When one variable of distribution is measured or predicted with the help of another variable of the distribution then that equation through which this result is acquired is called regression of equations. In this situation you’ll find:

  • Two variables of distribution involved in this equation.
  • There will be total of two regressions of equations, both of them predicted by the other. For example if it is X on Y then we’ll measure the value of X with the help of Y and if it is Y on X then we are going to measure Y with the help of X.
  • These two variables of distribution are called predictor variable and regressed variable.
  • In regression lines, any regression equation is found to be in straight line.

Now we come to regression line. First we’ll see X on Y:

Here X =c +dY,

Suppose the equations are moving forward using least square methods as (X1, Y1), (X1, Y2), (X3, Y3)…So to be able to calculate the values of both a andb, we are coming to:

X – X̅ = r (σx/ σy)(Y – Y̅)

X – X̅ = bxy(Y – Y̅)

Bxy = r(σx/ σy) = regression coefficient of x on y.

Second we’ll come to Y on X where Y = a + bX

Y – Y̅ = r σy/ σx(X – X̅)

Y – Y̅ = bxy(X – X̅)

Bxy = r(σy/ σx) = regression coefficient of y on x.

By following the method of regression equation, we are coming to few points:

  • The mean value of X and Y is something important for regression lines.
  • bxy.byx= r 2
  • r = ±√bxy.byx
  • Here r is represented with same importance when placing in regression coefficients.
  • Regression equations of two variables are different from each other and only remains identical with an exception of r = ± 1. We’ll find the angle with:

tanӨ= 1- r2/r . σxσy/ σ2x + σ2y

You’ll come across formulas of Regression coefficients in this case,

bxy=XY/ Y2, byx= XY/ X2

bxy=xy – (x.y/n)/ y2 – (y)2/n,       byx= xy – (x.y/n)/ x2 – (x)2/n

bxy = {dx.dy(∑dx.dy)/n}/ d2y – (dy)2/n

byx = {dx.dy(∑dx.dy)/n}/ d2x– (dx)2/n

Here n = number of observations.

If the measurement is based upon grouped data and N = f then we’ll consider:

bxy= {∑fdxdy– (∑fdx.(∑fdy )/N}/∑fd2y – {(∑fdy)2/N}

byx= {∑fdxdy– (∑fdx.(∑fdy )/N}/∑fd2x – {(∑fdx)2/N}

What is Standard Error of Estimation?

When studying for regression of equations, a standard error of estimation comes in front. This is the value of root mean square deviation that is obtained from the regression line of either X on Y or Y on X. In the first case we’ll use the equation,

Sx= σx1 – r2

In the second case we’ll use,

Sy= σy1 – r2

The necessities of this standard error are:

  • It presents a standard version of deviations of the predictions done on the values of either X from the equation of X on Y or Y from the equation of Y on X.
  • The size of the error is what predicted through this standard error.
  • The model used in this equation is also predicted with this standard error.

Next is Coefficient of Determination where one independent and one dependent variable are accounted and their percentage variation is what we’ll measure:

R2 = Explained variance / Total variance

For example if we suppose that X is the independent variable and Y is the dependent one and R2 = 0.85 then reduction is done 85% from the variation of Y on X.

If Y; = Go + al X; (i = 1, 2, …... n,) be the fitted values to the observation (xi , yi) i = 1, 2 … … n, then

R2 = {n ∑yi2 – (∑yi)2}/ n ∑ (yi)2 – (∑yi)2, 0 ≤ R2 ≤ 1

R 2 = 1 > all n observations lie on the fitted regression line.

Let us now measure regression lines and correlation coefficient from the data mentioned below:

Sales (x)1009878851109380
Purchase (y)85907072958174

Find out the value of y when x = 82.

Solution:

Here ∑x = 644,  ∑y = 567, n = 7

x̅ = ∑x/n = 92

y̅ = ∑y/n = 81

         xX = x – x̅      X2yY =  y – y̅         Y2         XY
100

98

78

85

110

95

80

8

6

-14

-7

18

3

-12

64

36

196

49

324

9

144

85

90

70

72

95

81

70

4

9

-11

-9

14

0

-11

16

81

121

81

196

0

121

32

54

154

63

252

0

132

822616687

Here regression coefficients are:

byx= ∑XY/ ∑X2 = 0.84

bxy= ∑XY/ ∑Y2 = 1.12

Regression equation of y on x:

Y-y’ = byx(x-x’)

y-81 = 0.84(x-92)

y=0.84x+ 3.72

Regression equation of x on y

x-x’ = bxy(y-y’)

x-92 = 1.12(y-81)

x=1.12y+ 1.28

The coefficient of correlation:

= √bxy.byx

= √(0.84). (1.12)    = 0.97

For x = 82, the value of y to be obtained from the regression equation of y on x. Hence

y = (0.84) (82) + 3.72 = 72.6.

 

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