Scatterplots and bivariate data - Year 10 Mathematics

What you will learn

explain what bivariate data is and why we use scatterplots,
construct scatterplots from paired data,
describe the association between two variables: strength, direction, and form,
draw a line of good fit by eye and use it for predictions,
distinguish between interpolation and extrapolation (and their limitations),
explain why correlation does not imply causation.

Worked example 0 Real-world example: study time and marks

A teacher records the hours studied and test marks (out of $50$ ) for $8$ students:

Hours	1	2	3	4	5	6	7	8
Mark	15	20	22	28	30	35	38	42

Plot each pair as a point: $(1, 15), (2, 20), \ldots, (8, 42)$ .
The points rise from left to right: positive association.
They lie close to a straight line: strong, linear association.
A line of good fit passes through approximately $(2, 20)$ and $(7, 38)$ .
Gradient $= \dfrac{38 - 20}{7 - 2} = \dfrac{18}{5} = 3.6$ . Equation: $y - 20 = 3.6(x - 2)$ , so $y = 3.6x + 12.8$ .

Key idea: we can use the line to predict marks for a given number of hours — but only within the data range.

1. What is bivariate data?

Bivariate data consists of pairs of measurements on two variables for each individual or observation. The explanatory variable (independent) is plotted on the horizontal axis, and the response variable (dependent) is plotted on the vertical axis.

Scatterplot with positive linear association and a line of good fit.

2. Describing association

When describing a scatterplot, address three features:

Feature	Options
Direction	Positive (upward trend) or negative (downward trend)
Strength	Strong (points close to a line), moderate, or weak (points widely scattered)
Form	Linear (straight-line pattern) or non-linear (curved pattern)

If there is no discernible pattern, we say there is no association.

Worked example 1 Describing a scatterplot

A scatterplot of car age (years) vs resale value ($) shows points falling from left to right. The points cluster tightly around a curve that flattens out for older cars.

Description: there is a strong, negative, non-linear association between car age and resale value. As car age increases, resale value decreases, but the rate of decrease slows for older cars.

3. Line of good fit by eye

A line of good fit (or trend line) is a straight line drawn through the data that best represents the overall pattern. When drawing by eye:

The line should pass through (or close to) the middle of the data cloud.
Roughly equal numbers of points should be above and below the line.
The line should follow the direction and slope of the data.

Once you have two points on the line, find the equation using gradient-intercept form:

Line equation

$y = mx + c$

where $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ is the gradient.

Worked example 2 Finding the equation of a line of good fit

A line of good fit passes through the points $(10, 45)$ and $(30, 25)$ .

Gradient: $m = \dfrac{25 - 45}{30 - 10} = \dfrac{-20}{20} = -1$ .
Using $(10, 45)$ : $45 = -1 \times 10 + c$ , so $c = 55$ .
Equation: $y = -x + 55$ .
Predict the value when $x = 20$ : $y = -20 + 55 = 35$ .

4. Interpolation vs extrapolation

Interpolation: predicting within the range of the data. This is generally reliable.
Extrapolation: predicting outside the range of the data. This is risky because the trend may not continue.

Worked example 3 Interpolation and extrapolation

Data covers hours studied from $1$ to $8$ , with the line $y = 3.6x + 12.8$ .

Predict the mark for $x = 5$ hours: $y = 3.6 \times 5 + 12.8 = 30.8$ . This is interpolation (within range) — reliable.
Predict the mark for $x = 15$ hours: $y = 3.6 \times 15 + 12.8 = 66.8$ . This is extrapolation (outside range) — unreliable. The test is out of $50$ , so $66.8$ is impossible.

Key idea: always check whether a prediction falls within the data range before trusting it.

5. Correlation vs causation

A strong association (correlation) between two variables does not mean one causes the other. There may be:

a confounding variable (a third variable that influences both),
reverse causation (the direction of influence is opposite to what was assumed),
coincidence (the association is purely by chance).

To establish causation, you need a properly designed experiment with a control group.

Worked example 4 Correlation is not causation

Data shows a strong positive correlation between ice-cream sales and drowning incidents.

Does ice cream cause drowning? No. The confounding variable is temperature: hot weather increases both ice-cream sales and swimming activity, which increases drowning risk. Ice cream and drowning are correlated but not causally linked.

Practice

Fluency

Tier 1: basic skills

Define bivariate data and give an example of two variables you might investigate.
Which variable goes on the horizontal axis: the explanatory variable or the response variable?
A scatterplot shows points rising steeply from left to right with little scatter. Describe the association.
A scatterplot shows points scattered randomly with no pattern. Describe the association.
A line of good fit passes through $(2, 10)$ and $(8, 28)$ . Find the gradient.
Using the line from Q5, find the equation and predict $y$ when $x = 5$ .
Is predicting $y$ for $x = 5$ (data range $2$ — $8$ ) interpolation or extrapolation?
Is predicting $y$ for $x = 12$ interpolation or extrapolation?
State whether each is positive or negative association: (a) height and shoe size, (b) altitude and temperature, (c) practice hours and error count.
True or false: a strong correlation between two variables proves that one causes the other.

Reasoning

Tier 2: mixed practice

Plot the following data on a scatterplot and describe the association:

$x$ 2 4 6 8 10 12
$y$ 35 30 24 20 14 10
Draw a line of good fit for the data in Q1, find its equation, and predict $y$ when $x = 7$ .
A researcher finds a strong positive correlation between the number of firefighters at a fire and the amount of damage caused. Does this mean firefighters cause damage? Explain.
Data on advertising spend ($‘000) and sales ($‘000) for $6$ months is:

Advertising 5 10 15 20 25 30
Sales 40 55 65 80 90 100

Find the equation of the line of good fit and predict sales for an advertising spend of $18,000.
Explain why extrapolating the line from Q4 to predict sales for $100,000 in advertising is unreliable.

$x$	2	4	6	8	10	12
$y$	35	30	24	20	14	10

Advertising	5	10	15	20	25	30
Sales	40	55	65	80	90	100

Reasoning

Tier 3: explain and apply

A scatterplot of study hours vs exam mark shows a strong positive linear association for $0$ — $6$ hours, but the points flatten out beyond $6$ hours. Explain this pattern and discuss the limitations of using a single straight line for the entire data set.
Two scatterplots are shown: (A) shows a strong linear pattern; (B) shows a moderate curved pattern. A student claims that (A) always provides better predictions. Evaluate this claim.
Explain the difference between an observed association, a confounding variable, and a causal relationship. Use a real-world example to illustrate all three concepts.
A line of good fit has the equation $y = -2.5x + 80$ . The data ranges from $x = 5$ to $x = 25$ . For what values of $x$ does the line predict negative $y$ values? Explain why these predictions are meaningless.

Challenge

Reasoning

Harder reasoning

Two students draw different lines of good fit for the same scatterplot. Student A’s line passes through $(5, 20)$ and $(15, 50)$ . Student B’s line passes through $(3, 14)$ and $(17, 56)$ . Show that both lines have the same gradient but different $y$ -intercepts. Which line would you trust more and why?
A data set of $12$ points has a strong positive linear association. One additional point is added far from the trend (an outlier). Describe how this outlier could affect (a) the position of the line of good fit, (b) the strength of the association, and (c) predictions made using the line.
A study finds that countries with more mobile phones per person also have higher life expectancy. A journalist writes “Mobile phones increase life expectancy.” Write a critique of this claim, identifying at least two confounding variables and explaining why a controlled experiment would be needed.
The residual for a data point is defined as $\text{residual} = \text{observed } y - \text{predicted } y$ . For the data $(3, 22), (5, 30), (7, 35), (9, 44)$ and the line $y = 3x + 12$ , calculate each residual. What does the pattern of residuals tell you about the fit?

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Tier 1

Bivariate data consists of pairs of measurements on two variables for each individual. Example: height (cm) and weight (kg) for each student in a class.
The explanatory (independent) variable goes on the horizontal axis.
Strong, positive, linear association.
No association (no pattern).
Gradient $= \dfrac{28 - 10}{8 - 2} = \dfrac{18}{6} = 3$ .
Using $(2, 10)$ : $10 = 3 \times 2 + c$ , so $c = 4$ . Equation: $y = 3x + 4$ . When $x = 5$ : $y = 3 \times 5 + 4 = 19$ .
Interpolation (5 is within the range 2—8).
Extrapolation (12 is outside the range 2—8).
(a) Positive. (b) Negative. (c) Negative.
False. Correlation does not prove causation.

Tier 2

The scatterplot shows a strong, negative, linear association. As $x$ increases, $y$ decreases steadily.
A line of good fit through approximately $(2, 35)$ and $(12, 10)$ gives gradient $= \dfrac{10 - 35}{12 - 2} = \dfrac{-25}{10} = -2.5$ . Using $(2, 35)$ : $35 = -2.5 \times 2 + c$ , so $c = 40$ . Equation: $y = -2.5x + 40$ . When $x = 7$ : $y = -2.5 \times 7 + 40 = -17.5 + 40 = 22.5$ .
No, firefighters do not cause damage. The confounding variable is the size of the fire. Larger fires cause more damage and also require more firefighters. The number of firefighters and the damage are both consequences of the fire’s severity.
Line through $(5, 40)$ and $(30, 100)$ : gradient $= \dfrac{100 - 40}{30 - 5} = \dfrac{60}{25} = 2.4$ . Using $(5, 40)$ : $40 = 2.4 \times 5 + c$ , so $c = 28$ . Equation: $y = 2.4x + 28$ . For $x = 18$ : $y = 2.4 \times 18 + 28 = 43.2 + 28 = 71.2$ . Predicted sales: $71,200.
At $x = 100$ : $y = 2.4 \times 100 + 28 = 268$ , predicting $268,000 in sales. This is extrapolation far beyond the data range ( $5$ — $30$ ). The linear trend may not continue: there could be diminishing returns on advertising, market saturation, or budget constraints. The prediction is unreliable.

Tier 3

The flattening suggests diminishing returns: beyond a certain point, additional study hours produce smaller improvements (perhaps due to fatigue or already knowing the material). A single straight line would overestimate marks at high hours and underestimate them in the middle range. A curve or two separate line segments would better fit the data.
The claim is not always correct. If the true relationship is curved, a straight line in (A) may give poor predictions at the extremes despite appearing strong. Plot (B), if fitted with an appropriate curve, could give better predictions than a straight line forced onto (A). The best model matches the form of the data, not just the apparent strength.
Observed association: data shows that students who eat breakfast tend to score higher on tests. Confounding variable: family income — wealthier families may provide both regular meals and better educational resources. Causal relationship: to establish that breakfast causes higher scores, you would need a controlled experiment where students are randomly assigned to eat or skip breakfast, with other factors held constant. Without this, the association may be driven by the confounding variable.
$y = 0$ when $-2.5x + 80 = 0$ , so $x = 32$ . For $x > 32$ , the line predicts negative $y$ values. Since $x = 32$ is outside the data range ( $5$ to $25$ ), these predictions are extrapolations. Negative values may be physically meaningless (e.g. you cannot have negative sales, negative height, etc.), confirming that extrapolation beyond the data range is unreliable.

Challenge

Student A: gradient $= \dfrac{50 - 20}{15 - 5} = \dfrac{30}{10} = 3$ . Equation: $y = 3x + 5$ . Student B: gradient $= \dfrac{56 - 14}{17 - 3} = \dfrac{42}{14} = 3$ . Equation: $y = 3x + 5$ . Both lines have gradient $3$ and $y$ -intercept $5$ — they are actually the same line. If the $y$ -intercepts differed, you would trust the line whose reference points are closer to the centre of the data cloud, as it is less influenced by extreme points.
(a) The outlier can “pull” the line of good fit toward it, tilting or shifting the line. (b) The strength of association decreases because the outlier increases the scatter around the line. (c) Predictions near the outlier become less reliable, and the line may give poorer predictions for the rest of the data if it has been pulled off course.
The claim confuses correlation with causation. Confounding variables include: (i) GDP per capita — wealthier countries can afford both more mobile phones and better healthcare, nutrition, and sanitation. (ii) Education levels — higher education leads to both greater technology adoption and healthier lifestyles. A controlled experiment (randomly assigning mobile phones and measuring life expectancy) is impractical and ethically complex. Without controlling for confounders, we cannot conclude that mobile phones increase life expectancy.
Predicted values: $y(3) = 3 \times 3 + 12 = 21$ ; $y(5) = 27$ ; $y(7) = 33$ ; $y(9) = 39$ . Residuals: $(22 - 21) = 1$ ; $(30 - 27) = 3$ ; $(35 - 33) = 2$ ; $(44 - 39) = 5$ . All residuals are positive and increasing, suggesting the line slightly underestimates $y$ values, and the underestimation grows for larger $x$ . This may indicate a slight curve (non-linearity) in the data, or that the gradient of the line is slightly too small.

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