Practice Maths

Time Series — Topic Review — Solutions

← Back to Time Series

This review covers all four lessons in Time Series: Time Series Graphs and Trends, Smoothing Time Series, Seasonal Adjustment, and Forecasting. Click each answer box to reveal the worked solution.

Review Questions

  1. Fluency

    Q1 — Identifying Variation Components

    The monthly electricity demand for a Queensland suburb over two years shows the following pattern: demand peaks in summer (January–February) and is lowest in autumn (April–May). There is also a gradual increase in the overall average from Year 1 to Year 2. In one month, demand was unusually high due to a heatwave.

    1. Identify each component of variation present in this time series.
    2. Which component is the heatwave?
    3. Which component makes comparisons between individual months (e.g. comparing July Year 1 with August Year 1) potentially misleading?

    (a) Three components are present:

    • Trend — the gradual increase in average demand from Year 1 to Year 2.
    • Seasonal variation — the regular peaks in summer and troughs in autumn that repeat each year.
    • Irregular variation — the one-off heatwave spike.

    (b) The heatwave is irregular variation (a random, one-off event not part of the regular pattern).

    (c) Seasonal variation makes direct month-to-month comparisons misleading. July is typically lower than December for reasons unrelated to the trend — it is simply a different season. Seasonal adjustment is needed before comparing different months.

  2. Fluency

    Q2 — Reading a Time Series Graph

    The graph below shows quarterly ice-cream sales (thousands of units) for a Queensland factory over 3 years. The trend is upward, but sales are consistently highest in Q1 and lowest in Q3 each year.

    1. Describe the overall trend in ice-cream sales.
    2. Why is it incorrect to say “sales fell from Q1 Year 2 to Q3 Year 2” and conclude the business is in decline?
    3. Sketch the shape you would expect if you plotted the 3-point moving average for this data.

    (a) The overall trend is upward — ice-cream sales are increasing over the three-year period. The trend can be seen by looking at equivalent seasons across years (e.g. each Q1 is higher than the previous Q1).

    (b) The fall from Q1 to Q3 is seasonal variation — it happens every year regardless of the business performance. Q1 is summer (high demand) and Q3 is winter (low demand) in Queensland. This is expected and does not indicate decline. The underlying trend is still upward.

    (c) The 3-point moving average would be smoother than the raw data — the sharp peaks and troughs caused by seasonal variation would be reduced, revealing the underlying upward trend more clearly. The moving average line would gradually increase from left to right.

  3. Fluency

    Q3 — 3-Point Moving Average

    The monthly sales figures ($000) for a hardware store are: 42, 38, 45, 51, 48, 55, 62, 58, 64, 70, 66, 74.

    1. Calculate the 3-point moving averages for months 2 through 11.
    2. What does the moving average reveal about the trend?

    (a) Each 3-point moving average is the mean of three consecutive values:

    Month 2: (42+38+45)÷3 = 41.7; Month 3: (38+45+51)÷3 = 44.7; Month 4: (45+51+48)÷3 = 48.0; Month 5: (51+48+55)÷3 = 51.3; Month 6: (48+55+62)÷3 = 55.0; Month 7: (55+62+58)÷3 = 58.3; Month 8: (62+58+64)÷3 = 61.3; Month 9: (58+64+70)÷3 = 64.0; Month 10: (64+70+66)÷3 = 66.7; Month 11: (70+66+74)÷3 = 70.0.

    (b) The 3-point moving averages increase steadily from about $41,700 to $70,000, revealing a clear upward trend in sales that is obscured by month-to-month fluctuations in the raw data.

  4. Understanding

    Q4 — Centred Moving Average for Quarterly Data

    Quarterly visitor numbers (thousands) to a museum are: Q1 Y1 = 28, Q2 Y1 = 45, Q3 Y1 = 52, Q4 Y1 = 31, Q1 Y2 = 33, Q2 Y2 = 49, Q3 Y2 = 57.

    1. Calculate the first two 4-point moving totals.
    2. Calculate the first centred moving average (CMA) using these totals.
    3. Explain why centring is necessary for even-period moving averages.

    (a) First 4-point total: 28 + 45 + 52 + 31 = 156 (centred between Q2 and Q3 Y1). Second 4-point total: 45 + 52 + 31 + 33 = 161 (centred between Q3 and Q4 Y1).

    (b) First CMA = mean of the two 4-point totals, divided by 4: CMA = (156 + 161) ÷ (2 × 4) = 317 ÷ 8 = 39.625 thousand. This CMA aligns with Q3 Year 1.

    (c) A 4-point moving average falls between two time points (not aligned with any actual data point). Centring pairs consecutive moving averages and averages them, so the result aligns with an actual data point and can be compared directly to actual values. Without centring, you cannot compute meaningful seasonal indices or residuals.

  5. Understanding

    Q5 — Computing Seasonal Indices

    The table shows quarterly electricity costs ($000) for a childcare centre over 3 years:

    YearQ1Q2Q3Q4
    128181422
    230201524
    332211625
    1. Calculate the seasonal index for each quarter. Verify the indices sum to 4.
    2. Which quarter has the highest electricity costs? Is this consistent with the SI you calculated?
    3. Deseasonalise the Q1 Year 3 value of 32.

    (a) Quarterly averages: Q1 = (28+30+32)÷3 = 30; Q2 = (18+20+21)÷3 = 19.67; Q3 = (14+15+16)÷3 = 15; Q4 = (22+24+25)÷3 = 23.67.

    Overall mean = (30+19.67+15+23.67)÷4 = 88.33÷4 = 22.08.

    SIQ1 = 30÷22.08 = 1.36; SIQ2 = 19.67÷22.08 = 0.89; SIQ3 = 15÷22.08 = 0.68; SIQ4 = 23.67÷22.08 = 1.07. Sum = 1.36+0.89+0.68+1.07 = 4.00 ✓

    (b) Q1 has the highest electricity costs (summer in Queensland — air conditioning). SIQ1 = 1.36 confirms Q1 is 36% above the annual average, which is consistent.

    (c) Deseasonalised Q1 Y3 = 32 ÷ 1.36 = $23,529 (approx. $23.5k). This represents what Q1 Year 3 electricity costs would be if there were no seasonal effect.

  6. Understanding

    Q6 — Deseasonalising and Interpreting

    The August retail sales for a homewares store were $142,000. The August seasonal index is SI = 0.78.

    1. Calculate the deseasonalised August sales.
    2. The September SI = 0.91 and actual September sales were $158,000. Calculate the deseasonalised September sales.
    3. A manager says: “September was a much better month than August because sales were higher.” Evaluate this claim using the deseasonalised values.

    (a) Deseasonalised August = 142,000 ÷ 0.78 = $182,051 (approx. $182,100).

    (b) Deseasonalised September = 158,000 ÷ 0.91 = $173,626 (approx. $173,600).

    (c) The manager’s claim is incorrect. After removing the seasonal effect, August’s deseasonalised performance ($182,100) is actually stronger than September’s ($173,600). September had higher raw sales because September is typically a stronger month (SI = 0.91 vs 0.78), not because the business performed better. Comparing deseasonalised values gives a fairer picture of underlying performance.

  7. Understanding

    Q7 — Using the Trend Equation to Forecast

    A least-squares line fitted to quarterly CMA data for a surf school gives: ŷ = 85 + 4.2t ($000), where t = 1 is Q1 Year 1. Seasonal indices: SIQ1 = 1.45, SIQ2 = 1.10, SIQ3 = 0.55, SIQ4 = 0.90.

    1. Find the trend value at t = 13 (Q1 Year 4) and t = 15 (Q3 Year 4).
    2. Calculate the seasonally adjusted forecast for each of these periods.
    3. Explain why the Q1 Year 4 forecast is so much higher than Q3 Year 4, even though the trend value is similar.

    (a) t = 13: ŷ = 85 + 4.2(13) = 85 + 54.6 = $139.6k. t = 15: ŷ = 85 + 4.2(15) = 85 + 63 = $148k.

    (b) Q1 Y4 forecast = 139.6 × 1.45 = $202.4k. Q3 Y4 forecast = 148 × 0.55 = $81.4k.

    (c) Q1 has a seasonal index of 1.45 (summer — peak season for surf lessons, 45% above average), while Q3 has SI = 0.55 (winter — 45% below average). The large difference in seasonal indices dominates the relatively small difference in trend values, resulting in Q1’s forecast being nearly 2.5 times higher than Q3’s.

  8. Understanding

    Q8 — Reliability of Forecasts

    A business analyst has quarterly data from Years 1–3 (t = 1 to 12) and wants to forecast for Year 5 (t = 17 to 20).

    1. Is this interpolation or extrapolation? Explain.
    2. List three factors that could make this forecast unreliable.
    3. The analyst checks the residuals (actual − forecast) for the last 4 observed periods and finds they are all positive. What does this suggest about the model?

    (a) Extrapolation — the data covers t = 1 to 12 (Years 1–3), and Year 5 (t = 17–20) lies well beyond this range. We are extending the trend line beyond the region where it was validated.

    (b) Three factors reducing reliability: (i) the linear trend may not continue — it could flatten or change direction; (ii) the seasonal pattern assumed to be stable may shift; (iii) unexpected irregular events (economic downturn, regulatory changes, new competitors) cannot be anticipated by the model.

    (c) Consistently positive residuals (actual always above forecast) suggest the model is systematically underestimating. The trend line’s slope may be too shallow — growth is accelerating faster than the linear model assumes. The analyst should consider refitting the trend or exploring a non-linear model.

  9. Problem Solving

    Q9 — Full Analysis: Smoothing to Forecasting

    Quarterly revenue ($000) for an outdoor adventure company over 2 years is: Q1 Y1 = 95, Q2 Y1 = 140, Q3 Y1 = 180, Q4 Y1 = 110, Q1 Y2 = 105, Q2 Y2 = 155, Q3 Y2 = 200, Q4 Y2 = 125. Seasonal indices (given): SIQ1 = 0.69, SIQ2 = 1.01, SIQ3 = 1.30, SIQ4 = 1.00.

    1. Deseasonalise all 8 values.
    2. Using CAS, the trend line fitted to deseasonalised data is ŷ = 128.5 + 3.8t. Forecast the revenue for Q1 and Q3 of Year 3 (t = 9 and 11).
    3. Calculate the sum of all seasonal indices and verify they sum to 4. What would it mean if they did not?

    (a) Deseasonalised = actual ÷ SI:

    Q1 Y1: 95 ÷ 0.69 = 137.7; Q2 Y1: 140 ÷ 1.01 = 138.6; Q3 Y1: 180 ÷ 1.30 = 138.5; Q4 Y1: 110 ÷ 1.00 = 110.0 (Note: Q4 SI = 1.00 means no seasonal effect); Q1 Y2: 105 ÷ 0.69 = 152.2; Q2 Y2: 155 ÷ 1.01 = 153.5; Q3 Y2: 200 ÷ 1.30 = 153.8; Q4 Y2: 125 ÷ 1.00 = 125.0.

    (b) t = 9 (Q1 Y3): ŷ = 128.5 + 3.8(9) = 162.7. Forecast = 162.7 × 0.69 = $112.3k. t = 11 (Q3 Y3): ŷ = 128.5 + 3.8(11) = 170.3. Forecast = 170.3 × 1.30 = $221.4k.

    (c) Sum = 0.69 + 1.01 + 1.30 + 1.00 = 4.00 ✓. If the sum were not 4.00, it would mean the seasonal indices are inconsistent — they imply an impossible situation where total annual expected sales are above or below the actual annual average. This would indicate a calculation error that must be corrected before proceeding.

  10. Problem Solving

    Q10 — Evaluating a Forecast

    An ice-cream manufacturer uses the trend equation ŷ = 2400 + 180t (cases) with seasonal indices SIQ1 = 1.65, SIQ2 = 1.10, SIQ3 = 0.52, SIQ4 = 0.73 to forecast production. The actual Q3 Year 3 production (t = 11) was 3,800 cases.

    1. Calculate the forecast for Q3 Year 3 (t = 11).
    2. Find the residual. Is the model over- or underestimating?
    3. The manufacturer’s accountant says the model is good because the forecast was “close to the actual.” How could you quantify how close the forecast is as a percentage?
    4. If the actual Q1 Year 4 production is 9,200 cases and the forecast for t = 13 gives 9,540, calculate the residual and state what it means for inventory planning.

    (a) t = 11: ŷ = 2400 + 180(11) = 2400 + 1980 = 4380. Forecast = 4380 × 0.52 = 2,277.6 ≈ 2,278 cases.

    (b) Residual = 3800 − 2278 = +1,522 cases. The model is underestimating significantly — actual production was much higher than forecast. This could reflect an unusually warm winter driving unexpected ice-cream demand (irregular variation).

    (c) Percentage error = |residual| ÷ actual × 100 = 1522 ÷ 3800 × 100 ≈ 40%. This is a large percentage error, suggesting the model was not close at all for this period — the accountant’s claim is questionable.

    (d) Residual = 9200 − 9540 = −340 cases. The actual was 340 cases below the forecast. For inventory planning, this means the manufacturer produced too many raw materials (over-prepared). This could lead to waste or excess storage costs. The negative residual suggests the model slightly overestimated Q1 demand for this year.

  11. Problem Solving

    Q11 — Complete Time Series Investigation

    Monthly cafe revenue ($000) data shows a 3-point moving average (MA3) value of $48.2k at Month 6. The actual Month 6 value is $44.5k and actual Month 7 value is $52.1k.

    1. Using the MA3 = 48.2 at Month 6 and actual Month 7 = 52.1, work out the MA3 for Month 7 (given that Month 5 actual = $48.0k). Show your working.
    2. Month 7 has a seasonal index of 1.08. Deseasonalise the Month 7 actual value.
    3. The trend line fitted to deseasonalised data is ŷ = 42 + 0.85t. Predict the deseasonalised revenue for Month 14.
    4. The July seasonal index is 1.08. Forecast the actual July revenue for Month 14 (assuming July corresponds to Month 14 in a future year).

    (a) MA3 at Month 7 = (Month 5 + Month 6 + Month 7) ÷ 3 = (48.0 + 44.5 + 52.1) ÷ 3 = 144.6 ÷ 3 = $48.2k.

    Note: The MA3 at Month 6 uses Months 5, 6, 7: (48.0 + 44.5 + 52.1) ÷ 3 = 48.2 ✓. The MA3 at Month 7 uses Months 6, 7, 8 — we would need Month 8 data, which is not given. Using the data as provided: MA3 at Month 7 cannot be calculated without Month 8. If the question intends Month 7’s MA3 from Months 5–7: it is $48.2k as computed above.

    (b) Deseasonalised Month 7 = 52.1 ÷ 1.08 = $48.2k. This is the revenue with the July seasonal boost removed.

    (c) t = 14: ŷ = 42 + 0.85(14) = 42 + 11.9 = $53.9k.

    (d) Forecast = trend × SI = 53.9 × 1.08 = $58.2k. This is the expected actual revenue for that July, accounting for both the upward trend and the typical July seasonal boost.

  12. Understanding

    Q12 — Interpreting Seasonal Indices

    A seasonal index of SI = 0.62 is calculated for Q3 (July–September) for a pool cleaning business in Queensland.

    1. Interpret this seasonal index in plain language.
    2. If the trend predicts $120,000 revenue for Q3 next year, what is the seasonally adjusted forecast?
    3. The actual Q3 revenue was $78,000. Calculate the residual and interpret it.

    (a) SI = 0.62 for Q3 means that Q3 revenue is typically 38% below the annual quarterly average. This makes sense for a pool cleaning business in Queensland — winter (Q3) is the off-peak season when fewer people use their pools.

    (b) Seasonally adjusted forecast = 120,000 × 0.62 = $74,400.

    (c) Residual = actual − forecast = 78,000 − 74,400 = +$3,600. The actual Q3 revenue exceeded the forecast by $3,600. This positive residual could reflect an irregular event (e.g., an unusually early heatwave in September prompting early pool use) or gradual improvement in business not fully captured by the trend.

  13. Understanding

    Q13 — 5-Point Moving Average

    Monthly rainfall (mm) for a coastal town is: 85, 92, 78, 65, 58, 70, 88, 95, 110, 98, 86, 74.

    1. Calculate the 5-point moving averages for months 3 through 10.
    2. Compare the 5-point moving average with a 3-point moving average in terms of how much smoothing is achieved.

    (a) Month 3: (85+92+78+65+58)÷5 = 378÷5 = 75.6; Month 4: (92+78+65+58+70)÷5 = 363÷5 = 72.6; Month 5: (78+65+58+70+88)÷5 = 359÷5 = 71.8; Month 6: (65+58+70+88+95)÷5 = 376÷5 = 75.2; Month 7: (58+70+88+95+110)÷5 = 421÷5 = 84.2; Month 8: (70+88+95+110+98)÷5 = 461÷5 = 92.2; Month 9: (88+95+110+98+86)÷5 = 477÷5 = 95.4; Month 10: (95+110+98+86+74)÷5 = 463÷5 = 92.6.

    (b) The 5-point moving average produces a smoother series than the 3-point moving average because it averages over more data points, so individual extreme values have less influence. However, it also loses more data points at each end (2 points lost at each end vs 1 for the 3-point MA) and responds more slowly to genuine changes in trend. A 3-point MA responds faster to trend changes but retains more of the irregular fluctuation.

  14. Fluency

    Q14 — Describing Trends

    A time series graph of monthly online sales shows the following pattern: from Month 1–6 the trend is roughly flat; from Month 7–12 there is a steep upward trend; from Month 13–18 the trend flattens again.

    1. Describe the trend in each phase.
    2. Would a single straight-line trend fit this data well? Explain.
    3. What should an analyst do before fitting a linear trend line?

    (a) Month 1–6: stationary (flat) trend — no clear growth or decline. Month 7–12: strong upward trend — sales growing rapidly. Month 13–18: stationary trend again — sales have plateaued at the new higher level.

    (b) No, a single straight-line trend would not fit well. It would underestimate sales in the middle phase (Months 7–12) and overestimate growth in the flat phases. The data suggests an S-curve or step change rather than a consistent linear trend.

    (c) Before fitting a linear trend, the analyst should: (i) inspect the time series graph to confirm the trend is approximately linear; (ii) check the residual plot after fitting to ensure random scatter (no remaining pattern); (iii) consider whether a smoothed/deseasonalised series should be used rather than raw data.

  15. Problem Solving

    Q15 — Extended Problem: Full Time Series Analysis

    A Queensland irrigation equipment retailer records quarterly sales ($000) over 3 years. Seasonal indices are: SIQ1 = 0.88, SIQ2 = 1.32, SIQ3 = 1.18, SIQ4 = 0.62. The trend line fitted to CMA data is ŷ = 215 + 7.5t (t = 1 is Q1 Year 1).

    1. Show that the seasonal indices are consistent.
    2. Forecast the revenue for each quarter of Year 4 (t = 13, 14, 15, 16).
    3. In Year 4, the actual Q2 revenue was $512,000. Calculate the residual for Q2 Year 4.
    4. The owner wants to plan staffing for Year 4. Based on the forecasts, in which quarter should they expect the most business? Which quarter should they minimise staffing?
    5. Comment on any limitations of using this model to plan Year 4 staffing levels.

    (a) Sum of SI = 0.88 + 1.32 + 1.18 + 0.62 = 4.00 ✓. The indices are consistent (they sum to 4.00 for quarterly data).

    (b) Forecasts for Year 4:

    t = 13 (Q1): ŷ = 215 + 7.5(13) = 312.5. Forecast = 312.5 × 0.88 = $275,000

    t = 14 (Q2): ŷ = 215 + 7.5(14) = 320. Forecast = 320 × 1.32 = $422,400

    t = 15 (Q3): ŷ = 215 + 7.5(15) = 327.5. Forecast = 327.5 × 1.18 = $386,450

    t = 16 (Q4): ŷ = 215 + 7.5(16) = 335. Forecast = 335 × 0.62 = $207,700

    (c) Q2 Y4 forecast = $422,400 (from above, in $000: $422.4k). Actual = $512,000. Residual = 512,000 − 422,400 = +$89,600. Actual revenue exceeded the forecast by $89,600 — the model underestimated Q2 Year 4 by approximately 21%.

    (d) Most business expected in Q2 (SI = 1.32, forecast $422,400) — the spring planting season drives peak irrigation equipment sales. Minimum staffing in Q4 (SI = 0.62, forecast $207,700) — summer heat and school holidays reduce farm equipment purchasing activity.

    (e) Limitations: (i) the model is based on only 3 years of data — the seasonal pattern may not be fully established; (ii) the linear trend may not continue — market saturation or drought policy changes could affect demand; (iii) Year 4 is one period beyond the data (extrapolation), so forecasts carry more uncertainty; (iv) the large residual in Q2 Year 4 suggests irregular events can significantly affect actual outcomes.