Correlation and regression

The idea

A regression line $y = a + bx$ predicts $y$ from $x$ . The gradient $b$ is the change in $y$ for each extra unit of $x$ , in context. Predictions inside the data range (interpolation) are reliable; outside it (extrapolation) they are not.

Substitute the $x$ -value into the line.
Interpret $b$ with units: 'each extra hour adds $b$ marks'.
Check whether $x$ is inside the range of the data.

Watch out Correlation does not show that one variable causes the other.

See it worked

Worked example

The regression line of score $s$ on hours $h$ is $s = 12.5 + 0.8h$ , for $h$ from $5$ to $40$ . Interpret $0.8$ and estimate $s$ when $h = 20$ .

$0.8$ : each extra hour is linked to $0.8$ more marks.
$s = 12.5 + 0.8 \times 20 = 28.5$ .
$20$ is inside the data range, so the estimate is reliable.

Answer $0.8$ marks per hour; $s = 28.5$

Your turn

Question 1Non-calculator3 marks

The regression line of $y$ on $x$ is $y = 3.2 + 1.5x$ . Estimate $y$ when $x = 10$ .

Question 2Non-calculator3 marks

The regression line of distance $d$ on time $t$ is $d = 2.4t + 1.1$ . By how much does $d$ increase when $t$ increases by $5$ ?

Question 3Non-calculator3 marks

A regression line was found from daily temperatures between $5^\circ$ C and $25^\circ$ C. Explain why using it to predict sales at $40^\circ$ C is unreliable.

Write your answer, then show the working

The data only covers $5^\circ$ C to $25^\circ$ C.
$40^\circ$ C is outside that range: this is extrapolation.
The relationship may not continue in the same way, so the prediction is unreliable.

Answer $40^\circ$ C is outside the data range, so it is extrapolation

In past papers

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