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`

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