2. Find Correlation Coefficient from two Regression line equations x+y=2, 2x+3y=4
Solution:
`x+y=2`
and `2x+3y=4`
`x+y=2 ->(1)`
`2x+3y=4 ->(2)`
equation`(1) xx 2 =>2x+2y=4`
equation`(2) xx 1 =>2x+3y=4`
Substracting `=>-y=0`
`=>y=0`
Putting `y=0` in equation`(1)`, we have
`x+0=2`
`=>x=2`
`:.x=2" and "y=0`
`:. bar x = 2, bar y = 0`
Suppose `x+y=2` is regression equation of `y` on `x`
`=>1 x +1 y -2 = 0`
`=>1 y = -1 x +2`
`=>y = (-1)/(1) x +2/(1)`
`=>y = -1 x +2`
`:. byx = -1`
Suppose `2x+3y=4` is regression equation of `x` on `y`
`=> 2 x +3 y -4 = 0`
`=> 2 x = -3 y +4`
`=> x = (-3)/(2) y +4/2`
`=> x = -1.5 y +2`
`:. bxy = -1.5`
`byx * bxy = 1.5` which is > 1
So, Our guessing is wrong so we interchange our guessing
Now, Suppose `2x+3y=4` is regression equation of `y` on `x`
`=>2 x +3 y -4 = 0`
`=>3 y = -2 x +4`
`=>y = (-2)/(3) x +4/(3)`
`=>y = -0.6667 x +1.3333`
`:. byx = -0.6667`
and Suppose `x+y=2` is regression equation of `x` on `y`
`=> 1 x +1 y -2 = 0`
`=> 1 x = -1 y +2`
`=> x = (-1)/(1) y +2/1`
`=> x = -1 y +2`
`:. bxy = -1`
`r = sqrt(byx * bxy)`
`=sqrt(-0.6667 * -1)`
`=sqrt(0.6667)`
`=-0.8165`
This material is intended as a summary. Use your textbook for detail explanation.
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