Are they really different?
Tonight's idea
Two classes take the same quiz. The mean mark (add them up, divide by how many) is 15 for class A, 13 for class B. Is A really stronger? Draw every mark as a dot.
The gap between the means is 2 marks in both pictures. On the left it shows: A's clump sits right of B's. On the right it is lost: the marks are so scattered that the classes cover almost the same ground.
So measure the gap against the spread. One measure of spread is the mean absolute deviation, or MAD: find how far each value is from the mean, then average those distances.
Gap in MADs = difference between the means ÷ MAD.
On the left the MAD is 1 mark, so the gap is \(2 \div 1 = 2\) MADs. On the right the MAD is 4, so the gap is 0.5 MADs. A rough guide, for groups with dot plots of similar shape: 2 MADs or more is a clear difference. Under 1 MAD, the groups mostly overlap. Odd shapes can fool the guide, so look at the dots too.
Watch one: tomato plants in sun and shade
Heights in cm. Sunny bed: 24, 27, 28, 29, 32. Shady bed: 20, 21, 23, 25, 26.
Sunny mean: \(140 \div 5 = 28\). Distances from 28 are 4, 1, 0, 1, 4. Their mean is \(10 \div 5 = 2\), so the MAD is 2 cm.
Shady mean: \(115 \div 5 = 23\). Distances from 23 are 3, 2, 0, 2, 3. MAD: \(10 \div 5 = 2\) cm.
Gap: \(28 - 23 = 5\) cm, which is \(5 \div 2 = 2.5\) MADs. A clear difference.
Mean and median are on the grade 6 sheet Lying with averages.
Your turn
- Find the mean and the MAD of 3, 5, 8, 12.
- Work out each gap in MADs. Say whether the difference is clear, or the groups mostly overlap, or it is in between. (a) Means 50 and 56, MAD 3 for both groups. (b) Means 50 and 56, MAD 12 for both. (c) Means 150 cm and 162 cm, MAD 8 cm for both.
- Minutes spent reading last night. Group P: 6, 7, 8, 8, 9, 10. Group Q: 9, 9, 9, 11, 11, 11. (a) Draw both dot plots, one above the other, on the same number line. (b) Find the mean and MAD of each group. (c) How many MADs apart are the means?
- A headline says: "Left-handers beat right-handers in memory test." The means were 71 and 70, and the MAD in each group was about 9. (a) Find the gap in MADs. (b) You meet one left-hander and one right-hander from the test. Can you be confident the left-hander scored higher? (c) How small would the MAD have to be for this 1-point gap to equal 2 MADs?
Stretch: change the units
Class A has mean 15 and class B has mean 13. Both have MAD 1. Every mark in both classes is doubled. Find the new means, the new MAD and the new gap in MADs. Then go back to the start and add 3 to every mark. What changes now?
Answers
- Mean: \(28 \div 4 = 7\). Distances from 7 are 4, 2, 1, 5. MAD: \(12 \div 4 = 3\).
- (a) \(6 \div 3 = 2\) MADs: clear. (b) \(6 \div 12 = 0.5\) MADs: mostly overlapping. (c) \(12 \div 8 = 1.5\) MADs: in between. A real shift, with plenty of overlap.
- (a) P has dots at 6, 7, 8, 8, 9, 10, with two stacked at 8. Q has three stacked at 9 and three at 11. (b) P: mean \(48 \div 6 = 8\); distances 2, 1, 0, 0, 1, 2; MAD \(6 \div 6 = 1\). Q: mean \(60 \div 6 = 10\); every distance is 1; MAD 1. (c) \(2 \div 1 = 2\) MADs.
- (a) \(1 \div 9 \approx 0.11\) MADs. (b) No. The gap is tiny compared with the spread. Means and MADs alone cannot tell you exactly how much the groups overlap; for that you need the dot plots. But nothing here says left-handers reliably score higher. The headline makes a tiny gap sound like a rule. (c) \(1 \div 2 = 0.5\) points.
Stretch. Doubled: the means are 30 and 26. Every distance from the mean doubles too, so the MAD is 2. The gap is \(4 \div 2 = 2\) MADs, as before. Adding 3: the means are 18 and 16. Every dot slides the same way, so the distances and the MAD stay at 1. The gap is still 2 MADs. Measured in MADs, the gap does not depend on the units.