REGRESSION ANALYSIS LECTURE NOTES

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REGRESSION ANALYSIS LECTURE NOTES

 


•   Regression analysis refers to statistical technique for estimating the relationship among variables.

•   Regression analysis is concerned with estimating the value of one variable when the value of the other variable is known.

• Regression- is a measure of the average relationship between two or more variables in terms of the original units of data.

REGRESSION LINES

•         Regression lines refer to graphical devices that describe the average relationship between two variables.

•         There are two regression lines namely:

-Regression line of Y on X

-Regression line of X on Y.

REGRESSION EQUATIONS

•          Regression equation refers to algebraic expressions of regression lines.

•         Since there are two regression lines, there are two regression equations namely:

1. Regression equation of Y on X.

This equation is expressed as Y= a+bx.

               Where Y= is the dependent variable to be estimated.

                          X= is the independent variable

     a = is the interception of Y axis

               b = slope on the Y axis,

•         To determine the values of a and b the following normal equations are to be solved.

     ∑Y = Na + b∑x

       ∑XY= a∑X + b∑X²

                              Or

 NB -   There is an alternative formula to find the value of a and b

-          This formula is only applicable on the regression equation of y on x.

               a     = Σy - bΣx

                                n

                b = nΣxy - Î£xΣy

                        nΣx²  - (Σx)²       

                                                          

2. Regression equation of X on Y

•         This equation is expressed by x = a + by

•         The value of a. and b in the equation are obtained by solving the following normal equations simultaneously.

ASSUMPTIONS OF REGRESSION ANALYSIS🙋

1.       The variance of the error terms is constant across all value of independent variables. This is called     homoscedasticity.

2.      The values of items are normally distributed.

3.        There is no correlation between the independent variables in a linear regression equation.

4.      Residual errors have a mean value of Zero.

5.     There is a linear relationship between the independent variables and dependent variables.

ILLUSTRATION

Use the following values of X and Y to find the regression equation of Y on X and X on Y.

X         Y

1                    80

2                    96

3                    83

4                    94

5                    99

6                    92

 

SOLUTION

For the equation of Y on X

 

Y = a + bx

 

Normal equations

        ∑Y = Na + b∑x

        ∑XY= a∑X + b∑X²

 

X               Y            X²     Y²             XY

1                80         1        6400           80

2                96          4     9216           192

3                92         9      8464           276

4                83        16     6889           332

5                94        25     8836          470

6                99        36    9801            594

7                92        49    8464            644     

∑x=        ∑y=     ∑x²=   ∑y² =       ∑xy=

28           636     140      50,070    2588

 

Hence                         636 = 7a + 28b

            N= 7                2588 = 28a + 140b 


                Use elimination or substitution method to find a and b

                        a = 84.58

                        b = 1.57 therefore Y = 84.58 

Use the alternative method to find a and b i.e


               a =636 – b28   = 84.58

                                7  


  b = 7 * 2588 - 28*636    = 1.57

              7 * 140   - 28²

                        ii) Regression equation of x on y

                                    x = a +by

                                    Normal equations

                

                    ∑X = Na + b∑Y

                     ∑XY= a∑Y + b∑Y²

                                           Hence      28 = 7a + 636b

                                                             2588 = 636a +58070b

                                    Thus                a = -9.63

                                                            b = 0.15

                                Therefore         x = 0.15x -9.63

 

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