Can r be fooled?
A weather report gives the average daily high temperature for each month of the year. If you compute r for month number and temperature, you get about 0.29: a weak correlation. Yet anyone in Ontario can tell you that temperature depends very strongly on the month! What went wrong?
Nothing is wrong with the arithmetic. The problem is that r only measures linear (straight-line) relationships. This lesson is about what r does and does not tell you.
The coefficient of determination, r²
Squaring r gives the coefficient of determination, r². It answers a useful question:
What fraction of the variation in y is explained by the straight-line relationship with x?
For the study-time data in the last lesson, r ≈ 0.979, so r² ≈ 0.958. About 96% of the variation in marks is explained by the linear relationship with study time. The other 4% comes from everything else (sleep, prior knowledge, luck on the test day...).
Notice that r² is always between 0 and 1, and that it is smaller than |r| (unless r is 0, 1 or −1). An r of 0.5 sounds moderate, but r² = 0.25 says the line explains only a quarter of the variation.
Worked exampleExample 2: One point, big effect
Show by calculation that adding the point (10, 10) to (1, 3), (2, 1), (3, 4), (4, 2) changes r from 0 to 0.9.
- Four points: Σx = 10, Σy = 10, Σxy = 3 + 2 + 12 + 8 = 25. Numerator: 4(25) − 10(10) = 0, so r = 0.
- Five points: Σx = 20, Σy = 20, Σxy = 25 + 100 = 125, Σx² = 30 + 100 = 130, Σy² = 30 + 100 = 130.
- Numerator: 5(125) − 20(20) = 625 − 400 = 225.
- Brackets: 5(130) − 20² = 250 and 5(130) − 20² = 250.
- r = 225 ÷ √(250 × 250) = 225 ÷ 250 = 0.9.
r goes from 0 to 0.9 because of one point. A report that only said "r = 0.9" would badly misrepresent the four ordinary points. Check for outliers and report r with and without them.