Question 1

Question 1: From the two regression equations find r, and . 4y = 9x + 15 and 25x = 4y + 17

Accepted Answer

Given 4y = 9x + 15 and 25x = 4y + 17 Since byx and bxy are positive. ∴ r = = 0.6 (, ) is the point of intersection of the regression lines 9x – 4y = -15 …….(i) 25x – 4y = 17 ……….(ii) -16x = -32 x = 2 ∴ = 2 Substituting x = 2 in equation (i) 9(2) – 4y = -15 18 + 15 = 4y 33 = 4y y = 33/4 = 8.25 ∴ = 8.

Question 2

Question 2: In a partially destroyed laboratory record of an analysis of regression data, the following data are legible: Variance of X = 9 Regression equations: 8x – 10y + 66 = 0 And 40x – 18

Accepted Answer

Given, (i) (, ) is the point of intersection of the regression lines 40x – 50y = -330 …….(i) 40x – 50y = +214 ………(ii) -32y = -544 y = 17 ∴ = 17 8x – 10(17) + 66 = 0 8x = 104 x = 13 ∴ = 13

Question 3

Question 3: For 50 students of a class, the regression equation of marks in statistics (X) on the marks in Accountancy (Y) is 3y – 5x + 180 = 0. The mean marks in accountancy is 44 and the var

Accepted Answer

Given, n = 50, = 44 ∴ Since (, ) is the point intersection of the regression line. ∴ (, ) satisfies the regression equation. 3 – 5 + 180 = 0 3(44) – 5 + 180 = 0 ∴ 5 = 132 + 180 = = 62.4 ∴ Mean marks in statistics is 62.4 Regression equation of X on Y is 3y – 5x + 180 = 0 ∴ 5x = 3y + 180

Question 4

Question 4: For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.

Accepted Answer

Step-by-step solution provided.

Question 5

Question 5: The equation of two regression lines are 2x + 3y – 6 = 0 and 3x + 2y – 12 = 0 Find (i) Correlation coefficient (ii)

Accepted Answer

Step-by-step solution provided.

Question 6

Question 6: For a bivariate data = 53, = 28, byx =-1.5 and bxy = -0.2. Estimate Y when X = 50.

Accepted Answer

Regression equation of Y on X is (Y – ) = byx (X – ) (Y – 28) = -1.5(50 – 53) Y – 28 = -1.5(-3) Y – 28 = 4.5 Y = 32.5

Question 7

Question 7: The equation of two regression lines are x – 4y = 5 and 16y – x = 64. Find means of X and Y. Also, find the correlation coefficient between X and Y.

Accepted Answer

Since (, ) is the point of intersection of the regression lines. x – 4y = 5 …..(i) -x + 16y = 64 …….(ii) 12y = 69 y = 5.75 Substituting y = 5.75 in equation (i) x – 4(5.75) = 5 x – 23 = 5 x = 28 ∴ = 28, = 5.75 x – 4y = 5 x = 4y + 5 ∴ bxy = 4 16y – x = 64 16y = x + 64 y = x + 4 byx = byx. bxy = × 4 =

Question 8

Question 8: In partially destroyed record, the following data are available variance of X = 25. Regression equation of Y on X is 5y – x = 22 and Regression equation of X on Y is 64x – 45y = 22

Accepted Answer

Given = 25, ∴ σx = 5 (i) Since (, ) is the point of intersection of regression lines -x + 5y = 22 …….(i) 64x – 45y = 22 ………..(ii) equation (i) becomes -9x + 45y = 198 64y – 45y = 22 55x = 220 x = 4 Substituting x = 4 in equation (i) -4 + 5y = 22 5y = 26 ∴ y = 5.2 ∴ = 4, = 5.2 Regression equation of

Question 9

Question 9: If the two regression lines for a bivariate data are 2x = y + 15 (x on y) and 4y – 3x + 25 (y on x) find (i) (ii) (iii) byx (iv) bxy (v) r [Given √0.375 = 0.61]

Accepted Answer

Since (, ) is the point of intersection of the regression line 2x = y + 15 4y = 3x + 25 2x – y = 15 …….(i) 3x – 4y = -25 ……..(ii) Multiplying equation (i) by 4 8x – 4y = 60 3x – 4y = -25 on Subtracting, 5x = 85 ∴ x = 17 Substituting x in equation (i) 2(17) – y = 15 34 – 15 = y ∴ y = 15 Since byx and

Question 10

Question 10: The two regression equation are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find , , r.

Accepted Answer

Since (, ) is the point of intersection of the regression lines 5x – 6y + 90 = 0 ……(i) 15x – 8y – 130 = 0 15x – 18y + 270 = 0 15x – 8y – 130 = 0 on subtracting, -10y + 400 = 0 y = 40 Substituting y = 40 in equation (i) 5x – 6(40) + 90 = 0 5x = 150 x = 30 ∴ = 30, = 40 Since byx and bxy are positive ∴

Question 11

Question 11: Two lines of regression are 10x + 3y – 62 = 0 and 6x + 5y – 50 = 0 Identify the regression equation equation of x on y. Hence find , , and r.

Accepted Answer

∴ Our assumption is correct. ∴ Regression equation of X on Y is 10x + 3y – 62 = 0 r2 = byx. bxy r2 = r = ± Since, byx and bxy are negative, r = – = -0.6 Also (, ) is the point of intersection of the regression lines 50x + 15y = 310 18x + 15y = 150 on subtracting 32x = 160 x = 5 Substituting x = 5 in

Question 12

Question 12: For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean value

Accepted Answer

Since byx and bxy are positive, r = = 0.6 Since, (, ) is the point of intersection of the regression lines 3x – 10y = -170 …….(i) 5x – 6y = -70 ………(ii) 9x – 30y = -510 25x – 30y = -350 on subtracting -16x = -160 x = 10 Substituting x = 10 in equation (i) 3(10) – 10y = -170 30 + 170 = 10y 200 = 10y y

Question 13

Question 13: Regression equation of two series are 2x – y – 15 = 0 and 4y + 25 = 0 and 3x- 4y + 25 = 0. Find , and regression coefficients, Also find coefficients of correlation. [Given √0.375

Accepted Answer

Since (, ) is the point of intersection of the regression line 2x = y + 15 4y = 3x + 25 2x – y = 15 ……(i) 3x – 4y = -15 ……..(ii) Multiply equation (i) by 4 8x – 4y = 60 3x – 4y = -25 on subtracting, 5x = 85 x = 17 Substituting x in equation (i) 2(17) – y = 15 34 – 15 = y y = 15 ∴ = 17, = 19 ∴ Our as

Question 14

Question 14: The two regression lines between height (X) in includes and weight (Y) in kgs of girls are 4y – 15x + 500 = 0 and 20x – 3y – 900 = 0. Find the mean height and weight of the group.

Accepted Answer

Since (, ) is the point intersection of the regression lines 15x – 4y = 500 ……(i) 20x – 3y = 900 …….(ii) 60x – 16y – 2000 60x – 9y = 2700 on subtracting, -7y = -700 y = 100 Substituting y = 100 in equation (i) 15x – 4(100) = 500 15x = 900 x = 60 ∴ Our assumption is correct ∴ Regression equation of Y

Chapter 3 Linear Regression Ex 3.3 Solutions