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942 algebraic methods i regression equation the two

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Unformatted text preview: . i.e., ∑(Y – Yc)2 < ∑ (Y – Ai)2 Where Ai = corresponding values of any other straight line. (iii) The lines of regression (best fit) intersect at the mean values of the variables X and Y, i.e., intersecting point is x, y . 9.4 Methods of Regression Analysis: The various methods can be represented in the form of chart given below: Regression methods Graphic (through regression lines) Algebraic (through regression equations) Scatter Diagram Regression Equations (through normal equations) Regression Equations (through regression coefficient) 221 9.4.1 Graphic Method: Scatter Diagram: Under this method the points are plotted on a graph paper representing various parts of values of the concerned variables. These points give a picture of a scatter diagram with several points spread over. A regression line may be drawn in between these points either by free hand or by a scale rule in such a way that the squares of the vertical or the horizontal distances (as the case may be) between the points and the line of regression so drawn is the least. In other words, it should be drawn faithfully as the line of best fit leaving equal number of points on both sides in such a manner that the sum of the squares of the distances is the best. 9.4.2 Algebraic Methods: (i) Regression Equation. The two regression equations for X on Y; X = a + bY And for Y on X; Y = a + bX Where X, Y are variables, and a,b are constants whose values are to be determined For the equation, X = a + bY The normal equations are ∑X = na + b ∑Y and ∑XY = a∑Y + b∑Y2 For the equation, Y= a + bX, the normal equations are ∑Y = na + b∑ X and ∑XY = a∑X + b∑X2 From these normal equations the values of a and b can be determined. Example 1: Find the two regression equations from the following data: X: Y: 6 9 2 11 10 5 222 4 8 8 7 Solution: Y X2 Y2 XY 9 36 81 54 11 4 121 22 5 100 25 50 8 16 64 32 7 64 49 56 40 220 340 214 Regression equation of Y on X is Y = a + bX and the normal equations are X 6 2 10 4 8 30 ∑Y = na + b∑X ∑XY = a∑X + b∑X2 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 26 = - 0.65 40 Now, substituting the value of ‘ b’ in equation (1) 40 = 5a –...
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