Question

The equations of two regression lines are \(10x-4y=80\) and \(10y-9x=-40\). Find (a) \(\bar{x}\) and \(\bar{y}\) (b) \(b_{yx}\) and \(b_{xy}\) (c) \(r\) (d) If Var(Y) = 36, obtain var (X).

Hint:

When identifying regression lines, calculate both possibilities for \(b_{yx}\) and \(b_{xy}\). The correct assignment is the one where their product is less than or equal to 1.

Answer 1

(a) Find \(\bar{x}\) and \(\bar{y}\):
The two regression lines intersect at the point of means \((\bar{x}, \bar{y})\). We solve the two equations simultaneously.
1) \(10x - 4y = 80 \implies 5x - 2y = 40\)
2) \(10y - 9x = -40 \implies -9x + 10y = -40\)
Multiply equation (1) by 5: \(25x - 10y = 200\).
Add this to equation (2): \[ (25x - 10y) + (-9x + 10y) = 200 - 40 \] \[ 16x = 160 \implies \bar{x} = 10 \] Substitute \(\bar{x}=10\) into \(5x - 2y = 40\): \[ 5(10) - 2y = 40 \implies 50 - 2y = 40 \implies 2y = 10 \implies \bar{y} = 5 \] So, \(\bar{x} = 10\) and \(\bar{y} = 5\). (b) Find \(b_{yx}\) and \(b_{xy}\):
Assume the first line is Y on X: \(10x - 80 = 4y \implies y = \frac{10}{4}x - 20\). So, \(b_{yx} = \frac{10}{4} = 2.5\).
Assume the second line is X on Y: \(10y + 40 = 9x \implies x = \frac{10}{9}y + \frac{40}{9}\). So, \(b_{xy} = \frac{10}{9}\).
Let's check the condition \(|r| \leq 1\), which means \(b_{yx} \cdot b_{xy} \leq 1\).
\( 2.5 \times \frac{10}{9} = \frac{25}{9} \approx 2.78 \).
This is greater than 1, so our assumption is wrong.
Let's switch the assumptions.
Line of Y on X: \(10y - 9x = -40 \implies 10y = 9x - 40 \implies y = \frac{9}{10}x - 4\). So, \(b_{yx} = \frac{9}{10} = 0.9\).
Line of X on Y: \(10x - 4y = 80 \implies 10x = 4y + 80 \implies x = \frac{4}{10}y + 8\). So, \(b_{xy} = \frac{4}{10} = 0.4\).
Check: \(b_{yx} \cdot b_{xy} = 0.9 \times 0.4 = 0.36\). This is \(\leq 1\), so the assumption is correct.
\(b_{yx} = 0.9\) and \(b_{xy} = 0.4\).
(c) Find \(r\):
The correlation coefficient \(r\) is the geometric mean of the regression coefficients. \[ r = \sqrt{b_{yx} \cdot b_{xy}} = \sqrt{0.36} = 0.6 \] Since both \(b_{yx}\) and \(b_{xy}\) are positive, \(r\) is also positive. So, \(r = 0.6\).
(d) Find Var(X):
We know that \(b_{yx} = r \frac{\sigma_y}{\sigma_x}\). The variance is the square of the standard deviation (\(\sigma^2\)). \[ b_{yx}^2 = r^2 \frac{\sigma_y^2}{\sigma_x^2} = r^2 \frac{\text{Var}(Y)}{\text{Var}(X)} \] \[ 0.9^2 = (0.6)^2 \frac{36}{\text{Var}(X)} \] \[ 0.81 = 0.36 \frac{36}{\text{Var}(X)} \] \[ \text{Var}(X) = \frac{0.36 \times 36}{0.81} = \frac{12.96}{0.81} = 16 \] The variance of X is 16.