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Worked Examples · Example 1
Q.

Example 1. Calculate the Karl Pearson's coefficient of correlation between the number of years of schooling of farmers and the annual yield per acre.

No. of years of schooling of farmersAnnual yield per acre in '000 (Rs)
04
24
46
610
810
108
127
Haryana BsehTextbookSubjective· 5mImportance★★★★★est
4% · 1/25 Questions
✓ Free question

Using deviations from the means (Xˉ=6\bar X=6, Yˉ=7\bar Y=7), ∑xy=42\sum xy = 42, ∑x2=112\sum x^{2}=112, ∑y2=38\sum y^{2}=38, giving r≈+0.644r\approx +0.644 — a moderately strong positive relationship between a farmer's years of schooling and the annual yield per acre.

Concept first

Karl Pearson's coefficient of correlation (rr) measures the direction and strength of the linear association between two quantitative variables. It is a pure number lying between −1-1 and +1+1. Taking deviations from the arithmetic means simplifies the arithmetic:

r=∑xy∑x2 ∑y2,x=X−Xˉ,    y=Y−Yˉr=\frac{\sum xy}{\sqrt{\sum x^{2}\,\sum y^{2}}},\qquad x=X-\bar X,\;\; y=Y-\bar Y

The working table

Xˉ=427=6,Yˉ=497=7.\bar X=\dfrac{42}{7}=6,\qquad \bar Y=\dfrac{49}{7}=7.

XXYYx=X−6x=X-6y=Y−7y=Y-7xyxyx2x^{2}y2y^{2}
04-6-318369
24-4-312169
46-2-1241
61003009
81023649
108414161
127600360
Σ004211238

Substituting

r=42112×38=424256=4265.24=0.6438r=\frac{42}{\sqrt{112\times 38}}=\frac{42}{\sqrt{4256}}=\frac{42}{65.24}=0.6438

Interpretation

The value +0.644+0.644 is positive and moderately high: schooling and yield move together, though not perfectly — other factors (soil, irrigation, inputs) also affect yield.

✓Final answer

r=∑xy∑x2 ∑y2=42112×38≈+0.644r=\frac{\sum xy}{\sqrt{\sum x^{2}\,\sum y^{2}}}=\frac{42}{\sqrt{112\times38}}\approx \mathbf{+0.644}

A moderately strong positive correlation between years of schooling and annual yield per acre.

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