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VICGeneral MathematicsSyllabus dot point

How do seasonal indices quantify a repeating pattern, and how do you deseasonalise and reseasonalise time series data?

Calculate seasonal indices from time series data, interpret an index as a percentage above or below the seasonal average, deseasonalise data by dividing by the index, and reseasonalise a forecast by multiplying by the index

A focused answer to the VCE General Mathematics Unit 3 Data analysis key-knowledge point on seasonal indices. Calculating seasonal indices that sum to the number of seasons, interpreting them as percentages, deseasonalising by dividing, and reseasonalising a forecast by multiplying.

Generated by Claude Opus 4.77 min answer

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  1. What this dot point is asking
  2. Calculating seasonal indices
  3. Interpreting a seasonal index
  4. Deseasonalising and reseasonalising
  5. Why this matters for the exams

What this dot point is asking

VCAA wants you to quantify and remove a repeating seasonal pattern from a time series. You calculate a seasonal index for each season (each quarter, month or day), interpret it as a percentage above or below the seasonal average, deseasonalise the raw data by dividing by the index, and later reseasonalise a forecast by multiplying by the index. This lets you fit a trend line to data with the seasonal wobble removed and then put the season back for a realistic forecast.

Calculating seasonal indices

For each cycle (for example each year of quarterly data), divide each season's value by that cycle's seasonal average (the mean of the four quarters). Average these ratios across all cycles for each season. If the resulting indices do not sum to the number of seasons, scale them so they do.

Interpreting a seasonal index

An index above 11 means the season is above the yearly average; below 11 means below it. Convert to a percentage by subtracting 11: an index of 1.151.15 is 15%15\% above average, an index of 0.920.92 is 8%8\% below average.

Deseasonalising and reseasonalising

To strip out the seasonal effect, divide each actual value by its seasonal index:

deseasonalised value=actual valueseasonal index.\text{deseasonalised value} = \frac{\text{actual value}}{\text{seasonal index}}.

To put the season back, for example when forecasting from a deseasonalised trend line, multiply:

reseasonalised value=deseasonalised value×seasonal index.\text{reseasonalised value} = \text{deseasonalised value} \times \text{seasonal index}.

Why this matters for the exams

Seasonal index questions are a recurring written-response feature and reward careful direction: divide to deseasonalise, multiply to reseasonalise. Markers also check the interpretation sentence, so state the percentage above or below average. This dot point combines with the trend line from time series work: you deseasonalise, fit the trend, forecast, then reseasonalise for the final answer.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 VCAA1 marksThe number of visitors each month to a zoo is seasonal. To correct the number of visitors in January for seasonality, the actual number of visitors, to the nearest percent, is increased by 35%. The seasonal index for that month is closest to A. 0.61 B. 0.65 C. 0.69 D. 0.74 E. 0.77
Show worked answer →

Deseasonalising divides the actual figure by the seasonal index: deseasonalised = actual / seasonal index.

Increasing the actual figure by 35% to correct it means deseasonalised = 1.35 x actual.

So 1.35 x actual = actual / seasonal index, which gives seasonal index = 1 / 1.35 = 0.74.

This is closest to 0.74, so the answer is D. An index below 1 means January is a below-average month, which is why the figure must be scaled up.

2025 VCAA1 marksThe seasonal index for the number of meat pie sales in winter is 1.75. To correct for seasonality, the actual number of meat pie sales for winter should be reduced, to the nearest whole percentage, by A. 25% B. 43% C. 57% D. 75%
Show worked answer →

Deseasonalising divides by the seasonal index: deseasonalised = actual / 1.75 = 0.5714 x actual.

So the deseasonalised figure is 57.14% of the actual figure, which is a reduction of 100% - 57.14% = 42.86%.

To the nearest whole percentage this is a reduction of 43%, so the answer is B. An index above 1 means winter is a peak season, so the raw figure is scaled down.

2023 VCAA2 marksTable 4 shows the average monthly ice cream consumption (litres/person) for 2011: Jan 0.156, Feb 0.150, Mar 0.158, Apr 0.180, May 0.200, Jun 0.210, Jul 0.183, Aug 0.172, Sep 0.162, Oct 0.145, Nov 0.134, Dec 0.154. Show that, when rounded to two decimal places, the seasonal index for July 2011 estimated from this data is 1.10.
Show worked answer →

The seasonal index for a month is that month's value divided by the monthly average for the year.

First find the monthly average: add the 12 values to get 2.004, then divide by 12, giving 0.167 litres/person (1 mark).

seasonal index for July = July value / monthly average = 0.183 / 0.167 = 1.0958.

Rounded to two decimal places this is 1.10, as required (1 mark).