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Chapter: Linear Regression· Mathematics & Statistics (Commerce) Part 2

The equations of the two regression lines are \(3x + 2y - 26 = 0\) and \(6x + y - 31 = 0\). Obtain the correlation coefficient between \(x\) and \(y\).Solution :To find correlation coefficient, we have to find the regression coefficients \(b_{yx}\) a…

Q.6 (C)(i) The equations of the two regression lines are \(3x + 2y - 26 = 0\) and \(6x + y - 31 = 0\). Obtain the correlation coefficient between \(x\) and \(y\).

Solution :

To find correlation coefficient, we have to find the regression coefficients \(b_{yx}\) and \(b_{xy}\).

Let \(3x + 2y = 26\) be equation of the line of regression of \(y\) on \(x\).

This gives \(y = \square\,x + 13\)

\(\therefore b_{yx} = -\dfrac{3}{2}\)

Now consider \(6x + y = 31\) as equation of the line of regression of \(x\) on \(y\).

This can be written as \(x = \square\,y + \dfrac{31}{6}\)

\(\therefore b_{xy} = -\dfrac{1}{6}\)

Now \(r^2 = \square = 0.25\)

As both \(b_{yx}\) and \(b_{xy}\) are negative,

\(\therefore r = \square\)

1 answer

Best answer

Answer: the boxes are \(-\dfrac{3}{2}\); \(-\dfrac{1}{6}\); \(b_{yx} \cdot b_{xy} = \left(-\dfrac{3}{2}\right)\left(-\dfrac{1}{6}\right)\); \(-0.5\).

\(3x + 2y = 26 \Rightarrow y = -\dfrac{3}{2}x + 13\), so \(b_{yx} = -\dfrac{3}{2}\).

\(6x + y = 31 \Rightarrow x = -\dfrac{1}{6}y + \dfrac{31}{6}\), so \(b_{xy} = -\dfrac{1}{6}\).

\(r^2 = b_{yx} \cdot b_{xy} = \left(-\dfrac{3}{2}\right)\left(-\dfrac{1}{6}\right) = \dfrac{1}{4} = 0.25\). (Since \(r^2 \le 1\), the assumed choice of lines is correct.)

Both regression coefficients are negative, so \(r = -\sqrt{0.25} = -0.5\).

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