Properties of coefficient of correlation
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The document discusses properties of the coefficient of correlation (r) including: 1) r always lies between -1 and 1 2) r is the geometric mean of the two regression coefficients 3) Several examples are shown calculating r from regression coefficients and comparing to Pearson's coefficient of correlation.
Properties of coefficient of correlation
- 2. Properties of coefficient of correlation:- 1) The coefficient of correlation always lies between -1 and +1 i.e, −1 ≤ 𝑟 ≤ +1 2) The correlation coefficient is symmetrical with respect to X and Y i.e 𝑟𝑥𝑦 = 𝑟𝑦𝑥 3) The coefficient of correlation is the geomatric mean of the two regression coefficient. r = √𝑏 × 𝑑 Or r = √ 𝑏 𝑦𝑥 × 𝑏 𝑥𝑦 4) It does not depend upon the units employed 5) It is independent of orgin and unit of measurement 6) The coefficient of cerrelation is unaffected by change of origin and scale i.e 𝑟𝑥𝑦 = 𝑟𝑢𝑣 7) The coefficient of cerrelation is a pure number.
- 3. Example-7: i) Calculate regression co-efficient by𝑥 and 𝑏𝑥𝑦 and calculate correlation with the help of regression coefficients for the following pairs of observations. ii) Calculate Karl Pearson’s coefficient of correlation and then verify that. X 1 2 3 4 5 6 7 8 Y 12 14 16 18 20 22 24 26 Solution: We know that the correlation coefficient is the geometric mean of the two regression coefficients. 𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦 𝑏𝑦𝑥 = 𝑛∑𝑥𝑦−∑𝑥∑𝑦 𝑛∑𝑥2−(∑𝑥)2 𝑏𝑥𝑦 = 𝑛∑𝑥𝑦−∑𝑥∑𝑦 𝑛∑𝑦2−(∑𝑦)2 The necessary calculations for regression coefficients are given below. 𝒙 𝒚 𝒙𝒚 𝒙 𝟐 𝒚 𝟐 1 12 12 1 144 2 14 28 4 196 3 16 48 9 256 4 18 72 16 324 5 20 100 25 400 6 22 132 36 484 7 24 168 49 576 8 26 208 64 676 ∑x=36 ∑y=152 ∑xy=768 ∑x2 =204 ∑y2 =3056
- 4. 𝑏𝑦𝑥 = 𝑛∑𝑥𝑦−∑𝑥∑𝑦 𝑛∑𝑥2−(∑𝑥)2 𝑏𝑦𝑥 = 8(768)−(36)(152) 8(204)−(36)2 𝑏𝑦𝑥 = 6144−5472 1632−1296 𝑏𝑦𝑥 = 672 336 𝑏𝑦𝑥 = 2 𝑏𝑥𝑦 = 𝑛∑𝑥𝑦−∑𝑥∑𝑦 𝑛∑𝑦2−(∑𝑦)2 𝑏𝑥𝑦 = 8(768)−(36)(152) 8(3056)−(152)2 𝑏𝑥𝑦 = 6144−5472 24448−23104 𝑏𝑥𝑦 = 672 1344 𝑏𝑥𝑦 = 0.5 We know that correlation coefficient is the geometric mean of the two regression coefficients i.e. 𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦 𝑟 = √2 × 0.5 𝒓 = 𝟏
- 5. (ii) Karl Pearson’s co-efficient of correlation. 𝑟 = 𝑛∑𝑥𝑦−∑𝑥∑𝑦 √(𝑛∑𝑥2−(∑𝑥)2)(𝑛∑𝑦2−(∑𝑦)2) 𝑟 = 8(768)−(36)(152) √(8(204)−(36)2)(8(3056)−(15)2) 𝑟 = 6144−5472 √(1632−1296)(24448−23104) 𝑟 = 672 √(336)(1344) 𝑟 = 672 √451584 𝑟 = 672 672 𝒓 = 𝟏 Hence Proved 𝒓 = √𝒃𝒚𝒙 × 𝒃𝒙𝒚
- 6. Example-8: If 𝑏 𝑦𝑥 = 51.9 and 𝑏 𝑥𝑦 = 0.019 Find coefficient of determination Solution: 𝑟𝑥𝑦 2 = 𝑏 𝑦𝑥 × 𝑏 𝑥𝑦 × 100 𝑟𝑥𝑦 2 = (51.9)(0.019) × 100 𝒓 𝒙𝒚 𝟐 = 𝟗𝟖. 𝟔𝟏% It means that 98.61% of the variation in the 𝑦-variable is explained or accounted for 𝑏𝑦 variation in the 𝑥-variable. Example-9: For the following two sets, the regression lines for each set are respectively. i) 𝑦 = 1.94𝑥 + 10.83 (𝑦 𝑜𝑛 𝑥) and 𝑥 = 0.15𝑦 + 6.18 (𝑥 𝑜𝑛 𝑦) ii) 𝑦 = −1.96𝑥 + 15 (𝑦 𝑜𝑛 𝑥) and 𝑥 = −0.45𝑦 + 7.16 (𝑥 𝑜𝑛 𝑦) Find coefficient of correlation in each case. Solution: i) Regression coefficient 𝑦 on 𝑥 (𝑏𝑦𝑥) = 1.94 Regression coefficient 𝑥 on 𝑦 (𝑏𝑥𝑦) = 0.15 𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦 𝑟 = √(1.94)(0.15) 𝒓 = 𝟎. 𝟓𝟒
- 7. ii) Regression coefficient 𝑦 on 𝑥 (𝑏𝑦𝑥) = −1.96 Regression coefficient 𝑥 on 𝑦 (𝑏𝑥𝑦) = −0.45 𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦 𝑟 = √(1.96)(0.45) 𝒓 = −𝟎. 𝟗𝟒 It is to be noted when both regression coefficients are negative then “𝑟” is also negative. Example-10: If 𝑛 = 18, ∑𝑥 = 638, ∑𝑦 = 41, ∑𝑥𝑦 = 1569.5, ∑𝑥2 = 25814, ∑𝑦2 = 101.45 i) Find simple coefficient of correlation. ii) If 𝑢 = 𝑥−3 5 and 𝑣 = 𝑦 20 then what would be the coefficient of correlation between 𝑢 and 𝑣. Solution: 𝑟 = 𝑛∑𝑥𝑦 − (∑𝑥)(∑𝑦) √{𝑛(∑𝑥2) − (∑𝑥)2}{𝑛(∑𝑦2) − (∑𝑦)2} 𝑟 = 18(1569.5) − (638)(41) √{18(25814) − (638)2}{18(101.45) − (41)2} 𝑟 = 28251 − 26158 √(464652 − 407044)(18261 − 1681)
- 8. 𝑟 = 2093 √(57608)(145.1) 𝑟 = 2093 √8358920.8 𝑟 = 2093 2891.18 𝒓 = 𝟎. 𝟕𝟐 (ii)Correlation is unaffected by the change of origin and scale. i.e. 𝑟𝑢𝑣 = 𝑟𝑥𝑦 𝒓 𝒖𝒗 = 𝟎. 𝟕𝟐