(i) Given equations of regression are 3x + 2y - 26 = 0 i.e., 3x + 2y = 26 ….(i) and 6x + y - 31 = 0 i.e., 6x + y = 31 ….(ii) By (i) - 2 × (ii), we get 3x + 2y = 26 12x + 2y = 62 - - - ∴ x = `(-36)/-9 = 4` Substituting x = 4 in (ii), we get 6 × 4 + y = 31 ∴ 24 + y = 31 ∴ y = 31 - 24 ∴ y = 7 Since the point of intersection of two regression lines is `(bar x, bar y)`, `bar x` = mean of X = 4, and `bar y` = mean of Y = 7 (ii) Let 3x + 2y 26 = 0 be the regression equation of Y on X. ∴ The equation becomes 2Y = 3X + 26 i.e., Y = `(-3)/2"X" + 26/2` Comparing it with Y = bYX X + a, we get `"b"_"YX" = (- 3)/2` Now, the other equation 6x + y - 31 = 0 is the regression equation of X on Y. ∴ The equation becomes 6X = - Y + 31 i.e., X = `(-1)/6 "Y" + 31/6` Comparing it with X = bXY Y+ a', we get `"b"_"XY" = (-1)/6` ∴ r = `+-sqrt("b"_"XY" * "b"_"YX")` `= +- sqrt((- 1)/6 xx (- 3)/2) = +- sqrt(1/4) = +- 1/2 = +- 0.5` Since the values of bXY and bYX are negative, r is also negative. ∴ r = - 0.5 (iii) The regression equation of Y on X is Y = `(- 3)/2 "X" + 26/2` For X = 2, we get Y = `(- 3)/2 xx 2 + 26/2 = - 3 + 13 = 10` (iv) Given, Var (Y) = 36, i.e., `sigma_"Y"^2` = 36 ∴ `sigma_"Y" = 6` Since `"b"_"XY" = "r" xx sigma_"X"/sigma_"Y"` `(- 1)/6 = - 0.5 xx sigma_"X"/6` ∴ `sigma_"X" = (-6)/(- 6 xx 0.5) = 2` ∴ `sigma_"X"^2` = Var(X) = 4 The equations given of the two regression lines are 2x + 3y - 6 = 0 and 5x + 7y - 12 = 0. Find: (a) Correlation coefficient (b) `sigma_x/sigma_y` We assume that 2x + 3y - 6 = 0 to be the line of regression of y on x. 2x + 3y - 6 = 0 ⇒ `x = - 3/2y + 3` ⇒ `"bxy" = - 3/2` 5x + 7y - 12 = 0 to be the line of regression of x on y. 5x + 7y - 12 = 0 ⇒ `y = - 5/7x + 12/7` ⇒ `"byx" = - 5/7` Now, r = `sqrt("bxy.byx") = sqrt(15/14)` byx = `(rσ_y)/(σ_x) = - 5/7, "bxy" = (rσ_x)/(σ_y) = - 3/2` ⇒ `(σ_x^2)/(σ_y^2) = (3/2)/(5/7)` ⇒ `(σ_x^2)/(σ_y^2) = 21/10` ⇒ `(σ_x)/(σ_y) = sqrt(21/10)`. Concept: Lines of Regression of X on Y and Y on X Or Equation of Line of Regression Is there an error in this question or solution? |