The two regression lines were found to be 4X – 5Y + 33 = 0 and 20X – 9Y – 107 = 0. Find the mean values and coefficient of correlation between X and Y.

Question Collected From Telegram Group.

Solution:

To get mean values we must solve the given lines. 

4X – 5Y = -33 … (1) 

20X – 9Y = 107 … (2) 

(1) × 5 ⇒ 20X – 25Y = -165 

20X – 9Y = 107 

Subtracting (1) and (2),

-16Y = -272  Y = \(\frac{272}{16}\) = 17 

i.e., \(\bar{Y}\) = 17 

Using Y = 17 in (1) we get, 4X – 85 = -33 

4X = 85 – 33 

4X = 52  X = 13 

i.e., \(\bar{X}\) = 13 

Mean values are \(\bar{X}\) = 13, \(\bar{Y}\) = 17, 

Let regression line of Y on X be 

4X – 5Y + 33 = 0 

5Y = 4X + 33 Y = (4X + 33) 

Y = \(\frac{1}{5}\)(4x + 33)

Y = \(\frac{4}{5}X+\frac{33}{5}\)

Y = 0.8X + 6.6 

∴ b_yx = 0.8 

Let regression line of X on Y be 20X – 9Y – 107 = 0 

20X = 9Y + 107 

X = \(\frac{1}{20}\)(9Y + 107) 

X = \(\frac{9}{20}Y+\frac{107}{20}\)

X = 0.45Y + 5.35 

∴ b_xy = 0.45 

Coefficient of correlation between X and Y is = ±0.6  = 0.6 

\[\begin{array}{l}r =  \pm \sqrt {{b_{yx}} \times {b_{xy}}} \\r =  \pm \sqrt {0.8 \times 0.45} \\r =  \pm 0.6\\r = 0.6\end{array}\]

Both b_yx and b_xy is positive take positive sign.