Single Regression Equation Dr. Anshul Singh Thapa

Regression Analysis It is clear from the definition of regression equation that it is a statistical device with the help of which we are in position to estimate (or predict) the unknown values of one variable (Y) from known values of another variable (X). The analysis used is called the simple liner regression analysis – simple because there is only one predictor or independent variable, and linear because of the assumed linear relationship between the dependent and dependent variable. The term linear means that an equation of a straight line of the form Y = a + b X, where ‘a’ and ‘b’ are constant, is used to describe the average relationship that exist between the two variables.

Regression Lines If we take the case of two variables X and Y, we shall have two regression lines as the regression of X and Y and the regression of Y on X. The regression line of Y on X gives the most probable values of X for given value of Y. However when there is either perfect positive or perfect negative correlation between the two variables (r = ± 1) the regression lines will coincide, i.e., we will have only one line. The farther the two regression line from each other, the lesser the degree of cancellation and the nearer the two regression lines to each other, the higher is the degree of correlation. If the variables are independent, r is zero and the lines of regression are at right angles i.e., parallel to OX and OY

Regression Equation Regression equation, also known as estimating equations, are algebraic expression of the regression lines. Since there are two regression lines, there are two regression equations – the regression equation of X on Y is used to describe the variations in the value of X for given changes in Y and the regression equation of Y on X is used to describe the variation in the values of Y for given changes in X.

Regression Equation of Y on X Regression Equation of X on Y The regression equation of Y on X is expressed as follows: Y = a+ b X It may be noted that in this equation ‘Y’ is a dependent variable i.e., its value depends on X. ‘X’ is independent variable, i.e., we can take a given value of X and compute the value of Y. ‘a’ is ‘Y-intersect” because its value is the point at which the regression line crosses the Y-axis that is the vertical axis ‘b’ is the ‘slope’ of line. It represent change in Y variable for a unit change in X variable. ‘a’ and ‘b’ in the equation are called numerical constants because for any given straight line, their value does not changed. The value of constants ‘a’ and ‘b’ are obtained by the method of Least Squares The regression equation of X on Y is expressed as follows: X = a+ b Y

Normal Equations X on Y (X = a + bY) Y on X (Y = a + bX) To determine the values of ‘a’ and ‘b’ the following two normal equations are to be solved simultaneously. ΣX = Na + bΣY ΣXY = aΣY + b ΣY2 To determine the values of ‘a’ and ‘b’ the following two normal equations are to be solved simultaneously. ΣY = Na + bΣX ΣXY = aΣX + b ΣX2

Regression equation X Y 6 9 2 11 10 5 4 8 7

Regression equation X Y XY 6 9 54 2 11 22 10 5 50 4 8 32 7 56

Regression equation X Y XY X2 6 9 54 36 2 11 22 4 10 5 50 100 8 32 16 7 56 64

Regression equation X Y XY X2 Y2 6 9 54 36 81 2 11 22 4 121 10 5 50 100 25 8 32 16 64 7 56 49

Regression equation X Y XY X2 Y2 6 9 54 36 81 2 11 22 4 121 10 5 50 100 25 8 32 16 64 7 56 49 ΣX = 30 ΣY = 40 ΣXY = 214 ΣX2 = 220 ΣY2= 340

X on Y Equations Solution ΣX = Na + bΣY ΣXY = aΣY + b ΣY2 ΣX = 30 ΣY = 40 ΣXY = 214 ΣX2 = 220 ΣY2= 340 30 = 5a + 40b …………….i 214 = 40a + 340b………..ii 240 = 40a + 320b…………..iii (Mult. 8) 214 = 40a + 340b…………….iv 26 = - 20b -20b = 26 b = 26/20 b = -1.3 ΣX = Na + bΣY 30 = 5a + (1.3) 40 5a = 30 + 52 a = 82/5 a= 16.4

Y on X Equations Solution ΣY = Na + bΣX ΣXY = aΣX + b ΣX2 ΣX = 30 ΣY = 40 ΣXY = 214 ΣX2 = 220 ΣY2= 340 40 = 5a + 30b …………….i 214 = 30a + 220b………..ii 240 = 30a + 180b………..iii (Mult. 6) 214 = 30a + 220b…………….iv 26 = - 40b -40b = 26 b = 26/40 b = 0.65 ΣY = Na + bΣX 40 = 5a + (0.65) 30 5a = 40 + 19.5 a = 59.5/5 a= 11.9

Calculate the regression equation of Y on X and If the value of X is 65 calculate the value of Y 50 30 60 40 80 70

Calculate the regression equation of Y on X and If the value of X is 65 calculate the value of Y XY 50 30 1500 60 3600 40 2000 3000 80 4800 70 5600 4200

Calculate the regression equation of Y on X and If the value of X is 65 calculate the value of Y XY X2 50 30 1500 2500 60 3600 40 2000 3000 80 4800 6400 70 5600 1600 4200 4900

Calculate the regression equation of Y on X and If the value of X is 65 calculate the value of Y XY X2 Y2 50 30 1500 2500 900 60 3600 40 2000 1600 3000 80 4800 6400 70 5600 4900 4200

Calculate the regression equation of Y on X and If the value of X is 65 calculate the value of Y XY X2 Y2 50 30 1500 2500 900 60 3600 40 2000 1600 3000 80 4800 6400 70 5600 4900 4200 ΣX=540 ΣY=450 ΣXY=28200 ΣX2 =34000 ΣY2=24100

Y on X 450 = 9a + 540b 28200 = 540a + 34000b 27000 = 540a + 32400b (Mult. 60) -1200 = -1600b 1600b = 1200 b = 1200/1600 b = 0.75 ΣY = Na + bΣX 450 = 9a + 540 (0.75) 9a + 405 = 450 9a = 450 – 405 a = 45/9 a = 5 Y = a +bX Y = 5 + (0.75) 65 Y = 53.75