Two lines of regression are 10x + 3y − 62 = 0 and 6x + 5y − 50 = 0. Identify the regression of x on y. Hence find `bar x, bar y` and r.

#### Solution

Given, two lines of regression are

10x + 3y – 62 = 0

i.e., 10x + 3y = 62 …(i)

and 6x + 5y – 50 = 0

i.e., 6x + 5y = 50 …(ii)

By (i) × 5 - (ii) × 3, we get

50x + 15y = 310

18x + 15y = 150

- - -

32x = 160

∴ x = 5

Substituting x = 5 in (i) we get,

10(5) + 3y = 62

∴ 50 + 3y = 62

∴ 3y = 62 - 50 = 12

∴ y = 4

Since the point of intersection of two regression lines is `(bar x, bar y)`,

`bar x = 5 and bar y = 4`

Now,

Let 10x + 3y - 62 = 0 be the regression equation of X on Y.

∴ The equation becomes 10x = –3y + 62

i.e., 10X = –3Y + 62

i.e., X = `- 3/10 "Y" + 62/10`

Comparing it with X = b_{XY} Y + a, we get

∴ `"b"_"XY" = - 3/10`

Now, other equation 6x + 5y – 50 = 0 be the regression equation of Y on X.

∴ The equation becomes 5y = – 6x + 50

i.e., 5Y = – 6X + 50

i.e., Y = `- 6/5 "x" + 50/5`

Comparing it with Y = b_{YX} X + a', we get

`"b"_"YX" = - 6/5`

Now, `"b"_"YX" * "b"_"XY" = (- 3/10)(- 6/5) = 9/25`

i.e., b_{XY} . b_{YX} < 1

∴ Assumption of regression equations is true.

∴ r = `+-sqrt("b"_"XY" * "b"_"YX") = +-sqrt(9/25) = +- 3/5`

Since b_{YX} and b_{XY} both are negative,

r is negative.

∴ r = `- 3/5 = - 0.6`