Question

b) For 25 army personnels, line of regression of weight of kidneys (Y) on weight of 

heart (X ) is Y = .0 399X + .6 934 and the line of regression of weight of heart on 

weight of kidney is X − .1 212Y + .2 461= .0 Find the correlation coefficient between 

X and Y and their mean values.

Answer 1

Rewrite the regression equations uniformly

$$Y = 0.0 399X + 0.6 934$$

$$X = 0 .1 212Y - 0.2 461$$

The basic formulas are

$$Y = r\dfrac{s_y}{s_x}X -r\dfrac{s_y}{s_x}\bar{x}+\bar{y}$$

$$X = r\dfrac{s_x}{s_y}Y -r\dfrac{s_x}{s_y}\bar{y}+\bar{x}$$

Then

$$r\dfrac{s_y}{s_x}=0.0399$$

$$r\dfrac{s_x}{s_y}=0.1212$$

Hence

$$r^2\dfrac{s_y}{s_x}\cdot\dfrac{s_y}{s_x}=0.0399(0.1212)$$

$$r^2=0.00483588$$

$$r=\sqrt{0.00483588}\approx0.06954049$$

$$\begin{cases} -0.0399\bar{x}+\bar{y}=0.6934 \\ -0.1212\bar{y}+\bar{x}=-0.2461 \end{cases}$$

$$\begin{cases} \bar{y}=0.6934+0.0399\bar{x} \\ -0.1212(0.6934+0.0399\bar{x} )+\bar{x}=-0.2461 \end{cases}$$

$$\begin{cases} \bar{y}=0.6934+0.0399\bar{x} \\ \bar{x}=-\dfrac{0.16205992}{0.99516412} \end{cases}$$

$$\begin{cases} \bar{y}=0.6869 \\ \bar{x}=-0.1628 \end{cases}$$

$$r=0.0695$$