Trend lines and forecasting in WACE Mathematics Applications Unit 4

Syllabus dot point

“Fit a least-squares trend line to a time series, forecast future values, and reseasonalise forecasts for seasonal data.”

WACEMathematics ApplicationsUnit 48 min readUpdated 2 Jun 2026

Quick answer

A trend line is a least-squares line fitted to a time series using a numerical time variable $t$ (such as $t = 1, 2, 3, \dots$ ). Its slope is the change per time period and its intercept is the value at $t = 0$ . Forecast by substituting a future $t$ . For seasonal data, fit the trend line to the deseasonalised values, forecast, then reseasonalise the forecast by multiplying by the target season's index.

Jump to a section

What this dot point is asking
Coding time as a number
Interpreting and forecasting
Seasonal data: deseasonalise, fit, reseasonalise
Why deseasonalise before fitting
Trend lines on non-seasonal data
Exam-style questions

What this dot point is asking

You must fit a least-squares trend line, forecast future values, and reseasonalise forecasts for seasonal data.

Coding time as a number

To fit a least-squares line, time must be a number, so code the periods as $t = 1, 2, 3, \dots$ in order.

Trend line

A trend line is the least-squares line $y = a + bt$ fitted to a time series, where $t$ is the coded time variable. The slope $b$ is the average change per time period; the intercept $a$ is the value when $t = 0$ .

Enter the coded times and the data values into the calculator's regression tool to get $a$ and $b$ , exactly as for any least-squares line.

Interpreting and forecasting

The slope and intercept have time-series meanings.

Slope $b$ : the average increase (or decrease) in the variable per time period.
Intercept $a$ : the value the line gives at $t = 0$ (often just before the data starts).

To forecast, substitute a future value of $t$ into the trend line. A forecast for a $t$ inside the data range is interpolation; beyond it is extrapolation and should be flagged as less reliable.

Forecasts assume the trend continues

A trend-line forecast assumes the straight-line pattern keeps going. The further past the data you forecast, the riskier this is, because trends can flatten or reverse. Always note that an extrapolated forecast depends on the trend continuing.

Seasonal data: deseasonalise, fit, reseasonalise

When the series is seasonal, the raw data swings up and down each cycle, which would distort the trend line. So you fit the line to the deseasonalised data, then restore the season for the final forecast.

Forecasting seasonal data

Deseasonalise the data (divide each value by its seasonal index).
Fit the trend line $y = a + bt$ to the deseasonalised values.
Forecast the deseasonalised value at the future $t$ .
Reseasonalise: actual forecast $=$ deseasonalised forecast $\times$ seasonal index of that season.

Worked example

Fitting a trend line and forecasting

Deseasonalised quarterly sales (units) for $t = 1$ to $6$ are $50, 54, 57, 61, 64, 68$ . Technology fits the trend line $y = 46.6 + 3.6t$ . The seasonal index for the target quarter ( $t = 10$ , a Quarter 2) is $1.15$ .

Interpret the slope

Slope $3.6$ : deseasonalised sales rise by about $3.6$ units per quarter on the underlying trend.

Forecast the deseasonalised value

At $t = 10$ : $y = 46.6 + 3.6(10) = 46.6 + 36 = 82.6$ units (deseasonalised).

Reseasonalise for Quarter 2

Multiply by the Q2 index: $82.6 \times 1.15 = 94.99$ , so about $95$ units actual.

Note the reliability

$t = 10$ is beyond the data ( $t = 1$ to $6$ ), so this is an extrapolation; the forecast assumes both the linear trend and the seasonal pattern continue, which may not hold.

Why deseasonalise before fitting

If you fit a trend line straight onto raw seasonal data, the regular peaks and dips pull the line about and the slope no longer measures the true trend. Deseasonalising first removes that swing so the line captures the genuine long-term movement; reseasonalising afterwards converts the trend forecast back into a realistic figure for the specific season.

Trend lines on non-seasonal data

When a time series has no seasonal component (for example, annual figures already aggregated over the year), there is nothing to deseasonalise: fit the least-squares trend line directly to the data against the coded time $t$ , forecast by substitution, and report without reseasonalising. The interpolation-versus-extrapolation judgement still applies, and the slope still reads as the change per period. Recognising whether the data is seasonal decides whether the deseasonalise-and-reseasonalise steps are needed at all, so check the series for a repeating pattern before choosing your approach.

Common errors

Fitting the trend line to raw seasonal data: Deseasonalise first so the seasonal swing does not distort the slope.
Forgetting to reseasonalise: The final forecast for a season must be multiplied by that season's index; otherwise it is only the deseasonalised value.
Wrong time code: Keep $t$ consistent (for example $t = 1$ for the first period) so the forecast time is correct.
Treating an extrapolated forecast as certain: Forecasts beyond the data assume the trend continues; state this limitation.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

Annual revenue (

\$

millions) for years coded

t = 1, 2, 3, 4, 5

8.0, 9.5, 10.5, 12.0, 13.5

. A least-squares trend line

y = 6.6 + 1.36t

is fitted. (a) Interpret the slope. (b) Forecast the revenue for

t = 8

. (c) State why this forecast should be treated with caution.

Show worked answer →

Read the slope as growth per period, then forecast.

(a) Slope $1.36$ : revenue is increasing by about $\$1.36$ million per year, on the linear trend. (2 marks)

(b) At $t = 8$ : $y = 6.6 + 1.36(8) = 6.6 + 10.88 = 17.48$ , so about $\$17.48$ million. (3 marks)

(c) $t = 8$ is beyond the data ( $t = 1$ to $5$ ), so the forecast is an extrapolation; it assumes the linear trend continues, which may not hold. (2 marks)

Markers reward the slope as annual growth, the trend substitution, and identifying extrapolation.

WACE 2024-style6 marks

Deseasonalised quarterly data gives a trend line

y = 120 + 5t

with

t = 1

the first quarter. (a) Forecast the deseasonalised value for the eighth quarter. (b) Reseasonalise the forecast if that quarter has a seasonal index of

0.75

. (c) Explain why deseasonalised data is used to fit the trend line.

Show worked answer →

Forecast on the trend, then restore the season.

(a) At $t = 8$ : $y = 120 + 5(8) = 120 + 40 = 160$ (deseasonalised). (2 marks)

(b) Reseasonalise by multiplying by the index: $160 \times 0.75 = 120$ . (2 marks)

(c) Deseasonalised data has the seasonal swing removed, so the trend line measures the genuine long-term movement rather than being distorted by regular seasonal peaks and dips. (2 marks)

Markers reward the trend substitution, multiplying by the index, and the distortion-free reason for deseasonalising.