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A
PRESENTATION
ON
REGRESSION
ANALYSIS
Presented By:
Sonia gupta
MEANING OF REGRESSION:
The dictionary meaning of the word Regression is
‘Stepping back’ or ‘Going back’. Regression is the
measures of the average relationship between two or
more variables in terms of the original units of the
data. And it is also attempts to establish the nature of
the relationship between variables that is to study
the functional relationship between the variables and
thereby provide a mechanism for prediction, or
forecasting.
REGRESSION ANALYSIS:
The statistical technique of estimating the unknown value
of one variable (i.e., dependent variable) from the known
value of other variable (i.e., independent variable) is
called regression analysis.
How the typical value of the dependent variable changes
when any one of the independent variables is varied,
while the other independent variables are held fixed.
Examples:
The effect of a price increase upon demand, for
example, or the effect of changes in the money supply
upon the inflation rate.
Factors that are associated with variations in earnings
across individuals—occupation, age, experience,
educational attainment, motivation, and ability. For
the time being, let us restrict attention to a single
factor—call it education. Regression analysis with a
single explanatory variable is termed “simple
regression.”
Price = f (Qty.)
Sales = f(advt.)
Yield = f(Fertilizer)
No. of students = f(Infrastructure)
Earning = f(Education)
Weight = f(Height)
Production = f(Employment)
Dependent
Variables
Independent
Variables
Importance of Regression Analysis
Regression analysis helps in three important ways :-
1. It provides estimate of values of dependent variables from
values of independent variables.
2. It can be extended to 2or more variables, which is known as
multiple regression.
3. It shows the nature of relationship between two or more
variable.
USE IN ORGANIZATION
In the field of business regression is widely used.
Businessman are interested in predicting future
production, consumption, investment, prices, profits,
sales etc. So the success of a businessman depends on
the correctness of the various estimates that he is
required to make. It is also use in sociological study and
economic planning to find the projections of population,
birth rates. death rates etc.
Regression Lines
A regression line is a line that best describes the linear
relationship between the two variables. It is expressed by
means of an equation of the form:
y = a + bx
The Regression equation of X on Y is:
X = a + bY
The Regression equation of Y on X is:
Y = a + bX
Regression Lines And Coefficient of Correlation
Perfect Positive Correlation Perfect Negative Correlation
r = + 1 r = -1
High Degree of Positive Correlation
High Degree of Negative Correlation
Low Degree of Positive Correlation
Low Degree of Positive Correlation
No Correlation
Y on X
X on Y
r = 0
REGRESSION
Through
Regression
Coefficient
DEVIATION
METHOD FROM
AIRTHMETIC
MEAN
DEVIATION
METHOD FORM
ASSUMED
MEAN
Through Normal
Equation
Through Normal Equation:
Least Square Method
The regression equation of X on Y is :
X= a+bY
Where,
X=Dependent variable
Y=Independent variable
The regression equation of Y on X is:
Y = a+bX
Where,
Y=Dependent variable
X=Independent variable
And the values of a and b in the above equations are found by the method of
least of Squares-reference . The values of a and b are found with the help of
normal equations given below:
(I ) (II )
 
 


2
XbXaXY
XbnaY
 
 


2
YbYaXY
YbnaX
Example1-:From the following data obtain the two regression equations using
the method of Least Squares.
X 3 2 7 4 8
Y 6 1 8 5 9
Solution-:
X Y XY X2 Y2
3 6 18 9 36
2 1 2 4 1
7 8 56 49 64
4 5 20 16 25
8 9 72 64 81
  24X   29Y  168XY 1422
X 2072
Y
  XbnaY    2
XbXaXY
Substitution the values from the table we get
29=5a+24b…………………(i)
168=24a+142b
84=12a+71b………………..(ii)
Multiplying equation (i ) by 12 and (ii) by 5
348=60a+288b………………(iii)
420=60a+355b………………(iv)
By solving equation(iii)and (iv) we get
a=0.66 and b=1.07
By putting the value of a and b in the Regression equation Y on X we get
Y=0.66+1.07X
Now to find the regression equation of X on Y ,
The two normal equation are
 
 


2
YbYaXY
YbnaX
Substituting the values in the equations we get
24=5a+29b………………………(i)
168=29a+207b…………………..(ii)
Multiplying equation (i)by 29 and in (ii) by 5 we get
a=0.49 and b=0.74
Substituting the values of a and b in the Regression equation X and Y
X=0.49+0.74Y
Through Regression Coefficient:
Deviations from the Arithmetic mean method:
The calculation by the least squares method are quit difficult when the values of
X and Y are large. So the work can be simplified by using this method.
The formula for the calculation of Regression Equations by this method:
Regression Equation of X on Y-
)()( YYbXX xy 
Regression Equation of Y on X-
)()( XXbYY yx 

 2
y
xy
bxy

 2
x
xy
byx
and
Where, xyb
yxband = Regression Coefficient
Example2-: From the previous data obtain the regression equations by
Taking deviations from the actual means of X and Y series.
X 3 2 7 4 8
Y 6 1 8 5 9
X Y x2 y2 xy
3 6 -1.8 0.2 3.24 0.04 -0.36
2 1 -2.8 -4.8 7.84 23.04 13.44
7 8 2.2 2.2 4.84 4.84 4.84
4 5 -0.8 -0.8 0.64 0.64 0.64
8 9 3.2 3.2 10.24 10.24 10.24
XXx  YYy 
  24X   29Y 8.26
2
x 8.28xy8.382
y  0x 0 y
Solution-:
Regression Equation of X on Y is
 
 
YX
YX
YX
y
xy
bxy
74.049.0
8.574.08.4
8.5
8.38
8.28
8.4
2






Regression Equation of Y on X is
)()( XXbYY yx 
 
XY
XY
XY
x
xy
byx
07.166.0
)8.4(07.18.5
8.4
8.26
8.28
8.5
2






………….(I)
………….(II)
)()( YYbXX xy 
It would be observed that these regression equations are same as those
obtained by the direct method .
Deviation from Assumed mean method-:
When actual mean of X and Y variables are in fractions ,the calculations
can be simplified by taking the deviations from the assumed mean.
The Regression Equation of X on Y-:
  
  


 22
yy
yxyx
xy
ddN
ddddN
b
The Regression Equation of Y on X-:
  
  


 22
xx
yxyx
yx
ddN
ddddN
b
)()( YYbXX xy 
)()( XXbYY yx 
But , here the values of and will be calculated by following formula:xyb yxb
Example-3: From the data given in previous example calculate regression
equations by assuming 7 as the mean of X series and 6 as the mean of Y series.
X Y
Dev. From
assu.
Mean 7
(dx)=X-7
Dev. From
assu. Mean
6 (dy)=Y-6 dxdy
3 6 -4 16 0 0 0
2 1 -5 25 -5 25 +25
7 8 0 0 2 4 0
4 5 -3 9 -1 1 +3
8 9 1 1 3 9 +3
Solution-:
2
xd 2
yd
  24X   29Y
  11xd   1yd 512
xd  392
yd  31yxdd
The Regression Coefficient of X on Y-:
  
  


 22
yy
yxyx
xy
ddN
ddddN
b
74.0
194
144
1195
11155
)1()39(5
)1)(11()31(5
2








xy
xy
xy
xy
b
b
b
b
8.5
5
29

 Y
N
Y
Y
The Regression equation of X on Y-:
49.074.0
)8.5(74.0)8.4(
)()(



YX
YX
YYbXX xy
8.4
5
24

 X
N
X
X
The Regression coefficient of Y on X-:
  
  


 22
xx
yxyx
yx
ddN
ddddN
b
07.1
134
144
121255
11155
)11()51(5
)1)(11()31(5
2








yx
yx
yx
yx
b
b
b
b
The Regression Equation of Y on X-:
)()( XXbYY yx 
66.007.1
)8.4(07.1)8.5(


XY
XY
It would be observed the these regression equations are same as those
obtained by the least squares method and deviation from arithmetic mean .
Difference Between
Correlation and
Regression Analysis
Regression analysis.

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Regression analysis.

  • 1.
  • 3. MEANING OF REGRESSION: The dictionary meaning of the word Regression is ‘Stepping back’ or ‘Going back’. Regression is the measures of the average relationship between two or more variables in terms of the original units of the data. And it is also attempts to establish the nature of the relationship between variables that is to study the functional relationship between the variables and thereby provide a mechanism for prediction, or forecasting.
  • 4. REGRESSION ANALYSIS: The statistical technique of estimating the unknown value of one variable (i.e., dependent variable) from the known value of other variable (i.e., independent variable) is called regression analysis. How the typical value of the dependent variable changes when any one of the independent variables is varied, while the other independent variables are held fixed.
  • 5. Examples: The effect of a price increase upon demand, for example, or the effect of changes in the money supply upon the inflation rate. Factors that are associated with variations in earnings across individuals—occupation, age, experience, educational attainment, motivation, and ability. For the time being, let us restrict attention to a single factor—call it education. Regression analysis with a single explanatory variable is termed “simple regression.”
  • 6. Price = f (Qty.) Sales = f(advt.) Yield = f(Fertilizer) No. of students = f(Infrastructure) Earning = f(Education) Weight = f(Height) Production = f(Employment) Dependent Variables Independent Variables
  • 7. Importance of Regression Analysis Regression analysis helps in three important ways :- 1. It provides estimate of values of dependent variables from values of independent variables. 2. It can be extended to 2or more variables, which is known as multiple regression. 3. It shows the nature of relationship between two or more variable.
  • 8. USE IN ORGANIZATION In the field of business regression is widely used. Businessman are interested in predicting future production, consumption, investment, prices, profits, sales etc. So the success of a businessman depends on the correctness of the various estimates that he is required to make. It is also use in sociological study and economic planning to find the projections of population, birth rates. death rates etc.
  • 9. Regression Lines A regression line is a line that best describes the linear relationship between the two variables. It is expressed by means of an equation of the form: y = a + bx The Regression equation of X on Y is: X = a + bY The Regression equation of Y on X is: Y = a + bX
  • 10. Regression Lines And Coefficient of Correlation Perfect Positive Correlation Perfect Negative Correlation r = + 1 r = -1
  • 11. High Degree of Positive Correlation High Degree of Negative Correlation
  • 12. Low Degree of Positive Correlation Low Degree of Positive Correlation
  • 13. No Correlation Y on X X on Y r = 0
  • 15. Through Normal Equation: Least Square Method The regression equation of X on Y is : X= a+bY Where, X=Dependent variable Y=Independent variable The regression equation of Y on X is: Y = a+bX Where, Y=Dependent variable X=Independent variable And the values of a and b in the above equations are found by the method of least of Squares-reference . The values of a and b are found with the help of normal equations given below: (I ) (II )       2 XbXaXY XbnaY       2 YbYaXY YbnaX
  • 16. Example1-:From the following data obtain the two regression equations using the method of Least Squares. X 3 2 7 4 8 Y 6 1 8 5 9 Solution-: X Y XY X2 Y2 3 6 18 9 36 2 1 2 4 1 7 8 56 49 64 4 5 20 16 25 8 9 72 64 81   24X   29Y  168XY 1422 X 2072 Y
  • 17.   XbnaY    2 XbXaXY Substitution the values from the table we get 29=5a+24b…………………(i) 168=24a+142b 84=12a+71b………………..(ii) Multiplying equation (i ) by 12 and (ii) by 5 348=60a+288b………………(iii) 420=60a+355b………………(iv) By solving equation(iii)and (iv) we get a=0.66 and b=1.07 By putting the value of a and b in the Regression equation Y on X we get Y=0.66+1.07X
  • 18. Now to find the regression equation of X on Y , The two normal equation are       2 YbYaXY YbnaX Substituting the values in the equations we get 24=5a+29b………………………(i) 168=29a+207b…………………..(ii) Multiplying equation (i)by 29 and in (ii) by 5 we get a=0.49 and b=0.74 Substituting the values of a and b in the Regression equation X and Y X=0.49+0.74Y
  • 19. Through Regression Coefficient: Deviations from the Arithmetic mean method: The calculation by the least squares method are quit difficult when the values of X and Y are large. So the work can be simplified by using this method. The formula for the calculation of Regression Equations by this method: Regression Equation of X on Y- )()( YYbXX xy  Regression Equation of Y on X- )()( XXbYY yx    2 y xy bxy   2 x xy byx and Where, xyb yxband = Regression Coefficient
  • 20. Example2-: From the previous data obtain the regression equations by Taking deviations from the actual means of X and Y series. X 3 2 7 4 8 Y 6 1 8 5 9 X Y x2 y2 xy 3 6 -1.8 0.2 3.24 0.04 -0.36 2 1 -2.8 -4.8 7.84 23.04 13.44 7 8 2.2 2.2 4.84 4.84 4.84 4 5 -0.8 -0.8 0.64 0.64 0.64 8 9 3.2 3.2 10.24 10.24 10.24 XXx  YYy    24X   29Y 8.26 2 x 8.28xy8.382 y  0x 0 y Solution-:
  • 21. Regression Equation of X on Y is     YX YX YX y xy bxy 74.049.0 8.574.08.4 8.5 8.38 8.28 8.4 2       Regression Equation of Y on X is )()( XXbYY yx    XY XY XY x xy byx 07.166.0 )8.4(07.18.5 8.4 8.26 8.28 8.5 2       ………….(I) ………….(II) )()( YYbXX xy 
  • 22. It would be observed that these regression equations are same as those obtained by the direct method . Deviation from Assumed mean method-: When actual mean of X and Y variables are in fractions ,the calculations can be simplified by taking the deviations from the assumed mean. The Regression Equation of X on Y-:          22 yy yxyx xy ddN ddddN b The Regression Equation of Y on X-:          22 xx yxyx yx ddN ddddN b )()( YYbXX xy  )()( XXbYY yx  But , here the values of and will be calculated by following formula:xyb yxb
  • 23. Example-3: From the data given in previous example calculate regression equations by assuming 7 as the mean of X series and 6 as the mean of Y series. X Y Dev. From assu. Mean 7 (dx)=X-7 Dev. From assu. Mean 6 (dy)=Y-6 dxdy 3 6 -4 16 0 0 0 2 1 -5 25 -5 25 +25 7 8 0 0 2 4 0 4 5 -3 9 -1 1 +3 8 9 1 1 3 9 +3 Solution-: 2 xd 2 yd   24X   29Y   11xd   1yd 512 xd  392 yd  31yxdd
  • 24. The Regression Coefficient of X on Y-:          22 yy yxyx xy ddN ddddN b 74.0 194 144 1195 11155 )1()39(5 )1)(11()31(5 2         xy xy xy xy b b b b 8.5 5 29   Y N Y Y The Regression equation of X on Y-: 49.074.0 )8.5(74.0)8.4( )()(    YX YX YYbXX xy 8.4 5 24   X N X X
  • 25. The Regression coefficient of Y on X-:          22 xx yxyx yx ddN ddddN b 07.1 134 144 121255 11155 )11()51(5 )1)(11()31(5 2         yx yx yx yx b b b b The Regression Equation of Y on X-: )()( XXbYY yx  66.007.1 )8.4(07.1)8.5(   XY XY It would be observed the these regression equations are same as those obtained by the least squares method and deviation from arithmetic mean .