Question 10
Which of the following graphs represents a time series plot with a three-point moving average?
Not quite. (A) has only three smoothed points. “Three-point” describes how each smoothed value is made, from three values in a row. It doesn’t mean three points on the graph.
Not quite. (B) also has only three smoothed points, and they don’t sit on the averages. The first average belongs at 2004 and is $\frac{11+12+9}{3}\approx10.7$, but (B) plots its first square up at 12.
Not quite. (C)’s squares sit right on top of every third original value: 2003, 2006, 2009, 2012 and 2015. Picking out every third value isn’t averaging. A moving average combines three values in a row for every point.
That’s it. Step through the hints below to see why.
Slide the three-year window along the data. Each stop averages the three orange values and plots the result at the middle year.
(A) has three blue squares, at 2003, 2009 and 2015. A three-point moving average can’t have a value at 2003 or 2015, because those years have a neighbour on one side only.
(B) has three blue squares, at 2004, 2009 and 2014. The years are possible, but none of the squares sits on an average. Slide to 2004 and compare.
(C)’s blue squares sit exactly on the black dots at 2003, 2006, 2009, 2012 and 2015. Those are original values, picked out every third year. Nothing has been averaged.
(D) has a blue square for every year from 2004 to 2014. Slide the window all the way to 2014 and every orange square lands on a blue one.
A three-point moving average replaces each value with the mean of three values in a row: that year, the year before and the year after. The result is plotted at the middle year.
“Three-point” is about how each smoothed value is made. It says nothing about how many smoothed points end up on the graph.
The data starts in 2003, so the first full window is 2003, 2004 and 2005, centred on 2004:
$$\frac{11+12+9}{3}\approx10.7$$
Then the window moves on one year at a time, all the way to the last window centred on 2014. That gives a smoothed point for every year from 2004 to 2014, eleven in all. Only 2003 and 2015 miss out.
Now count the squares in each option. (A) and (B) have three, and (C) has five. Only (D) has eleven, from 2004 to 2014.
Check one to be sure. (D)’s first square sits a little under 11 at 2004, which matches 10.7. The smoothed line also does what smoothing should: it follows the ups and downs of the original but with smaller swings.
Each smoothed value is the mean of three consecutive years, plotted at the middle year:
$$\begin{aligned}\text{2004: }&\frac{11+12+9}{3}\approx10.7\\\text{2005: }&\frac{12+9+15}{3}=12\\&\ \ \vdots\\\text{2014: }&\frac{11+16+12}{3}=13\end{aligned}$$
That is eleven smoothed points, 2004 to 2014, and only (D) has them, each on its average.
On the QCAA paper the 2011 value has no dot of its own, but the original line passes through 8 there and (D)’s averages for 2010 to 2012 use 8, so it is drawn here as a dot.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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