---
title: "Partial r12.3 and multiple R1.23 (0.69, 0.64, 0.85)"
subject: "Probability and Statistics"
chapter: 5
key: "semester_5/probability_and_statistics:partial-multiple-corr-r12-069"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/partial-multiple-corr-r12-069
site: BCT Engineering
---

# Partial r12.3 and multiple R1.23 (0.69, 0.64, 0.85)

Probability and Statistics · Chapter 5: Correlation and regression

- Asked in: 2069 Chaitra, 4 marks
- 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-r12-069
- 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 = 0.69, r13 = 0.64 and r23 = 0.85. Calculate the partial correlation r12.3 and multiple correlations R1.23.

## Short answer

Partial correlation r12.3 = (r12 − r13 r23)/√[(1−r13²)(1−r23²)] = 0.361. Multiple correlation R1.23 = √[(r12²+r13²−2r12 r13 r23)/(1−r23²)] = 0.697.

## Answer

Given: r₁₂ = 0.69, r₁₃ = 0.64, r₂₃ = 0.85 (X₁ = fertilizer, X₂ = seeds, X₃ = productivity).

#### Partial correlation r₁₂.₃

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

- Numerator = 0.69 − (0.64)(0.85) = 0.69 − 0.5440 = 0.1460
- 1 − r₁₃² = 0.5904; 1 − r₂₃² = 0.2775
- Denominator = √(0.5904 × 0.2775) = 0.4048

r₁₂.₃ = 0.1460 / 0.4048 = **0.3607**

#### Multiple correlation R₁.₂₃

R₁.₂₃² = (r₁₂² + r₁₃² − 2r₁₂r₁₃r₂₃) / (1 − r₂₃²)

- r₁₂² + r₁₃² = 0.4761 + 0.4096 = 0.8857
- 2r₁₂r₁₃r₂₃ = 2(0.69)(0.64)(0.85) = 0.7507
- Numerator = 0.1350; denominator = 1 − 0.7225 = 0.2775

R₁.₂₃² = 0.4864, so R₁.₂₃ = **0.6974**

**Answer: r₁₂.₃ ≈ 0.361, R₁.₂₃ ≈ 0.697.** R₁.₂₃ exceeds both r₁₂ and r₁₃, as it must.

## Check yourself

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

1. Which numerator is used for r12.3?
   A. r12 − r13r23
   B. r12 + r13r23
   C. r13 − r12r23
   D. r12r13 − r23
   Answer: A. r12.3 = (r12 − r13r23)/√[(1 − r13²)(1 − r23²)]. (Key point: Partial formula)

2. What is r12.3?
   A. 0.69
   B. 0.146
   C. 0.361
   D. 0.697
   Answer: C. 0.1460/0.4048 = 0.3607. (Key point: Partial value)

3. What is the denominator in R1.23²?
   A. 1 − r12²
   B. 1 − r13²
   C. 1 − r23²
   D. 1 − r12r13
   Answer: C. For R1.23 the denominator is 1 − r23², the correlation between predictors. (Key point: Multiple denominator)

4. R1.23 ≈ 0.697 is expected to be:
   A. Smaller than r12 and r13
   B. Larger than or equal to r12 and r13
   C. Negative
   D. Equal to r12.3
   Answer: B. The multiple correlation is never smaller than any simple correlation of the dependent variable with a predictor. (Key point: Result)

## Close variants set in other exams

- [Partial r12.3 and multiple R1.23 (0.59, 0.46, 0.77)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apartial-multiple-corr-r12-059) — Probability and Statistics, chapter 5, asked 3×

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