Question

The equations of two regression lines are 10x − 4y = 80 and 10y − 9x = − 40 Find:

1. `bar x and bar y`
2. b_YX and b_XY
3. If var (Y) = 36, obtain var (X)
4. r

Answer 1

(i) Given equations of regression are

10x − 4y = 80

i.e., 5x − 2y = 40        .....(i)

and 10y − 9x = −40

i.e., − 9x + 10y = −40         .....(ii)

By 5 × (i) + (ii), we get

 25x − 10y = 200

− 9x + 10y = − 40 
16x             = 160

∴ x = 10

Substituting x = 10 in (i), we get

5(10) − 2y = 40

∴ 50 − 2y = 40

∴ −2y = 40 − 50

∴ −2y = − 10

∴ y = 5

Since the point of intersection of two regression lines is `(bar x, bar y)`, `bar x = 10  and bar y = 5`

(ii) Let 10y − 9x = −40 be the regression equation of Y on X.

∴ The equation becomes 10Y = 9X − 40

i.e., Y = `9/10X − 40/10`

Comparing it with Y = b_YX X + a, we get

`b_(YX) = 9/10 = 0.9`

Now, the other equation 10x − 4y = 80 be the regression equation of X on Y.

∴ The equation becomes 10X = 4Y + 80

i.e., X = `4/10 Y + 80/10`

i.e., X = `2/5 Y + 8`

Comparing it with X = b_XY Y + a', we get

`b_(XY) = 2/5 = 0.4`

(iii) Given, Var (Y) = 36, i.e., `sigma_Y^2` = 36

∴ σY = 6

Since `b_(XY) = r xx sigma_X/sigma_Y`

`2/5 = 0.6 xx sigma_X/6`

∴ `2/5 = 0.1 xx sigma_X`

∴ `2/(5 xx 0.1) = sigma_X`

∴ `sigma_X` = 4

∴ `sigma_X^2 = 16` i.e., Var(X) = 16

(iv) r = `+-sqrt(b_(XY) *b_(YX)`

`= +-sqrt(2/5 xx 9/10) +- sqrt(9/25)`

`= +- 3/5`

 `= +- 0.6`

Since b_YX and b_XY are positive,

r is also positive.

∴ r = 0.6