In a partially destroyed laboratory record of an analysis of
regression data, the following data are legible:
Variance of X = 9
Regression equations:
8x − 10y + 66 = 0
and 40x − 18y = 214.
Find based on above information
- The
mean values of X and Y.
- Correlation
coefficient between X and Y.
- Standard
deviation of Y.
Given, ${\sigma _X}^2$ = 9$
∴ σX = 3
(i) The two regression equations are
8x - 10y + 66 = 0
i.e., 8x - 10y = - 66 ...(i)
and 40x - 18y = 214 ....(ii)
By 5 × (i) - (ii), we get
40x - 50y = - 330
40x - 18y = 214
(-) (+) (-)
- 32y = - 544
∴ y = $\frac{544}{32}\ = 17$
Substituting y = 17 in (i), we get
8x - 10 × 17 = - 66
∴ 8x - 170 = - 66
∴ 8x = - 66 + 170
∴ 8x = 104
∴ x = $\frac{104}{8}\ = 13$
Since the point of intersection of two regression lines is ($\overline x $,$\overline y $)
$\overline x $ = mean value of X = 13, and
$\overline y $ = mean value of X = 17.
(ii) Let 8x - 10y + 66 = 0 be the regression
equation of Y on X.
∴ The equation becomes 10Y = 8X + 66
i.e., Y = $\frac{8}{{10}}X$ + $\frac{{66}}{{10}}$
i.e., Y = $\frac{4}{{5}}X$ + $\frac{{33}}{{5}}$
Comparing it with Y = bYX X + a, we get
${b_{YX}}$ = $\frac{4}{5}$
Now, the other equation, i.e., 40x - 18y = 214 is the
regression equation of X on Y.
∴ The equation becomes X =$X{\rm{ = }}\frac{{18}}{{40}}Y + \frac{{214}}{{40}}$
i.e.,$X{\rm{ = }}\frac{9}{{20}}Y + \frac{{107}}{{20}}$
Comparing it with X = bXY Y + a', we get
$\begin{array}{l}{b_{xy}} = \frac{9}{{20}}\\r = \pm \sqrt {{b_{xy}} \times {b_{yx}}} \\\therefore r = \pm \sqrt {\frac{9}{{20}} \times \frac{4}{5}} = \pm \sqrt {\frac{9}{{25}}} = \pm \frac{3}{5} = \pm 0.6\end{array}$
Since bYX and bXY both are
positive,
r is also positive.
∴ r = 0.6
(iii) ${b_{YX}} = r\frac{{{\sigma _Y}}}{{{\sigma _X}}}$
$\therefore \frac{4}{5} = 0.6 \times \frac{{{\sigma _Y}}}{3}$
$\therefore {\sigma _Y} = 4$