe the strength and direction of the association. (c) What percentage of the variation in resting heart rate is **not** explained by exercise?

Show worked answer →

(a) $|r| = \sqrt{0.64} = 0.8$ . The association is negative, so $r = -0.8$ . (1 mark)

(b) A strong, negative association: people who exercise more tend to have lower resting heart rates. (1 mark)

(c) $1 - 0.64 = 0.36$ , so 36% of the variation is due to other factors (such as age, fitness history or genetics) or random variation. (1 mark)

Original4 marks

A farmer records annual rainfall (mm), fertiliser used (kg/ha) and crop yield (t/ha) for eight seasons. Rainfall: 410, 520, 380, 610, 450, 560, 490, 630. Fertiliser: 120, 90, 150, 110, 160, 100, 140, 130. Yield: 3.2, 3.4, 3.1, 3.8, 3.6, 3.5, 3.3, 4.1. (a) Calculate

r

for rainfall against yield and for fertiliser against yield. (b) Which variable is the better explanatory variable for yield? Justify using

R^2

Show worked answer →

(a) Rainfall: $r \approx 0.864$ . Fertiliser: $r \approx -0.106$ . (2 marks)

(b) Rainfall. $R^2 \approx 0.746$ , so about 75% of the variation in yield is explained by rainfall, compared with $R^2 \approx 0.011$ (about 1%) for fertiliser, which shows almost no linear association in these data. (2 marks)

Note that this does not show fertiliser has no effect in general; the farmer varied it only over a small range, and other variables changed at the same time.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Identify the explanatory and response variables: (a) the number of hours a candle has burned and its remaining height; (b) a person's height and their arm span, when using height to predict arm span.

Show worked solution →

(a) Explanatory: hours burned. Response: remaining height. (1 mark)

(b) Explanatory: height. Response: arm span. (1 mark)

foundation2 marks

r = -0.7

. Find

R^2

and interpret it for a study of temperature (explanatory) and heating costs (response).

Show worked solution →

$R^2 = (-0.7)^2 = 0.49$ . (1 mark)

About 49% of the variation in heating costs can be explained by the variation in temperature. (1 mark)

foundation2 marks

R^2 = 0.81

and the scatterplot slopes upward from left to right. Find

r

and describe the association.

Show worked solution →

$r = +\sqrt{0.81} = 0.9$ (positive because the scatterplot slopes upward). (1 mark)

A strong, positive association. (1 mark)

core3 marks

Daily screen time after 6 pm (hours) and sleep (hours) for ten students: screen time 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 5.0, 5.5, 6.0; sleep 8.9, 8.6, 8.8, 8.1, 7.9, 8.2, 7.4, 7.3, 6.8, 6.9. Find

r

and

R^2

and describe the association.

Show worked solution →

$r \approx -0.961$ and $R^2 \approx 0.923$ . (2 marks)

A strong, negative, linear association: students with more evening screen time tend to sleep less. About 92% of the variation in sleep is explained by screen time. (1 mark)

core2 marks

A data set has

r = 0.08

, but the scatterplot shows the points lying close to a U-shaped curve. A student says "there is no relationship". Evaluate this claim.

Show worked solution →

$r$ only measures linear association, so $r \approx 0$ means there is almost no linear association. (1 mark)

The U-shaped pattern shows a strong non-linear relationship, so the claim is wrong: there is a relationship, but it is not linear and $r$ should not be used to describe it. (1 mark)

exam3 marks

For the ten students in the screen-time question,

r \approx -0.961

. An eleventh student reports 2 hours of screen time and 5 hours of sleep (they were unwell). With this student included,

r \approx -0.401

. Explain what has happened and what an analyst should do.

Show worked solution →

The new point is an outlier: far below the pattern of the other points. (1 mark)

A single outlier has weakened the association from strong ( $r \approx -0.961$ ) to weak ( $r \approx -0.401$ ), because $r$ is sensitive to outliers. (1 mark)

The analyst should investigate it. If it is unrepresentative (the student was unwell), report the analysis with and without it, or justify excluding it; do not just delete it silently. (1 mark)

Practise this

Sources & how we know this

General Mathematics General senior syllabus 2025 — QCAA

What this dot point is asking

The answer

Explanatory and response variables

Constructing a scatterplot

Describing an association

Pearson's correlation coefficient, rrr

The coefficient of determination, R2R^2R2

Converting between rrr and R2R^2R2

1. Using the calculator

2. Describing the association in context

3. Interpreting R2R^2R2

4. From R2R^2R2 to rrr

5. The effect of an outlier

6. When rrr misleads

Exam-style questions

Practice questions

Practise this

Sources & how we know this

Pearson's correlation coefficient, $r$

The coefficient of determination, $R^2$

Converting between $r$ and $R^2$

3. Interpreting $R^2$

4. From $R^2$ to $r$

6. When $r$ misleads