Pearson’s correlation coefficient
Answers are for parents and teachers.
1
Six pupils: x = hours of football practice, y = minutes to run 1 km (table). Find r by the five sums and name the type.
| x | 2 | 4 | 5 | 7 | 9 | 11 |
|---|---|---|---|---|---|---|
| y | 14 | 13 | 10 | 9 | 6 | 5 |
Σx = 38, Σy = 57, Σxy = 302, Σx² = 296, Σy² = 607. Numerator 6 × 302 − 38 × 57 = −354; roots √332 and √393. r = −354 ÷ 361.21 = −0.9800: a strong inverse correlation.
2
Find r when n = 7, Σx = 113, Σy = 281, Σxy = 4735, Σx² = 1933, Σy² = 11675, and describe the correlation.
Numerator 7 × 4735 − 113 × 281 = 33145 − 31753 = 1392; roots √(13531 − 12769) = √762 and √(81725 − 78961) = √2764. r = 1392 ÷ 1451.26 = 0.9592: strong and direct.
3
For n = 5, Σx = 21, Σy = 47, Σxy = 129, Σx² = 111, Σy² = 647 show that the correlation is perfect and say which type.
Numerator 5 × 129 − 21 × 47 = −342. Roots √(555 − 441) = √114 and √(3235 − 2209) = √1026; the product √116964 = 342. So r = −342 ÷ 342 = −1: perfect inverse.
4
A student says: “For x = 2, 4, 6 the value of Σx² is 144, because 2 + 4 + 6 = 12 and 12² = 144.” Find the mistake and give the right value.
He squared the sum, which is (Σx)². The value of Σx² is 4 + 16 + 36 = 56.