Question #349510

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


1
Expert's answer
2022-06-13T15:54:50-0400

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

r12.3=r12r13r231r1321r232r_{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)

r12.3=r12r13r231r1321r232=0.80.60.410.6210.42=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}}=


=0.80.2410.3610.160.560.80.91650.76=\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)

r13.2=r13r12r231r1221r232=0.60.80.410.8210.42=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}}=


=0.60.3210.6410.160.280.60.91650.51=\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)

r23.1=r23r12r131r1221r132=0.40.80.610.8210.62=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}}=


=0.40.4810.6410.36=0.080.60.80.17=\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:

R1.23=r122+r1322r12r13r231r232=0.82+0.6220.80.60.410.42=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}}=


=0.64+0.360.38410.16=0.6160.840.86=\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


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