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Exercises · Q19
Q.

Calculate the correlation coefficient between X and Y and comment on their relationship:

X–3–2–1123
Y941149

(Ans. r = 0)

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Here ∑X=0\sum X=0 and ∑XY=0\sum XY=0, so the numerator is 00 and r=0r=0. Yet Y=X2Y=X^{2} exactly — a perfect non-linear relationship. This is the classic case where zero correlation does not mean no relationship.

Concept first

r=N∑XY−∑X ∑Y[N∑X2−(∑X)2][N∑Y2−(∑Y)2]r=\frac{N\sum XY-\sum X\,\sum Y}{\sqrt{[N\sum X^{2}-(\sum X)^{2}][N\sum Y^{2}-(\sum Y)^{2}]}}

The working table (N=6N=6)

XXYYXYXYX2X^{2}Y2Y^{2}
-39-27981
-24-8416
-11-111
11111
248416
3927981
028028196

Substituting

N∑XY−∑X∑Y=6(0)−(0)(28)=0N\sum XY-\sum X\sum Y = 6(0)-(0)(28)=0

Since the numerator is 00,

r=0[6(28)−0][6(196)−282]=0r=\frac{0}{\sqrt{[6(28)-0][6(196)-28^{2}]}}=0

Comment on the relationship …

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