3x+2y−26=06x+y−31=0Solving these equations simultaneously, we get
6x+4y−52=06x+y−31=0
3x+2y−26=03y−21=0
3x=−2(7)+26y=7Then x=4,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=−32y+326=>bxy=−32
y=−6x+31=>byx=−6 But r2=bxy⋅byx=4 which cannot be true, because −1≤r≤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=−61y+631=>bxy=−61
y=−23x+13=>byx=−23
r2=bxy⋅byx=(−61)(−23)=41 As the regression coefficients have negative signs, we have to consider negative value of r.
r=−41=−21
a) x=4,y=7b)\ \ \ r=-0.5 \ \ \ \ \
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