Lines of Regression of X on Y and Y on X Or Equation of Line of Regression

Calculate the Spearman’s rank correlation coefficient for the following data and interpret the result: 

If `bar"X"` = 40, `bar"Y"` = 6, σ_x = 10, σ_y = 1.5 and r = 0.9 for the two sets of data X and Y, then the regression line of X on Y will be:

For a bivariate data `barx = 10`, `bary = 12`, V(X) = 9, σ_y = 4 and r = 0.6
Estimate y when x = 5

Solution: Line of regression of Y on X is

`"Y" - bary = square ("X" - barx)`

∴ Y − 12 = `r.(σ_y)/(σ_x)("X" - 10)`

∴ Y − 12 = `0.6 xx 4/square ("X" - 10)`

∴ When x = 5

Y − 12 = `square(5 - 10)`

∴ Y − 12 = −4

∴ Y = `square`

Find graphical solution for following system of linear inequations :
3x + 2y ≤ 180; x+ 2y ≤ 120, x ≥ 0, y ≥ 0
Hence find co-ordinates of corner points of the common region.

If Σ_x1 = 56 Σ_y1 = 56, Σ`x_1^2` = 478,
Σ`y_1^2` = 476, Σ_x1y1 = 469 and n = 7, Find
(a) the regression equation of y on x.
(b) y, if x = 12.

For the given lines of regression, 3x – 2y = 5 and x – 4y = 7, find:
(a) regression coefficients b_yx and b_xy
(b) coefficient of correlation r (x, y)

The regression equation of x on y is 40x – 18y = 214  ......(i)

The regression equation of y on x is 8x – 10y + 66 = 0  ......(ii)

Solving equations (i) and (ii),

`barx = square`

`bary = square`

∴ b_yx = `square/square`

∴ b_xy = `square/square`

∴ r = `square`

Given variance of x = 9

∴ b_yx = `square/square`

∴ `sigma_y = square`

Information on v:ehicles [in thousands) passing through seven different highways during a day (X) and number of accidents reported (Y) is given as follows :   

`Sigmax_i` = 105, `Sigmay_i` = 409, n = 7, `Sigmax_i^2` = 1681, `Sigmay_i^2` = 39350 `Sigmax_iy_i` = 8075

  Obtain the linear regression of Y on X.