--- title: "Regression: operator experience and performance" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:experience-performance-regression-r2" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/experience-performance-regression-r2 site: BCT Engineering --- # Regression: operator experience and performance Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2076 Asoj, 5 marks - Video: [Regression equation || How to find regression equation](https://www.youtube.com/watch?v=sKfAmFK6u8A) — Civil learning online - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/experience-performance-regression-r2 - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question The following data gives the experience of machine operators in years and their performance as given by the number of good parts turned out per 100 pieces. | Experience (X) | 16 | 12 | 18 | 4 | 3 | 10 | 5 | 12 | |---|---|---|---|---|---|---|---|---| | Performance (Y) | 87 | 88 | 89 | 68 | 78 | 80 | 75 | 83 | a) Fit the regression equation of performance rating on experience and estimate the probable performance of an operator had 8 years experience. b) Determine coefficient of determination and interpret it. ## Short answer Least squares fit gives Y = 69.670 + 1.133X. For X = 8 years, predicted performance is about 78.7 good parts per 100. r squared = 0.7605, so about 76% of variation in performance is explained by experience. ## Answer Let X = experience (years), Y = performance (good parts per 100). #### a) Regression of Y on X Sums (n = 8): ```text x y xy x² y² 16 87 1392 256 7569 12 88 1056 144 7744 18 89 1602 324 7921 4 68 272 16 4624 3 78 234 9 6084 10 80 800 100 6400 5 75 375 25 5625 12 83 996 144 6889 --------------------------------------------- 80 648 6727 1018 52856 ``` Σx = 80, Σy = 648, Σxy = 6727, Σx² = 1018, Σy² = 52856 Means: x̄ = 10.0000, ȳ = 81.0000 Sxx = Σx² − (Σx)²/n = 218.0000 Sxy = Σxy − (Σx)(Σy)/n = 247.0000 Syy = Σy² − (Σy)²/n = 368.0000 Slope b = Sxy/Sxx = 247.0000/218.0000 = 1.1330 Intercept a = ȳ − bx̄ = 81.0000 − (1.1330)(10.0000) = 69.6697 **Regression equation: Ŷ = 69.670 + 1.133X** For X = 8 years: Ŷ = 69.6697 + 1.1330(8) = 78.73 ≈ **78.7 good parts per 100**. #### b) Coefficient of determination r = Sxy/√(Sxx·Syy) = 247.0000/√(218.0000 × 368.0000) = 0.8721 **r² = 0.7605** Interpretation: about 76.0% of the variation in operators' performance is explained by differences in their experience; the remaining 24.0% is due to other factors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The slope b = Sxy/Sxx equals: A. 218/247 = 0.88 B. 247/368 = 0.67 C. 247/218 = 1.133 D. 81/10 = 8.1 Answer: C. b = 247/218 = 1.133 good parts per extra year of experience. (Key point: Slope) 2. Predicted performance for 8 years of experience (Ŷ = 69.67 + 1.133X): A. 78.7 B. 81.0 C. 69.7 D. 8.1 Answer: A. 69.67 + 1.133(8) = 78.73. (Key point: Prediction) 3. r² = 0.7605 means: A. 76% of operators are experienced B. Performance rises 76% per year C. About 76% of the variation in performance is explained by experience; 24% by other factors D. r = 0.7605 Answer: C. r² is the proportion of variation in Y explained by X. 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