Bivariate Data and Regression - TCE Mathematics Applications (Tasmania)

Syllabus dot point

“Analyse bivariate data using scatterplots, correlation, and least-squares regression lines.”

TCEMathematics ApplicationsUnit 48 min readUpdated 2 Jun 2026

Quick answer

You must describe the relationship between two numerical variables from a scatterplot, quantify it with the correlation coefficient $r$ and the coefficient of determination $r^2$ , and fit a least-squares line to make and judge predictions.

What this dot point is asking

Bivariate data is paired data: each subject gives two numbers, such as a person's height and weight. The whole topic is about whether one variable helps predict the other, and how strongly.

Explanatory and response variables

The explanatory (independent) variable is the one used to predict; it goes on the horizontal axis. The response (dependent) variable is the one being predicted; it goes on the vertical axis.

Describing a scatterplot

When you read a scatterplot, comment on four features: form (linear or non-linear), direction (positive or negative), strength (how tightly the points cluster about a line), and any outliers. A positive direction means $y$ tends to rise as $x$ rises.

Correlation coefficient

Pearson's correlation coefficient $r$ measures the strength and direction of a linear relationship. It always lies between $-1$ and $+1$ . Values near $\pm 1$ mean a strong linear pattern; values near $0$ mean little or no linear relationship.

Interpreting r

$|r| = 1$ is a perfect straight line. Roughly: $0.75 \le |r| < 1$ strong, $0.5 \le |r| < 0.75$ moderate, $0.25 \le |r| < 0.5$ weak. The sign of $r$ always matches the slope of the line.

Coefficient of determination

The coefficient of determination is simply $r^2$ . It gives the proportion of the variation in the response variable that is explained by the linear relationship with the explanatory variable. If $r = 0.8$ then $r^2 = 0.64$ , so about $64\%$ of the variation in $y$ is explained by $x$ (and $36\%$ is due to other factors).

Least-squares regression line

The least-squares line is the straight line $\hat{y} = a + bx$ that minimises the sum of the squared vertical distances from the points to the line. In a TCE exam you usually read $a$ (intercept) and $b$ (slope/gradient) from technology, then interpret them.

Interpreting the line

\hat{y} = a + bx

Slope

b

: the predicted change in

y

for each one-unit increase in

x

. Intercept

a

: the predicted value of

y

when

x = 0

(only meaningful if

x = 0

is realistic).

Fitting and using a regression line

A study of 8 students records hours studied ( $x$ ) and test score ( $y$ ). Technology gives the least-squares line $\hat{y} = 42 + 5.5x$ with $r = 0.86$ .

Interpret the slope and intercept

Slope $5.5$ : each extra hour of study is associated with a predicted increase of $5.5$ marks. Intercept $42$ : a student who studies $0$ hours is predicted to score $42$ .

Predict and judge

For $x = 6$ hours: $\hat{y} = 42 + 5.5(6) = 42 + 33 = 75$ marks.

Since $6$ hours lies within the data range, this is interpolation and is reliable. The coefficient of determination is $r^2 = 0.86^2 = 0.74$ , so about $74\%$ of the variation in scores is explained by study time.

Interpolation and extrapolation

Predicting inside the range of the data (interpolation) is reasonably safe. Predicting outside the range (extrapolation) is risky because the linear pattern may not continue. State which you are doing whenever you make a prediction.

Confusing correlation with causation

A strong $r$ does not prove that $x$ causes $y$ . A third (lurking) variable may drive both, or the link may be coincidence. Always describe a relationship as an association, and never claim cause from correlation alone.

Always check the sign and the range

The sign of the slope $b$ must match the sign of $r$ . Before trusting a prediction, confirm the $x$ value sits inside the data range; if not, flag it as extrapolation.

Computing the line from sums

When the question gives the summary statistics rather than the raw data, use the least-squares formulae directly. The slope is

b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2},

and the intercept is

a = \dfrac{\sum y - b\sum x}{n} = \bar{y} - b\bar{x}

. The arithmetic must be careful: compute the numerator and denominator of

b

first, then use that value of

b

in the intercept formula. The information sheet supplied in the TCE exam carries these formulae, so the marks are for substituting correctly, not for memorising them.

Slope and intercept from sums

For five data points, $n = 5$ , $\sum x = 15$ , $\sum y = 40$ , $\sum xy = 145$ , $\sum x^2 = 55$ .

Step 1: Compute the slope

The slope formula requires you to substitute the five summary statistics into the numerator and denominator separately before dividing, because mixing up the order is the most common arithmetic error. Compute the numerator first, then the denominator, then divide.

b = \frac{5 \times 145 - 15 \times 40}{5 \times 55 - 15^2} = \frac{725 - 600}{275 - 225} = \frac{125}{50} = 2.5.

Step 2: Compute the intercept

Once the slope is known, the intercept is most reliably found using $a = \bar{y} - b\bar{x}$ , which requires the sample means. Divide each sum by $n$ to obtain the means, then substitute the slope you just calculated.
$\bar{x} = \frac{15}{5} = 3$ , $\bar{y} = \frac{40}{5} = 8$ , so $a = 8 - 2.5 \times 3 = 8 - 7.5 = 0.5$ .

Step 3: Write the equation and interpret it

Combining the intercept and slope gives the least-squares equation. Interpreting the slope in context is essential because the examiner awards marks for the practical meaning, not just the number. Each one-unit rise in $x$ predicts a rise of $2.5$ in $y$ , and the predicted value of $y$ when $x = 0$ is $0.5$ .

The least-squares line is $\hat{y} = 0.5 + 2.5x$ .

Reporting good practice: quote $r$ to two decimals, state the form, direction and strength of the relationship, give the equation of the line, and interpret slope and intercept in the words of the context.

Exam-style questions

Questions in the style of TASC 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.

Practice6 marks

A table shows blood pressure for a random selection of people of different ages. a) Taking age as the independent variable

A

, use your calculator's regression function to find a linear equation, using

A

(age) and

B

(blood pressure) to three decimal places. b) Find

r

and

r^2

to four decimal places and interpret

r

. c) Predict the blood pressure of a 22 year old.

Show worked answer →

a) (2 marks) Enter the eight (age, blood pressure) pairs into linear regression and read slope and intercept to 3 dp, giving a model of the form $B = 0.755A + 102.439$ (use the values your calculator returns).

b) (3 marks) From the same regression, $r \approx 0.6049$ and $r^2 \approx 0.3659$ (4 dp). Interpret $r$ : there is a weak-to-moderate positive linear correlation between age and blood pressure, so blood pressure tends to rise as age increases, but the association is not strong.

c) (1 mark) Substitute $A = 22$ : $B = 0.755(22) + 102.439 \approx 119$ mmHg. Markers reward including the units mmHg.

Practice2 marks

A table gives the women's Olympic 400m freestyle winning time (seconds). Given

\sum x = 44

\sum y = 2581

\sum xy = 13658

\sum x^2 = 324

and

n = 8

, use the regression formulae to find

a

and

b

(intercept and slope) to two decimal places.

Show worked answer →

(2 marks) Use the least-squares formulae from the information sheet.

Slope $b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{8 \times 13658 - 44 \times 2581}{8 \times 324 - 44^2} = \frac{109264 - 113564}{2592 - 1936} = \frac{-4300}{656} = -6.55$ (2 dp).

Intercept $a = \frac{\sum y - b\sum x}{n} = \frac{2581 - (-6.55)(44)}{8} = \frac{2581 + 288.20}{8} = 358.65$ (2 dp). So the line is $\text{Time} = 358.65 - 6.55x$ (small differences arise from rounding $b$ ).

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC