Answer in Statistics and Probability for Krishnakant Thakre #90220
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 “-1leq rleq 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 “
