The regression lines of a sample are x +6y=6 and 3x+2y=10

Given data :

regression Lines :   x +6y=6  __(1) and 3x+2y=10   __(2)

To Find : (i) sample means x and y = ? (ii) coefficient of correlation between x and y = ?

Solution : (i) sample means x and y : x +6y=6  __(1) 3x+2y=10   __(2)  https://www.youtube.com/watch?v=OPDTazrPk_Y

(i) \(\bar{x} = 3,\ \bar{y} = 0.5\); (ii) \(r = -\tfrac{1}{3} \approx -0.33\)

(i) Means. Both regression lines pass through \((\bar{x}, \bar{y})\), so solve them together:

\[x + 6y = 6 \quad (1)\]

\[3x + 2y = 10 \quad (2)\]

From (1), \(x = 6 - 6y\). Substituting in (2):

\[18 - 18y + 2y = 10 \Rightarrow y = 0.5\]

\[x = 6 - 3 = 3\]

(ii) Correlation. Decide which line is which. Try (1) as y on x and (2) as x on y:

\[(1):\ y = 1 - \tfrac{1}{6}x \Rightarrow b_{yx} = -\tfrac{1}{6}\]

\[(2):\ x = \tfrac{10}{3} - \tfrac{2}{3}y \Rightarrow b_{xy} = -\tfrac{2}{3}\]

\[b_{yx}\,b_{xy} = \tfrac{1}{9} \le 1\quad\text{(valid)}\]

(The other choice gives \(b_{xy} = -6,\ b_{yx} = -\tfrac{3}{2}\), product 9 > 1, which is impossible.)

\[r^2 = \tfrac{1}{9} \Rightarrow r = \pm\tfrac{1}{3}\]

Both regression coefficients are negative, so \(r\) is negative:

\[r = -\tfrac{1}{3} \approx -0.333\]