--- title: "Regression of wear on oil viscosity" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:fretting-wear-regression-viscosity" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/fretting-wear-regression-viscosity site: BCT Engineering --- # Regression of wear on oil viscosity Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2072 Chaitra, 5 marks; 2072 Kartik, 5 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/fretting-wear-regression-viscosity - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question An article in Wear (Vol.152, 1992, pp. 171-181) presents data on the fretting wear of mild steel and oil viscosity. Representative data follow, with x = oil viscosity and y = wear volume (10^-4 cubic millimeters). | y | 240 | 181 | 193 | 155 | 172 | 110 | 113 | 75 | 94 | |---|---|---|---|---|---|---|---|---|---| | x | 1.6 | 9.4 | 15.5 | 20.0 | 22.0 | 35.5 | 43.0 | 40.5 | 33.0 | i) Fit the sample linear regression model using least squares. ii) Predict fretting wear when viscosity x = 30. ## Short answer Least squares gives ŷ = 234.071 − 3.509x. At x = 30, predicted wear volume is 128.8 × 10⁻⁴ mm³. ## Answer Take x = oil viscosity (independent) and y = wear volume (dependent). Model: ŷ = a + bx, with least squares estimates b = Sxy/Sxx, a = ȳ − b·x̄ #### i) Fitting the regression line Sums (n = 9): ```text x y xy x² y² 1.6 240 384 2.56 57600 9.4 181 1701.4 88.36 32761 15.5 193 2991.5 240.25 37249 20 155 3100 400 24025 22 172 3784 484 29584 35.5 110 3905 1260.25 12100 43 113 4859 1849 12769 40.5 75 3037.5 1640.25 5625 33 94 3102 1089 8836 --------------------------------------------- 220.5 1333 26864.4 7053.67 220549 ``` Σx = 220.5, Σy = 1333, Σxy = 26864.4, Σx² = 7053.67, Σy² = 220549 Means: x̄ = 24.5000, ȳ = 148.1111 Sxx = Σx² − (Σx)²/n = 1651.4200 Sxy = Σxy − (Σx)(Σy)/n = -5794.1000 Syy = Σy² − (Σy)²/n = 23116.8889 Slope b = Sxy/Sxx = -5794.1000/1651.4200 = -3.5086 Intercept a = ȳ − bx̄ = 148.1111 − (-3.5086)(24.5000) = 234.0707 Fitted line: **ŷ = 234.071 − 3.509 x** (Correlation r = -0.938, strong negative: wear falls as viscosity rises.) #### ii) Prediction at x = 30 ŷ = 234.0707 + (-3.5086)(30) = 234.0707 − 105.2567 = 128.81 **Answer: ŷ = 234.07 − 3.509x; predicted wear at x = 30 is about 128.8 × 10⁻⁴ mm³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Least squares estimates for ŷ = a + bx are: A. b = Sxx/Sxy, a = x̄ − bȳ B. b = Sxy/Sxx, a = ȳ − b·x̄ C. b = Sxy/Syy, a = ȳ + b·x̄ D. b = r, a = ȳ Answer: B. The slope is Sxy/Sxx and the intercept is ȳ − b·x̄. (Key point: Least squares formulas) 2. The slope is −3.509. What does it mean? A. Wear rises by 3.509 per unit viscosity B. Wear is 3.509 when viscosity is zero C. The correlation is −3.509 D. Wear falls by about 3.509 units for each unit rise in viscosity Answer: D. Negative slope means wear decreases as oil viscosity increases, by about 3.509 (10^-4 mm³) per unit. (Key point: Meaning of the slope) 3. Predicted wear at x = 30 using ŷ = 234.07 − 3.509x is about: A. 128.8 B. 105.3 C. 234.1 D. 339.3 Answer: A. 234.07 − 3.5086(30) = 234.07 − 105.26 = 128.8. 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