Comparing Data Sets
Comparing data sets means using centre, spread and shape together to make a fair judgement. The mean and standard deviation suit symmetric data, while the median and IQR suit skewed data or data with outliers. In NSW Year 10 Advanced maths you compare sets shown in back-to-back stem-and-leaf plots, dot plots, histograms and box plots (NESA outcome MA5‑DAT‑C‑01).
You will compare homework times, sport hours, help-line waiting times, seedlings grown with and without fertiliser, earnings at a cafe and a supermarket, and pizza delivery times. You will choose the better measure when there is an outlier, compare times before and after a swim program, and choose a bus route to catch a train. Harder questions find unknown scores so two sets share a median, range or mean. It is part of Year 10 Advanced Single Variable and Bivariant Statistics.
Every question with a fully worked solution.
- Comparing Data Sets - Video - Back to Back Stem Plots Watch
Theory
To compare data sets, compare their centre, spread and shape with suitable statistics, then draw a conclusion in context. This Year 10 Advanced maths page covers every common display, how to choose between the mean and standard deviation and the median and IQR, what outliers do, and why a difference between groups does not prove a cause.
A good comparison has three parts: centre (which group is generally higher), spread (which is more consistent) and shape (symmetrical or skewed, outliers, clusters). Quote statistics for each.
For roughly symmetrical data without outliers, the mean and standard deviation use all the data. For skewed data or data with outliers, the median and IQR describe typical values better.
Shifting every score by the same amount changes the centre but not the spread. An outlier or a correction to one score can change the mean and SD noticeably while hardly moving the median and IQR.
Data that show two groups differ (an association) do not prove that one thing causes another. Other factors may explain the difference.
| Display | Read off |
|---|---|
| back-to-back stem-and-leaf | ordered scores for both groups: medians, quartiles, ranges |
| parallel dot plots | every score: medians, modes, shape, and mean or SD with a calculator |
| histograms | modal class, shape, the class containing the median |
| parallel box plots | five-number summaries: medians, IQRs, ranges, skewness |
| summary table | mean vs median for skew; SD vs IQR for the effect of outliers |
Comparing two data sets
- Get the data from the display (read back-to-back leaves outwards from the stem).
- Choose statistics: look at the shape first; use median and IQR if skewed or with outliers, mean and SD if roughly symmetrical.
- Calculate them for both groups (statistics mode for the mean and \(\sigma_x\)).
- Write one sentence each for centre, spread and shape, with numbers and units.
- Conclude in context, answering the question asked (which is better, faster, more reliable).
Careful conclusions
- Before and after: compare the means (did it change?) and SDs (did it become more consistent?).
- Adding or correcting a score: recalculate; a score close to the mean lowers the SD, a score far from it raises the SD.
- Cause: say "is associated with", not "causes", unless the groups were set up as a fair experiment.
| Class P | Stem | Class Q |
|---|---|---|
| \(\) | \(3\) | \(5\) |
| \(8\ \ 2\) | \(4\) | \(6\) |
| \(7\ \ 5\ \ 1\) | \(5\) | \(8\) |
| \(7\ \ 3\ \ 2\ \ 0\) | \(6\) | \(4\ \ 6\ \ 8\) |
| \(1\) | \(7\) | \(0\ \ 3\ \ 5\ \ 9\) |
| \(4\) | \(8\) | \(2\) |
Key: \(2\,|\,4\,|\,6\) means \(42\) for class P and \(46\) for class Q
(i) Find the median, IQR and range of each class.
(ii) Compare the classes.
(i) With \(11\) marks, the median is the \(6\)th, \(Q_1\) the \(3\)rd and \(Q_3\) the \(9\)th. Read class P's leaves from the stem outwards (right to left).
| \(\text{P}:\ \) | \(\) | \(42,\ 51,\ 60,\ 67,\ 84:\quad \text{IQR} = \mathbf{16},\ \text{range} = \mathbf{42}\) |
| \(\text{Q}:\ \) | \(\) | \(35,\ 58,\ 68,\ 75,\ 82:\quad \text{IQR} = \mathbf{17},\ \text{range} = \mathbf{47}\) |
Medians: P \(\mathbf{60}\), Q \(\mathbf{68}\).
(ii) Centre: Q's median is \(8\) marks higher, so Q generally did better. Spread: the IQRs are almost equal (\(16\) and \(17\)); Q's larger range comes from one low mark, \(35\). Shape: Q is negatively skewed (tail of low marks); P is fairly symmetrical with one high mark, \(84\).
| Prices (\(\$\) million) | Suburb A | Suburb B |
|---|---|---|
| Mean | \(1.02\) | \(0.95\) |
| Median | \(0.88\) | \(0.94\) |
| Standard deviation | \(0.31\) | \(0.12\) |
| IQR | \(0.15\) | \(0.14\) |
(i) Describe the shape of each suburb's prices.
(ii) In which suburb is a typical house cheaper?
(iii) The standard deviations are very different but the IQRs are almost equal. Explain.
(i) A: mean \(>\) median by \(\$0.14\)m, so a few expensive sales pull the mean up: positively skewed. B: mean \(\approx\) median: roughly symmetrical.
(ii) For skewed data use the median. A's median, \(\$0.88\)m, is lower than B's \(\$0.94\)m: a typical house is cheaper in A, even though A's mean is higher.
(iii) The middle halves of the two suburbs are equally spread (IQRs \(\$0.15\)m and \(\$0.14\)m). The standard deviation uses every price, and A's few very expensive houses, far above the mean, make its SD large. The IQR ignores those extremes.
(i) Find \(x\) and \(y\).
(ii) Which set has the larger population standard deviation?
(i) With \(5\) scores, \(Q_1\) is the average of the \(1\)st and \(2\)nd, and \(Q_3\) of the \(4\)th and \(5\)th. Set A: \(Q_1 = 6.5\), \(Q_3 = 17.5\), IQR \(= 11\); total \(60\).
| \(\dfrac{y + 18}{2} - \dfrac{3 + x}{2}\) | \(=\) | \(11 \quad\Rightarrow\quad y - x = 7\) |
| \(3 + x + 12 + y + 18\) | \(=\) | \(60 \quad\Rightarrow\quad x + y = 27\) |
| \(2y\) | \(=\) | \(34\) |
| \(y\) | \(=\) | \(\mathbf{17},\quad x = \mathbf{10}\) |
Check order: \(3, 10, 12, 17, 18\ \checkmark\)
(ii) Statistics mode: \(\sigma_A = 5.253\ldots \approx 5.25\) and \(\sigma_B = 5.403\ldots \approx 5.40\). Set B is slightly more spread out by standard deviation, even though the means and IQRs match.
| Person | Before | After |
|---|---|---|
| \(1\) | \(72\) | \(70\) |
| \(2\) | \(80\) | \(76\) |
| \(3\) | \(68\) | \(67\) |
| \(4\) | \(75\) | \(72\) |
| \(5\) | \(84\) | \(79\) |
| \(6\) | \(70\) | \(69\) |
| \(7\) | \(77\) | \(74\) |
| \(8\) | \(79\) | \(75\) |
| \(9\) | \(66\) | \(66\) |
| \(10\) | \(74\) | \(71\) |
(i) Find the mean and population standard deviation before and after, correct to \(2\) decimal places.
(ii) Describe the change.
(iii) Does this show that the app caused the change?
(i) Statistics mode:
| \(\text{Before}:\ \bar{x}\) | \(=\) | \(\dfrac{745}{10} = \mathbf{74.5},\quad \sigma \approx \mathbf{5.37}\) |
| \(\text{After}:\ \bar{x}\) | \(=\) | \(\dfrac{719}{10} = \mathbf{71.9},\quad \sigma \approx \mathbf{3.91}\) |
(ii) The mean fell by \(2.6\) beats per minute and no one's rate went up. The standard deviation also fell, so the rates became lower and more consistent.
(iii) No. The data show an association, not a cause. There is no comparison group that did not use the app; the season, other exercise, or simply being measured again could explain the change.
Common pitfalls
Frequently asked questions
How do you compare two sets of data?
Compare the centre using the means or medians, the spread using the standard deviations or interquartile ranges, and the shape. Quote the values and finish with a conclusion in context.
When should I use the median and IQR instead of the mean and standard deviation?
When the data are skewed or contain outliers. Extreme values pull the mean and standard deviation, but they barely affect the median and interquartile range.
How do you read a back-to-back stem-and-leaf plot?
The stems are in the middle. Leaves on the right belong to one group and read outwards as usual; leaves on the left belong to the other group and also read outwards from the stem, so they are written in reverse order.
What does it mean if the standard deviation is much larger than the IQR suggests?
There are probably a few scores far from the mean, such as outliers or a long tail, which increase the standard deviation but do not affect the middle half of the data.
Does a difference between two groups prove a cause?
No. It shows an association. Other factors could explain the difference unless the groups were set up as a fair experiment.