CORRELATION ANALYSIS NDIM

CORRELATION

ANALYSIS

Introduction



_{Correlation a LINEAR association between two}

random variables



Correlation analysis show us how to determine

both the nature and strength of relationship

between two variables



When variables are dependent on time

correlation is applied



A zero correlation indicates that there is no

relationship between the variables



A correlation of –1 indicates a perfect negative

correlation



A correlation of +1 indicates a perfect positive

correlation

Types of Correlation



There are three types of correlation

Types

Type1

Positive

Negative

No

Perfect



If two related variables are such that when

one increases (decreases), the other also

increases (decreases).



If two variables are such that when one

increases (decreases), the other decreases

(increases)



When plotted on a graph it tends to be a perfect

line



When plotted on a graph it is not a straight line

Type 2

Type 3

Simple Multiple

Partial

Two independent and one dependent variable

One dependent and more than one independent

variables

One dependent variable and more than one

independent variable but only one independent

variable is considered and other independent

variables are considered constant

Methods of Studying Correlation



Scatter Diagram Method



Karl Pearson Coefficient Correlation of

Method

180 160 140 120 100 80 60 40 20 0 0 5 0 100 150 2 0 0 2 5 0 Drug A (dose in mg) S ym pt om In de x 160 140 120 100 80 60 40 20 0 0 50 100 150 200 250 Drug B (dose in mg) S y m p to m In d ex

Very good fit Moderate fit

Correlation:

Linear

Relationship

s

Strong relationship = good linear fit

Points clustered closely around a line show a strong correlation. The line is a good predictor (good fit) with the data. The more spread out the points, the weaker the correlation, and the less good the fit. The line is a REGRESSSION line (Y = bX + a)

Coefficient of Correlation



A measure of the strength of the linear relationship

between two variables that is defined in terms of the

(sample) covariance of the variables divided by their

(sample) standard deviations



Represented by “r”



r lies between +1 to -1



-1 <

r

< +1



The + and – signs are used for positive linear

correlations and negative linear

n Y

2

(

Y

)

2

X

2

n

(

X

)

2

n XY

X Y

r

xy

Shared variability of X and Y variables on the

top

Individual variability of X and Y variables on the

bottom

Interpreting Correlation

Coefficient

r



strong correlation:

r > .70 or r < –.70



moderate correlation:

r is between .30 &

.70

or

r is between –.30

and –.70



weak correlation:

r is between 0 and

Spearmans rank coefficient



A method to determine correlation when the data

is not available in numerical form and as an

alternative the method, the method of rank

correlation is used. Thus when the values of the

two variables are converted to their ranks, and

there from the correlation is obtained, the

Computation of Rank

Correlation

Spearman’s rank correlation coefficient

ρ can be calculated when



Actual ranks given



Ranks are not given but grades are given but not

repeated



Ranks are not given and grades are given and

repeated

Algebraically method 1.Least Square Method-:

The regression equation of X on Y is : X= a+bX

Where,

X=Dependent variable and Y=Independent variable The regression equation of Y on X is:

Y = a+bX Where,

Y=Dependent variable X=Independent variable

Simple Linear Regression Independent variable (x) De p e n d e n t v a ri a b le (y )

The output of a regression is a function that predicts the dependent variable based upon values of the independent variables.

y = a + bX ± є

a (y intercept)

b = slope = ∆y/ ∆x є

Example1-: From the following data obtain the 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



X 24



Y  29



XY 168



X2 142



Y2 207



Y



na



b



X











2 X b X a XY

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 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 Y b Y a XY Y b na X

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

Substituting the values of a and b in the Regression equation X and Y

X=0.49+0.74Y

2.Deaviation from the Arithmetic mean method:

The calculation by the least squares method are quit cumbersome 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-

₍

_X

_X

₎

_b

₍

_Y

_Y

₎

xy







Regression Equation of Y on

X-)

(

)

(

Y



Y



b

_yx

X



X





xy

Where,

b

_xy yx

b

and _{= 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

X

X

x





y Y Y



X 24



Y  29



x 0



y 0



x2  26.8



y2 38.8



xy28.8 Solution-:

Regression Equation of X on Y is









49

.

0

74

.

0

8

.

5

74

.

0

8

.

4

8

.

5

8

.

38

8

.

28

8

.

4

2























Y

X

Y

X

Y

X

y

xy

b

_xy Regression Equation of Y on X is ) ( ) (Y Y b_yx X  X





) 8 . 4 ( 07 . 1 8 . 5 8 . 4 8 . 26 8 . 28 8 . 5 2       





X Y X Y x xy b_yx ………….(I)

)

(

)

(

X



X



b

_xy

Y



Y

It would be observed that these regression equations are same as those obtained by the direct method .

3.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-:

 

 



 







₂ 2 y x y x xy

d

d

N

d

d

d

d

N

b

The Regression Equation of Y on X-:

 







 







₂ 2 y x y x yx

d

d

N

d

d

d

d

N

b

)

(

)

(

X



X



b

_xy

Y



Y

)

(

)

(

Y



Y



b

_yx

X



X

But , here the values of and will be calculated by following formula:

b

xy yx

Example-: 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

(d_x)=X-7

Dev. From assu. Mean 6 (d_y)=Y-6 d_xd_y 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 x

d

2 y

d

The Regression Coefficient of X on Y-:

 







 







₂ 2 y y y x y x xy

d

d

N

d

d

d

d

N

b

74

.

0

194

144

1

195

11

155

)

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-:

)

8

.

5

(

74

.

0

)

8

.

4

(

)

(

)

(













Y

X

Y

Y

b

X

X

_xy

8

.

4

5

24











X

N

X

X

The Regression coefficient of Y on X-:











 

   ₂ 2 x x y x y x yx d d N d d d d N b

07

.

1

134

144

121

255

11

155

)

11

(

)

51

(

5

)

1

)(

11

(

)

31

(

5

2























yx yx yx yx

b

b

b

b

The Regression Equation of Y on X-:

)

(

)

(

Y



Y



b

_yx

X



X

66 . 0 07 . 1 ) 8 . 4 ( 07 . 1 ) 8 . 5 (      X Y X Y