---
title: "Partial r12.3 and multiple R1.23 (0.59, 0.46, 0.77)"
subject: "Probability and Statistics"
chapter: 5
key: "semester_5/probability_and_statistics:partial-multiple-corr-r12-059"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/partial-multiple-corr-r12-059
site: BCT Engineering
---

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

Probability and Statistics · Chapter 5: Correlation and regression

- Asked in: 2075 Asoj, 4 marks; 2068 Chaitra, 5 marks (asked 2 times)
- Video: [3.4 - Multiple and Partial Correlation](https://www.youtube.com/watch?v=6N8P_mr0s2k) — SIAM VIT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/partial-multiple-corr-r12-059
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5

## Question

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.

## Short answer

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

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The numerator of r12.3 is:
   A. r12 + r13 r23 = 0.944
   B. r12 − r13 r23 = 0.2358
   C. r12 r13 r23
   D. r12 − r13 = 0.13
   Answer: B. 0.59 − (0.46)(0.77) = 0.59 − 0.3542 = 0.2358. (Key point: Numerator of partial correlation)

2. r12.3 equals:
   A. 0.590
   B. 0.348
   C. 0.416
   D. 0.566
   Answer: C. 0.2358 divided by √(0.7884 × 0.4071) = 0.5665 gives 0.4162. (Key point: Value of r12.3)

3. R1.23 = 0.590 and R² = 0.348 means:
   A. R is 34.8%, the average of r12 and r13
   B. 34.8% of variation in X1 is explained by X2 and X3 together
   C. X2 and X3 are 34.8% correlated
   D. R is smaller than r12
   Answer: B. R² is the coefficient of multiple determination: the proportion of variation in X1 explained by X2 and X3 jointly. (Key point: Multiple correlation meaning)

4. Which check confirms R1.23 = 0.590 is plausible?
   A. R must be below r12
   B. R must equal r12.3
   C. R must be at least as large as r12 (0.59) and r13 (0.46)
   D. R must be negative
   Answer: C. Multiple correlation is never smaller than the simple correlations of X1 with X2 or X3. (Key point: Check against simple correlations)

## Close variants set in other exams

- [Partial r12.3 and multiple R1.23 (0.69, 0.64, 0.85)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apartial-multiple-corr-r12-069) — Probability and Statistics, chapter 5, asked 3×

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