Fitting, predicting and interpreting models | Year 11

01 / LEARN

Understand the idea.

Choose and fit a rule

Use the input spacing when fitting a model. An increase of twenty-two over two input units gives gradient eleven, not twenty-two. For a positive exponential rule, a time gap of three means the output ratio equals $b^{3}$ .

After fitting parameters, check all supplied observations and state the input range they cover. Matching two observations determines parameters but leaves the model’s wider reliability to be tested with further evidence.

Residuals show what is missed

A residual measures what the model missed at one input: observed minus predicted. Positive residuals place observations above the fitted rule; negative residuals place them below. Keep this subtraction order consistent.

Examine several residuals for patterns. Increasing positive residuals can mean the model underestimates later growth. Small scattered residuals may support a model over the observed interval, but do not establish reliable forecasts indefinitely beyond it.

residual = observed - predicted

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A line and a new measurement

Data are $(0, 40), (2, 62), (4, 84)$ . Fit a line, predict at 6, then find the residual if the new observation is 104.

Worked solution

1
Find the rate per input unit.
$m = \frac{62 - 40}{2 - 0} = \frac{22}{2} = 11$
Both successive two-unit intervals add twenty-two, supporting a constant rate.
2
Fit and check the linear rule.
$\begin{aligned} y & = 40 + 11 t \\ y (4) & = 40 + 11 (4) = 84 \end{aligned}$
The time-zero observation supplies the intercept, and the third observation agrees.
3
Predict at the new input.
$y (6) = 40 + 11 (6) = 106$
The same fitted rate is extended beyond the observed zero-to-four range.
4
Calculate the new residual.
$e = 104 - 106 = - 2$
The actual new observation lies two units below the prediction.

Answer

$y = 40 + 11 t$ ; prediction 106; residual −2.

EXAMPLE 02

Recover an exponential factor

A quantity is 150 at time 0 and 259.2 at time 3. Fit $P = A b^{t}$ and predict at 5.

Worked solution

1
Recover the initial amount.
$A = 150$
At time zero the exponential factor is one.
2
Use the later observation.
$\begin{aligned} 259.2 & = 150 b^{3} \\ b^{3} & = \frac{259.2}{150} = 1.728 \end{aligned}$
Three time intervals contribute three factors of the unknown multiplier.
3
Take the positive cube root.
$b = \sqrt[3]{1.728} = 1.2$
A positive exponential multiplier is required for this quantity model.
4
State the fitted model and predict.
$\begin{aligned} P (t) & = 150 {(1.2)}^{t} \\ P (5) & = 150 {(1.2)}^{5} = 373.248 \end{aligned}$
The fitted twenty-percent growth factor is applied for five periods.

Answer

$P = 150 {(1.2)}^{t}$ ; prediction 373.248.

03 / YOUR TURN

Try it yourself.

Check your understanding

A model predicts 72 and the observation is 75. Find the residual.

Show answer & reasoning

The residual is $75 - 72 = 3$ . The observed output is three units above the model prediction.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Fit a line to $(0, 12), (3, 27), (6, 42)$ .
2 marks
A prediction is 80 and the observation is 77. Find the residual.
2 marks
Fit $P = A b^{n}$ to $P (0) = 200$ and $P (2) = 288$ .
2 marks
For $y = 7 + 4 x$ , observations at $x = 1, 2, 3$ are 12,14,20. Find residuals.
2 marks
Does accurate fitting between times 2 and 8 alone justify a prediction at 80?
2 marks
An exponential quantity is 60 at time 1 and 135 at time 3. Find its positive factor and value at zero.
3 marks

Read the worked solutions

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

Fit a line to $(0, 12), (3, 27), (6, 42)$ .
2 marks
Each 3-unit input step adds 15, so gradient is 5.
$y = 12 + 5 x$ .
A prediction is 80 and the observation is 77. Find the residual.
2 marks
$77 - 80 = - 3$ .
$- 3$ .
Fit $P = A b^{n}$ to $P (0) = 200$ and $P (2) = 288$ .
2 marks
$b^{2} = \frac{288}{200} = 1.44$ ; use positive factor 1.2.
$P = 200 {(1.2)}^{n}$ .
For $y = 7 + 4 x$ , observations at $x = 1, 2, 3$ are 12,14,20. Find residuals.
2 marks
Predictions are 11,15,19.
Subtract each prediction from its matching observation.
1, $- 1$ , 1.
Does accurate fitting between times 2 and 8 alone justify a prediction at 80?
2 marks
80 is far beyond the observed range.
The process may change; local fit does not establish long-range validity.
No.
An exponential quantity is 60 at time 1 and 135 at time 3. Find its positive factor and value at zero.
3 marks
The two-step ratio is 2.25, so $b = 1.5$ .
Initial value $= \frac{60}{1.5} = 40$ .
Factor 1.5; initial value 40.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

In this lesson

Understand the idea.

Choose and fit a rule

Residuals show what is missed

See the method in action.

A line and a new measurement

Find the rate per input unit.

Fit and check the linear rule.

Predict at the new input.

Calculate the new residual.

Recover an exponential factor

Recover the initial amount.

Use the later observation.

Take the positive cube root.

State the fitted model and predict.

Try it yourself.

Check your understanding

Keep practising