Skip to main content

Partial r12.3 and multiple R1.23 (0.59, 0.46, 0.77)

Short answer

Partial correlation r12.3 = 0.416, and multiple correlation R1.23 = 0.590, computed from the given simple correlations using standard formulas.

The simple correlation coefficient between fertilizer (x1), seeds (x2) and productivity (x3) are r12 = 0.59, r13 = 0.46, r23 = 0.77. Calculate the partial correlation coefficient r12.3 and multiple correlation R1.23.

Given: r₁₂ = 0.59, r₁₃ = 0.46, r₂₃ = 0.77 (X₁ = fertilizer, X₂ = seeds, X₃ = productivity).

Partial correlation r₁₂.₃

r₁₂.₃ = (r₁₂ − r₁₃r₂₃) / √[(1 − r₁₃²)(1 − r₂₃²)]

  • Numerator = 0.59 − (0.46)(0.77) = 0.59 − 0.3542 = 0.2358
  • 1 − r₁₃² = 1 − 0.2116 = 0.7884
  • 1 − r₂₃² = 1 − 0.5929 = 0.4071
  • Denominator = √(0.7884 × 0.4071) = 0.5665

r₁₂.₃ = 0.2358 / 0.5665 = 0.4162

Multiple correlation R₁.₂₃

R₁.₂₃ = √[(r₁₂² + r₁₃² − 2r₁₂r₁₃r₂₃) / (1 − r₂₃²)]

  • r₁₂² + r₁₃² = 0.3481 + 0.2116 = 0.5597
  • 2r₁₂r₁₃r₂₃ = 2(0.59)(0.46)(0.77) = 0.4180
  • Numerator = 0.1417; denominator = 1 − 0.5929 = 0.4071

R₁.₂₃ = √(0.1417/0.4071) = √0.3482 = 0.5901

Answer: r₁₂.₃ ≈ 0.416, R₁.₂₃ ≈ 0.590. Here X₁ (fertilizer) is the dependent variable, so seeds and productivity together explain about 34.8% of the variation in fertilizer (R² = 0.348). As expected, R₁.₂₃ is not smaller than r₁₂ or r₁₃.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 5 questions →