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How do we fit the best straight line to data and use it to predict?

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

How to find and interpret the least-squares line y = a + bx, use it for prediction, distinguish interpolation from extrapolation, and read residuals to judge the fit.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-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 20232 marksCalculator-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 20212 marksCalculator-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.