Seasonal indices and deseasonalisation: VCE General Mathematics Unit 3 Data analysis

Syllabus dot point

“Seasonal adjustment including the use and interpretation of seasonal indices and their calculation using seasonal and yearly means, and modelling trend by fitting a least squares line to a time series with time as the explanatory variable (data de-seasonalised where necessary), with forecasts re-seasonalised where necessary”

13 min readUpdated 23 Sep 2026

Quick answer

A seasonal index compares a season with the average season: $1.2$ means $20\%$ above average and $0.85$ means $15\%$ below. For one year, index $=$ value for the season $\div$ the mean of all the seasons that year; for several years, average the indices from each year. A full set of indices sums to the number of seasons (4 for quarters, 12 for months). Deseasonalise with actual $\div$ index and reseasonalise with deseasonalised $\times$ index. To forecast, fit a least squares line to the deseasonalised values against time, forecast, then multiply by the season's index.

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What this dot point is asking
The answer
Why seasonal adjustment matters
Exam-style questions
Practice questions

What this dot point is asking

In "Investigating and modelling time series data" the Study Design lists "seasonal adjustment including the use and interpretation of seasonal indices and their calculation using seasonal and yearly means" and "modelling trend by fitting a least squares line to a time series with time as the explanatory variable (data de-seasonalised where necessary), and the use of the model to make forecasts (with re-seasonalisation where necessary)". You need to:

interpret a seasonal index as a percentage above or below the average season;
calculate seasonal indices from one year or several years of data;
deseasonalise data (divide by the index) and reseasonalise a forecast (multiply by the index), including the percentage change this causes;
fit a trend line to deseasonalised data with time as the explanatory variable, forecast with it, and reseasonalise.

Seasonal indices are among the least well answered Data analysis questions. Exam 1 Question 16 was about seasonal correction in both 2023 ( $43\%$ ) and 2025 ( $50\%$ ), and in Examination 2 only $21\%$ (2023 Question 4c) and $12\%$ (2025 Question 6b) scored full marks for calculating an index.

The answer

What a seasonal index measures

A time series with seasonality repeats a calendar pattern: ice cream sales peak in summer, heating bills in winter. A seasonal index measures how a season compares with the average season:

\text{seasonal index} = \frac{\text{value for the season}}{\text{seasonal average}}.

Index	Meaning
1.00	an average season
1.35	$35\%$ above the average season
0.74	$26\%$ below the average season

Because the indices compare each season with the average of all of them, a full set of seasonal indices averages 1, so they sum to the number of seasons: 4 for quarterly data, 12 for monthly data, 7 for days of the week. Use this to find a missing index or to check your working.

Calculating seasonal indices

From one year of data:

Find the seasonal average for the year: the yearly total divided by the number of seasons.
Divide each season's value by that average.

For the 2011 ice cream data (2023 Examination 2 Question 4c), the total is 2.004, the monthly mean is $\frac{2.004}{12} = 0.167$ , and July's index is $\frac{0.183}{0.167} = 1.10$ .

From several years of data (the usual method):

For each year, find the seasonal average.
Divide every value by its own year's average (this gives one ratio per season per year).
Average the ratios for each season across the years.

Each year's ratios sum to the number of seasons, so the averages do too. The 2025 Examination 2 report also accepted a pooled version: the mean value for the season over all years divided by the overall seasonal mean.

Deseasonalising and reseasonalising

\text{deseasonalised value} = \frac{\text{actual value}}{\text{seasonal index}}, \qquad \text{actual (reseasonalised) value} = \text{deseasonalised value} \times \text{seasonal index}.

Deseasonalising removes the seasonal effect: a peak season (index above 1) is scaled down and a quiet season (index below 1) is scaled up. What remains shows the trend and the irregular fluctuations.

Percentage corrections. Dividing by an index changes the value by the factor $\frac{1}{\text{index}}$ :

Index	$\frac{1}{\text{index}}$	Deseasonalising
0.74	1.35	increases the actual value by $35\%$
0.8	1.25	increases it by $25\%$
1.25	0.8	reduces it by $20\%$
1.75	0.571	reduces it by $43\%$

The change is not the distance of the index from 1: an index of 1.75 does not mean "reduce by $75\%$ " (2025 Examination 1 Question 16) and an increase of $35\%$ does not mean an index of 0.65 (2023 Examination 1 Question 16).

Trend lines on deseasonalised data

Seasonal swings make a least squares line on the raw data almost useless. Deseasonalise first, fit the line with time as the explanatory variable, forecast the deseasonalised value, then reseasonalise by multiplying by the index for the season being forecast.

Formula

Seasonal index $= \dfrac{\text{value for the season}}{\text{seasonal average}}$ ; indices sum to the number of seasons. Deseasonalised $= \dfrac{\text{actual}}{\text{seasonal index}}$ . Actual $=$ deseasonalised $\times$ seasonal index. Deseasonalising multiplies a value by $\dfrac{1}{\text{index}}$ .

Worked examples: calculating, correcting and forecasting

Seasonal indices from three years of quarterly data

A kayak hire business records hires each quarter:

Year	Q1	Q2	Q3	Q4	Yearly mean
1	52	30	22	44	37
2	58	33	25	48	41
3	63	37	27	53	45

Step 1: yearly means. Year 1: $\frac{148}{4} = 37$ ; year 2: $\frac{164}{4} = 41$ ; year 3: $\frac{180}{4} = 45$ .

Step 2: divide each value by its year's mean.

Year	Q1	Q2	Q3	Q4
1	1.4054	0.8108	0.5946	1.1892
2	1.4146	0.8049	0.6098	1.1707
3	1.4000	0.8222	0.6000	1.1778

Step 3: average each column. Q1: 1.407, Q2: 0.813, Q3: 0.601, Q4: 1.179.

Step 4: check. $1.407 + 0.813 + 0.601 + 1.179 = 4.000$ . Quarter 1 runs about $41\%$ above an average quarter and quarter 3 about $40\%$ below.

Deseasonalise, fit a trend line and forecast

Step 1: deseasonalise. Divide each value by its quarter's index. Year 1: $\frac{52}{1.407} = 36.96$ , $\frac{30}{0.813} = 36.90$ , $\frac{22}{0.601} = 36.61$ , $\frac{44}{1.179} = 37.32$ ; year 2: 41.22, 40.59, 41.60, 40.71; year 3: 44.78, 45.51, 44.93, 44.95.

Step 2: fit the trend line with time period $t = 1, 2, \ldots, 12$ as the explanatory variable:

\text{deseasonalised hires} = 35.11 + 0.9065t \qquad (r = 0.942).

On the raw data the line would have $r = 0.13$ : the seasonal swings hide a clear upward trend of about 0.9 hires per quarter.

Step 3: forecast. Quarter 1 of year 4 is $t = 13$ : $35.11 + 0.9065 \times 13 = 46.90$ (deseasonalised).

Step 4: reseasonalise. $46.90 \times 1.407 = 65.99$ , about 66 hires. Quarter 3 of year 4 ( $t = 15$ ): $48.71 \times 0.601 = 29.3$ , about 29 hires.

A percentage correction (2025 Examination 1 Question 16)

Step 1: deseasonalised $= \frac{\text{actual}}{1.75} = 0.571 \times \text{actual}$ .

Step 2: $0.571$ keeps $57\%$ , so the reduction is $43\%$ . Answer: B.

Finding the index from a correction (2023 Examination 1 Question 16)

Step 1: "increased by $35\%$ " means deseasonalised $= 1.35 \times$ actual.

Step 2: $\frac{1}{\text{index}} = 1.35$ , so index $= \frac{1}{1.35} = 0.74$ . Answer: D.

A monthly index over several years (2025 Examination 2 Question 6b)

Step 1: monthly means: $\frac{361}{12} = 30.083$ , $\frac{354}{12} = 29.5$ , $\frac{358}{12} = 29.833$ , $\frac{357}{12} = 29.75$ .
Step 2: September ratios: $\frac{15}{30.083} = 0.4986$ , $\frac{20}{29.5} = 0.6780$ , $\frac{20}{29.833} = 0.6704$ , $\frac{15}{29.75} = 0.5042$ .
Step 3: average: $\frac{2.3512}{4} = 0.588$ (to three decimal places).

Common traps

Subtracting instead of dividing: An index of 1.75 does not mean "remove $75\%$ ": $\frac{1}{1.75} = 0.57$ , a $43\%$ reduction.
Using $1 -$ the percentage for the index: A $35\%$ increase gives an index of $\frac{1}{1.35} = 0.74$ , not $0.65$ .
Multiplying to deseasonalise: Deseasonalise by dividing; multiply only to put the season back into a forecast.
Dividing by the wrong average: The index uses that year's seasonal (monthly or quarterly) mean, not the yearly total.
Leaving out steps in a "show that": Show the total, the mean and the division (2023 Examination 2 Question 4c).
Forecasting from raw seasonal data: Fit the trend line to deseasonalised values, then reseasonalise the forecast.
Comparing two seasons through their indices as percentages: 1.407 is $40.7\%$ above the average season, not $40.7\%$ above another season.

Exam technique

For percentage-correction multiple choice, try an actual value of 100, as the 2023 report suggests: deseasonalise it and see how far it moved. For index calculations, set out a small table (value, yearly mean, ratio) so every step is visible, and check that a full set of indices sums to the number of seasons. Round only at the end.

Why seasonal adjustment matters

Many real series, from unemployment to retail sales, are published "seasonally adjusted" for the same reason as in this dot point: a raw monthly figure mixes the season with the underlying change. A seasonally adjusted rise means the series grew by more than the usual seasonal amount. In exam terms, the seasonal index answers "is this month typically high or low?", deseasonalising answers "what would this month look like without its season?", and the trend line on deseasonalised data answers "where is the series heading?". The time series page covers identifying seasonality and smoothing.

Note

Some things go up and down with the seasons. A seasonal index says how busy a season usually is compared with an average season: 1.2 means $20\%$ busier than average, 0.8 means $20\%$ quieter. To take the season out of a number, divide it by the index. To put the season back into a forecast, multiply by the index. The indices always add up to the number of seasons, so you can check your working or find a missing one.

Exam-style questions

Questions in the style of VCAA 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.

2023 VCAA Exam 1 Q16Past paper1 mark

The 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 →

Correcting for seasonality means deseasonalising, and an increase of $35\%$ means

\text{deseasonalised} = 1.35 \times \text{actual}.

But $\text{deseasonalised} = \dfrac{\text{actual}}{\text{seasonal index}}$ , so $\dfrac{1}{\text{seasonal index}} = 1.35$ and

\text{seasonal index} = \frac{1}{1.35} = 0.7407\ldots \approx 0.74.

The answer is D. Only $43\%$ were correct and $39\%$ chose B (0.65), from $1 - 0.35$ . Check: dividing by 0.65 would increase the figure by $54\%$ , not $35\%$ . The report suggests trying an actual value of 100: deseasonalised $= 135$ , so the index is $\frac{100}{135}$ .

Source: VCAA 2023 General Mathematics Examination 1, Question 16, and the 2023 examination report.

2025 VCAA Exam 1 Q16Past paper1 mark

The 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 →

\text{deseasonalised} = \frac{\text{actual}}{1.75} = 0.5714\ldots \times \text{actual}.

Multiplying by 0.57 keeps $57\%$ of the value, a reduction of $100\% - 57\% = 43\%$ . The answer is B ( $50\%$ correct).

$23\%$ chose D (75%), treating the index as a $75\%$ excess to be removed. Subtracting $75\%$ of a value is not the same as dividing by 1.75. Option C (57%) is what remains, not the reduction.

Source: VCAA 2025 General Mathematics Examination 1, Question 16, and the 2025 examination report.

2023 VCAA Exam 2 Q4bPast paper1 mark

The time series plot below shows the average monthly ice cream consumption recorded over three years, from January 2010 to December 2012. [The plot shows April 2010 (month 4) at 0.18 litres/person.] The long-term seasonal index for April is 1.05. Determine the deseasonalised value for average monthly ice cream consumption in April 2010 (month 4). Round your answer to two decimal places.

Show worked answer →

\text{deseasonalised} = \frac{\text{actual}}{\text{seasonal index}} = \frac{0.18}{1.05} = 0.1714\ldots \approx 0.17 \text{ litres/person}.

$72\%$ correct. April is a slightly above-average month (index above 1), so its deseasonalised value is a little lower than the actual value.

Source: VCAA 2023 General Mathematics Examination 2, Question 4, and the 2023 examination report.

2023 VCAA Exam 2 Q4cPast paper2 marks

Table 4 below shows the average monthly ice cream consumption for 2011. Consumption (litres/person), 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 →

Step 1: the monthly mean for 2011. The 12 values total 2.004, so

\text{monthly mean} = \frac{2.004}{12} = 0.167.

Step 2: the index.

\text{seasonal index (July)} = \frac{0.183}{0.167} = 1.0958\ldots \approx 1.10.

Only $21\%$ scored both marks and $50\%$ scored zero. Both calculations were required: many showed $\frac{0.183}{0.167}$ without showing where 0.167 came from. In a "show that" question, every step to the given value must appear.

Source: VCAA 2023 General Mathematics Examination 2, Question 4, and the 2023 examination report.

2025 VCAA Exam 2 Q6bPast paper2 marks

The time series plot shows the number of homes sold in a town each month over a four-year period. Month 1 is January 2016 and month 48 is December 2019. [September values read from the plot: 15 (2016), 20 (2017), 20 (2018), 15 (2019).] The total number of sales in each of the four years is given in the table below. 2016: 361, 2017: 354, 2018: 358, 2019: 357. A seasonal index can be calculated for each month based on the four-year period. Calculate this seasonal index for September, the ninth month in the calendar year. Round your answer to three decimal places.

Show worked answer →

Method 1 (one ratio per year, then average). Divide each September value by that year's monthly mean (total $\div 12$ ):

\frac{15}{361/12} = 0.4986, \quad \frac{20}{354/12} = 0.6780, \quad \frac{20}{358/12} = 0.6704, \quad \frac{15}{357/12} = 0.5042.

\text{seasonal index} = \frac{0.4986 + 0.6780 + 0.6704 + 0.5042}{4} = 0.588.

Method 2 (pooled), also accepted. Mean September value $= \frac{15 + 20 + 20 + 15}{4} = 17.5$ ; overall monthly mean $= \frac{361 + 354 + 358 + 357}{48} = 29.79\ldots$ ; index $= \frac{17.5}{29.79\ldots} = 0.587$ .

Only $12\%$ scored both marks ( $74\%$ scored zero). September sales run about $41\%$ below an average month.

Source: VCAA 2025 General Mathematics Examination 2, Question 6, and the 2025 examination report.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

The seasonal index for summer sales at a surf shop is 1.22. (a) Interpret this index. (b) Summer sales were 480. Find the deseasonalised sales, to the nearest whole number.

Show worked solution →

(a) Summer sales are typically $22\%$ above the average for a season. (1 mark)

(b) $\frac{480}{1.22} = 393.4\ldots \approx 393$ . (1 mark)

foundation1 mark

Three of the quarterly seasonal indices for a business are 1.22, 0.95 and 0.78. Find the fourth.

Show worked solution →

Quarterly indices sum to 4: $4 - (1.22 + 0.95 + 0.78) = 4 - 2.95 = 1.05$ . (1 mark)

core2 marks

A cafe served 140, 95, 80 and 125 hot meals in quarters 1 to 4 of one year. Calculate the seasonal index for each quarter from this year's data, to three decimal places.

Show worked solution →

Quarterly mean $= \frac{140 + 95 + 80 + 125}{4} = \frac{440}{4} = 110$ . (1 mark)

Indices: $\frac{140}{110} = 1.273$ , $\frac{95}{110} = 0.864$ , $\frac{80}{110} = 0.727$ , $\frac{125}{110} = 1.136$ . (They sum to 4.000.) (1 mark)

core1 mark

A trend line fitted to deseasonalised data forecasts a deseasonalised value of 520 for a quarter whose seasonal index is 0.78. Find the actual forecast.

Show worked solution →

Reseasonalise by multiplying: $520 \times 0.78 = 405.6$ . (1 mark)

core2 marks

(a) A month has seasonal index 0.8. By what percentage must actual values for that month be increased to deseasonalise them? (b) Another month has index 1.25. By what percentage must its actual values be reduced?

Show worked solution →

(a) $\frac{1}{0.8} = 1.25$ , so increase by $25\%$ . (1 mark)

(b) $\frac{1}{1.25} = 0.8$ , so reduce by $20\%$ . (1 mark) The percentage change is not simply the distance of the index from 1.

exam2 marks

A least squares line fitted to deseasonalised monthly sales is deseasonalised sales

= 212 + 3.4 \times

month number, where month 1 is January 2024. The seasonal index for June is 0.92. Forecast the actual sales for June 2026 (month 30).

Show worked solution →

Deseasonalised forecast $= 212 + 3.4 \times 30 = 314$ . (1 mark)

Reseasonalise: $314 \times 0.92 = 288.88 \approx 289$ . (1 mark)

exam3 marks

A kayak hire business has quarterly seasonal indices 1.407, 0.813, 0.601 and 1.179 (quarters 1 to 4). (a) Explain why these indices must add to 4. (b) Quarter 3 hires were 27. Find the deseasonalised figure, to one decimal place, and explain why it is larger than 27. (c) A student says "Quarter 1 hires are 40.7% above quarter 3 hires." Explain the error.

Show worked solution →

(a) Each index compares a quarter with the average quarter, and the average of the four indices must be 1, so they sum to 4. (1 mark)

(b) $\frac{27}{0.601} = 44.9$ . Quarter 3 is a below-average season (index less than 1), so removing the seasonal effect raises the value. (1 mark)

(c) 1.407 means quarter 1 is $40.7\%$ above the average quarter, not above quarter 3. Compared with quarter 3, quarter 1 is typically $\frac{1.407}{0.601} = 2.34$ times as large. (1 mark)

Practise this

Sources & how we know this

VCE General Mathematics Examination 1 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 2 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 1 (2025) and examination report — VCAA (2025)
VCE General Mathematics Examination 2 (2025) and examination report — VCAA (2025)
VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
VCE General Mathematics: examination specifications, past examinations and reports — VCAA