--- title: "22 pairs: means, consistency, correlation" subject: "Probability and Statistics" chapter: 5 key: "semester_5/probability_and_statistics:sample-22-pairs-xy-correlation" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/sample-22-pairs-xy-correlation site: BCT Engineering --- # 22 pairs: means, consistency, correlation Probability and Statistics · Chapter 5: Correlation and regression - Asked in: 2075 Chaitra, 6 marks - Video: [Correlation Coefficient](https://www.youtube.com/watch?v=11c9cs6WpJU) — The Organic Chemistry Tutor - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/sample-22-pairs-xy-correlation - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-5 ## Question Following data reveals the sample of 22 pairs of observation (X,Y) drawn from large population. X: 46 61 56 68 58 45 50 59 45 66 57 59 66 62 57 57 45 50 61 55 47 51 Y: 49 46 43 32 26 27 29 47 37 30 43 32 27 37 24 43 49 48 29 37 32 26 i) Find the sample mean for each variable X and Y. ii) Which series is more consistent and why? iii) Find the standard error of the difference of mean. iv) Find the coefficient of Karl Pearson correlation. ## Short answer x̄ = 55.5, ȳ = 36.0. X is more consistent (CV 13.0% vs 23.8%). SE of the difference of means = 2.39. Karl Pearson r = -0.234, a weak negative correlation. ## Answer n = 22 pairs. Σx = 1221, Σy = 793, Σx² = 68857, Σy² = 30125, Σxy = 43708. #### i) Sample means x̄ = Σx/n = 1221/22 = **55.500** ȳ = Σy/n = 793/22 = **36.045** #### ii) Consistency (coefficient of variation) Sample standard deviations: s = √[(Σx² − n·x̄²)/(n − 1)] - s_x = 7.209, s_y = 8.566 - CV(X) = s_x/x̄ × 100 = 7.209/55.500 × 100 = **12.99%** - CV(Y) = s_y/ȳ × 100 = 8.566/36.045 × 100 = **23.76%** **Series X is more consistent**, because its coefficient of variation is smaller (less relative variation). #### iii) Standard error of the difference of means For independent samples of size 22 each, SE(x̄ − ȳ) = √(s_x²/n₁ + s_y²/n₂) = √(51.976/22 + 73.379/22) = √5.6980 = **2.387** (x̄ − ȳ = 19.455.) #### iv) Karl Pearson's correlation coefficient Sxx = 1091.500, Syy = 1540.955, Sxy = Σxy − n·x̄·ȳ = -303.500 r = Sxy/√(Sxx·Syy) = -303.500/√(1091.500 × 1540.955) = **-0.2340** The two series have a weak-to-moderate negative linear correlation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With Σx = 1221 and n = 22, x̄ is: A. 55.5 B. 36.0 C. 61.1 D. 27.7 Answer: A. 1221/22 = 55.5. (Key point: Mean) 2. Which series is more consistent, and why? A. Y, because its standard deviation is larger B. X, because its mean is larger C. Y, because its CV is 23.76% D. X, because its coefficient of variation (12.99%) is smaller than Y's (23.76%) Answer: D. Consistency is compared with the coefficient of variation s/x̄ × 100; smaller CV means more consistent. (Key point: Consistency measure) 3. The standard error of the difference of means for independent samples is: A. √(s_x²/n1 + s_y²/n2) = 2.387 B. (s_x + s_y)/n C. √(s_x² − s_y²) D. x̄ − ȳ = 19.455 Answer: A. SE = √(51.976/22 + 73.379/22) = √5.698 = 2.387. (Key point: SE of difference of means) 4. r = −0.234 indicates: A. A strong positive relationship B. A weak negative linear relationship C. A perfect negative relationship D. No data Answer: B. The sign is negative and the magnitude is small, so the linear relationship is weak and negative. 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