3 x + 2 y − 26 = 0 3x+2y-26=0 3 x + 2 y − 26 = 0 6 x + y − 31 = 0 6x+y-31=0 6 x + y − 31 = 0 Solving these equations simultaneously, we get
6 x + 4 y − 52 = 0 6x+4y-52=0 6 x + 4 y − 52 = 0 6 x + y − 31 = 0 6x+y-31=0 6 x + y − 31 = 0
3 x + 2 y − 26 = 0 3x+2y-26=0 3 x + 2 y − 26 = 0 3 y − 21 = 0 3y-21=0 3 y − 21 = 0
3 x = − 2 ( 7 ) + 26 3x=-2(7)+26 3 x = − 2 ( 7 ) + 26 y = 7 y=7 y = 7 Then x ‾ = 4 , y ‾ = 7. \overline{x}=4, \overline{y}=7. x = 4 , y = 7.
Let 3 x + 2 y − 26 = 0 3x+2y-26=0 3 x + 2 y − 26 = 0 be the regression line of X on Y and the other line as Y on X. Then
x = − 2 3 y + 26 3 = > b x y = − 2 3 x=-{2 \over 3}y+{26 \over 3}=>b_{xy}=-{2 \over 3} x = − 3 2 y + 3 26 => b x y = − 3 2
y = − 6 x + 31 = > b y x = − 6 y=-6x+31=>b_{yx}=-6 y = − 6 x + 31 => b y x = − 6 But r 2 = b x y ⋅ b y x = 4 r^2=b_{xy}\cdot b_{yx}=4 r 2 = b x y ⋅ b y x = 4 which cannot be true, because − 1 ≤ r ≤ 1. -1\leq r\leq 1. − 1 ≤ r ≤ 1.
So we change our assumptions i.e., the line 6 x + y − 31 = 0 6x+y-31=0 6 x + y − 31 = 0 be the regression line of X on Y and the other line as Y on X. Then
x = − 1 6 y + 31 6 = > b x y = − 1 6 x=-{1 \over 6}y+{31 \over 6}=>b_{xy}=-{1 \over 6} x = − 6 1 y + 6 31 => b x y = − 6 1
y = − 3 2 x + 13 = > b y x = − 3 2 y=-{3 \over 2}x+13=>b_{yx}=-{3 \over 2} y = − 2 3 x + 13 => b y x = − 2 3
r 2 = b x y ⋅ b y x = ( − 1 6 ) ( − 3 2 ) = 1 4 r^2=b_{xy}\cdot b_{yx}=(-{1 \over 6})(-{3 \over 2})={1 \over 4} r 2 = b x y ⋅ b y x = ( − 6 1 ) ( − 2 3 ) = 4 1 As the regression coefficients have negative signs, we have to consider negative value of r . r. r .
r = − 1 4 = − 1 2 r=-\sqrt{{1 \over 4}}=-{1 \over 2} r = − 4 1 = − 2 1
a ) x ‾ = 4 , y ‾ = 7 a)\ \ \ \overline{x}=4, \overline{y}=7 a ) x = 4 , y = 7 b)\ \ \ r=-0.5 \ \ \ \ \
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