📈 Time Series Analysis Quiz

Model Answer:
A time series is a set of data values recorded at successive, equally-spaced points in time. The horizontal (x) axis must always represent time (e.g. year, quarter, month), and the vertical (y) axis represents the measured variable. Points are usually joined with line segments to show how the variable changes over time.

Model Answer:
(a) Trend: Overall increasing long-term trend — values in Year 2 are consistently higher than the corresponding quarters in Year 1.

(b) Seasonality: Clear seasonal pattern — Q1 is the highest each year (summer peak) and Q3 is the lowest (winter trough). This repeats in a predictable, calendar-related cycle.

(c) Irregular fluctuations: No obvious irregular fluctuations — the data follows a consistent seasonal pattern without unexpected spikes or dips.

Model Answer:
Seasonality refers to systematic, calendar-related movements that repeat predictably at the same time each year — e.g. ice-cream sales peaking every summer and falling every winter.

Irregular fluctuations are unsystematic, random, short-term variations caused by unpredictable one-off events (e.g. a natural disaster, a strike, or unusual weather). They do not repeat in a regular pattern and cannot be forecast from the data.

Model Answer:
Trend: Overall increasing long-term trend — electricity consumption generally rises across the full period, suggesting growing demand over time.

Irregular fluctuation: The sharp drop in 2017 is an irregular fluctuation — an unsystematic, short-term deviation from the trend likely caused by a one-off event (e.g. unusually mild weather, a major energy efficiency initiative, or an economic shock).

Seasonality: Cannot be confirmed from annual data alone — sub-yearly observations would be needed to reveal a seasonal pattern.

Model Answer:
A decreasing trend with seasonality means the overall long-term direction of the data is downward (values generally fall over time), but within each year there are still predictable, calendar-related peaks and troughs that repeat.

Example: DVD rental store revenue from 2008–2015. Overall revenue declines year-on-year as streaming services grow (decreasing trend), but there are still predictable seasonal peaks in winter (people stay home more) and troughs in summer — a repeating seasonal pattern superimposed on the declining trend.

Model Answer:
A 3-point moving mean averages each group of 3 consecutive values, placed at the centre period:

Period 2: (18 + 24 + 15) ÷ 3 = 57 ÷ 3 = 19.0
Period 3: (24 + 15 + 21) ÷ 3 = 60 ÷ 3 = 20.0
Period 4: (15 + 21 + 27) ÷ 3 = 63 ÷ 3 = 21.0
Period 5: (21 + 27 + 18) ÷ 3 = 66 ÷ 3 = 22.0
Period 6: (27 + 18 + 24) ÷ 3 = 69 ÷ 3 = 23.0
Period 7: (18 + 24 + 30) ÷ 3 = 72 ÷ 3 = 24.0

No smoothed values for periods 1 or 8 — insufficient neighbours on one side.

Model Answer:
Order each group of 5 consecutive values and select the middle (3rd) value:

Period 3 — values: 10, 14, 8, 20, 12 → ordered: 8, 10, 12, 14, 20 → Median = 12
Period 4 — values: 14, 8, 20, 12, 16 → ordered: 8, 12, 14, 16, 20 → Median = 14
Period 5 — values: 8, 20, 12, 16, 11 → ordered: 8, 11, 12, 16, 20 → Median = 12

Model Answer:
Smoothing removes (or reduces) short-term irregular fluctuations and seasonal variation from the data, leaving a clearer picture of the long-term trend. By averaging consecutive values, extreme highs and lows are dampened, so the overall direction of the data — whether increasing, decreasing, or stationary — becomes much more visible and easier to analyse or model.

Model Answer:
With an odd number of data points, the smoothed value can be placed at the exact centre of the group — e.g. a 3-point average sits at the 2nd value, a 5-point at the 3rd. This gives a clearly defined time period to assign the smoothed value to.

With an even number of data points there is no single middle value, so the smoothed value falls between two time periods — making it impossible to assign it to a specific period without an additional centring step (which is beyond the scope of this course).

Model Answer:
Period 2: (5 + 9 + 3) ÷ 3 = 5.67
Period 3: (9 + 3 + 42) ÷ 3 = 18.0
Period 4: (3 + 42 + 7) ÷ 3 = 17.33
Period 5: (42 + 7 + 11) ÷ 3 = 20.0
Period 6: (7 + 11 + 4) ÷ 3 = 7.33

Median vs Mean: The median is preferred when the data contains outliers (extreme values). Here, the value 42 is a clear outlier that heavily inflates the mean-smoothed values at periods 3, 4, and 5, masking the true underlying pattern. The median is resistant to outliers and would give more representative smoothed values in this case.

Model Answer:
A seasonal index measures how much a particular season (quarter or month) typically differs from the overall average. A value of 1 means the season is exactly at the average level.

A seasonal index of 1.25 means that season is typically 25% above the overall seasonal average — it is a high/peak season.

A seasonal index of 0.80 means that season is typically 20% below the overall seasonal average — it is a low/off-peak season.

Model Answer:
Step 1 — Overall mean of all 8 values:
40 + 60 + 80 + 50 + 44 + 66 + 88 + 54 = 482  →  Mean = 482 ÷ 8 = 60.25

Step 2 — Average for each quarter:
Q1: (40 + 44) ÷ 2 = 42  |  Q2: (60 + 66) ÷ 2 = 63  |  Q3: (80 + 88) ÷ 2 = 84  |  Q4: (50 + 54) ÷ 2 = 52

Step 3 — Seasonal index = Quarter average ÷ Overall mean:
SI(Q1) = 42 ÷ 60.25 ≈ 0.697
SI(Q2) = 63 ÷ 60.25 ≈ 1.046
SI(Q3) = 84 ÷ 60.25 ≈ 1.394
SI(Q4) = 52 ÷ 60.25 ≈ 0.863

Check: 0.697 + 1.046 + 1.394 + 0.863 = 4.000 ✓ (quarterly indices must sum to 4)

Model Answer:
Deseasonalised value = Actual value ÷ Seasonal index

Q1: 40 ÷ 0.697 ≈ 57.4
Q2: 60 ÷ 1.046 ≈ 57.4
Q3: 80 ÷ 1.394 ≈ 57.4
Q4: 50 ÷ 0.863 ≈ 57.9

Observation: All four deseasonalised values are nearly identical (~57.4). This confirms that Year 1 revenue is essentially flat once the seasonal effects are removed — the large variation between quarters was entirely due to seasonality, not any underlying growth.

Model Answer:
For quarterly data, the four seasonal indices must always sum to 4 (since there are 4 seasons and the average index = 1).

SI(Q4) = 4 − (0.84 + 1.12 + 1.30) = 4 − 3.26 = 0.74

A seasonal index of 0.74 means Q4 is typically 26% below the quarterly average — it is an off-peak season.

Model Answer:
Deseasonalised value = Actual ÷ Seasonal index = $87,000 ÷ 1.45 ≈ $60,000

This represents the underlying sales performance for December after removing the typical Christmas-season boost (which is 45% above the monthly average). The deseasonalised figure of $60,000 is directly comparable to other months' deseasonalised values, allowing a fair assessment of the business's true performance independent of seasonal effects.

Model Answer:
The least-squares trend line: ŷ = a + b·t

ŷ = the predicted value of the variable at time t
t = the time period (coded as 1, 2, 3, …)
a (y-intercept) = the predicted value when t = 0; represents the baseline starting point of the model
b (slope) = the average change in the variable per unit of time. A positive b indicates an increasing trend; a negative b indicates a decreasing trend. For example, b = 2.5 means the variable increases by 2.5 units per time period on average.

Model Answer:
n = 5  |  Σt = 15  |  Σy = 85  |  Σt² = 55
Σty = (1×12)+(2×15)+(3×17)+(4×20)+(5×21) = 12+30+51+80+105 = 278

b = [Σty − (Σt · Σy)/n] ÷ [Σt² − (Σt)²/n]
   = [278 − (15×85)/5] ÷ [55 − 225/5]
   = [278 − 255] ÷ [55 − 45] = 23 ÷ 10 = 2.3

a = ȳ − b·t̄ = (85/5) − 2.3×(15/5) = 17 − 6.9 = 10.1

∴ Trend line: ŷ = 10.1 + 2.3t
The deseasonalised value increases by an average of 2.3 units per time period.

Model Answer:
(a) For Year 8, t = 8:
ŷ = 4.2 + 1.8 × 8 = 4.2 + 14.4 = $18.6 million

(b) The slope of 1.8 means the company's deseasonalised annual sales increase by an average of $1.8 million per year. This indicates a consistent positive long-term trend in underlying sales performance.

Model Answer:
If a trend line is fitted to raw seasonal data, the recurring seasonal peaks and troughs distort the line of best fit — pulling it up or down rather than tracking the true underlying direction.

Fitting to deseasonalised data removes the seasonal effects first, so the trend line accurately models the long-term direction only. This produces more reliable predictions, and seasonal effects can then be re-applied separately to generate actual forecasts for specific seasons.

Model Answer:
(a) Decreasing trend — the variable falls by an average of 3.2 units per time period (negative slope).

(b) ŷ = 50 − 3.2 × 10 = 50 − 32 = 18

(c) Set ŷ = 0:   0 = 50 − 3.2t  →  3.2t = 50  →  t = 50 ÷ 3.2 ≈ 15.6
The model predicts the value reaches zero at approximately t = 15.6. In practice, the trend may change before this point.

Model Answer:
Q3 of Year 3 = t = 11  (Year 1: t=1–4; Year 2: t=5–8; Year 3 Q1=9, Q2=10, Q3=11)

Step 1 — Deseasonalised forecast:
ŷ = 18 + 0.5 × 11 = 18 + 5.5 = 23.5 thousand

Step 2 — Re-seasonalise (multiply by seasonal index for Q3):
Actual forecast = 23.5 × 2.20 = 51.7 thousand visitors

This large value is expected — the Q3 seasonal index of 2.20 indicates winter visits are 120% above the quarterly average, reflecting the peak skiing season.

Model Answer:
Deseasonalised = Actual ÷ Seasonal index

This year: $42,000 ÷ 1.40 = $30,000
Last year: $38,000 ÷ 1.40 ≈ $27,143

The underlying (deseasonalised) sales did increase by ~$2,857 (~10.5%), confirming genuine growth. However, the shop owner overstates the improvement — the raw $4,000 difference is partly inflated by the fact that April is a seasonally strong month (40% above average). The deseasonalised figures give a much more accurate picture of true business performance.

Conclusion: Business has improved, but comparing raw figures across seasons without deseasonalising is misleading.

Model Answer:
For Year 7, t = 7:
ŷ = 8.4 − 0.6 × 7 = 8.4 − 4.2 = 4.2%

Reliability: Year 7 is an extrapolation — 2 years beyond the data range (t = 1 to 5). The historical trend is very consistent (close to linear), which provides moderate confidence. However:
• Extrapolation always carries more uncertainty than interpolation within the data range
• Economic conditions can shift unexpectedly (recession, government policy changes) that could reverse the trend

Overall: Cautiously reliable given the strong, consistent trend — but should be treated as an estimate only.

Model Answer:
A seasonal index of 0.42 for July means revenue is typically 58% below the monthly average — it is indeed a very slow month. However, a low seasonal index alone does not justify closure:

1. The overall trend matters: If deseasonalised revenue is positive and growing, the business is fundamentally healthy
2. Fixed costs continue: Rent, insurance, and staff costs don't necessarily disappear during closure
3. Customer loyalty: Closing may damage relationships with regular customers who return in peak season
4. Low SI ≠ unprofitable: July may still generate revenue that exceeds variable costs

The seasonal index tells us July is below average — not whether July is profitable. The manager would need to compare July revenue against July-specific costs before drawing this conclusion.

Model Answer:
(a) Positive (increasing) long-term trend. Deseasonalised quarterly profit increases by an average of 1.2 units per quarter.

(b) Year 4 Q2 = t = 14  (Y1: t=1–4; Y2: t=5–8; Y3: t=9–12; Y4 Q1=13, Q2=14)
Deseasonalised forecast: ŷ = 12.5 + 1.2 × 14 = 12.5 + 16.8 = 29.3
Actual forecast = Deseasonalised × SI(Q2) = 29.3 × 1.10 = 32.23

(c) Any one of:
• This is an extrapolation beyond observed data — unexpected events (irregular fluctuations) could significantly alter the actual result
• The linear model assumes a constant rate of growth, which may not continue
• The seasonal indices are based on past patterns and may not remain stable in the future