Substituting the values, we get 40 = 5a + 30b (1) …… 214 = 30a + 220b (2) ……. Multiplying (1) by 6 240 = 30a + 180b……. (3) (2) (3) - 26 = 40b – b = = - 0.65 Now, substituting the value of in equation (1) 40 = 5a 19.5 ‘b’ – 5a = 59.5 = = 11.9 a Hence, required regression line Y on X is Y = 11.9 0.65 X. Again, regression – equation of X on Y is X = + and a bY The normal equations are X = na + and bY = + XY aY bY2 Now, substituting the corresponding values from the above table, we get 30 = 5a + 40b (3) …. 214 = 40a + 340b (4) …. Multiplying (3) by 8, we get 240 = 40a + 320 (5) b …. (4) (5) gives – -26 = 20b b = = - 1.3 Substituting = - 1.3 in equation (3) gives 30 = 5a 52 b – 5a = 82 16.4 a = = Hence, required regression line of X on Y is X = 16.4 1.3Y – Regression Co-efficient: The regression equation of Y on X is y = x) ȳ+r(x-
Substituting the values, we get 40 = 5a + 30b …… (1) 214 = 30a + 220b ……. (2) Multiplying (1) by 6 240 = 30a + 180b……. (3) (2) – (3) - 26 = 40b b = = - 0.65 Now, substituting the value of ‘b’ in equation (1) 40 = 5a – 19.5 5a = 59.5 a = = 11.9 Hence, required regression line Y on X is Y = 11.9 – 0.65 X. Again, regression equation of X on Y is X = a + bY and The normal equations are X = na + bY and XY = aY + bY2 Now, substituting the corresponding values from the above table, we get 30 = 5a + 40b …. (3) 214 = 40a + 340b …. (4) Multiplying (3) by 8, we get 240 = 40a + 320 b …. (5) (4) – (5) gives -26 = 20b b = = - 1.3 Substituting b = - 1.3 in equation (3) gives 30 = 5a – 52 5a = 82 a = = 16.4 Hence, required regression line of X on Y is X = 16.4 – 1.3Y Regression Co-efficient: The regression equation of Y on X is y = ȳ+r(x- x)
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