Interpreting slope/intercept | Year 12

01 / LEARN

Understand the idea.

Translate the slope into a rate

In $\hat{y} = a + b x$ , the slope $b$ is the predicted response change for one additional unit of $x$ . Its units are response units per predictor unit. A negative slope predicts a decrease; it does not say every individual observation changes by that exact amount.

Always attach units and the phrase “the model predicts” when interpreting a fitted rate. Changing the predictor from hours to minutes changes the numerical coefficient, because one minute is a smaller increment; it must still represent the same physical rate.

Check whether the intercept has practical meaning

The intercept $a$ is the predicted response at $x = 0$ . Check whether zero belongs to the situation and lies within a useful observed range. An intercept can position the fitted line correctly without representing an actual observed starting measurement.

Interpret a base charge or initial amount only when the context supports that interpretation. If zero is outside the applicable range, state this limitation instead of treating the model as evidence of what would really happen there.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A delivery model

Delivery cost C dollars is modelled by C=6.5+1.8m, where m is parcel mass in kg. Interpret both coefficients.

Worked solution

1
Evaluate the model at zero mass.
$C (0) = 6.5 + 1.8 (0) = 6.5$
The intercept is the modelled base cost of 6.50 dollars, subject to whether zero mass is a meaningful service case.
2
Compare predictions one kilogram apart.
$\begin{aligned} C (m + 1) - C (m) & = [6.5 + 1.8 (m + 1)] \\ - [6.5 + 1.8 m] \\ = 1.8 \end{aligned}$
The predicted cost increases by 1.80 dollars for each extra kilogram.
3
State both coefficient meanings with units.
$a = 6.50 dollars, b = 1.80 dollars per kg$
The base amount and rate have different units. Confirm the applicable mass range before treating the intercept as an actual charge.

Answer

Intercept: modelled base cost 6.50 dollars; slope: 1.80 dollars per kg.

EXAMPLE 02

A different time unit

A model is d=12+30t, with d in km and t in hours. Rewrite it using u minutes.

Worked solution

1
Express the same time in hours.
$t = \frac{u}{60}$
Sixty minutes make one hour, so a time of u minutes is u divided by 60 hours.
2
Substitute into the original model.
$\begin{aligned} d & = 12 + 30 (\frac{u}{60}) \\ = 12 + 0.5 u \end{aligned}$
Substitution changes the input unit while preserving the distance predicted at the same physical time.
3
Interpret the new coefficient.
$30 km/h = \frac{30}{60} km/min = 0.5 km/min$
The intercept remains 12 km; the rate is now written per minute rather than per hour.

Answer

d=12+0.5u.

03 / YOUR TURN

Try it yourself.

Check your understanding

A fitted line predicts fuel remaining F=50-0.08d, with d in km. Interpret the slope.

Show answer & reasoning

Predicted fuel falls by 0.08 L for each additional km.

The slope is $- 0.08 L/km$ . The model predicts that fuel remaining falls by $0.08 L$ for each additional kilometre; the negative sign indicates a decrease.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

A fitted crop yield is $Y = 2 + 0.4 R$ , with $Y$ in tonnes and $R$ in mm rainfall. Interpret 0.4.
3 marks
A model $T = 90 - 2.5 x$ predicts test time in minutes from weeks of training. Interpret $- 2.5$ .
3 marks
A tree model $h = 1.2 + 0.7 t$ was fitted only for ages 5–12 years. Is 1.2 m a reliable measured height at birth?
3 marks
Change $y = 4 + 12 x$ from $x$ hours to $t$ minutes.
3 marks
The model $y = 5 + 0.03 x$ uses $x$ in dollars. Express it with $z$ in hundreds of dollars.
3 marks
A regression gives an intercept of $- 8$ for mass predicted from length. Must the entire model be useless?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A fitted crop yield is $Y = 2 + 0.4 R$ , with $Y$ in tonnes and $R$ in mm rainfall. Interpret 0.4.
3 marks
State response units per explanatory unit.
Qualify the statement as a model prediction.
The model predicts 0.4 more tonnes for each additional mm of rainfall.
A model $T = 90 - 2.5 x$ predicts test time in minutes from weeks of training. Interpret $- 2.5$ .
3 marks
The negative sign indicates a decrease.
Each extra training week predicts a 2.5-minute decrease in test time.
A tree model $h = 1.2 + 0.7 t$ was fitted only for ages 5–12 years. Is 1.2 m a reliable measured height at birth?
3 marks
Zero is outside the observed age range.
No. It is the extrapolated prediction at age zero, not a measured birth height.
Change $y = 4 + 12 x$ from $x$ hours to $t$ minutes.
3 marks
Substitute $x = \frac{t}{60}$ .
$\frac{12}{60} = 0.2$ .
$y = 4 + 0.2 t$ .
The model $y = 5 + 0.03 x$ uses $x$ in dollars. Express it with $z$ in hundreds of dollars.
3 marks
$x = 100 z$ .
$0.03 \times 100 = 3$ .
$y = 5 + 3 z$ .
A regression gives an intercept of $- 8$ for mass predicted from length. Must the entire model be useless?
3 marks
A negative extrapolated zero-length prediction does not alone invalidate a local fitted relationship.
No. The intercept may be outside the relevant length range; inspect fit and predictions within that range.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: statistical models. Original lessons and assessment practice; not an official SACE assessment.

Official curriculum source ↗

In this lesson

Understand the idea.

Translate the slope into a rate

Check whether the intercept has practical meaning

See the method in action.

A delivery model

Evaluate the model at zero mass.

Compare predictions one kilogram apart.

State both coefficient meanings with units.

A different time unit

Express the same time in hours.

Substitute into the original model.

Interpret the new coefficient.

Try it yourself.

Check your understanding

Keep practising