Given two regression lines 5x+7y 22 and 6x+2y 20 find mean values of x and y

A) 3 and 4

B) 4 and 3

C) 6 and 8

D) 8 and 6

Description for Correct answer:
The regression lines are

\( \Large 5x-y=22\qquad ...(1)\)

\( \Large 64x-5y=24\qquad ...(2)\)

\( \Large (1) \times 45\Rightarrow 225x-45y=990\qquad ...(3)\)

\( \Large (2)\Rightarrow 64x-45y=24\qquad ...(4)\)

\( \Large (3)-(4)4\Rightarrow 161x=966\qquad ...(5)\)

\( \Large x=\frac{966}{161}=6\)

\( \Large \therefore (1)\Rightarrow 5(6)-y=22\)

\( \Large y=8\)

\( \Large \therefore\) The mean values of X and Y are 6 and 8.

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(i) Given equations of regression are

3x + 2y - 26 = 0

i.e., 3x + 2y = 26       ….(i)

and 6x + y - 31 = 0

i.e., 6x + y = 31       ….(ii)

By (i) - 2 × (ii), we get

3x + 2y = 26

12x + 2y = 62

-      -       -
- 9x = - 36

∴ x = `(-36)/-9 = 4`

Substituting x = 4 in (ii), we get

6 × 4 + y = 31

∴ 24 + y = 31

∴ y = 31 - 24

∴ y = 7

Since the point of intersection of two regression lines is `(bar x, bar y)`, 

`bar x` = mean of X = 4, and

`bar y` = mean of Y = 7

(ii) Let 3x + 2y 26 = 0 be the regression equation of Y on X.

∴ The equation becomes 2Y = 3X + 26

i.e., Y = `(-3)/2"X" + 26/2`

Comparing it with Y = bYX X + a, we get

`"b"_"YX" = (- 3)/2`

Now, the other equation 6x + y - 31 = 0 is the regression equation of X on Y.

∴ The equation becomes 6X = - Y + 31

i.e., X = `(-1)/6 "Y" + 31/6`

Comparing it with X = bXY Y+ a', we get

`"b"_"XY" = (-1)/6`

∴ r = `+-sqrt("b"_"XY" * "b"_"YX")`

`= +- sqrt((- 1)/6 xx (- 3)/2) = +- sqrt(1/4) = +- 1/2 = +- 0.5`

Since the values of bXY and bYX are negative,

r is also negative.

∴ r = - 0.5

(iii) The regression equation of Y on X is

Y = `(- 3)/2 "X" + 26/2`

For X = 2, we get

Y = `(- 3)/2 xx 2 + 26/2 = - 3 + 13 = 10`

(iv) Given, Var (Y) = 36, i.e., `sigma_"Y"^2` = 36

∴ `sigma_"Y" = 6`

Since `"b"_"XY" = "r" xx sigma_"X"/sigma_"Y"`

`(- 1)/6 = - 0.5 xx sigma_"X"/6`

∴ `sigma_"X" = (-6)/(- 6 xx 0.5) = 2`

∴ `sigma_"X"^2` = Var(X) = 4

The line of regression of y on x is given by: \(y - \;\bar y = {b_{yx}}\;\left( {x - \;\bar x} \right)\) where byx is called the regression coefficient of y on x.

Similarly, the line of regression of x on y is given by: \(x - \;\bar x = {b_{xy}}\;\left( {y - \bar y} \right)\) wherebxy is called the regression coefficient of x on y.

The correlation coefficient r2 = byx × bxy

The two lines of regression intersect each other at \(\left( {\bar x\;,\;\bar y} \right)\)

Calculation:

Given: Two regression lines are 6x + y = 30 and 3x + 2y = 25.

As we know that, the two lines of regression intersect each other at \(\left( {\bar x\;,\;\bar y} \right)\)

By solving these two equations: 6x + y = 30 and 3x + 2y = 25

We get \(\left( {\bar x\;,\;\bar y} \right) = \left( {\frac{{35}}{9},\frac{{20}}{3}} \right)\)

We can write 6x + y = 30 as line of regression of x on y: \(x - \frac{{35}}{9} = \; - \frac{1}{6} \times \left( {y - \frac{{20}}{3}} \right)\)   ------(1)

By comparing equation (1), with line of regression of x on y which is given by: \(x - \;\bar x = {b_{xy}}\;\left( {y - \bar y} \right)\) we get \({b_{xy}} = \; - \frac{1}{6}\) 

Similarly, we can write 3x + 2y = 25 as line of regression of y on x: \(y - \frac{{20}}{3} = \; - \frac{3}{2}\;\left( {y - \frac{{35}}{9}} \right)\)    ------(2)

By comparing equation (2), with line of regression of x on y which is given by \(\;y - \;\bar y = {b_{yx}}\;\left( {x - \;\bar x} \right)\): we get \({b_{yx}} = \; - \frac{3}{2}\)