Least-squares regression (SACE Stage 2 General Mathematics)

Syllabus dot point

“Determine and interpret the least-squares regression line, use it to make predictions, and assess fit using residuals.”

8 min readUpdated 2 Jun 2026

Quick answer

The least-squares regression line is the straight line $y = a + bx$ that minimises the sum of the squared vertical gaps (residuals) between the data and the line. You interpret the slope and intercept in context, then substitute to predict.

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What this dot point is asking
The least-squares line
Prediction, interpolation and extrapolation
Using residuals to assess fit
What "least squares" actually minimises
Choosing a non-linear model
Linking back to correlation
Exam-style questions

What this dot point is asking

You must find the regression line (usually with a calculator), interpret its slope and intercept, use it to predict, and judge its fit using residuals.

The least-squares line

The least-squares line is written $y = a + bx$ , where $b$ is the slope and $a$ is the vertical intercept. It is the unique line that makes the total of the squared vertical distances from the points to the line as small as possible.

Residual

A residual is the vertical gap between an observed data point and the value the line predicts: $\text{residual} = y_{\text{actual}} - y_{\text{predicted}}$ . The least-squares line minimises the sum of the squared residuals.

In practice you read $a$ and $b$ from a calculator, but you must interpret both:

The slope $b$ is the predicted change in $y$ for each one-unit increase in $x$ .
The intercept $a$ is the predicted value of $y$ when $x = 0$ .

Worked example

A regression of weekly sales $y$ (thousands of dollars) against advertising spend $x$ (hundreds of dollars) gives $y = 5.2 + 1.8x$ . Interpret the slope and intercept in context, then predict weekly sales when $x = 6$ .

Step 1: Interpret the slope in context

The slope $b = 1.8$ is the predicted change in $y$ for each one-unit increase in $x$ . Because $x$ is measured in hundreds of dollars and $y$ in thousands of dollars, a one-unit rise in $x$ represents an extra $\$100$ of advertising. So the slope tells us that each additional $\$100$ spent on advertising is associated with a predicted rise of $\$1800$ in weekly sales.

Step 2: Interpret the intercept in context

The intercept $a = 5.2$ is the predicted value of $y$ when $x = 0$ , that is, when nothing is spent on advertising. This predicts weekly sales of $\$5200$ with zero advertising spend. In practice this figure may reflect sales from loyal customers or other channels, though it should be used cautiously if $x = 0$ is outside the range of the data.

Step 3: Substitute to make the prediction

To predict sales when $x = 6$ (that is, $\$600$ of advertising), substitute $x = 6$ into the regression equation. The multiplication comes first, then the addition:

y = 5.2 + 1.8(6) = 5.2 + 10.8 = 16.0.

Final answer: Predicted weekly sales are $\$16\,000$ when $\$600$ is spent on advertising.

Prediction, interpolation and extrapolation

Substituting an $x$ -value inside the range of the data is interpolation, which is usually reliable. Substituting beyond the data range is extrapolation, which is risky because the linear pattern may not continue.

Interpolation versus extrapolation

Predicting within the range of the collected data is interpolation and is generally trustworthy. Predicting outside that range is extrapolation; treat it with caution because there is no evidence the relationship holds there.

Using residuals to assess fit

After fitting the line, residuals tell you how good the fit is. A residual plot graphs each residual against $x$ .

If the residuals are scattered randomly above and below zero with no pattern, a linear model is appropriate.
If the residual plot shows a clear curve or pattern, the relationship is not really linear and a straight line is the wrong model.

Worked example

Using the regression line $y = 5.2 + 1.8x$ , suppose that at $x = 4$ the actual recorded sales were $\$13\,500$ , that is $y_{\text{actual}} = 13.5$ (thousands). Find the residual at this point and explain what it means.

Step 1: Find the predicted value from the regression line

The residual compares what the line predicts against what actually happened. We first calculate what the line says sales should be at $x = 4$ by substituting into the equation:

y_{\text{predicted}} = 5.2 + 1.8(4) = 5.2 + 7.2 = 12.4.

So the model predicts weekly sales of $\$12\,400$ when $\$400$ is spent on advertising.

Step 2: Compute the residual

The residual is defined as the actual value minus the predicted value. Taking the actual minus the predicted preserves the sign: a positive residual means the model underestimated, and a negative residual means it overestimated.

\text{residual} = y_{\text{actual}} - y_{\text{predicted}} = 13.5 - 12.4 = 1.1.

Step 3: Interpret the sign and size of the residual

The residual of $1.1$ (thousands of dollars) is positive, which means the actual sales of $\$13\,500$ were $\$1100$ above the value the line predicted. The model underestimated sales at this point. A positive residual places the data point above the regression line on a scatterplot.

Final answer: The residual is $1.1$ (that is, $\$1100$ ), indicating actual sales were $\$1100$ higher than the line's prediction.

What "least squares" actually minimises

The line is called "least squares" because, among all possible straight lines, it makes the sum of the squared residuals as small as possible. Squaring the residuals serves two purposes: it stops positive and negative gaps from cancelling, and it penalises large misses more heavily than small ones, so the line is pulled toward the bulk of the data. The line always passes through the mean point $(\bar{x}, \bar{y})$ , which is a useful checkpoint and explains why a single far-off outlier can swing the line noticeably - it drags the squared-distance total up until the line tilts toward it.

Choosing a non-linear model

When a residual plot shows a clear curved pattern, the straight-line model is wrong even if its $r^2$ looks high. SACE often compares a linear fit with an exponential one for growth data such as subscriber numbers or populations. The better model is the one whose residual plot is patternless and whose $r^2$ is closer to $1$ . Citing both pieces of evidence - the residual pattern and the $r^2$ value - is what a full-mark justification requires; relying on $r^2$ alone is not enough, because a curved relationship can still produce a deceptively high $r^2$ .

Linking back to correlation

The coefficient of determination $r^2$ from the correlation work tells you the proportion of variation in $y$ explained by the line. A high $r^2$ together with a patternless residual plot means the linear model is a strong fit.

Round only at the end

Keep full calculator accuracy for $a$ and $b$ while predicting, and round the final answer to a sensible number of figures. Rounding the coefficients early can throw a prediction off by a noticeable amount.

Common mistake

Do not swap the explanatory and response variables; the regression of $y$ on $x$ is different from the regression of $x$ on $y$ . Also do not trust a far extrapolation, and do not call a fit "good" from $r^2$ alone; always check the residual plot for a leftover pattern.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style2 marks

Calculator-assumed. The area of Arctic sea ice (

S

million km squared) is recorded against the year of observation (

D

) from 1980 to 2020. State the equation of the least-squares regression line using the context variables.

Show worked answer →

Enter the $(D, S)$ pairs into a calculator's linear regression. With $D$ explanatory and $S$ response, the calculator gives approximately slope $-0.0872$ and intercept $178$ .

So $S = -0.0872D + 178$ (small rounding differences accepted). (1 mark for slope and intercept, 1 mark for using the context variables $S$ and $D$ rather than $x$ and $y$ .)

The negative slope reflects the steady decline in sea-ice area over time.

SACE 2023-style2 marks

Calculator-assumed. Using the least-squares line

N = 2.01t + 79.5

for the number of potoroos

N

after

t

months, predict when the population reaches 300, and comment on the reliability.

Show worked answer →

Substitute $N = 300$ : $300 = 2.01t + 79.5$ , so $220.5 = 2.01t$ and $t = \dfrac{220.5}{2.01} \approx 109.7$ months. (1 mark)

So about 110 months. This is an extrapolation well beyond the 60-month data range, so it should be treated with caution. (1 mark)

SACE 2021-style2 marks

Calculator-assumed. A regression of weekly sales

y

(thousands of dollars) on advertising spend

x

(hundreds of dollars) gives

y = 5.2 + 1.8x

. At

x = 4

the actual sales were 13.5 (thousand dollars). Find the predicted value and the residual.

Show worked answer →

Predicted value: $y = 5.2 + 1.8(4) = 12.4$ (thousand dollars). [1 mark]

Residual $= y_{\text{actual}} - y_{\text{predicted}} = 13.5 - 12.4 = 1.1$ (thousand dollars), so the actual sales were 1100 dollars above the line. [1 mark]

A positive residual means the model underestimated this point.

Practise this

Sources & how we know this

Stage 2 General Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 General Mathematics: external assessment — SACE Board