Pearson Correlation — Formula & Worked Example (n = 10)
1) Definition (authentic)
Population correlation (rho): ρ = Cov(X,Y) / (σX σY)
Sample Pearson correlation (r):
r = Σ (xi − x̄)(yi − ȳ) / √[ Σ(xi − x̄)² · Σ(yi − ȳ)² ]
Sample Pearson correlation (r):
r = Σ (xi − x̄)(yi − ȳ) / √[ Σ(xi − x̄)² · Σ(yi − ȳ)² ]
2) Computational (shortcut) formula
For calculations using raw sums:
r = [ n·Σ(xy) − (Σx)(Σy) ] / √{ [ n·Σ(x²) − (Σx)² ] · [ n·Σ(y²) − (Σy)² ] }
r = [ n·Σ(xy) − (Σx)(Σy) ] / √{ [ n·Σ(x²) − (Σx)² ] · [ n·Σ(y²) − (Σy)² ] }
Worked example (complete steps) — n = 10
Dataset (given)
X = 5, 6, 7, 8, 9, 4, 3, 2, 1, 10
Y = 12, 14, 15, 18, 20, 10, 8, 7, 5, 22
Step A — Construct table of intermediate values
| i | Xi | Yi | Xi·Yi | Xi² | Yi² |
|---|---|---|---|---|---|
| 1 | 5 | 12 | 60 | 25 | 144 |
| 2 | 6 | 14 | 84 | 36 | 196 |
| 3 | 7 | 15 | 105 | 49 | 225 |
| 4 | 8 | 18 | 144 | 64 | 324 |
| 5 | 9 | 20 | 180 | 81 | 400 |
| 6 | 4 | 10 | 40 | 16 | 100 |
| 7 | 3 | 8 | 24 | 9 | 64 |
| 8 | 2 | 7 | 14 | 4 | 49 |
| 9 | 1 | 5 | 5 | 1 | 25 |
| 10 | 10 | 22 | 220 | 100 | 484 |
| Σ | ΣX = 55 | ΣY = 131 | ΣXY = 876 | ΣX² = 385 | ΣY² = 2011 |
Step B — Compute means
n = 10
x̄ = ΣX / n = 55 / 10 = 5.5
ȳ = ΣY / n = 131 / 10 = 13.1
Step C — Compute centered products (alternate view)
We can also compute Σ (Xi − x̄)(Yi − ȳ), Σ (Xi − x̄)² and Σ (Yi − ȳ)².
| i | Xi | Yi | Xi−x̄ | Yi−ȳ | (X−x̄)(Y−ȳ) | (X−x̄)² | (Y−ȳ)² |
|---|---|---|---|---|---|---|---|
| 1 | 5 | 12 | -0.5 | -1.1 | 0.55 | 0.25 | 1.21 |
| 2 | 6 | 14 | 0.5 | 0.9 | 0.45 | 0.25 | 0.81 |
| 3 | 7 | 15 | 1.5 | 1.9 | 2.85 | 2.25 | 3.61 |
| 4 | 8 | 18 | 2.5 | 4.9 | 12.25 | 6.25 | 24.01 |
| 5 | 9 | 20 | 3.5 | 6.9 | 24.15 | 12.25 | 47.61 |
| 6 | 4 | 10 | -1.5 | -3.1 | 4.65 | 2.25 | 9.61 |
| 7 | 3 | 8 | -2.5 | -5.1 | 12.75 | 6.25 | 26.01 |
| 8 | 2 | 7 | -3.5 | -6.1 | 21.35 | 12.25 | 37.21 |
| 9 | 1 | 5 | -4.5 | -8.1 | 36.45 | 20.25 | 65.61 |
| 10 | 10 | 22 | 4.5 | 8.9 | 40.05 | 20.25 | 79.21 |
| Σ (centered) | Σ (X−x̄)(Y−ȳ) = 155.50 | Σ (X−x̄)² = 82.50 | Σ (Y−ȳ)² = 294.90 | ||||
Step D — Compute Pearson r using centered sums
r = Σ (X−x̄)(Y−ȳ) / √[ Σ (X−x̄)² · Σ (Y−ȳ)² ]
Substitute numbers: r = 155.50 / √(82.50 × 294.90)
√(82.50 × 294.90) = √24359.25 = 155.981...
⇒ r = 155.50 / 155.981... = 0.996933 (approx)
Substitute numbers: r = 155.50 / √(82.50 × 294.90)
√(82.50 × 294.90) = √24359.25 = 155.981...
⇒ r = 155.50 / 155.981... = 0.996933 (approx)
Step E — (Optional) compute r using shortcut formula (same result)
r = [ n·ΣXY − (ΣX)(ΣY) ] / √{ [ n·ΣX² − (ΣX)² ] · [ n·ΣY² − (ΣY)² ] }
Numerator = 10×876 − 55×131 = 8760 − 7205 = 1555
Denominator = √[ (10×385 − 55²) × (10×2011 − 131²) ] = √[ 825 × 2949 ] = √2435925 = 1559.7836...
r = 1555 / 1559.7836... = 0.996933 (same as above)
Numerator = 10×876 − 55×131 = 8760 − 7205 = 1555
Denominator = √[ (10×385 − 55²) × (10×2011 − 131²) ] = √[ 825 × 2949 ] = √2435925 = 1559.7836...
r = 1555 / 1559.7836... = 0.996933 (same as above)
Step F — Test significance (t-test for r)
Test statistic: t = r · √[ (n−2) / (1 − r²) ] , df = n − 2
r = 0.996933 , n = 10 ⇒ t ≈ 36.03 , df = 8
This t is extremely large ⇒ two-tailed p ≪ 0.001 (highly significant).
r = 0.996933 , n = 10 ⇒ t ≈ 36.03 , df = 8
This t is extremely large ⇒ two-tailed p ≪ 0.001 (highly significant).
Step G — 95% Confidence Interval (Fisher z method)
1) z' = 0.5 · ln((1+r)/(1−r)) = 0.5·ln((1.996933)/(0.003067)) ≈ 3.2579
2) SE(z') = 1 / √(n−3) = 1 / √7 ≈ 0.37796
3) 95% CI on z': z' ± 1.96·SE ⇒ [ 3.2579 − 0.7408 , 3.2579 + 0.7408 ] = [2.5171 , 3.9987 ]
4) Back-transform bounds to r: lower ≈ 0.98658 , upper ≈ 0.99930
⇒ 95% CI for r ≈ [0.9866 , 0.9993]
2) SE(z') = 1 / √(n−3) = 1 / √7 ≈ 0.37796
3) 95% CI on z': z' ± 1.96·SE ⇒ [ 3.2579 − 0.7408 , 3.2579 + 0.7408 ] = [2.5171 , 3.9987 ]
4) Back-transform bounds to r: lower ≈ 0.98658 , upper ≈ 0.99930
⇒ 95% CI for r ≈ [0.9866 , 0.9993]
English interpretation: r ≈ 0.997 indicates an extremely strong positive linear relationship between X and Y. The correlation is statistically significant (t ≈ 36, p ≪ 0.001). The 95% CI [0.9866, 0.9993] shows the population correlation is also very likely to be very high.
पियरसन सहसंबंध — सूत्र और पूरा उदाहरण (n = 10)
1) परिभाषा (सटीक)
जनसंख्या सहसंबंध (ρ): ρ = Cov(X,Y) / (σX σY)
नमूना पियरसन सहसंबंध (r):
r = Σ (xi − x̄)(yi − ȳ) / √[ Σ(xi − x̄)² · Σ(yi − ȳ)² ]
नमूना पियरसन सहसंबंध (r):
r = Σ (xi − x̄)(yi − ȳ) / √[ Σ(xi − x̄)² · Σ(yi − ȳ)² ]
2) गणनात्मक (shortcut) सूत्र
r = [ n·Σ(xy) − (Σx)(Σy) ] / √{ [ n·Σ(x²) − (Σx)² ] · [ n·Σ(y²) − (Σy)² ] }
पूर्ण उदाहरण (चरण-दर-चरण)
दिए गए आंकड़े
X = 5, 6, 7, 8, 9, 4, 3, 2, 1, 10
Y = 12, 14, 15, 18, 20, 10, 8, 7, 5, 22
चरण A — मध्यवर्ती तालिका
(ऊपर दी गयी तालिका देखें — वही X, Y, XY, X², Y² और सम हैं: ΣX=55, ΣY=131, ΣXY=876, ΣX²=385, ΣY²=2011)
चरण B — माध्य निकालना
n = 10
x̄ = 55 / 10 = 5.5
ȳ = 131 / 10 = 13.1
चरण C — केंद्रित गुणन और वर्ग जमा
Σ (X−x̄)(Y−ȳ) = 155.50 ; Σ (X−x̄)² = 82.50 ; Σ (Y−ȳ)² = 294.90
चरण D — पियरसन r गणना
r = 155.50 / √(82.50 × 294.90) = 155.50 / 155.981... = 0.996933
चरण E — महत्व परीक्षण (t)
t ≈ 36.03, df = 8 ⇒ p बहुत छोटा (p ≪ 0.001) — सार्थक (statistically significant)
चरण F — 95% विश्वसनीयता अंतर (Fisher z)
95% CI for r ≈ [0.9866 , 0.9993]
व्याख्या: r ≈ 0.997 बहुत मजबूत धनात्मक रैखिक सम्बन्ध दर्शाता है। यह सांख्यिकीय रूप से महत्वपुर्ण है और 95% CI भी ऊँचा है।
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