Seasonal indices, deseasonalising and trend forecasts: QCE General Mathematics Unit 3 Time series analysis

Syllabus dot point

“Deseasonalise a time series by calculating seasonal indices using the average percentage method, fit a least-squares line to model the long-term trend, and solve practical problems that involve forecasting with time series data”

QCEGeneral MathematicsUnit 3: Bivariate data and time series…16 min readUpdated 23 Sep 2026

Quick answer

With the average percentage method, divide each value by its year's average and average these proportions for each season to get the seasonal indices, which sum to the number of seasons. Deseasonalise by dividing actual values by the index, fit a least-squares line to the deseasonalised values against time $t$ , forecast by substituting a future $t$ and multiplying by the season's index, and remember that forecasts are extrapolations.

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Practice questions

What this dot point is asking

Many time series rise and fall with the seasons: ice cream sales peak in summer, ski resort visits peak in winter, electricity use peaks in both. That regular, calendar-related pattern hides the long-term trend. The QCAA General Mathematics 2025 syllabus asks you to measure the seasonal pattern with seasonal indices calculated by the average percentage method, to deseasonalise the data (remove the seasonal effect), to fit a least-squares line to model the long-term trend, and to solve practical problems with the results, which usually means forecasting and then commenting on how reliable the forecast is.

This is the second half of the Time series analysis topic (the first half, plotting, describing and smoothing, is on the time series page). It is ideal for the problem-solving and modelling task and appears regularly in the external assessment.

The answer

What a seasonal index measures

A seasonal index compares a season with the average season.

An index of 1 means the season is exactly average.
An index of 1.35 means the season is typically $35\%$ above average.
An index of 0.70 means the season is typically $30\%$ below average.

Because they compare each season with the average, the indices for one complete cycle always add to the number of seasons: four quarterly indices add to 4, twelve monthly indices add to 12, seven daily indices add to 7. Indices are sometimes written as percentages (135%, 70%), in which case quarterly indices add to 400%.

The average percentage method

The syllabus method builds the indices from the data in four steps.

Find the average for each year (each complete cycle): add the seasons and divide by the number of seasons.
Express each value as a percentage (or proportion) of its own year's average.
Average these percentages for each season across the years. These averages are the seasonal indices.
Check that the indices add to the number of seasons (allowing for rounding).

Here is a tourist park's quarterly visitor numbers (thousands) for three years:

Year	$Q_1$	$Q_2$	$Q_3$	$Q_4$	Yearly average
1	42	58	75	45	55
2	46	62	81	51	60
3	50	69	86	55	65

Each value divided by its yearly average:

Year	$Q_1$	$Q_2$	$Q_3$	$Q_4$
1	0.7636	1.0545	1.3636	0.8182
2	0.7667	1.0333	1.3500	0.8500
3	0.7692	1.0615	1.3231	0.8462
Seasonal index (mean of column)	0.767	1.050	1.346	0.838

Check: $0.767 + 1.050 + 1.346 + 0.838 = 4.001$ , which is 4 apart from rounding. $Q_3$ (the holiday quarter) is typically $34.6\%$ above average and $Q_1$ is $23.3\%$ below.

A spreadsheet makes this quick (the syllabus explicitly mentions spreadsheets), but in an examination you work with a scientific calculator, so keep the tables tidy and round only at the end of each step.

Key fact

The seasonal index for a season is the average, over the years, of $\dfrac{\text{value}}{\text{that year's average}}$ . Indices sum to the number of seasons. Deseasonalised value $= \dfrac{\text{actual value}}{\text{seasonal index}}$ , and a forecast actual value is the deseasonalised forecast multiplied by the seasonal index.

Deseasonalising

To remove the seasonal effect, divide each actual value by its season's index:

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

A strong quarter (index above 1) is scaled down; a weak quarter (index below 1) is scaled up. What is left shows how the series is moving once the "normal" seasonal swing is taken out. For the tourist park:

$t$	1	2	3	4	5	6	7	8	9	10	11	12
Actual	42	58	75	45	46	62	81	51	50	69	86	55
Deseasonalised	54.8	55.2	55.7	53.7	60.0	59.0	60.2	60.9	65.2	65.7	63.9	65.6

For example, $\frac{75}{1.346} \approx 55.7$ and $\frac{45}{0.838} \approx 53.7$ . The deseasonalised series climbs steadily, revealing the upward trend that the zig-zag of the actual data disguised.

Deseasonalised values also allow fair comparisons between seasons. A ski resort with 8000 visitors in June (index 1.60) and 3300 in October (index 0.55) had deseasonalised values of 5000 and 6000: October was the stronger month relative to its normal level.

Fitting a least-squares trend line

With the seasonal effect removed, model the trend by fitting a least-squares line with time as the explanatory variable. Number the periods $t = 1, 2, 3, \ldots$ in order (do not restart at each year), and enter $t$ and the deseasonalised values into your calculator's linear regression.

For the tourist park, the least-squares line is

\text{deseasonalised visitors} = 52.53 + 1.149t \quad (r \approx 0.934).

The slope tells you that, with the seasonal effect removed, visitor numbers are increasing by about 1.149 thousand per quarter on average.

Forecasting

To forecast an actual future value:

Find the value of $t$ for the future period (continue the numbering: $Q_1$ of year 4 is $t = 13$ , $Q_3$ of year 4 is $t = 15$ ).
Substitute into the trend line to get a deseasonalised forecast.
Reseasonalise: multiply by that season's index.

For $Q_3$ of year 4: $52.53 + 1.149 \times 15 = 69.765$ , and $69.765 \times 1.346 \approx 93.9$ , so about 93 900 visitors. For $Q_1$ of year 4: $52.53 + 1.149 \times 13 = 67.467$ , and $67.467 \times 0.767 \approx 51.7$ , so about 51 700 visitors.

How reliable is a forecast?

Every time series forecast is an extrapolation: it predicts outside the range of the data. It relies on the trend continuing in a straight line and the seasonal pattern staying the same. Forecasts one or two periods ahead are usually reasonable; forecasts many years ahead are not, because conditions change (new competitors, economic changes, one-off events). In the problem-solving and modelling task and in complex unfamiliar questions, you are expected to comment on this.

Worked examples: indices, deseasonalising and forecasting

Seasonal indices from two years of data

Quarterly sales (hundreds): Year 1: 36, 60, 84, 60. Year 2: 44, 66, 99, 71. Find the seasonal indices.

Yearly averages: $\frac{36 + 60 + 84 + 60}{4} = 60$ and $\frac{44 + 66 + 99 + 71}{4} = 70$ .
Proportions of the yearly average: Year 1: $\frac{36}{60} = 0.600$ , $\frac{60}{60} = 1.000$ , $\frac{84}{60} = 1.400$ , $\frac{60}{60} = 1.000$ . Year 2: $\frac{44}{70} \approx 0.629$ , $\frac{66}{70} \approx 0.943$ , $\frac{99}{70} \approx 1.414$ , $\frac{71}{70} \approx 1.014$ .
Average by quarter: $Q_1 \approx 0.614$ , $Q_2 \approx 0.971$ , $Q_3 \approx 1.407$ , $Q_4 \approx 1.007$ .
Check: Sum $= 3.999$ , which is 4 within rounding.

Marker's note: always divide by the average of the same year, not the overall average of all the data.

Deseasonalising and comparing

Using those indices, deseasonalise the Year 2 values.

$Q_1$ : $\frac{44}{0.614} \approx 71.7$ . $Q_2$ : $\frac{66}{0.971} \approx 68.0$ . $Q_3$ : $\frac{99}{1.407} \approx 70.4$ . $Q_4$ : $\frac{71}{1.007} \approx 70.5$ .

Interpret. Once seasonality is removed, Year 2 sales are fairly steady at about 70 (hundred) per quarter; the large $Q_3$ figure is mostly the normal seasonal peak, not unusual growth.

Marker's note: deseasonalised values of a single year should cluster around that year's average. If one is far away, check your division.

A forecast from the trend line

The tourist park's trend line is deseasonalised visitors $= 52.53 + 1.149t$ with $Q_1$ of year 1 as $t = 1$ . Forecast $Q_4$ of year 4 and comment.

$t$ value: $Q_4$ of year 4 is $t = 16$ .
Deseasonalised forecast: $52.53 + 1.149 \times 16 = 52.53 + 18.384 = 70.914$ .
Reseasonalise: $70.914 \times 0.838 \approx 59.4$ , about 59 400 visitors.
Comment: This is an extrapolation one year beyond the data; it assumes the steady trend and the seasonal pattern continue, which is reasonable for a short-term forecast.

Marker's note: the two most common errors are restarting $t$ at each year and forgetting to reseasonalise.

Finding an actual value from deseasonalised data

A surf shop's deseasonalised autumn sales were 36.4 thousand dollars and the autumn seasonal index is 0.85. Find the actual autumn sales.

Rearrange. Actual $=$ deseasonalised $\times$ index $= 36.4 \times 0.85 = 30.94$ .

Answer. About $30 900.

Marker's note: if the other three indices are given instead, first find the missing index from the sum (quarterly indices sum to 4).

Common traps

Dividing by the overall average instead of each year's average: The average percentage method compares each value with its own year.
Multiplying to deseasonalise: Deseasonalise by dividing by the index; reseasonalise by multiplying.
Indices that do not sum correctly: Quarterly indices sum to 4 and monthly to 12. A different total signals an error (small differences from rounding are fine).
Restarting the time count: Number the periods continuously: the first quarter of year 2 is $t = 5$ , not $t = 1$ .
Fitting the trend line to actual data with strong seasonality: Deseasonalise first, or the line is distorted by the seasonal swing.
Forgetting to reseasonalise the forecast: The trend line gives a deseasonalised value; multiply by the season's index for the actual forecast.
Ignoring the limitation: Forecasts are extrapolations and become unreliable far into the future.

Exam technique

Set out the average percentage method as a table: one row per year, a column for the yearly average, then a row of seasonal indices. Show one example calculation in full so the marker can follow your method, and check the indices' sum. For a forecast, write three labelled lines: " $t = \ldots$ ", "deseasonalised forecast $= \ldots$ ", "actual forecast $= \ldots \times$ index $= \ldots$ ". Finish with one sentence on reliability.

Note

Some numbers go up and down with the seasons, like ice cream sales. A seasonal index tells you how much busier or quieter a season normally is: 1.35 means 35% busier than an average season. If you divide by the index, you take out the normal seasonal bump and can see whether things are really growing. Then you draw a straight trend line through those adjusted numbers to predict the future, and finally multiply by the index again to put the seasonal bump back in.

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 ice cream sales is 1.35 and for winter is 0.70. Interpret each index.

Show worked solution →

Summer, 1.35: summer sales are typically $35\%$ above the average for a season. (1 mark)

Winter, 0.70: winter sales are typically $30\%$ below the average for a season. (1 mark)

foundation2 marks

Quarterly seasonal indices for a business are

Q_1

0.85,

Q_2

1.10,

Q_3

1.20 and

Q_4

unknown. (a) Find the

Q_4

index. (b) Deseasonalise a

Q_3

figure of 540.

Show worked solution →

(a) Quarterly indices sum to 4: $Q_4 = 4 - (0.85 + 1.10 + 1.20) = 4 - 3.15 = 0.85$ . (1 mark)

(b) Deseasonalised $= \frac{540}{1.20} = 450$ . (1 mark)

foundation2 marks

A trend line fitted to deseasonalised data predicts a deseasonalised value of 320 for a future winter. The winter seasonal index is 0.75. (a) Find the forecast actual value. (b) Explain why you multiply rather than divide.

Show worked solution →

(a) Reseasonalise: $320 \times 0.75 = 240$ . (1 mark)

(b) The trend line gives values with the seasonal effect removed. Winter figures are typically only $75\%$ of the seasonal average, so the seasonal effect is put back by multiplying by 0.75. Dividing would move the value the wrong way. (1 mark)

core3 marks

A café's quarterly sales (thousands of dollars) for one year are 42, 58, 75 and 45. Using this year alone, calculate each quarter's value as a percentage of the yearly average, to one decimal place.

Show worked solution →

Yearly average: $\frac{42 + 58 + 75 + 45}{4} = \frac{220}{4} = 55$ . (1 mark)

Percentages of the average:

$Q_1$ : $\frac{42}{55} \times 100\% \approx 76.4\%$
$Q_2$ : $\frac{58}{55} \times 100\% \approx 105.5\%$
$Q_3$ : $\frac{75}{55} \times 100\% \approx 136.4\%$
$Q_4$ : $\frac{45}{55} \times 100\% \approx 81.8\%$

(2 marks) With several years of data, these percentages are averaged across the years for each quarter to give the seasonal indices (the average percentage method).

core3 marks

Monthly seasonal indices are given for a ski resort. The June index is 1.60 and the October index is 0.55. In one year the resort had 8000 visitors in June and 3300 in October. (a) Deseasonalise both figures. (b) Which month was busier once the seasonal effect is removed?

Show worked solution →

(a) June: $\frac{8000}{1.60} = 5000$ . October: $\frac{3300}{0.55} = 6000$ . (2 marks)

(b) October. Although October's actual number is much lower, its deseasonalised value (6000) is higher than June's (5000), so October performed better relative to what is normal for that month. (1 mark)

core3 marks

Two years of quarterly data (in hundreds) are: Year 1: 36, 60, 84, 60; Year 2: 44, 66, 99, 71. Use the average percentage method to find the seasonal indices, to three decimal places.

Show worked solution →

Yearly averages. Year 1: $\frac{240}{4} = 60$ . Year 2: $\frac{280}{4} = 70$ . (1 mark)

Percentages of each yearly average.

	$Q_1$	$Q_2$	$Q_3$	$Q_4$
Year 1	0.600	1.000	1.400	1.000
Year 2	0.629	0.943	1.414	1.014

(1 mark)

Average for each quarter. $Q_1 = \frac{0.600 + 0.629}{2} \approx 0.614$ , $Q_2 = \frac{1.000 + 0.943}{2} \approx 0.971$ , $Q_3 = \frac{1.400 + 1.414}{2} = 1.407$ , $Q_4 = \frac{1.000 + 1.014}{2} = 1.007$ . (1 mark)

Check: $0.614 + 0.971 + 1.407 + 1.007 = 3.999$ , which is 4 apart from rounding.

exam4 marks

Seasonal indices for a tourist park are

Q_1

0.767,

Q_2

1.050,

Q_3

1.346 and

Q_4

0.838. The least-squares line fitted to the deseasonalised visitor numbers (thousands) against the quarter number

t

(with

t = 1

for

Q_1

of year 1) is deseasonalised visitors

= 52.53 + 1.149t

. (a) Interpret the slope. (b) Forecast the actual number of visitors in

Q_3

of year 4. (c) Comment on the reliability of a forecast for

Q_3

of year 10.

Show worked solution →

(a) After removing the seasonal effect, visitor numbers increase by about 1.149 thousand (about 1150 visitors) per quarter on average. (1 mark)

(b) $Q_3$ of year 4 is $t = 15$ . Deseasonalised forecast: $52.53 + 1.149 \times 15 = 52.53 + 17.235 = 69.765$ . Reseasonalise: $69.765 \times 1.346 \approx 93.9$ . About 93 900 visitors. (2 marks)

(c) Year 10 is $t = 39$ , far beyond the three years of data used ( $t = 1$ to 12). This is extrapolation a long way into the future, and it assumes the trend and seasonal pattern stay exactly the same for many years, so the forecast is unreliable. (1 mark)

exam3 marks

A company's deseasonalised sales for

Q_2

were 450 units and the

Q_2

seasonal index is 0.90. The manager says actual

Q_2

sales were "500 units, because 450 is 90% of 500". (a) Determine the actual sales. (b) Explain the manager's error. (c) State what a seasonal index of 0.90 tells you about

Q_2

Show worked solution →

(a) Actual $=$ deseasonalised $\times$ seasonal index $= 450 \times 0.90 = 405$ units. (1 mark)

(b) The manager has reversed the relationship. Deseasonalised $= \frac{\text{actual}}{\text{index}}$ , so the actual value is the smaller one when the index is below 1: $405 \div 0.90 = 450$ . The manager divided when they should have multiplied. (1 mark)

(c) $Q_2$ sales are typically $10\%$ below the average quarter. (1 mark)

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Sources & how we know this

General Mathematics General senior syllabus 2025 — QCAA