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Partial and multiple correlation from sums

Short answer

Partial correlation r12.3 = -0.654, multiple correlation R1.23 = 0.767, using corrected sums of squares and cross‑products from the given sums.

A sample of 10 values of three variables X1, X2 and X3 were obtained as ΣX1 = 10, ΣX2 = 20, ΣX3 = 30, ΣX1² = 20, ΣX2² = 68, ΣX3² = 170, ΣX1X2 = 10, ΣX1X3 = 15, ΣX2X3 = 64. Find (i) Partial correlation between X1 and X2 eliminating the effect of X3 (ii) Multiple correlation between X1, X2 and X3 assuming X1 as dependent variable.

n = 10. Corrected sums of squares and products Sij = ΣXiXj − (ΣXi)(ΣXj)/n:

  • S₁₁ = 20 − 10²/10 = 10.00
  • S₂₂ = 68 − 20²/10 = 28.00
  • S₃₃ = 170 − 30²/10 = 80.00
  • S₁₂ = 10 − (10)(20)/10 = -10.00
  • S₁₃ = 15 − (10)(30)/10 = -15.00
  • S₂₃ = 64 − (20)(30)/10 = 4.00

Simple correlations, rij = Sij/√(Sii·Sjj):

  • r₁₂ = -10/√(10×28) = -0.5976
  • r₁₃ = -15/√(10×80) = -0.5303
  • r₂₃ = 4/√(28×80) = 0.0845

i) Partial correlation r₁₂.₃

r₁₂.₃ = (r₁₂ − r₁₃r₂₃)/√[(1 − r₁₃²)(1 − r₂₃²)] = (-0.5976 − (-0.5303)(0.0845)) / √[(1 − 0.2812)(1 − 0.0071)] = -0.5528 / 0.8448 = -0.6544

ii) Multiple correlation R₁.₂₃ (X₁ dependent)

R₁.₂₃² = (r₁₂² + r₁₃² − 2r₁₂r₁₃r₂₃)/(1 − r₂₃²) = (0.3571 + 0.2812 − 2(-0.5976)(-0.5303)(0.0845)) / (1 − 0.0071) = 0.5848 / 0.9929 = 0.5890

R₁.₂₃ = √0.5890 = 0.7675

Answer: r₁₂.₃ = -0.654 and R₁.₂₃ = 0.767. After removing the effect of X₃, X₁ and X₂ are negatively related; X₂ and X₃ together explain about 58.9% of the variation in X₁.

✓ Answer checked by Claude Sonnet 5.5

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