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Find coefficient of correlation and standard deviation

The lines of regression of bivariate population are 8x-10y+66=0 and 40x-18y = 214. The variance of x is 9.

Find,

a. coefficient of correlation r

b. the standard deviation y

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Given

$\begin{align} 8x-10y+66 &=0 \\ 8x-10y & =-66 - - - -1 \\ 40x-18y& = 214 - - - - 2 \\ \sigma x& = 9\\ \end{align}$

To find x&y solve equation 1 & 2, simultaneously, we get,

x=13

y=17

To find r: Suppose, the first equation represent the line of regression of x on y

$\begin{align} \therefore 8x &=10y-66\\ x&= \frac{10y}{8} - \frac{66}{8}\\ x&= \frac54 y - \frac{33}{4} \ \therefore we \ get \ b_{xy} = \frac54 \\ \end{align}$

Suppose the second equation represents the line of regression of y on x.

$\begin{align} \therefore 18y &=40x-214 \\ y&= \frac{40}{18} x - \frac{214}{18} \\ y&= \frac{20}{9} x - \frac{107}{9} \\ we \ get, \ b_{yx} = \frac{20}{9} \\ r&= \sqrt{b_{xy}.b_{yx} } = \sqrt{\frac54 \times \frac{20}{9}}\\ r&=1.667\\ \end{align}$

but the value of r can never be greater than 1. Hence our supposition is wrong.

Now, treating the first equation as representing, the line of regression on of y on x, we write it as

$\begin{align} 10y &= 8x+66 \\ y&= \frac{8}{10} x+ \frac{66}{10} \\ y&= frac45 x +6.6 \\ b_{yx} & = \frac45 \\ \end{align}$

Treating the second equation as representing the line of regression of x on y

$\begin{align} 40x&=18y+214 \\ x&=\frac{18}{40} y+ \frac{214}{40} \\ x&= \frac{9}{20} y + \frac{107}{20} \\ b_{xy} & = \frac{9}{20} \\ r&=\sqrt{b_{yx} .b_{xy}}\\ r&=\sqrt{\frac45 \times \frac{9}{20}} \\ r&= 0.6 \\ \end{align}$

To find $\sigma y$,

Consdier, $b_{yx} = r \frac{\sigma y}{\sigma x} $

but $\sigma x = 9, \to given$

$\begin{align} \sigma y & = \frac{b_{yx} . \sigma x}{r} \\ \sigma y & = \frac{(\frac45)( 9)}{0.6} \\ \sigma y & = \frac{7.2}{0.6} \\ \sigma y & = 12 \\ \end{align}$

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