Answer to Question #90220 in Statistics and Probability for Krishnakant Thakre

Question

The equation of two variables x and y are follows: 3x+2y-26=0;6x+y-31=0
Find a)mean
b) regression correlation

Answer 1

$$3x+2y-26=0$$ $$6x+y-31=0$$

Solving these equations simultaneously, we get 

$$6x+4y-52=0$$ $$6x+y-31=0$$

$$3x+2y-26=0$$ $$3y-21=0$$

$$3x=-2(7)+26$$ $$y=7$$

Then $\overline{x}=4, \overline{y}=7.$

Let $3x+2y-26=0$ be the regression line of X on Y and the other line as Y on X. Then

$$x=-{2 \over 3}y+{26 \over 3}=>b_{xy}=-{2 \over 3}$$

$$y=-6x+31=>b_{yx}=-6$$

But $r^2=b_{xy}\cdot b_{yx}=4$ which cannot be true, because $-1\leq r\leq 1.$

So we change our assumptions i.e., the line $6x+y-31=0$ be the regression line of X on Y and the other line as Y on X. Then

$$x=-{1 \over 6}y+{31 \over 6}=>b_{xy}=-{1 \over 6}$$

$$y=-{3 \over 2}x+13=>b_{yx}=-{3 \over 2}$$

$$r^2=b_{xy}\cdot b_{yx}=(-{1 \over 6})(-{3 \over 2})={1 \over 4}$$

As the regression coefficients have negative signs, we have to consider negative value of $r.$

$$r=-\sqrt{{1 \over 4}}=-{1 \over 2}$$

$$a)\ \ \ \overline{x}=4, \overline{y}=7$$ $$b)\ \ \ r=-0.5 \ \ \ \ \$$