written 5.2 years ago by | modified 4.0 years ago by |
Find i) $\bar {x}$, $\bar{y}$
ii) correlation coefficient r
--
written 5.2 years ago by | modified 4.0 years ago by |
Find i) $\bar {x}$, $\bar{y}$
ii) correlation coefficient r
--
written 5.2 years ago by | modified 5.2 years ago by |
i) To find $\bar {x}$, $\bar{y}$ 15x = 8y+130.
Given, 6y= 5x +90 -----------1
-5x + 6y =90 -------------A
and 15x = 8y+130. ------------2
15x - 8y= 130 ------------B
To solve both the equation A and B simultaneously, We get
x = 30 and y = 40
$\bar {x}$= x = 30 and $\bar{y}$ = y = 40
ii) To find correlation coefficeint 'r'
Let first equation represents the line of regression of $\bar {x}$, $\bar{y}$
by using equation 1, we get
$\phi$ X = $\frac{6}{5}$ y= 18
$\therefore$ we find fxy = $\frac{6}{5}$
the second equation represents the line of regression of x only
by using equation 2 we get
y= $\frac{15}{8}$$\times$x - $\frac{130}{8}$ $\therefore$ we find **fyx = $\frac{15}{8}$** r = $\sqrt{fxy\times fyx}$ = $\sqrt{\frac{6}{5}\times \frac{15}{8}}$ - $\sqrt{\frac{9}{4}}$ r= 1.5 \gt 1 -------- failed $\therefore$ the value r can be never grater than 1 numerically, hence our supposition is wrong Now, let the first equaion representing the line of regression of y on x $\therefore$ by using equation 1 we get, y= $\frac{5}{6}\times x + \frac{90}{6}$ fyx = $\frac{5}{6}$ the second equation representing the line of regression of x on y $\therefore$ by using equation 2 we get x= $\frac{8}{15}\times y + \frac{130}{15}$ fxy = $\frac{8}{15}$ r = $\sqrt{fxy\times fyx}$ = $\sqrt{\frac{5}{6}\times \frac{8}{15}}$ r= $\sqrt{\frac{40}{90}}$ = $\sqrt{\frac{4}{9}}$ = $\frac{2}{3}$
r= 0.667 which is less than 1