The equations of two Regressions lines are 7𝑥 − 16𝑦 + 9 = 0 and 5𝑦 − 4𝑥 − 3 = 0. Find the coefficient of correlation and the means of x and y

Solution:

$7𝑥 − 16𝑦 + 9 = 0 \ ...(i) \\5𝑦 − 4𝑥 − 3 = 0\ ...(ii)$

From (i)

$16y=7x+9 \\\Rightarrow y=\dfrac7{16}x+\dfrac9{16}$ ...(iii)

From (ii)

$5y=4x+3 \\\Rightarrow y=\dfrac45x+\dfrac35$ ...(iv)

On equating these values of y,

$\dfrac7{16}x+\dfrac9{16}=\dfrac45x+\dfrac35 \\ \Rightarrow x=-\dfrac{3}{29}$

Put this in (iii), we get,

$y=\dfrac{15}{29}$

So, mean of x $=-\dfrac3{29}$

And mean of y $=\dfrac{15}{29}$

Next, slope of 1st line $=b_{yx}=\dfrac7{16}$ [From (iii)]

And slope of 2nd line $=b_{xy}=\dfrac4{5}$ [From (iv)]

Then, $r^2=b_{yx}b_{xy}=\dfrac7{16}\times \dfrac4{5}=\dfrac7{20}$

So, $r=\pm\sqrt{\dfrac7{20}}=\pm0.592$

But r has same sign as $b_{yx},b_{xy}$ , so $r=0.592$