--- title: "Regression of rice yield on tillers" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:rice-yield-tillers-regression" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/rice-yield-tillers-regression site: BCT Engineering --- # Regression of rice yield on tillers Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2068 Jestha (old course), 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/rice-yield-tillers-regression - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question The international rice research institute in the Philippines wants to relate the grain yield of rice varieties, y, to the tiller number, x. They conducted experiments for some rice varieties and tillers. Below are the results obtained for the rice variety Milfor 6. | Grain Yield, kg/ha | 4,862 | 5,244 | 5,128 | 5,052 | 5,298 | 5,410 | |---|---|---|---|---|---|---| | Tillers, no./m² | 160 | 175 | 192 | 195 | 238 | 240 | i) Fit a linear regression line to predict grain yield with respect to Tillers. ii) Estimate the grain yield if Tillers value is 210. ## Short answer Regression line: ŷ = 4200.38 + 4.826x (r = 0.811). At tillers x = 210, predicted grain yield ≈ 5214 kg/ha. ## Answer Let x = tillers (no./m²), the independent variable, and y = grain yield (kg/ha), the dependent variable. n = 6. #### i) Fitting the regression line ```text x y xy x² y² 160 4862 777920 25600 23639044 175 5244 917700 30625 27499536 192 5128 984576 36864 26296384 195 5052 985140 38025 25522704 238 5298 1260924 56644 28068804 240 5410 1298400 57600 29268100 -------------------------------------------------- 1200 30994 6224660 245358 160294572 ``` - x̄ = 1200/6 = 200.000, ȳ = 30994/6 = 5165.667 - Sxx = Σx² − (Σx)²/n = 5358.000 - Sxy = Σxy − (Σx)(Σy)/n = 25860.000 b = Sxy/Sxx = 25860.000/5358.000 = 4.8264 a = ȳ − b·x̄ = 5165.667 − (4.8264)(200.000) = 4200.381 **Regression line: ŷ = 4200.38 + 4.826 x** (r = 0.811) #### ii) Estimate at x = 210 ŷ = 4200.381 + 4.8264 × 210 = **5213.9 kg/ha** **Answer: grain yield ≈ 5214 kg/ha when tillers = 210/m².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the independent variable? A. Grain yield B. Tillers per m² C. Variety D. Both Answer: B. Yield is predicted from tillers, so x = tillers. (Key point: Variables) 2. What is the slope b? A. Sxy/Sxx = 25860/5358 ≈ 4.83 B. Sxx/Sxy C. Sxy/Syy D. ȳ/x̄ Answer: A. b = 25860/5358 = 4.8264 kg/ha per tiller. (Key point: Slope) 3. What is a? A. 200.0 B. 5165.7 C. 4200.4 D. 4.83 Answer: C. a = 5165.667 − 4.8264(200) = 4200.38. (Key point: Intercept) 4. Yield at 210 tillers/m²? A. About 5214 kg/ha B. About 5040 kg/ha C. About 4200 kg/ha D. About 5400 kg/ha Answer: A. ŷ = 4200.38 + 4.8264(210) ≈ 5213.9. 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