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Exercises · Q20
Q.

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

X134578
Y268101416

(Ans. r = 1)

Puducherry CbseNCERTSubjective· 3mImportance★★★★★est
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∑X=28, ∑Y=56, ∑XY=328, ∑X2=164, ∑Y2=656\sum X=28,\ \sum Y=56,\ \sum XY=328,\ \sum X^{2}=164,\ \sum Y^{2}=656 give r=400200×800=+1r=\frac{400}{\sqrt{200\times800}}=+1. The data satisfy Y=2XY=2X exactly — a perfect positive linear 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}
12214
3618936
48321664
5105025100
7149849196
81612864256
2856328164656

Substituting

N∑XY−∑X∑Y=6(328)−(28)(56)=1968−1568=400N\sum XY-\sum X\sum Y=6(328)-(28)(56)=1968-1568=400

N∑X2−(∑X)2=6(164)−282=984−784=200N\sum X^{2}-(\sum X)^{2}=6(164)-28^{2}=984-784=200

N∑Y2−(∑Y)2=6(656)−562=3936−3136=800N\sum Y^{2}-(\sum Y)^{2}=6(656)-56^{2}=3936-3136=800 …

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