Question

If the regression lines of a bivariate distribution are 4x − 5y + 33 = 0 and 20x − 9y − 107 = 0, then

1. Calculate the arithmetic mean of x and y.
2. Estimate the value of x when y = 7.
3. Find the variance of y when σx = 3.

Answer 1

(a) The two regression lines always intersect at the point of the means `(barx, bary)`. We solve the given equations simultaneously:

4x − 5y = −33   ...[1]

20x − 9y = 107   ...[2]

Multiply the first equation by 5:

20x − 25y = − 165      ...[3]

Subtract Equation 3 from Equation 2:

(20x − 9y) − (20x − 25y) = 107 − (−165)

⇒ 16y = 272 

⇒ `bary` = 17

Substitute y = 17 back into the first equation:

4x − 5(17) + 33 = 0

4x − 85 + 33 = 0

⇒ 4x = 52

⇒ `barx = 13`

(b) To estimate x, we must use the regression line of x on y We first identify which line is which by checking the condition `b_(yx) × b_(yx)` ≤ 1.

Line 1 as y on x: 5y = 4x + 33 

⇒ y = 0.8x + 6.6 (so, b_yx = 0.8)

Line 2 as x on y: 20x = 9y + 107

⇒ x = 0.45y + 5.35 (so, b_xy = 0.45)

r² = 0.8 × 0.45 = 0.36

Since 0.36 ≤ 1, this identification is correct.

Now, substitute y = 7 into the x on y line:

x = 0.45(7) + 5.35

x = 3.15 + 5.35

x = 8.5

(c) Find the variance of y when σx = 3

We use the relationship between the regression coefficients and standard deviations:

`b_(yx)/b_(xy) = (r (σy)/(σx))/(r (σx)/(σy)) = σ_y^2/σ_x^2`

Substitute the known values:

`0.8/0.45 = σ_y^2/3^2`

`16/9 = σ_y^2/9`

`σ_y^2 = 16`

The variance of y = 16