Question #173480

8(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.


Expert's answer

Rewrite the regression equations uniformly


Y=.0399X+.6934Y = .0 399X + .6 934

X=.1212Y−.2461X = .1 212Y - .2 461


Denote mean values as x‾\overline{x} and y‾\overline{y}, correlation coefficient as rr, standard deviations as sxs_x and sys_y. Then the basic formulas are


Y=rsysxX−rsysxx‾+y‾Y=r\dfrac{s_y}{s_x}X-r\dfrac{s_y}{s_x}\overline{x}+\overline{y}


X=rsxsyY−rsxsyy‾+x‾X=r\dfrac{s_x}{s_y}Y-r\dfrac{s_x}{s_y}\overline{y}+\overline{x}


So find correlation coefficient from following


{rsysx=.0399rsxsy=.1212\left\{\begin{array}{rr}r\dfrac{s_y}{s_x}=.0 399\\ \\ r\dfrac{s_x}{s_y}= .1 212\end{array}\right.


Multiply the equations


rsysx⋅rsxsy=.0399⋅.1212r\dfrac{s_y}{s_x}\cdot r\dfrac{s_x}{s_y}=.0 399 \cdot .1 212 ⇒\Rightarrow r2=.00483588r^2=.00483588


Thus

r=.00483588=.069540492r=\sqrt{.00483588}=.069540492


Further, find mean values from


−rsysxx‾+y‾=.6934-r\dfrac{s_y}{s_x}\overline{x}+\overline{y}= .6 934


−rsxsyy‾+x‾=−.2461-r\dfrac{s_x}{s_y}\overline{y}+\overline{x}=- .2 461


or


−.0399x‾+y‾=.6934- .0 399\overline{x}+\overline{y}= .6 934

−.1212y‾+x‾=−.2461-.1 212\overline{y}+\overline{x}=- .2 461


Construct the system of linear equations and solve it using Cramer's rule


{−.0399 x‾+ y‾=.6934x‾−.1212 y‾=−.2461\left\{\begin{array}{rrr}- .0 399\ \overline{x}&+\ \overline{y}=& .6 934\\ \overline{x}&-.1 212\ \overline{y}=&- .2 461\end{array}\right.


det⁡(A)=∣−.039911−.1212∣=\det(A)= \begin{vmatrix} - .0 399 & 1 \\ 1 & -.1 212 \end{vmatrix}= −.0399⋅(−.1212)−1⋅1=-.0 399\cdot( -.1 212)-1\cdot1= −.99516412-.99516412


det⁡(Ax)=∣.69341−.2461−.1212∣=\det(A_x)= \begin{vmatrix} .6 934 & 1 \\ - .2 461 & -.1 212 \end{vmatrix}= .6934⋅(−.1212)−1⋅(−.2461)=.6934\cdot( -.1 212)-1\cdot(-.2461)= .16205992.16205992


det⁡(Ay)=∣−.0399.69341−.2461∣=\det(A_y)= \begin{vmatrix} - .0 399 & .6 934 \\ 1 & - .2 461 \end{vmatrix}= −.0399⋅(−.2461)−.6934⋅1=-.0 399\cdot( -.2461)-.6934\cdot1= −.68358061-.68358061


x‾=det⁡(Ax)det⁡(A)=.16205992−.99516412=−.162847431\overline{x}=\dfrac{\det(A_x)}{\det(A)}=\dfrac{.16205992}{-.99516412}=-.162847431


y‾=det⁡(Ay)det⁡(A)=−.68358061−.99516412=.686902388\overline{y}=\dfrac{\det(A_y)}{\det(A)}=\dfrac{-.68358061}{-.99516412}=.686902388


Answer


correlation coefficient

r=.0695r=.0695


mean values

x‾=−.1628\overline{x}=-.1628

y‾=.6869\overline{y}=.6869


Remark. Weight must be positive. Perhaps condition is incorrect.



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