Course pack unit 5

RAI UNIVERSITY, AHMEDABAD 1
Course: MCA
Subject: Computer Oriented Numerical
Statistical Methods
Unit-5
RAI UNIVERSITY, AHMEDABAD

Unit-V-Regression
Sr.
No.
Name of the Topic Page
No.
1 Introduction and Definition of Regression Analysis 2
2 Regression lines ,Properties and its explanation 3
3 Regression coefficients and its Properties 5
4 Difference between Regression and Correlation 6
5 Example based on the Regression line and Regression Co-
efficients
7
6 Example based on the fitting of regression line and estimation for
bivariate frequency distribution
13
7 Advantage and limitations of Regression Analysis 17
8 References 18
9 Exercise 19

Unit-V-Regression
1.1 Introduction:
If two variables are significantly correlated, and if there is some theoretical basis
for doing so, it is possible to predict (estimate) values of one variable from the
other. This observation leads to a very important conceptknown as ‘Regression
Analysis’.
For example, if we know that the advertising and sales are correlated we find out
expected amount of sales for a given advertising expenditure for attaining a given
amount of sales. Similarly if we know the yield of rice and rainfall are closely
related we may find out the amount of rain is required to achieve a certain
production figure.
In general Regression analysis means the estimation or prediction of the unknown
value of one variable from the known value of the other variable. It is one of the
most important statistical tools which is extensively used in almost all sciences –
Natural, Social and Physical.
1.2 Definition:
The dictionary meaning of ‘Regression’ is returning or going back. The term
‘Regression’ is first used by Sir Francis Galton (1822-1911) in 1877 while
studying the relationship between the height of father and sons. This term was
introduced by him in the paper of “Regression towards Mediocrity in healthcare
structure”.
Regression analysis was explained by M. M. Blair as follows:
“Regressionanalysis is a mathematical measure of the average relationship
betweentwo or more variables in terms of the original units of the data”.

2.1 RegressionLine:
Regressionline is the line which gives the bestestimate of one variable from the
value of any other given variable.
The regressionline gives the average relationship between the two variables in
mathematical form.
2.2 The Regressionwouldhave the following properties:
a) ∑(𝑦 – 𝑦𝑐) = 0 and b) ∑(𝑦 – 𝑦𝑐)
2
= Minimum
• For two variables X and Y, there are always two lines of regression –
(A)Regressionline of 𝒙 on 𝒚 :
It gives the bestestimate for the value of X for any specific given values of Y
𝒙 = 𝒂 + 𝒃 𝒚
where 𝑎 = 𝑥 - intercept
𝑏 = Slope of the line
𝑥 = Dependent variable
𝑦 = Independent variable
(B) Regressionline of 𝒀 on :

• It gives the best estimate for the value of 𝑌 for any specific given values of
𝑋
𝒚 = 𝒂 + 𝒃𝒙
Where 𝑎 = 𝑦 - intercept
𝑏 = Slope of the line
𝑦 = Dependent variable
𝑥= Independent variable
2.3 The Explanation of RegressionLine
• In case of perfect correlation (positive or negative ) the two line of
regression coincide.
• If the two R. line are far from each other then degree of correlation is less,
& vice versa.
• The mean values of 𝑥 & 𝑦 can be obtained as the point of intersection of the
two regression line.
• The higher degree of correlation between the variables, the angle between
the lines is smaller & vice versa.
2.4 RegressionEquation / Line & Method of LeastSquares
2.4.1 RegressionEquationof 𝒚 on 𝒙
𝒚 = 𝒂 + 𝒃𝒙
• In order to obtain the values of ‘𝑎’ & ‘𝑏’
∑𝑦 = 𝑛𝑎 + 𝑏∑𝑥
∑𝑥𝑦 = 𝑎∑𝑥 + 𝑏∑𝑥2
2.4.2 RegressionEquationof 𝒙 on
𝒙 = 𝒄 + 𝒅𝒚

• In order to obtain the values of ‘𝑐’ & ‘𝑑’
∑𝑥 = 𝑛𝑐 + 𝑑∑𝑦
∑𝑥𝑦 = 𝑐∑𝑦 + 𝑑∑𝑦2
3.1 RegressionCoefficients:
The regression coefficient between two variables is a numerical measure showing
the change in the value of one variable for a unit change in the value of the other
variable.
3.1.1 Formula for finding regressioncoefficient 𝒃 𝒚𝒙:
Regression Equation of y on x:
𝑦 – 𝑦̅ = 𝑏 𝑦𝑥 (𝑥 – 𝑥̅)
𝑏 𝑦𝑥 =
∑𝑋𝑌
∑𝑋2
𝑏 𝑦𝑥 = 𝑟 (
𝜎 𝑦
𝜎𝑥
)
Also by using the formula of 𝑟, 𝜎𝑥 𝑎𝑛𝑑 𝜎 𝑦 we get 𝑏 𝑦𝑥 =
𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥2−(∑ 𝑥)
2
3.1.2 Formula for finding regressioncoefficient 𝒃 𝒙𝒚:
Regression Equation of x on y:
𝑥 – 𝑥̅ = 𝑏𝑥𝑦 (𝑦 – 𝑦̅)
𝑏𝑥𝑦 =
∑𝑋𝑌
∑𝑌2
𝑏𝑥𝑦 = 𝑟 (
𝜎𝑥
𝜎 𝑦
)
Also by using the formula of 𝑟, 𝜎𝑥 𝑎𝑛𝑑 𝜎 𝑦 we get 𝑏𝑥𝑦 =
𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑦2−(∑ 𝑦)
2
3.2 Properties of Regressionco-efficients:

(1) The productof regression co-efficients is equal to the square of the
correlation co-efficient.
Since 𝑏 𝑦𝑥 = 𝑟 (
𝜎 𝑦
𝜎 𝑥
) and 𝑏𝑥𝑦 = 𝑟 (
𝜎 𝑥
𝜎 𝑦
)
𝑏 𝑦𝑥 × 𝑏𝑥𝑦 = 𝑟2
√ 𝑏 𝑦𝑥 × 𝑏𝑥𝑦 = 𝑟2
Thus regression coefficient is the geometric mean between two regression
coefficients.
(2) 𝑏 𝑦𝑥 , 𝑏𝑥𝑦 and 𝑟 have always the same sign.
𝑆𝑖𝑛𝑐𝑒 ∑𝑋2
𝑎𝑛𝑑 ∑𝑌2
is always positive ,the signs of 𝑏𝑥𝑦, 𝑏 𝑦𝑥 and 𝑟 depend
upon the sign of ∑𝑋𝑌. If ∑𝑋𝑌 is positive then 𝑏𝑥𝑦, 𝑏 𝑦𝑥 and 𝑟 are positive
and
If ∑𝑋𝑌is negative then 𝑏𝑥𝑦, 𝑏 𝑦𝑥 and 𝑟 are negative. Thus all the three 𝑏𝑥𝑦, 𝑏 𝑦𝑥
and 𝑟 have always the same sign.
(3) If two variables have perfect relationship one regression co-efficient is
reciprocal of the other.
For perfect relationship 𝑟 = ± 1
Now 𝑏 𝑦𝑥 × 𝑏𝑥𝑦 = 𝑟2
𝑏 𝑦𝑥 × 𝑏𝑥𝑦 = (± 1)2
= 1
∴ 𝑏 𝑦𝑥 =
1
𝑏 𝑥𝑦
(4)The productof regression co-efficients is 𝑟2
which can not exceed 1.Hence
if one regression co-efficient is greater than 1, the other regression co-
efficient is must be less than 1.
(5) The regression co-efficients are independent of change of origin but not of
scale.

4.1 Difference betweencorrelationand regression:
Correlation Regression
It gives a numerical measure of the
linear relationship between the
variables.
It gives functional relationship between
the variables, and this relationship helps
us in estimating the value of one
variable for a given value of another
variable.
Correlation co-efficient is always
between -1 and +1.
One regression co-efficient can be
greater than 1.
Correlation co-efficient is independent
of change of origin and scale.
Regression co-efficients are independent
of change of origin but not of scale.
Correlation co-efficient can be obtained
from regression co-efficients.
Regression co-efficient can not be
obtained from only correlation co-
efficient.
5.1 Example:-
From the following data obtain the two regressionequation and calculate the
regressionequationtaking deviation of items from mean of x and y series.
X 6 2 10 4 8
Y 9 11 5 8 7
Solution:-
Method-1
OBTAINING REGRESSION EQUATION
𝒙 𝒚 𝒙𝒚 𝐱 𝟐
𝐲 𝟐
6 9 54 36 81
2 11 22 4 121
10 5 50 100 25

4 8 32 16 64
8 7 56 64 49
∑𝒙 = 𝟑𝟎 ∑𝑦 = 40 ∑𝑥𝑦 = 214 ∑x2
= 220 ∑y2
= 340
Regression equation of y on x: 𝑦 = 𝑎 + 𝑏𝑥
∑𝑦 = 𝑛𝑎 + 𝑏∑𝑥
∑𝑥𝑦 = 𝑎∑𝑥 + 𝑏∑𝑥2
Substituting the values
40 = 5𝑎 + 30𝑏⋯(𝑖)
214 = 30𝑎 + 220𝑏⋯( 𝑖𝑖)
Multiplying equation (𝑖) by 6,
240 = 30𝑎 + 180𝑏⋯ ( 𝑖𝑖𝑖)
214 = 30𝑎 + 220𝑏⋯( 𝑖𝑣)
Subtracting equation ( 𝑖𝑣) from ( 𝑖𝑖𝑖) we get
−40𝑏 = 26 𝑜𝑟 𝑏 = −0.65
Substituting the value of b in equation( 𝑖)
40 = 5𝑎 + 30(−0.65) 𝑜𝑟 5𝑎 = 40 + 19.5 = 59.5 𝑜𝑟 𝑎 = 11.9
Putting the values of a and b in equation, the regression of 𝑦 on 𝑥 is
𝑦 = 11.9− 0.65𝑥
Regression equation of 𝑥 on 𝑦: 𝑥 = 𝑎 + 𝑏𝑦
∑𝑥 = 𝑛𝑎 + 𝑏∑𝑦
∑𝑥𝑦 = 𝑎∑𝑦 + 𝑏∑𝑦2
30 = 5𝑎 + 40𝑏⋯(𝑖)
214 = 40𝑎 + 340𝑏⋯(𝑖𝑖)
Multiplying equation ( 𝑖) by 8:

240 = 40𝑎 + 320𝑏⋯(𝑖𝑖𝑖)
214 = 40𝑎 + 340𝑏⋯(𝑖𝑣)
From equation ( 𝑖𝑖𝑖) and ( 𝑖𝑣)
−20𝑏 = 26 𝑜𝑟 𝑏 = −13
Substituting the value of b in equation ( 𝑖);
30 = 5𝑎 + 40(−1.3) 𝑜𝑟 5𝑎 = 30 + 52 = 82 𝑎 = 16.4
Putting the value of a and b in the equation, the regression line of 𝑥 on 𝑦 is
𝑥 = 16.4− 1.3𝑦
Now we find the Regression line by using the second method.
Method-2
Here, we use the formula of regression line which contain regression coefficients.
CALCULATION OF REGRESSION EQUATIONS
𝒙 𝒙 − 𝒙̅
= 𝑿
𝑿 𝟐 Y 𝒚 − 𝒚̅ = 𝒀 𝒀 𝟐 𝑿𝒀
6 0 0 9 +1 1 0
2 -4 16 11 +3 9 -12
10 +4 16 5 -3 9 -12
4 -2 4 8 0 0 0
8 +2 4 7 -1 1 -2
∑𝒙 = 𝟑𝟎 ∑𝑋 = 0 ∑𝑋2
= 40
∑𝑦 = 40 ∑𝑌 = 0 ∑𝑌2
= 20
∑𝑋𝑌
= −26
𝑥̅ =
30
5
= 6 ; 𝑦̅ =
40
5
= 8

The line of regression 𝑥 on 𝑦 is
( 𝑥 − 𝑥̅) = 𝑟
𝜎𝑥
𝜎 𝑦
( 𝑦 − 𝑦̅)
𝑟
𝜎𝑥
𝜎 𝑦
=
∑𝑋𝑌
∑𝑌2
=
−26
20
= −1.3
𝑥 − 6 = −1.3( 𝑦 − 8) = −1.3𝑦 + 10.4
𝑥 = −1.3𝑦 + 10.4+ 6
𝑥 = 16.4− 1.3𝑦
The line of regression 𝑦 on 𝑥 is
( 𝑦 − 𝑦̅) = 𝑟
𝜎 𝑦
𝜎𝑥
( 𝑥 − 𝑥̅)
𝑟
𝜎 𝑦
𝜎𝑥
=
∑𝑋𝑌
∑𝑋2
=
−26
40
= −0.65
𝑦 − 8 = −0.65( 𝑥 − 6) = −0.65𝑥 + 3.9
𝑦 = −0.65𝑥 + 3.9+ 8
𝑦 = 11.9− 0.65𝑥
Thus we find the same answer what obtained earlier. However, the calculations are
very much simplified without the use of the normal equation.
5.2 Example—
The following Information is obtained from result of an examination.
Marks in Mathematics
(x)
Marks in English (y)
Average 39.5 47.5
S.D. 10.8 16.8
Correlationco-efficientbetween 𝒙 and 𝒚 = 0.42

Obtain the two regressionlines and hence estimate 𝒚 for 𝒙 = 𝟓𝟎 and 𝒙 for 𝒚 =
𝟑𝟎.
Solution:
The equation of regression line of 𝑦 on 𝑥 is:
𝑦 = 𝑎 + 𝑏 𝑦𝑥 𝑥
Where 𝑏 𝑦𝑥 = 𝑟
𝜎 𝑦
𝜎 𝑥
= 0.42
16.8
10.8
= 0.653
And 𝑎 = 𝑦̅ − 𝑏 𝑦𝑥 𝑥̅
𝑎 = 47.5− 0.653(39.5)
𝑎 = 47.5− 25.79
𝑎 = 21.71
∴ 𝑦 = 21.71+ 0.653𝑥 is the regression line of 𝑦 on 𝑥.
The equation of regression line of 𝑥 on 𝑦 is
𝑥 = 𝑎 + 𝑏𝑥𝑦 𝑦
Where 𝑏𝑥𝑦 = 𝑟
𝜎 𝑥
𝜎 𝑦
= 0.42
10.8
16.8
= 0.27
And 𝑎 = 𝑥̅ − 𝑏𝑥𝑦 𝑦̅
𝑎 = 39.5− 0.27(47.5)
𝑎 = 39.5− 12.82
𝑎 = 26.68
∴ 𝑥 = 26.68+ 0.27𝑦 is regression line of 𝑥 on 𝑦.
When 𝑥 = 50, the estimated value of 𝑦 is
𝑦 = 21.71+ 0.653(50)
𝑦 = 21.71+ 32.65

𝑦 = 54.36
When 𝑦 = 30, the estimated value of 𝑥 is
𝑥 = 26.68+ 0.27(30)
𝑥 = 26.68+ 8.10
𝑥 = 34.78
5.3 Example—
The following information is obtained for two variables 𝒙 and 𝒚. Find
regressionequationof 𝒚 on 𝒙. 𝒏 = 𝟏𝟎; ∑ 𝒙 = 𝟏𝟑𝟎,∑ 𝒚 = 𝟐𝟐𝟎, ∑ 𝒙 𝟐
=
𝟐𝟐𝟖𝟖; ∑ 𝒙𝒚 = 𝟑𝟒𝟔𝟕.
Solution:
Supposethe regression line of 𝑦 on 𝑥 is 𝑦 = 𝑎 + 𝑏 𝑦𝑥 𝑥
Here 𝑥̅ =
∑ 𝑥
𝑛
=
130
10
= 13
𝑦̅ =
∑ 𝑦
𝑛
=
220
10
= 22
𝑏 𝑦𝑥 =
𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥2−(∑ 𝑥)
2
𝑏 𝑦𝑥 =
10(3467)−(130)(220)
10(2288)−(130)2
𝑏 𝑦𝑥 =
34670−28600
22880−16900
𝑏 𝑦𝑥 =
6070
5980
𝑏 𝑦𝑥 = 1.015
And 𝑎 = 𝑦̅ − 𝑏 𝑦𝑥 𝑥̅
𝑎 = 22 − 1.015(13)
𝑎 = 8.805

∴ 𝑦 = 8.805+ 1.015𝑥 Is the regression line of 𝑦 on 𝑥.
6. Fitting of regressionlines and estimation for bivariate frequency
Distribution:
6.1 Example:
Find two lines of regressionfrom the following bivariate table:
Age of Wife
Age
Of
Husband
Solution:
↓ → 𝒙
Y
10-20 20-30 30-40 40-50 50-60 𝒇 𝒚 M.V.
𝒚
𝒗 𝒗𝒇 𝒗 𝒗 𝟐
𝒇 𝒗 𝒇𝒖𝒗
15-25 (24)
6
(6)
3 9 20 -2 -18 36 30
25-35 (6)
3
(16)
16
(0)
10 29 30 -1 -29 29 22
10-20 20-30 30-40 40-50 50-60
15-25 6 3 - - -
25-35 3 16 10 - -
35-45 - 10 15 7 -
45-55 - - 7 10 4
55-65 - - 4 5

35-45 (0)
10
(0)
15
(0)
7 32 40 0 0 0 0
45-55 (0)
7
(10)
10
(8)
4 21 50 1 21 21 18
55-65 (8)
4
(20)
5 9 60 2 18 36 28
𝒇 𝒙 9 29 32 21 9 100 -8 122 98
M.V. 𝒙 15 25 35 45 55
𝒖 -2 -1 0 1 2
𝒖𝒇 𝒖 -18 -29 0 21 18 -8
𝒖 𝟐
𝒇 𝒖 36 29 0 21 36 122
𝒇𝒖𝒗 30 22 0 18 28 98
Here 𝑋̅ = 𝐴 +
∑ 𝑢𝑓𝑢
𝑛
× 𝐶𝑥
𝑋̅ = 35 +
−8
100
× 10
𝑋̅ = 34.2
𝑌̅ = 𝐵 +
∑ 𝑣𝑓𝑣
𝑛
× 𝐶𝑦
𝑌̅ = 40 +
−8
100
× 10
𝑌̅ = 39.2
Supposethe regression equation of 𝑦 on 𝑥 is
𝑦 = 𝑎 + 𝑏 𝑦𝑥 𝑥
Where, 𝑏 𝑦𝑥 =
𝑛 ∑ 𝑓𝑢𝑣−∑ 𝑢𝑓𝑢 ∑ 𝑣𝑓𝑣
𝑛 ∑ 𝑢2 𝑓𝑢−(∑ 𝑢𝑓𝑢)
2 ×
𝐶 𝑦
𝐶 𝑥

𝑏 𝑦𝑥 =
100×98−(−8)(−8)
100×122−(−8)2
×
10
10
𝑏 𝑦𝑥 =
9800−64
12200−64
× 1
𝑏 𝑦𝑥 =
9736
12136
𝑏 𝑦𝑥 = 0.802
And, 𝑎 = 𝑦̅ − 𝑏 𝑦𝑥 𝑥̅
𝑎 = 39.2− 0.802(34.2)
𝑎 = 39.2− 27.43
𝑎 = 11.77
∴ 𝑦 = 11.77+ 0.802𝑥 is the regression line of 𝑦 on 𝑥.
Now supposethe regression line of 𝑥 on 𝑦 is
𝑥 = 𝑎 + 𝑏𝑥𝑦 𝑦
Where 𝑏𝑥𝑦 =
𝑛 ∑ 𝑓𝑢𝑣−∑ 𝑢𝑓𝑢 ∑ 𝑣𝑓𝑣
𝑛 ∑ 𝑣2 𝑓𝑣−(∑ 𝑣𝑓𝑣)
2 ×
𝐶 𝑥
𝐶 𝑦
𝑏𝑥𝑦 =
9736
100×122−(−8)2
×
10
10
𝑏𝑥𝑦 = 0.802
And 𝑎 = 𝑥̅ − 𝑏𝑥𝑦 𝑦̅
𝑎 = 34.2− 0.802(39.2)
𝑎 = 34.2− 31.44
𝑎 = 2.76
∴ 𝑥 = 2.76 + 0.802𝑦 is the regression line of 𝑥 on 𝑦.
6.2 Example—

Calculate 𝒃 𝒚𝒙, 𝒃 𝒙𝒚& 𝑟 using the following:
Given Value Estimated Value
𝒙 = 𝟏𝟎 𝑦 = 22
𝒙 = 𝟐𝟎 𝑦 = 34
𝒚 = 𝟑𝟎 𝑥 = 17
𝒚 = 𝟓𝟎 𝑥 = 23
Solution:
Let the regression equation of 𝑦 on 𝑥 be 𝑦 = 𝑎 + 𝑏 𝑦𝑥 𝑦
∴ 22 = 𝑎 + 𝑏 𝑦𝑥 . 10
34 = 𝑎 + 𝑏 𝑦𝑥 . 20
_____________________
−12 = −10𝑏 𝑦𝑥
∴ 𝑏 𝑦𝑥 = 1.2
Let the regression equation of 𝑥 on 𝑦 be 𝑥 = 𝑎 + 𝑏𝑥𝑦 𝑦
∴ 17 = 𝑎 + 𝑏𝑥𝑦. 30
23 = 𝑎 + 𝑏𝑥𝑦. 50
_______________________
−6 = −𝑏𝑥𝑦. 20
𝑏𝑥𝑦 =
6
20
= 0.3
Now 𝑟 = √ 𝑏 𝑦𝑥. 𝑏𝑥𝑦
𝑟 = √(1.2)(0.3)
𝑟 = √0.36
𝑟 = 0.6

7.1Advantages of RegressionAnalysis:
1. The estimates of the unknown parameters obtained from linear least squares
regression are the optimal.
2. Estimates from a broad class of possibleparameter estimates under the usual
assumptions are used for process modeling.
3. It uses data very eﬃciently. Good results can be obtained with relatively
small data sets.
4. The theory associated with linear regression is well-understood and allows
for construction of diﬀerent types of easily-interpretable statistical intervals
for predictions, calibrations, and optimizations.
7.2 Limitations of RegressionAnalysis:
1. In making estimate from a regression equation, it is important to remember
that the assumption is being made that relationship has not changed since the
regression equation was computed. Another point worth remembering is that
the relationship shown by the scatter diagram may not be the same if the
equation is extended beyond the values used in computing the equation.
 For example there may be a close linear relationship between the yield of a
crop and the amount of fertilizer applied, with the yield increasing as the
amount of fertilizer is increased. It would not be logical, however, to extend
this equation beyond the limits of the experiment for it is quite likely that if
the amount of fertilizer were increased indefinitely, the yield would
eventually decline as too much fertilizer was applied.

Reference BookandWebsite Name:
1. Statistical Methods by S.P.Gupat
2. Business statistics by R.S.Bardwaj
3. Business statistics ( B.S. Shah Prakashan)
4. www.answers.com/Q/What_is_the_advantages_and_disadvantages_of_mult
iple_regression_analysis
5. http://www.biomedware.com/files/documentation/spacestat/Statistics/Multiv
ariate_Modeling/Regression/regression_line.png

EXERCISE
Q-1. Evaluate the following Questions:
1. Find the equations of regression lines from the following data and also
estimate 𝑦 for 𝑥 = 1 and 𝑥 for 𝑦 = 4.
𝒙: 3 2 -1 6 4 -2 5 7
𝒚: 5 13 12 -1 2 20 0 -3
2. Find regression co-efficients from the following data:
𝒙: 21 22 23 24 25 26 27 28 29 30
𝒚: 17 19 19 20 23 24 27 26 28 27
3. Obtain two regression lines from the following bivariate table:
Height Weight
90-100 100-110 110-120 120-130
50-55 4 7 5 2
55-60 6 10 7 4
60-65 6 12 10 7
65-70 3 8 6 3
4. The two regression lines are 𝑥 + 2𝑦 − 5 = 0 and 2𝑥 + 3𝑦 − 8 = 0and
𝜎𝑥
2
=12, find 𝑥̅ ,𝑦̅,𝜎 𝑦
2
𝑎𝑛𝑑 𝑟.

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