--- title: "Twists regression; CI for slope" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:twists-regression-with-ci-slope" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/twists-regression-with-ci-slope site: BCT Engineering --- # Twists regression; CI for slope Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2071 Chaitra, 5 marks; 2075 Asoj, 4 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/twists-regression-with-ci-slope - 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 | i) 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. ii) Find 99% confidence interval for the regression coefficient (i.e. slope). ## Short answer Fitted equation: ŷ = -21.091 + 5.523x; at 20% element A, predicted twists ≈ 89. The 99% CI for slope is (-2.959, 14.004) using t(0.005,8)=3.355 and SE=2.5277. ## Answer Let x = percentage of element A (independent), y = number of twists (dependent). #### i) Regression of y on x 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) = 89.36 ≈ **89 twists**. (Note: x = 20 lies outside the observed range 10–15, so this is an extrapolation and less reliable.) #### ii) 99% confidence interval for the slope β β̂ ± t(α/2, n−2) · SE(b), where SE(b) = s/√Sxx and s² = SSE/(n − 2). - SSE = Syy − b·Sxy = 1436.4000 − (5.5227)(97.2000) = 899.5909 - s² = SSE/(n − 2) = 899.5909/8 = 112.4489, s = 10.6042 - SE(b) = √(112.4489/17.6000) = 2.5277 - t(0.005, 8) = 3.355 b ± t·SE(b) = 5.5227 ± (3.355)(2.5277) = 5.5227 ± 8.4813 **99% CI for β: (-2.959, 14.004).** Since the interval contains 0, the slope is not significantly different from zero at the 1% level. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Predicting twists at x = 20 (observed x is 10 to 15) is: A. Interpolation and very reliable B. Extrapolation, so less reliable C. Impossible with regression D. Reliable because r = 1 Answer: B. 20 lies outside the data range 10–15, so the prediction (about 89) is an extrapolation. (Key point: Prediction and extrapolation) 2. How is s² for the slope's standard error obtained? A. SSE/n B. Syy/(n − 1) C. SSE/(n − 2) with SSE = Syy − b·Sxy D. Sxy/Sxx Answer: C. s² = SSE/(n − 2) = 899.59/8 = 112.45, with SSE = Syy − b·Sxy. (Key point: Residual variance) 3. For a 99% CI for the slope with n = 10, the t value is: A. t(0.005, 9) = 3.250 B. t(0.01, 10) = 2.764 C. t(0.005, 8) = 3.355 D. z = 2.576 Answer: C. The slope CI uses n − 2 = 8 degrees of freedom and α/2 = 0.005, giving 3.355. (Key point: t value for CI) 4. The 99% CI for the slope is (−2.959, 14.004). What does it show? A. The slope is significantly positive B. The slope is exactly 5.52 C. The slope is negative D. It contains 0, so the slope is not significantly different from zero at the 1% level Answer: D. Since 0 lies inside the interval, we cannot conclude a linear effect at the 1% level. (Key point: Interpreting the CI) ## Close variants set in other exams - [Twists regression; predict at 20%](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Atwists-regression-predict-20) — Probability and Statistics, chapter 5, asked 3×