The equations of the two lines of regression are $4x + 3y + 7 = 0\displaystyle and $3x + 4y + 8 = 0. Find the correlation coefficient between x and y?
Given the two regression equations:
1) $4x + 3y + 7 = 0$
2) $3x + 4y + 8 = 0$
We need to determine the regression lines of y on x and x on y.
First, let's assume equation (1) is the regression line of y on x:
4x+3y+7=0⟹3y=−4x−7⟹y=−34x−37
This gives the regression coefficient of y\displaystyle yy on x\displaystyle xx as:
byx=−43≈−1.33b_{yx} = -43\displaystyle \frac{4}{3}34≈−1\displaystyle \approx -1≈−1.33byx=−34≈−1.33
Now, let's assume equation (2) is the regression line of x\displaystyle xx on y\displaystyle yy:
3x+4y+8=0⟹3x=−4y−8⟹x=−43y−833x + 4y + 8 = 0 \implies 3x = -4y - 8 \implies x = -43\displaystyle \frac{4}{3}34y - 83\displaystyle \frac{8}{3}383x+4y+8=0⟹3x=−4y−8⟹x=−34y−38
This gives the regression coefficient of x\displaystyle xx on y\displaystyle yy as:
bxy=−43≈−1.33b_{xy} = -43\displaystyle \frac{4}{3}34≈−1\displaystyle \approx -1≈−1.33bxy=−34≈−1.33
Now, let's check the product of these regression coefficients, which is r2\displaystyle r^2r2:
r2=byx×bxy=(−43)×(−43)=169≈1.78>1r^2 = b_{yx} ×\displaystyle \times× b_{xy} = \left(-43\displaystyle \frac{4}{3}34\right) ×\displaystyle \times× \left(-43\displaystyle \frac{4}{3}34\right) = 169\displaystyle \frac{16}{9}916≈\displaystyle \approx≈ 1.78 > 1r2=byx×bxy=(−34)×(−34)=916≈1.78>1
Since the coefficient of determination r2\displaystyle r^2r2 must lie between 0\displaystyle 00 and 1\displaystyle 11 (0≤r2≤1\displaystyle 0 ≤\displaystyle \le≤ r^2 ≤\displaystyle \le≤ 10≤r2≤1), this assumption is invalid.
Therefore, we reverse the assumptions:
Assume equation (1) is the regression line of x\displaystyle xx on y\displaystyle yy:
4x+3y+7=0⟹4x=−3y−7⟹x=−34y−744x + 3y + 7 = 0 \implies 4x = -3y - 7 \implies x = -34\displaystyle \frac{3}{4}43y - 74\displaystyle \frac{7}{4}474x+3y+7=0⟹4x=−3y−7⟹x=−43y−47
Thus:
bxy=−34=−0.75b_{xy} = -34\displaystyle \frac{3}{4}43 = -0.75bxy=−43=−0.75
Assume equation (2) is the regression line of y\displaystyle yy on x\displaystyle xx:
3x+4y+8=0⟹4y=−3x−8⟹y=−34x−23x + 4y + 8 = 0 \implies 4y = -3x - 8 \implies y = -34\displaystyle \frac{3}{4}43x - 23x+4y+8=0⟹4y=−3x−8⟹y=−43x−2
Thus:
byx=−34=−0.75b_{yx} = -34\displaystyle \frac{3}{4}43 = -0.75byx=−43=−0.75
Now, let's calculate r2\displaystyle r^2r2:
r2=byx×bxy=(−0.75)×(−0.75)=0.5625r^2 = b_{yx} ×\displaystyle \times× b_{xy} = (-0.75) ×\displaystyle \times× (-0.75) = 0.5625r2=byx×bxy=(−0.75)×(−0.75)=0.5625
This is valid since $0.5625 ≤\displaystyle \le≤ 1$.
The correlation coefficient r\displaystyle rr is:
r=±r2=±0.5625=±0.75r = ±\displaystyle \pm±r2=±\displaystyle \sqrt{r^2} = \pmr2=±0.5625=±\displaystyle \sqrt{0.5625} = \pm0.5625=± 0.75r=±r2=±0.5625=±0.75
Since both byx\displaystyle b_{yx}byx and bxy\displaystyle b_{xy}bxy are negative, the correlation coefficient r\displaystyle rr must also be negative.
Therefore, r=−0.75\displaystyle r = -0.75r=−0.75.
Hence, **Option A** is the correct answer.
About This Chapter: Correlation and Regression
Paper
Paper 3: Quantitative Aptitude
Weightage
4-6 Marks
Key Topics
Correlation Coefficient, Regression Equations
This chapter covers Correlation Coefficient, Regression Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.