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How do we fit the best straight line to bivariate data and read meaning from it?

Fit a least-squares line using technology, interpret the slope and intercept in context, and predict while distinguishing interpolation from extrapolation.

How to fit the least-squares regression line with technology, interpret its slope and intercept in real units, predict with the equation, and judge interpolation against extrapolation.

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
What "least squares" means
Interpreting slope and intercept
Interpolation versus extrapolation

What this dot point is asking

You must fit the line with technology, interpret the slope and intercept in context, predict, and distinguish interpolation from extrapolation.

What "least squares" means

For any straight line, each point has a vertical gap to the line called a residual. The least-squares line is the unique line that makes the sum of the squared residuals as small as possible, which is why it is the standard line of best fit.

Enter the paired data into the statistics menu, run the linear regression, and write the equation with the response variable on the left.

Interpreting slope and intercept

The whole point of fitting the line is to read meaning from $a$ and $b$ in the variables of the problem.

Slope $b$ : the predicted change in the response variable for each one-unit increase in the explanatory variable. Always state the direction (increase or decrease) and the units.
Intercept $a$ : the predicted response when the explanatory variable is zero. Sometimes this is meaningful, sometimes (for example "weight at age $0$ ") it is not, and you should say so.

Worked example

Fitting and using a least-squares line

A cafe records the daily maximum temperature ( $x$ , degrees Celsius) and the number of iced coffees sold ( $y$ ) over data from $x = 18$ to $x = 35$ . Technology gives $y = -30 + 6x$ .

Interpret the slope

The slope $6$ means each $1$ degree rise in temperature is associated with $6$ more iced coffees sold per day.

Interpret the intercept

The intercept $-30$ predicts sales at $0$ degrees Celsius. This is meaningless here (you cannot sell a negative number, and $0$ degrees is far below the data), so the intercept is not interpreted in context.

Predict at 28 degrees (interpolation)

$x = 28$ lies inside the data range, so this is interpolation: $y = -30 + 6(28) = -30 + 168 = 138$ iced coffees.

Comment on a prediction at 45 degrees

$x = 45$ is well outside the data range ( $18$ to $35$ ), so predicting there is extrapolation and unreliable; the linear pattern may break down at extreme heat.

Interpolation versus extrapolation

A prediction is only as trustworthy as the data behind it.

Always check whether the prediction value lies inside the minimum-to-maximum range of the explanatory variable before judging reliability. Markers expect this label on every prediction.

The reliability of an interpolated prediction also depends on $r^2$ : a line explaining $85\%$ of the variation gives tighter predictions than one explaining $40\%$ . So a complete answer to "how reliable is this prediction" mentions both whether it is interpolation or extrapolation and how much variation the model explains.

Exam-style practice questions

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

WACE 20216 marksFor ten students, the least-squares line relating exam mark (

y

, %) to hours studied (

x

) is

y = 41.0 + 3.5x

, with data ranging over

x = 2

x = 12

hours. (a) Interpret the slope and intercept. (b) Predict the mark for

9

hours. (c) Comment on predicting the mark for

20

hours.

Show worked answer →

Read the slope as the change per unit and the intercept as the value at $x = 0$ .

(a) Slope $3.5$ : each extra hour of study is associated with a $3.5$ percentage-point increase in exam mark. Intercept $41.0$ : a student who studies $0$ hours is predicted to score $41\%$ . (3 marks)

(b) $x = 9$ lies inside the data range ( $2$ to $12$ ), so this is interpolation: $y = 41.0 + 3.5(9) = 41.0 + 31.5 = 72.5\%$ . (2 marks)

(c) $x = 20$ is far outside the data range, so predicting there is extrapolation and unreliable; the linear trend may not continue and a mark cannot exceed $100\%$ . (1 mark)

Markers reward slope-as-rate and intercept-in-context, a correct interpolation, and identifying extrapolation as unreliable.

WACE 20235 marksA least-squares line of weekly sales

y

(units) on advertising spend

x

(

\

hundreds) is

y = 120 + 8x

. (a) Predict the sales for a spend of

1500

. (b) The actual sales at that spend were

260

units. Calculate the residual and state what it shows.

Show worked answer →

Use the line for the prediction, then compare with the observed value.

(a) A spend of $\$1500$ is $x = 15$ (hundreds), so $y = 120 + 8(15) = 120 + 120 = 240$ units predicted. (2 marks)

(b) Residual $=$ observed $-$ predicted $= 260 - 240 = 20$ . The positive residual shows the actual sales were $20$ units above the line, so the model underpredicted at that point. (3 marks)

Markers reward consistent units for $x$ , the prediction, and a residual defined as observed minus predicted.