What is wrong time code?

Keep t consistent (for example t = 1 for the first period) so the forecast time is correct.

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How do we fit a trend line to a time series and forecast future values?

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

How to fit a least-squares trend line to a time series using a numerical time variable, forecast future values, and reseasonalise a forecast when the data is seasonal.

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
Coding time as a number
Interpreting and forecasting
Seasonal data: deseasonalise, fit, reseasonalise
Why deseasonalise before fitting
Trend lines on non-seasonal data

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.

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.

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.

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.

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 20227 marksAnnual revenue (

\

millions) for years coded

t = 1, 2, 3, 4, 5

is

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 20246 marksDeseasonalised 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.