Question #90220
The equation of two variables x and y are follows: 3x+2y-26=0;6x+y-31=0
Find a)mean
b) regression correlation
1
Expert's answer
2019-05-27T13:13:16-0400
3x+2y26=03x+2y-26=06x+y31=06x+y-31=0

Solving these equations simultaneously, we get 


6x+4y52=06x+4y-52=06x+y31=06x+y-31=0

3x+2y26=03x+2y-26=03y21=03y-21=0

3x=2(7)+263x=-2(7)+26y=7y=7

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

Let 3x+2y26=03x+2y-26=0 be the regression line of X on Y and the other line as Y on X. Then


x=23y+263=>bxy=23x=-{2 \over 3}y+{26 \over 3}=>b_{xy}=-{2 \over 3}

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

But r2=bxybyx=4r^2=b_{xy}\cdot b_{yx}=4 which cannot be true, because 1r1.-1\leq r\leq 1.

So we change our assumptions i.e., the line 6x+y31=06x+y-31=0 be the regression line of X on Y and the other line as Y on X. Then


x=16y+316=>bxy=16x=-{1 \over 6}y+{31 \over 6}=>b_{xy}=-{1 \over 6}

y=32x+13=>byx=32y=-{3 \over 2}x+13=>b_{yx}=-{3 \over 2}

r2=bxybyx=(16)(32)=14r^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.


r=14=12r=-\sqrt{{1 \over 4}}=-{1 \over 2}

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

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