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Year 10 Maths - 10 Statistics

Time-Series Data

20 practice questions 2 video lessons Theory + worked examples
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Theory

A time series records the same quantity at regular times, such as each month or each year. This Year 10 maths page shows how to read time-series graphs and tables, describe the trend and any seasonal pattern, find the biggest changes, use percentage change, fit a linear trend and use it to predict, and judge when a prediction is reliable.

A time series is a set of data collected at equal time intervals. Time is the independent variable, so it goes on the horizontal axis, and the points are joined in order to make a line graph. Data with one variable for each item (such as heights) are univariate; a time series pairs each value with a time, so it is bivariate.

The trend is the long-term direction: increasing, decreasing or steady. A seasonal pattern repeats at the same time every year, such as heating costs peaking each winter. A cycle repeats over a longer, less regular period, and irregular changes are one-off jumps or dips.

If a series rises by the same amount each time period, the trend is linear and can be modelled by \(y = a + bt\), where \(a\) is the value at \(t = 0\) and \(b\) is the change per time period. If it rises by the same percentage each period, the increases keep growing and the trend is non-linear.

Using a trend to predict inside the time span of the data is interpolation; predicting the future is extrapolation, which assumes the trend continues and is less reliable the further ahead you go.

Trend and seasonal patternQuarterly sales over two years: 18, 32, 26, 12 then 26, 41, 34, 20. The sales peak every second quarter and dip every fourth quarter (the seasonal pattern), while a dashed trend line rises slowly. Y1 Q1Y2 Q101020304050QuarterSales ($ thousand) trend
The seasonal pattern repeats every four quarters. The dashed trend line shows the slow long-term rise through the ups and downs.
Linear and non-linear growthTwo series start at 100. The linear one rises by 20 each year: 100, 120, 140, 160, 180. The non-linear one rises by 20% each year: 100, 120, 144, 172.8, 207.4, curving upwards. 01234050100150200YearsValue +20 a year+20% a year
Equal increases give a straight line. Equal percentage increases give a curve that gets steeper, so a straight trend line does not fit it.
To findUse
change between two timeslater value \(-\) earlier value
biggest rise or fallthe change for each pair of neighbouring times; the largest is the steepest segment
rate of change\(\dfrac{\text{change in value}}{\text{change in time}}\)
linear model\(y = a + bt\) with \(t\) = time since the start
\[\text{percentage change} = \dfrac{\text{new} - \text{old}}{\text{old}} \times 100\%\]
\[\begin{aligned}&\text{percentage error}\\ &\quad= \dfrac{|\text{predicted} - \text{actual}|}{\text{actual}} \times 100\%\end{aligned}\]
Compare like with like. With seasonal data, compare the same season in different years (Quarter 3 with Quarter 3) to see the trend. A fall from winter to spring every year is seasonal, not a real change.
Units of time. A gradient of \(0.03\) per year is \(10 \times 0.03 = 0.3\) per decade. Let \(t\) be the number of years after the first year so the numbers stay small.
Where two trends meet. For \(A = 120 + 8t\) and \(B = 200 - 2t\), solve \(120 + 8t = 200 - 2t\): \(10t = 80\), so \(t = 8\) and both equal \(184\). Check that \(t = 8\) is inside the time span of the data.
Checking a prediction. A trend predicted \(14.4\) but the actual value was \(13.5\). The percentage error is \(\dfrac{0.9}{13.5} \times 100\% \approx 6.7\%\).

How to describe a time series

  1. Read the axes: the time interval and the units of the values.
  2. Trend: look past the ups and downs. Is the series rising, falling or steady overall?
  3. Season: is there a pattern that repeats each year? Name when the peaks and dips happen and explain them.
  4. Changes: subtract neighbouring values to find the biggest rise or fall; use percentage change to compare the same season in different years.
  5. Unusual values: note any one-off jump or dip.

How to predict from a trend

  1. Check the form: equal increases (linear) or growing increases (for example a constant percentage).
  2. Linear: find \(b\) from two points, \(b = \dfrac{\text{change in value}}{\text{change in time}}\), then \(y = a + bt\). Substitute \(t\) to predict, or set \(y\) and solve for \(t\).
  3. Percentage growth: multiply by the growth factor once for each period.
  4. Comment: a prediction far beyond the data is extrapolation and may fail if conditions change.
Example 1 β€” Trend and season
The graph shows the number of visitors, in thousands, to a ski resort each quarter from \(2023\) to \(2025\). (Quarter \(3\) is July to September.)
Ski resort visitors each quarterQuarterly visitors to a ski resort, 2023 to 2025: 12, 30, 95, 20; 14, 33, 101, 22; 15, 36, 108, 25. There is a tall peak in Quarter 3 of every year. 202320242025020406080100120Quarter (Q1 of each year marked)Visitors (thousands)
(i) In which quarter are visitor numbers highest each year? Explain why.
(ii) Is there a long-term trend?
(iii) Find the percentage change in Quarter \(3\) visitors from \(2024\) to \(2025\), to \(1\) decimal place.
Solution

(i) Quarter 3 every year (\(95\), \(101\), \(108\) thousand). It is winter, the ski season, so this is a seasonal pattern.

(ii) Compare the same quarter each year. Quarter 3: \(95 \to 101 \to 108\); Quarter 1: \(12 \to 14 \to 15\). Every quarter is higher than a year before, so the trend is increasing.

\(\text{(iii)}\ \dfrac{108 - 101}{101} \times 100\%\)\(=\)\(\dfrac{7}{101} \times 100\%\)
\(\)\(\approx\)\(\mathbf{6.9\%}\)
Example 2 β€” The biggest changes
The table shows the number of cars registered in a town, in thousands.
Year\(2015\)\(2016\)\(2017\)\(2018\)
Cars\(48\)\(51\)\(55\)\(54\)
Year\(2019\)\(2020\)\(2021\)
Cars\(60\)\(57\)\(62\)
(i) Between which two years was the greatest increase?
(ii) Between which two years did the number fall the most?
(iii) Describe the trend.
Solution

Subtract each year's value from the next year's.

\(2015 \to 2016\)\(:\)\(+3\)
\(2016 \to 2017\)\(:\)\(+4\)
\(2017 \to 2018\)\(:\)\(-1\)
\(2018 \to 2019\)\(:\)\(+6\)
\(2019 \to 2020\)\(:\)\(-3\)
\(2020 \to 2021\)\(:\)\(+5\)

(i) The greatest increase was 2018 to 2019 (\(6\) thousand): the steepest rising segment of the graph.

(ii) The biggest fall was 2019 to 2020 (\(3\) thousand).

(iii) Despite two small dips, the overall trend is increasing: from \(48\) thousand to \(62\) thousand in six years.

Example 3 β€” Predicting with a trend line
The average weekly hours of streaming per household has a linear trend line through \((2015, 8.4)\) and \((2023, 11.6)\). The data cover \(2015\) to \(2023\).
(i) Find the gradient in hours per year, and per decade.
(ii) Write a rule for \(H\), with \(t\) the number of years after \(2015\).
(iii) Predict \(H\) in \(2030\).
(iv) In which year does the trend give \(13.2\) hours? How reliable is it?
Solution
\(\text{(i) } b\)\(=\)\(\dfrac{11.6 - 8.4}{2023 - 2015} = \dfrac{3.2}{8}\)
\(\)\(=\)\(\mathbf{0.4}\text{ h per year}\)

That is \(10 \times 0.4 = \mathbf{4}\) hours per decade.

\(\text{(ii) } H\)\(=\)\(\mathbf{8.4 + 0.4t}\)
\(\text{(iii) } t = 15:\ H\)\(=\)\(8.4 + 0.4(15) = \mathbf{14.4}\)
\(\text{(iv) } 8.4 + 0.4t\)\(=\)\(13.2\)
\(0.4t\)\(=\)\(4.8\)
\(t\)\(=\)\(12 \Rightarrow \mathbf{2027}\)

Both \(2027\) and \(2030\) are after the data, so these are extrapolations. Viewing habits can change quickly, so the predictions become less reliable the further ahead they go.

Example 4 β€” Linear or not?
The number of subscribers to a video channel, in thousands, at the start of each year was \(256\), \(320\), \(400\), \(500\), \(625\).
(i) Is the growth linear?
(ii) Find the percentage increase each year.
(iii) If the pattern continues, predict the next value.
Solution
\(\text{(i) increases}\)\(:\)\(64,\ 80,\ 100,\ 125\)

The increases are not equal, so the growth is not linear. A straight trend line would under-predict.

\(\text{(ii)}\ \dfrac{320}{256}\)\(=\)\(1.25\)
\(\dfrac{400}{320} = \dfrac{500}{400} = \dfrac{625}{500}\)\(=\)\(1.25\)

Each value is \(1.25\) times the one before: a \(25\%\) increase each year.

\(\text{(iii)}\ 625 \times 1.25\)\(=\)\(\mathbf{781.25}\text{ thousand}\)

About \(781\,000\) subscribers, assuming the \(25\%\) growth continues.

Common pitfalls

Compare the same season. A fall from summer to autumn may happen every year. Compare a quarter with the same quarter a year earlier before claiming a real change.
Steepest segment, not highest point. The greatest increase is the biggest difference between neighbouring values, not the year with the largest value.
Check the form before drawing a straight line. Equal percentage increases curve upwards; a linear trend would under-predict.
Extrapolation is a guess. A trend line can predict impossible values, such as a negative price, far outside the data. Say why the trend might not continue.

Frequently asked questions

What is a time series?

A set of data recorded at equal time intervals, such as monthly rainfall or yearly population. Time goes on the horizontal axis and the points are joined to make a line graph.

What is the difference between a trend and a seasonal pattern?

The trend is the long-term direction of the data. A seasonal pattern is a rise and fall that repeats at the same time every year.

How do you find the biggest increase in a time series?

Subtract each value from the next one. The largest positive difference is the biggest increase, and it shows as the steepest rising part of the graph.

How do you predict from a linear trend?

Find the change per time period from two points, write y = a + bt, and substitute the time you want. To find when a value is reached, set y equal to it and solve for t.

Why are predictions far into the future unreliable?

They are extrapolations. They assume the trend carries on unchanged, but conditions such as prices, weather or habits can change.