Answer to Question #349510 in Statistics and Probability for Anu

Question

Given the following data: r12 = 0' 8 , r13 = 0.6 and r23= 0.4 then find

(i) r12.3

(ii) r13.2

(iii) r23.1

(iv) R1.23

Answer 1

The formula for calculating the partial correlation coefficient between x₁ and x₂ when controlling x₃:

$r_{12.3}=\frac{r_{12}-r_{13}\cdot r_{23}}{\sqrt{1-r_{13}^2}\cdot \sqrt{1-r_{23}^2}}$ .

So we have:

(i)

$r_{12.3}=\frac{r_{12}-r_{13}\cdot r_{23}}{\sqrt{1-r_{13}^2}\cdot \sqrt{1-r_{23}^2}}=\frac{0.8-0.6\cdot 0.4}{\sqrt{1-0.6^2}\cdot \sqrt{1-0.4^2}}=$

$=\frac{0.8-0.24}{\sqrt{1-0.36}\cdot \sqrt{1-0.16}} \approx \frac{0.56}{0.8\cdot 0.9165}\approx 0.76$

(ii)

$r_{13.2}=\frac{r_{13}-r_{12}\cdot r_{23}}{\sqrt{1-r_{12}^2}\cdot \sqrt{1-r_{23}^2}}=\frac{0.6-0.8\cdot 0.4}{\sqrt{1-0.8^2}\cdot \sqrt{1-0.4^2}}=$

$=\frac{0.6-0.32}{\sqrt{1-0.64}\cdot \sqrt{1-0.16}} \approx \frac{0.28}{0.6\cdot 0.9165}\approx 0.51$

(iii)

$r_{23.1}=\frac{r_{23}-r_{12}\cdot r_{13}}{\sqrt{1-r_{12}^2}\cdot \sqrt{1-r_{13}^2}}=\frac{0.4-0.8\cdot 0.6}{\sqrt{1-0.8^2}\cdot \sqrt{1-0.6^2}}=$

$=\frac{0.4-0.48}{\sqrt{1-0.64}\cdot \sqrt{1-0.36}} =-\frac{0.08}{0.6\cdot 0.8}\approx -0.17$

(iv)

The multiple correlation coefficient:

$R_{1.23}=\sqrt{\frac{r_{12}^2+r_{13}^2-2r_{12}r_{13}r_{23}}{1-r_{23}^2}}=\sqrt{\frac{0.8^2+0.6^2-2\cdot 0.8\cdot 0.6\cdot 0.4}{1-0.4^2}}=$

$=\sqrt{\frac{0.64+0.36-0.384}{1-0.16}}=\sqrt{\frac{0.616}{0.84}}\approx 0.86$

Answer: (i) 0.76 (ii) 0.51 (iii) -0.17 (iv) 0.86