Publish On: 2019-04-13

# Jason steev

Total Post: 475

## Question: From the following data obtain the two regression equations

Reply On: 2013-07-10

# Gaurabh Raghav

Total Post: 33

## ANS: From the following data obtain the two regression equations

Regression equation of

To determine the values of a and b the following two normal equations are to be solved.

Let

Differentiating partially with respect to and b,

∂S/∂a = Σ (Y – a – b X) (-1) = 0 and ∂ S / ∂ b = Σ (Y – a – b X) (-X) = 0

Or Σ (Y – a – b A) = 0 or Σ X = N a + b Σ X and Σ XY = a Σ X + b Σ X

Σ Y = Na + b Σ X

Σ X Y = a Σ X + b Σ X

Substitution the values 40 = 5 a + 30 b

214 = 30 a + 220 b

Multiplying equation (i) by 6,240 = 30 a + 180 b

214 = 30 a + 220 b

Deduction equation (iv) from (iii) 40 b = 26 or b = - 0.56

Substituting the value of b in equation (i)

40 = 5a + 30 (- 0.56) or 5a = 40 + 19.5 = 59.5 or a = 11.9

Putting the values of a and b in the equation the regression of Y on X is

Y = 11.9 – 0.56 X

Regression line of X on Y; X c = a + by

And the two normal equations are

Σ X = Na + b Σ Y

Σ X Y = a Σ Y + b Σ Y

30 = 5 + 40 B

214 = 40 A + 340 b

Multiplying equation (i) by 8: 240 = 40 a + 320 b

214 = 40 a + 340 b

From eqn (iii) and (iv) - 20 b = 26 or b = - 1.3

Substituting the value of b in equation 30 = 5 a + 40 (-1.3)

5 a = 30 + 52 = 82 ∴ a = 16.4

Putting the values of a and b in the equation the regression line of X on Y is X = 16.4 – 1.3 Y.

**Y on X: Yc = a + bx**To determine the values of a and b the following two normal equations are to be solved.

**Σ (Y – Yc)**should be minimum or^{2}**Σ (Y – a – vx)**should be minimum**(since Yc = a + bx)**Let

**S = Σ (Y – a – b X)**^{2}Differentiating partially with respect to and b,

∂S/∂a = Σ (Y – a – b X) (-1) = 0 and ∂ S / ∂ b = Σ (Y – a – b X) (-X) = 0

Or Σ (Y – a – b A) = 0 or Σ X = N a + b Σ X and Σ XY = a Σ X + b Σ X

^{2}Σ Y = Na + b Σ X

Σ X Y = a Σ X + b Σ X

^{2}Substitution the values 40 = 5 a + 30 b

214 = 30 a + 220 b

Multiplying equation (i) by 6,240 = 30 a + 180 b

214 = 30 a + 220 b

Deduction equation (iv) from (iii) 40 b = 26 or b = - 0.56

Substituting the value of b in equation (i)

40 = 5a + 30 (- 0.56) or 5a = 40 + 19.5 = 59.5 or a = 11.9

Putting the values of a and b in the equation the regression of Y on X is

Y = 11.9 – 0.56 X

Regression line of X on Y; X c = a + by

And the two normal equations are

Σ X = Na + b Σ Y

Σ X Y = a Σ Y + b Σ Y

^{2}30 = 5 + 40 B

214 = 40 A + 340 b

Multiplying equation (i) by 8: 240 = 40 a + 320 b

214 = 40 a + 340 b

From eqn (iii) and (iv) - 20 b = 26 or b = - 1.3

Substituting the value of b in equation 30 = 5 a + 40 (-1.3)

5 a = 30 + 52 = 82 ∴ a = 16.4

Putting the values of a and b in the equation the regression line of X on Y is X = 16.4 – 1.3 Y.

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