Correlation & Regression - 2 Presentation - Unitedworld School of Business

7/28/2019 Correlation & Regression - 2 Presentation - Unitedworld School of Business

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Correlation-Regression


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It deals with association between two ormore variables

Correlation analysis deals with

covariation between two or more

variables

Types

1. Positive or negative

Simple or multipleLinear or non-linear


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Karl Pearsons Coefficient of Correlation

dx dy ( Gamma) = -------------------------

dx2 dy2

dx dy= -------------------------

N xydx = x-xbardy = y- ybar

dx dy = sum of products of deviations from respective

arithmetic means of both series


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After calculating assumed or working mean Ax & Ay dx dy ( dx) x( dy) ( Gamma) = --------------------------------

[ N dx2 - ( dx)2 x [ Ndy2 - ( dy)2 ]

dx dy = total of products of deviation from assumedmeans of x and y series

dx = total of deviations of x series dy = total of deviations of y series

dx2 = total of squared deviations of x series dy2 = total of squared deviations of y series

N= No. of items ( no. of paired items


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After calculating assumed or working mean Ax & Ay dx x dy

dx dy - ----------------N

( Gamma) = -------------------------( dx)2 ( dy)2

[ dx2 - --------- ] x [ dy2 - ------------]N N


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Assumptions of Karl Pearsons Coefficient of Correlation

1. Linear relationship exists between the variablesProperties of Karl Pearsons Coefficient of Correlation1.value lies between +1 & - 1

2.Zero means no correlation

3. ( Gamma) = bxy X byxWhere bxy X byx are regression coefficicent

Merit

Convenient for accurate interpretation as it gives degree &

direction of relationship between two variables


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Limitations

1. Assumes linear relationship , even though it

may not be

2. Method & process of calculation is difficult &

time consuming3. Affected by extreme values in distribution


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Probable Error of Karl Pearsons Coefficient of

Correlation

1- 2

Probable Error of ( Gamma) = 0.6745 -------- N


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Q7.Calculate coefficient of correlation for following data

X65 63 67 64 68 62 70 66 68 67 69 71

Y 68 66 68 65 69 66 68 65 71 67 68 70

Ans dx dy ( Gamma) = ------------------------- dx2 dy2

dx dy= -------------------

N xy


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1 2 3 4 5 6 7 8 9 10 11 12

Su

mX Xbar

X 65 63 67 64 68 62 70 66 68 67 69 71 800 66.67

Y 68 66 68 65 69 66 68 65 71 67 68 70 811 67.58

dx=x-xbar -1.67 -3.67 0.33 -2.67 1.33 -4.67 3.33 -0.67 1.33 0.33 2.33 4.33

dx2 2.78 13.44 0.11 7.11 1.78 21.78 11.11 0.44 1.78 0.11 5.44 18.78

84.

67

dx.dy -0.69 5.81 0.14 6.89 1.89 7.39 1.39 1.72 4.56 -0.19 0.97 10.47

40.

33

dy=y-ybar 0.42 -1.58 0.42 -2.58 1.42 -1.58 0.42 -2.58 3.42 -0.58 0.42 2.42

dy2 0.17 2.51 0.17 6.67 2.01 2.51 0.17 6.67 11.67 0.34 0.17 5.84

38.

92

dx dysum dx2*

sumdy2

3294.

9

dx2 dy2 57.40

coeff of

correlation = 0.70


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Q8. following information about age of husbands

& wives. Find correlation coefficient

Husband

23 27 28 29 30 31 33 35 36 39

Wife 18 22 23 24 25 26 28 29 30 32

( Gamma) =0.99


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1 2 3 4 5 6 7 8 9 10

Sum

X Xbar

X 23 27 28 29 30 31 33 35 36 39 311 31.10

Y 18 22 23 24 25 26 28 29 30 32 257 25.70

dx=x-

xbar -8.10 -4.10 -3.10 -2.10 -1.10 -0.10 1.90 3.90 4.90 7.90

dx2 65.61 16.81 9.61 4.41 1.21 0.01 3.61 15.21 24.01 62.41

202.

9

dx.dy 62.37 15.17 8.37 3.57 0.77 -0.03 4.37 12.87 21.07 49.77178.

3

dy=y-

ybar -7.70 -3.70 -2.70 -1.70 -0.70 0.30 2.30 3.30 4.30 6.30

dy2 59.29 13.69 7.29 2.89 0.49 0.09 5.29 10.89 18.49 39.69

158.

1

dx dy sum dx2* sumdy232078.4

9

dx2 dy2 179.10

coeff of correlation

= 1.00


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Rank Correlation : some times variable are not

quantitative in nature but can be arranged inserial order.

Specially while eading with attributes like

honesty , beauty , character , morality etcTo deal with such situations , Charles Edward

Spearman , in 1904 developed a formula for

obtaining correlation coefficient between ranks

of n individuals in two attributes under study , or

ranks given by two or three judges


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Rank coefficient of correlation

6 d2

(rho) = 1 - -------------------

N3-N

6 d2 (rho) = 1 - -------------------

N(N2-1)

d2 = total of squared differenceN = number of items


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Q9. ten competitors in a cooking competition are ranked

by three judges in the following way .by using rank

coorelation method find out which pair of judges havenearest approach

P Q R

1 1 3 6

2 6 5 43 5 8 9

4 10 4 8

5 3 7 1

6 2 10 27 4 2 3

8 9 1 10

9 7 6 5

10 8 9 7


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P Q R Rp-Rq dpq2 Rq-Rr dqr2 Rp-Rr dpr2

1 1 3 6 -2 4 -3 9 -5 25

2 6 5 4 1 1 1 1 2 4

3 5 8 9 -3 9 -1 1 -4 16

4 10 4 8 6 36 -4 16 2 4

5 3 7 1 -4 16 6 36 2 4

6 2 10 2 -8 64 8 64 0 0

7 4 2 3 2 4 -1 1 1 1

8 9 1 10 8 64 -9 81 -1 1

9 7 6 5 1 1 1 1 2 4

10 8 9 7 -1 1 2 4 1 1

1000 200 214 0 60

6Sigma d2 1200 1284 360

N3-N 990 6Sigma d2/N3-N 1.21 1.297 0.3636

(rho) -0.21 -0.297 0.636364


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Regression Analysis is the process of

developing a statistical model which is usedto predict the value of a dependant variable

by an independent variable

Application

Advertising v/s sales revenue

First used by Sir Francis Gatton in 1877 for

study of height of sons w.r.t height of fathers


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Regression Analysisgoing back or to revert to

the former condition or returnRefers to functional relationship between x & y

and estimates of value of depebdent variable y

for given values of independeny variable x

Relationship between income of employees and

savings

Regression coefficients can be used to calculate ,

correlation coeffecient. ( Gamma) = bxy Xbyx


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Types of Regression

1. Simple & Multiple Regression2. Total or Partial

3. Linear / Non-linear

Methods of Regression Analysis

1. Scatter Diagram

2. Regression Equations

3. Regression LinesRegression of x on y y= a + bx

Regression of y on x x= a + by


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Regression coefficients coefficient of regressionof x on y = coefficient of regression of x on y =

( x- x-) (y- y-) dx dy

bxy= ------------------= ------- (y- y-)2 dy2

coefficient of regression of y on x

( x- x-) (y- y-) dx dybyx= ------------------= ---------- (x- x-)2 dx2


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Q2.From the data given below find

two regression coefficientstwo regression equations

coefficient of correlation between marks in

Economics & statistics

most likely marks in statistics when marks in

Economics are 30

let marks in Economics be x and that in statistics

be yMarks in Eco 25 28 35 32 31 36 29 38 34 32

Marks in Stat 43 46 49 41 36 32 31 30 33 39


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Marks in

Eco

25 28 35 32 31 36 29 38 34 32 x 320 x- 32

Marks in

Stat

43 46 49 41 36 32 31 30 33 39 y 380 y- 38


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Marks in

Eco

25 2

8

35 3

2

3

1

3

6

2

9

3

8

3

4

3

2

x 320 x- 3

2

Marks in

Stat

43 4

6

49 4

1

3

6

3

2

3

1

3

0

3

3

3

9

y 380 y- 3

8

dx=x- x-

=x-32

-7 -4 3 0 -1 4 -3 6 2 0 dx 0 3

3

3

3

dy=y- y

-

=x-38 5 8 11 3 -2 -6 -7 -8 -5 1dy

0


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Marks in

Eco

25 28 35 32 31 36 29 38 34 32 x 320 x- 32

Marks in

Stat

43 46 49 41 36 32 31 30 33 39 y 380 y- 38

dx=x- x-=x-

32

-7 -4 3 0 -1 4 -3 6 2 0 dx 0 33 33

dy=y- y-=x-

38

5 8 11 3 -2 -6 -7 -8 -5 1 dy 0

dx2 49 16 9 0 1 16 9 36 4 0 dx2 140

dy2 25 64 121 9 4 36 49 64 25 1 dy2 398

dx dy -35 -

32

33 0 2 -

24

21 -

48

-

10

0 dxd

y

-93


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Regression coefficients coefficient of regressionof x on y = coefficient of regression of x on y =

( x- x-) (y- y-) dx dy -93

bxy= ------------------= ------- = ------ = -0.2337 (y- y-)2 dy2 398

coefficient of regression of y on x =

( x- x-) (y- y-) dx dy -93byx= ------------------= ---------- = --------= -0.6643

(x- x-)2 dx2 140


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regression of x on yx-x- = bxy (y-y-)

x-32 = -0.2337(y-38)

= - 0.2337 y +0.2337 *38= -0.2337y + 8.8806

x = -0.2337y +32 + 8.8806

x = -0.2337y +40.8806


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Correlation Coefficient = bxy *byx

= -0.2337 *-0.6643 = 0.1552 = -0.394

Since byx & bxy are both negative


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regression of y on x

y-y- = bxy (x-x-)

y-38 = -0.6643(x-32)y -38= -0.6643x+0.6643*32

y = -0.6643x+38+0.6643*32

y = -0.6643x+38+21.2576y = -0.6643x+59.2576


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In order to estimate most likely marks in statistics

(y) when Economics (x) are 30 , we shall use the

line regression of y x viz

The required estimate is given by

y = -0.6643* 30+59.2576= -19.929+59.2576 =

=39.3286


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Sum of Squares- x&y

(x )*(y)SSxy = ( x-x- ) ( y-y- )= = xy - --------------

nSum of Squares xx

(x )

SSxx = ( x-x- )2=x2 - -------------n

Sales &advt expenses in Rs 1000 Develop a regression model


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advt sales

92 930

94 900

97 1020

98 990

100 1100102 1050

104 1150

105 1120

105 1130

107 1200

107 1250

110 1220

Sales &advt expenses in Rs.1000. Develop a regression model


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SSxy

b = ------------SSxx

y=a+bx

y= a+b x y= n* a+b x

n* a = b x - y y - b x y b x

a = ----------- = ------- - -------

n n n

xi= yi= ed residual


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xi yi ed residual

advt sales x2 xy (yi-y-) (yi-y-)2 y^=fits yi-y^ (yi-y^)2 y^-y- (y^-y-)2

92 930 8464 85560 = 902.4 27.6

94 900 8836 84600 940.54 -40.54

97 1020 9409 98940 997.75 22.25

98 990 9604 97020

1016.8

2 -26.82

100 1100 10000 110000

1054.9

6 45.04

102 1050 10404 107100 1093.1 -43.1

104 1150 10816 119600

1131.2

4 18.76

105 1120 11025 117600

1150.3

1 -30.31

105 1130 11025 118650

1150.3

1 -20.31

107 1200 11449 128400

1188.4

5 11.55

107 1250 11449 133750

1188.4

5 61.55

110 1220 12100 134200

1245.6

6 -25.66

1221 13060 124581 1335420 0

13059.

99 0.01x y x2 xy (yi-yc)


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xi= yi= predicted residual

advt sales x2 xy (yi-y-) (yi-y-

)2

y^=fits yi-y^ (yi-y^)2 y^-y-

(y^-y-

)2

92 930 8464 85560 -158.33

25069.44

902.4 27.6

761.76 -

185.9334571.20

94 900 8836 84600 -188.33

35469.44

940.54 -40.54

1643.49 -

147.79

21842.87

97 1020 9409 98940 -68.33 4669.44 997.75 22.25 495.06 -90.58 8205.34

98 990 9604 97020 -98.33 9669.44 1016.82 -26.82 719.31 -71.51 5114.16

100 1100 10000 110000 11.67 136.11

1054.96 45.04 2028.60

-33.37

1113.78

102 1050 10404 107100 -38.33 1469.44 1093.1 -43.1 1857.61 4.77 22.72

104 1150 10816 119600 61.67 3802.78 1131.24 18.76 351.94 42.91 1840.98

105 1120 11025 117600 31.67 1002.78 1150.31 -30.31 918.70 61.98 3841.11

105 1130 11025 118650 41.67 1736.11 1150.31 -20.31 412.50 61.98 3841.11

107 1200 11449 128400 111.67 12469.44 1188.45 11.55 133.40 100.12 10023.35

107 1250 11449 133750 161.67 26136.11 1188.45 61.55 3788.40 100.12 10023.35

110 1220 12100 134200 131.67 17336.11 1245.66 -25.66 658.44 157.33 24751.68

1221 13060 124581 1335420 0.00 138966.667 13059.99 0.01 13769.21 -0.01 125191.6

x y x2 xy (yi-yc)


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1221x- = ------------- = 101.75

12

(x *y) 1221*13060

SSxy = xy - ------------= 1335420 - -------------- =6565n 12

(x )2 ( 1221)2

SSxx = x2 - -------------= 124581 - ------- = = 344.25

n 12


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SSxy 6565

b = ------------- = ----------------= 19.0704

SSxx 344.25y=a+bx

y= a+b x y= n* a+b xn* a = b x - y

y - b x y b x 13060 19.0704*1221a = ----------- = ------- - ------- = ---------- - --------------

n n n 12 12

= - 852.08


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equation for simple regression line

y= a+bxy= -852.08+ 19.0704 x

for regression of y on x


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For testing the Fit

yi = yi- value of yrecorded value in the given datay- = Mean ( Average )of y

y^ = Predicted Values from regression line

deviation = (yi- y-) = difference in actual value of y from

meanResiduals = (yi- y^)= gap ( error , difference ) between

actual value of y & predicted value calculated from

regression line

Deviation of predicted value from mean = (y^- y-

)a = intercept on y -axis

b= slope of regression line


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total sum of squares = SST = (yi-y-

)2

regression sum of squares = SSR = (y^- y-)2

Error sum of squares = SSE = (yi-y^)2

SSR

coefficient of determination = 2= -------SST


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SSEStandard Error of Estimate =Syx= ----------------

n-2

In order to to determine whether a significant linear relationship

exists between independent variable x and dependent variable y weperform whether population slope is zero

b - t= ----------

Sb

Syx

Sb = Standard error of b= -----------

SSxx


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H0:Slope of thr regression line is zero

H1-Slope of the regression line is not zero


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SSE

Syx= Standard Error of Estimate =--------

n-2 (yi-y^)2 13769.21= -------- = ------------ = 1376.92 = 37.1068

n-2 10-2

(x )2 (1221)2SSxx = x2 - -------- = 124581 - -------= 344.25

n 12

Syx


SSxx


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Syx


SSxxb- 19.07-0

t= ---------- = ------------------------------- = 9.53Sb 37.1068/( 344.25)

As calculated value of t is more than table

value of t for 12-2 = 10 degrees of freedomNull hypothesis is rejected


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Coefficient of Determination Definition

The Coefficient of Determination, also known as R Squared,

is interpreted as the goodness of fit of a regression.

The higher the coefficient of determination, the better the

variance that the dependent variable is explained by theindependent variable.

The coefficient of determination is the overall measure of

the usefulness of a regression.

For example,r2 is given at 0.95. This means that thevariation in the regression is 95% explained by the

independent variable. That is a good regression.


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The Coefficient of Determination can be

calculated as the Regression sum of squares,SSR, divided by the total sum of squares, SST

SSR

Coefficient of Determination 2 = ----------SST

Campus Overview


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Campus Overview

907/A Uvarshad,

Gandhinagar

Highway, Ahmedabad

382422.

Ahmedabad Kolkata

Infinity Benchmark,

10th Floor, Plot G1,Block EP & GP,

Sector V, Salt-Lake,

Kolkata 700091.

Mumbai

Goldline Business Centre

Linkway Estate,Next to Chincholi Fire

Brigade, Malad (West),

Mumbai 400 064.


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Correlation & Regression - 2 Presentation - Unitedworld School of Business

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