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Worked Examples · Example 1
Q.

The advertisement expenditure (X, in Rs. lakh) and the corresponding sales (Y, in Rs. lakh) of a firm over 5 years are given below. Compute Karl Pearson's Coefficient of Correlation by the actual mean (deviation) method.

Year12345
Advertisement Expenditure (X)246810
Sales (Y)367910
Gujarat GsebTextbookSubjectiveImportance★★★★★
12% · 5/42 Questions
✓ Free question

Step 1 — Compute the means.

X‾=2+4+6+8+105=305=6Y‾=3+6+7+9+105=355=7\overline{X}=\dfrac{2+4+6+8+10}{5}=\dfrac{30}{5}=6 \qquad \overline{Y}=\dfrac{3+6+7+9+10}{5}=\dfrac{35}{5}=7

Step 2 — Compute the deviations and the required products.

XXYYx=X−6x=X-6y=Y−7y=Y-7xyxyx2x^2y2y^2
23−4−4161616
46−2−1241
6700000
8922444
10104312169
Total344030

Step 3 — Apply the formula.

r=∑xy∑x2∑y2=3440×30=341200=3434.641=0.9815r = \dfrac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} = \dfrac{34}{\sqrt{40\times 30}} = \dfrac{34}{\sqrt{1200}} = \dfrac{34}{34.641} = 0.9815

Verification by the direct (raw-score) method: ∑X=30\sum X=30, ∑Y=35\sum Y=35, ∑XY=6+24+42+72+100=244\sum XY=6+24+42+72+100=244, ∑X2=220\sum X^2=220, ∑Y2=275\sum Y^2=275, N=5N=5.

r=N∑XY−∑X∑Y[N∑X2−(∑X)2][N∑Y2−(∑Y)2]=5(244)−30(35)[5(220)−900][5(275)−1225]=1220−1050200×150=17030000=170173.21=0.9815r=\dfrac{N\sum XY-\sum X\sum Y}{\sqrt{[N\sum X^2-(\sum X)^2][N\sum Y^2-(\sum Y)^2]}}=\dfrac{5(244)-30(35)}{\sqrt{[5(220)-900][5(275)-1225]}}=\dfrac{1220-1050}{\sqrt{200\times150}}=\dfrac{170}{\sqrt{30000}}=\dfrac{170}{173.21}=0.9815

Both methods agree exactly at r≈0.9815r\approx 0.9815.

✓Final answer

r≈0.981r \approx 0.981 — a very high degree of positive correlation between advertisement expenditure and sales.

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