Answer to Question #349509 in Statistics and Probability for Anu

Question

In a statistical study relating to the prices (in T) of two shares, X and Y, the following two regression lines were found as 8X - 10Y + 70 = 0 and 20X - 9Y - 65 = 0. The standard deviation of X = 3, then find (i) the values of X and Y, (ii) r(X, Y), and (iii) standard deviation of Y.

Answer 1

(i) The given equation of the lines of regression are

$$8X - 10Y + 70 = 0$$ $$20X - 9Y - 65 = 0$$

$$40X - 50Y + 350 = 0$$ $$40X - 18Y - 130 = 0$$

$$-32Y+480=0$$

$$Y=15$$

$$8X - 10(15) + 70 = 0$$

$$X=10$$

$X=10, Y=15$

(ii) Let the equation $8X - 10Y + 70 = 0$ be the regression equation of $Y$ on $X.$ Then

$$Y=0.8X+7$$

Comparing it with $Y=b_{YX}X+a,$ we get $b_{YX}=0.8.$

Let the equation $20X - 9Y - 65 = 0$ be the regression equation of $X$ on $Y.$ Then

$$X=0.45X+3.25$$

Comparing it with $X=b_{XY}Y+a',$ we get $b_{XY}=0.45.$

$$r=\pm\sqrt{b_{XY}\cdot b_{YX}}$$

Since $b_{XY},b_{YX}$ both are positive, then $r$ is also positive:

$$r=\sqrt{0.8\cdot 0.45}=0.6$$

(iii)

$$b_{YX}=r\dfrac{\sigma_Y}{\sigma_X}$$

Then

$$\sigma_Y=b_{YX}\dfrac{\sigma_X}{r}$$

$$\sigma_Y=0.8(\dfrac{3}{0.6})=4$$