--- title: "Multiple correlation: expenditure, income, family size" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:multiple-correlation-food-expenditure" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/multiple-correlation-food-expenditure site: BCT Engineering --- # Multiple correlation: expenditure, income, family size Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2073 Shrawan, 5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/multiple-correlation-food-expenditure - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question A house survey on monthly expenditure on food yield following data: | Monthly expenditure (100 Rs.) | 10 | 15 | 20 | 25 | 30 | 35 | 40 | |---|---|---|---|---|---|---|---| | Monthly income (1000 Rs.) | 2 | 4 | 5 | 7 | 6 | 6 | 5 | | Size of the family | 4 | 5 | 7 | 10 | 8 | 11 | 4 | Obtain the multiple correlation coefficient. ## Short answer R_y.x1x2 = 0.718. Calculated from simple correlations: r_yx1=0.6614, r_yx2=0.3546, r_x1x2=0.7939 using formula R² = (r_yx1² + r_yx2² - 2r_yx1 r_yx2 r_x1x2)/(1 - r_x1x2²). ## Answer Let Y = monthly expenditure on food (dependent), X₁ = monthly income, X₂ = family size. n = 7. #### Step 1: Sums ```text y x1 x2 x1y x2y x1x2 y² x1² x2² 10 2 4 20 40 8 100 4 16 15 4 5 60 75 20 225 16 25 20 5 7 100 140 35 400 25 49 25 7 10 175 250 70 625 49 100 30 6 8 180 240 48 900 36 64 35 6 11 210 385 66 1225 36 121 40 5 4 200 160 20 1600 25 16 ------------------------------------------------------ 175 35 49 945 1290 267 5075 191 391 ``` #### Step 2: Simple correlation coefficients r = Sab/√(Saa·Sbb), with Sab = Σab − ΣaΣb/n. - r_yx₁: Sx₁y = 70.000, Sx₁x₁ = 16.000, Syy = 700.000 → **0.6614** - r_yx₂: Sx₂y = 65.000, Sx₂x₂ = 48.000 → **0.3546** - r_x₁x₂: Sx₁x₂ = 22.000 → **0.7939** #### Step 3: Multiple correlation R_y.x₁x₂ = √[(r_yx₁² + r_yx₂² − 2·r_yx₁·r_yx₂·r_x₁x₂) / (1 − r_x₁x₂²)] - r_yx₁² + r_yx₂² = 0.4375 + 0.1257 = 0.5632 - 2·r_yx₁·r_yx₂·r_x₁x₂ = 0.3724 - Numerator = 0.1908; denominator = 1 − 0.6302 = 0.3698 R² = 0.1908 / 0.3698 = 0.5161 **Answer: R_y.x₁x₂ = 0.7184 ≈ 0.718.** Interpretation: income and family size together explain about 51.6% of the variation in monthly food expenditure; the relationship is strong and positive. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which variable is dependent here? A. Monthly income B. Family size C. Monthly food expenditure D. Month Answer: C. Food expenditure Y is explained by income X1 and family size X2. (Key point: Dependent variable) 2. Which formula is used for R_y.x1x2²? A. (r_yx1² + r_yx2² − 2 r_yx1 r_yx2 r_x1x2)/(1 − r_x1x2²) B. (r_yx1² + r_yx2²)/2 C. (r_yx1 − r_yx2)/(1 − r_x1x2) D. r_yx1 · r_yx2 · r_x1x2 Answer: A. R² combines the two simple correlations while correcting for the correlation between predictors. (Key point: Multiple correlation formula) 3. What value of R_y.x1x2 is obtained? A. 0.516 B. 0.718 C. 0.794 D. 0.661 Answer: B. R² = 0.1908/0.3698 = 0.5161, so R = 0.718. (Key point: Result) 4. What does R² = 0.516 say? A. Income and family size together explain about 51.6% of variation in expenditure B. 51.6% of families spend on food C. Income alone explains 51.6% D. The relation is weak and negative Answer: A. R² is the proportion of variation in Y explained jointly by both predictors. 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