--- title: "Twists regression; predict at 20%" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:twists-regression-predict-20" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/twists-regression-predict-20 site: BCT Engineering --- # Twists regression; predict at 20% Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2071 Shrawan, 5 marks - Video: [Finding the Regression Equation/Regression Line by Hand (Formula)](https://www.youtube.com/watch?v=hKocQvDuSvs) — HelpYourMath - Statistics - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/twists-regression-predict-20 - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question The following data gives the number of twists required to break a certain kind of forged alloy bar and percentage of alloying element A present in the metal. | Number of twists | 41 | 49 | 69 | 65 | 40 | 50 | 58 | 57 | 31 | 36 | |---|---|---|---|---|---|---|---|---|---| | Percentage of element A | 10 | 12 | 14 | 15 | 13 | 12 | 13 | 14 | 13 | 12 | Fit the regression equation of number of twists on percentage of element A. Determine the predicted number of twists required to break an alloy when percentage of element is 20. ## Short answer The regression equation of twists on percentage of element A is y = -21.091 + 5.523x. For 20% element, predicted twists are about 89. Since 20% is outside the observed range, the prediction is an extrapolation and should be used cautiously. ## Answer Let x = percentage of element A (independent) and y = number of twists (dependent). The regression equation of y on x is ŷ = a + bx, with b = Sxy/Sxx and a = ȳ − b·x̄. Sums (n = 10): ```text x y xy x² y² 10 41 410 100 1681 12 49 588 144 2401 14 69 966 196 4761 15 65 975 225 4225 13 40 520 169 1600 12 50 600 144 2500 13 58 754 169 3364 14 57 798 196 3249 13 31 403 169 961 12 36 432 144 1296 --------------------------------------------- 128 496 6446 1656 26038 ``` Σx = 128, Σy = 496, Σxy = 6446, Σx² = 1656, Σy² = 26038 Means: x̄ = 12.8000, ȳ = 49.6000 Sxx = Σx² − (Σx)²/n = 17.6000 Sxy = Σxy − (Σx)(Σy)/n = 97.2000 Syy = Σy² − (Σy)²/n = 1436.4000 Slope b = Sxy/Sxx = 97.2000/17.6000 = 5.5227 Intercept a = ȳ − bx̄ = 49.6000 − (5.5227)(12.8000) = -21.0909 **Regression equation: ŷ = -21.091 + 5.523 x** #### Prediction at x = 20% ŷ = -21.0909 + 5.5227 × 20 = -21.0909 + 110.4545 = 89.36 **Answer: about 89 twists.** Since 20% is far outside the sample range (10% to 15%), this extrapolated value should be used with caution. The correlation r = 0.611 is only moderate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which variable is dependent in 'regression of twists on percentage of A'? A. Percentage of element A B. Number of twists C. Both D. Neither Answer: B. Twists is regressed on percentage, so y = twists is dependent and x = percentage is independent. (Key point: Variables) 2. How is the slope b computed? A. Sxy/Syy B. Sxy/Sxx C. Sxx/Sxy D. Σxy/Σx Answer: B. For y on x, b = Sxy/Sxx = 97.2/17.6 = 5.5227. (Key point: Slope formula) 3. How is intercept a found? A. a = ȳ − b·x̄ = -21.09 B. a = x̄ − b·ȳ C. a = b·x̄ − ȳ D. a = ȳ/x̄ Answer: A. a = 49.6 − 5.5227(12.8) = -21.0909. (Key point: Intercept) 4. Why should the prediction at 20% be treated with caution? A. 20% is far outside the sample range 10%–15% B. The slope is negative C. Sample size is too large D. r equals 1 Answer: A. Extrapolating well beyond observed data is unreliable; also r = 0.611 is only moderate. (Key point: Extrapolation) ## Close variants set in other exams - [Twists regression; CI for slope](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Atwists-regression-with-ci-slope) — Probability and Statistics, chapter 5, asked 3×