Correlation and Regression Example

 

Numerical Example – Karl Pearson’s Correlation Coefficient (r)

🔹 Question

Find the coefficient of correlation between X and Y.

X	2	4	6	8	10
Y	1	3	4	6	8

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🔹 Step 1: Formula

$$r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$$

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🔹 Step 2: Prepare Calculation Table

X	Y	X²	Y²	XY
2	1	4	1	2
4	3	16	9	12
6	4	36	16	24
8	6	64	36	48
10	8	100	64	80

Now compute totals:

$$\sum X = 30$$ $$\sum Y = 22$$ $$\sum X^2 = 220$$ $$\sum Y^2 = 126$$ $$\sum XY = 166$$ $$n = 5$$

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🔹 Step 3: Substitute in Formula

Numerator:

$$5(166) - (30)(22)$$ $$= 830 - 660 = 170$$

Denominator:

$$\sqrt{[5(220) - 30^2][5(126) - 22^2]}$$ $$= \sqrt{[1100 - 900][630 - 484]}$$ $$= \sqrt{(200)(146)}$$ $$= \sqrt{29200}$$ $$= 170.88$$

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🔹 Step 4: Final Answer

$$r = \frac{170}{170.88}$$ $$r ≈ 0.995$$

✅ Interpretation:

There is a very strong positive correlation between X and Y.

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2️⃣ Numerical Example – Spearman’s Rank Correlation (ρ)

Developed by Charles Spearman.

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🔹 Question

Two judges ranked 5 students as follows:

Student	A	B	C	D	E
Judge 1	1	2	3	4	5
Judge 2	2	1	4	3	5

Find Spearman’s rank correlation coefficient.

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🔹 Step 1: Formula

$$\rho = 1 - \frac{6\sum d^2}{n(n^2 - 1)}$$

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🔹 Step 2: Compute d and d²

Student	R₁	R₂	d = R₁ − R₂	d²
A	1	2	-1	1
B	2	1	1	1
C	3	4	-1	1
D	4	3	1	1
E	5	5	0	0

$$\sum d^2 = 4$$ $$n = 5$$

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🔹 Step 3: Substitute in Formula

$$\rho = 1 - \frac{6(4)}{5(25 - 1)}$$ $$= 1 - \frac{24}{5(24)}$$ $$= 1 - \frac{24}{120}$$ $$= 1 - 0.2$$ $$\rho = 0.8$$

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✅ Interpretation:

There is a strong positive agreement between the two judges.