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17-2

CORRELATION ANALYSISAND REGRESSION

ANALYSIS

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Correlation

Correlation A measure of association between

two numerical variables.

Example (positive correlation)Typically, in the summer as the

temperature increases people arethirstier.

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Scatter Diagram

Scatter diagrams provide therelationship between two variables ina graphical form

The diagram summarizes the natureof relationship between two variables

Whether the relationship is positiveor negative

The diagram also explains themagnitude of the relationship

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Scatter Diagrams with varied rvalues

r2 = 1, r2 = 1,

r2 = .81, r2 = 0,

Y

X

Y

X

Y

X

Y

X

r = +1 r = -1

r = +0.

9 r = 0

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17-6

Specific Example

For sevenrandomsummer days, a

person recordedthe temperatureand their waterconsumption,

during a three-hour periodspent outside.

Temperature(F)

Water

Consumption

(ounces)

75 16

83 20

85 25

85 27

92 32

97 48

99 48

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How would you describe the graph?

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How strong is the linearrelationship?

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17-9

Correlation Analysis Correlation Analysis is statistical

technique used to measure themagnitude of linear relationshipbetween two variables

Correlation can be used along withregression analysis to determine thenature of the relationship betweenvariables

The prominent correlation coefficientsare

1.The Pearson product moment

correlation coefficient

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Measuring the Relationship

Pearsons SampleCorrelation

Coefficient, r

measures the direction and

the strength of the linearassociation between twonumerical paired variables.

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Direction of Association

Positive Correlation NegativeCorrelation

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Strength of LinearAssociation

rvalue Interpretation1 perfect positive linear

relationship

0 no linear relationship-1 perfect negative linear

relationship

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Strength of LinearAssociation

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Other Strengths ofAssociation

r value Interpretation0.9 strong association

0.5 moderate association

0.25 weak association

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Other Strengths ofAssociation

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17-16

Product Moment CorrelationThe product moment correlation,

r, summarizes the strength ofassociation between two metric(interval or ratio scaled) variables,say X and Y.

As it was originally proposed by Karl

Pearson, it is also known as thePearson correlation coefficient. It isalso referred to as simple correlation,bivariate correlation, or merely the

correlation coefficient.

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Product Moment Correlation

From a sample ofn observations,Xand Y,the product moment correlation, r, can becalculated as:

rvaries between -1.0 and +1.0.

( ) ( )

( ) ( )

1

2 2

1 1

n

i i

i

n n

i i

i i

X X Y Y

r

X X Y Y

=

= =

=

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Ad Spending and Corresponding Sales ofRoyal Products

Company Adver t is ingE xp(X )

S ale s(Y )

1 6 10

2 9 12

3 8 12

4 3 4

5 1 0 1 2

6 4 6

7 5 8

8 2 2

9 1 1 1 8

1 0 9 9

1 1 1 0 1 7

1 2 2 2

Ad Ex(in Crores)Sales(inThousands)

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Product Moment CorrelationThe correlation coefficient may be calculated as follows:

X= (10 + 12 + 12 + 4 + 12 + 6 + 8 + 2 + 18 + 9 + 17 + 2)/12= 9.333

Y= (6 + 9 + 8 + 3 + 10 + 4 + 5 + 2 + 11 + 9 + 10 + 2)/12= 6.583

= (10 -9.33)(6-6.58) + (12-9.33)(9-6.58)+ (12-9.33)(8-6.58) + (4-9.33)(3-6.58)+ (12-9.33)(10-6.58) + (6-9.33)(4-6.58)+ (8-9.33)(5-6.58) + (2-9.33) (2-6.58)+ (18-9.33)(11-6.58) + (9-9.33)(9-6.58)

+ (17-9.33)(10-6.58) + (2-9.33)(2-6.58)= -0.3886 + 6.4614 + 3.7914 + 19.0814+ 9.1314 + 8.5914 + 2.1014 + 33.5714+ 38.3214 - 0.7986 + 26.2314 + 33.5714= 179.6668

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Product Moment Correlation1

= (10-9.33)2 + (12-9.33)2 + (12-9.33)2 + (4-9.33)2

+ (12-9.33)

2

+ (6-9.33)

2

+ (8-9.33)

2

+ (2-9.33)

2

+ (18-9.33)2 + (9-9.33)2 + (17-9.33)2 + (2-9.33)2

= 0.4489 + 7.1289 + 7.1289 + 28.4089+ 7.1289+ 11.0889 + 1.7689 + 53.7289+ 75.1689 + 0.1089 + 58.8289 + 53.7289= 304.6668

= (6-6.58)2 + (9-6.58)2 + (8-6.58)2 + (3-6.58)2+ (10-6.58)2+ (4-6.58)2 + (5-6.58)2 + (2-6.58)2

+ (11-6.58)2 + (9-6.58)2 + (10-6.58)2 + (2-6.58)2

= 0.3364 + 5.8564 + 2.0164 + 12.8164

+ 11.6964 + 6.6564 + 2.4964 + 20.9764+ 19.5364 + 5.8564 + 11.6964 + 20.9764= 120.9168

Thus, r= 179.6668

(304.6668) (120.9168)= 0.9361

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Product Moment CorrelationThe correlation coefficient may be calculated as follows:

X= (10 + 12 + 12 + 4 + 12 + 6 + 8 + 2 + 18 + 9 + 17 + 2)/12= 9.333

Y= (6 + 9 + 8 + 3 + 10 + 4 + 5 + 2 + 11 + 9 + 10 + 2)/12= 6.583

= (10 -9.33)(6-6.58) + (12-9.33)(9-6.58)+ (12-9.33)(8-6.58) + (4-9.33)(3-6.58)+ (12-9.33)(10-6.58) + (6-9.33)(4-6.58)+ (8-9.33)(5-6.58) + (2-9.33) (2-6.58)+ (18-9.33)(11-6.58) + (9-9.33)(9-6.58)

+ (17-9.33)(10-6.58) + (2-9.33)(2-6.58)= -0.3886 + 6.4614 + 3.7914 + 19.0814+ 9.1314 + 8.5914 + 2.1014 + 33.5714+ 38.3214 - 0.7986 + 26.2314 + 33.5714= 179.6668

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Rank correlation

Researchers often face situationswhere they have to take decisionsbased on data measured on ordinal

scale scales in such casesSpearmans rank correlation isappropriate to relationship between

variables.It can be calculated using following

formula

rs = 1 (( 6 D 2 )/( N(N2 -1))

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h ki f l i i

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17 23

The ranking of televisionModels

Television Models Existing System New system

A 3 1

B 5 5

C 10 9

D 2 3

E 7 2F 6 4

G 4 6

H 1 7

I 8 10J 9 8

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Calculation of Rank correlationcoefficient

Television

Models

Existing

System(X)

New

system(Y)

D =(R1 - R2 ) D2

A 3 12 4

B 5 50 0

C 10 91 1D 2 3-1 1

E 7 25 25

F 6 42 4

G 4 6-2 4

H 1 7-6 36

I 8 10-2 4

J 9 81 1

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rs = 1 (( 6 D2 )/( N(N2 -1))

= 1-((6X80) /(10(100-1)))

= 1-(480/990)

= 1-0.48

= 0.52

This indicates that there is a positivecorrelation between two variables.This means the both the systems are

giving similar results

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6

Regression

Regression

Specific statistical methods for

finding the line of best fit for oneresponse (dependent) numericalvariable based on one or more

explanatory (independent)variables.

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Regression: 3 MainPurposes

To describe (or model)

To predict (or estimate)

To control (or administer)

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Regression AnalysisRegression analysis examines

associative relationships between ametric dependent variable and oneor more independent variables in thefollowing ways:

Determine whether the independentvariables explain a significantvariation in the dependent variable

Determine how much of thevariation in the dependent variablecan be explained by theindependent variables: strength ofthe relationship.

Predict the values of the dependent

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Example

Plan an outdoor party.

Estimate number of softdrinks to buy per person, basedon how hot the weather is.

Use Temperature/Water dataand regression.

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Real Life Applications

Estimating Seasonal Sales forDepartment Stores (Periodic)

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Real Life Applications

Predicting Student Grades Based onTime Spent Studying

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Practice Problems

Can the number of pointsscored in a basketball game bepredicted by

The time a player plays inthe game?

By the players height?

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Types of Regression Models

Positive Linear Relationship

Negative Linear Relationship

Relationship NOT Linear

No Relationship

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Least square method

The equation for regression line assumedby Least Squares method is

Y=a+bx+ei Where ei =Yi-i Where Y is the dependent variable X is the independent variable a is the Y-intercept

b is the slope of the line b=( (n (XY)-( X Y))/ ((n (X 2)-( X) 2) a=Y-bX

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Calculations for determining

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Calculations for determiningconstants a and b

Man Hours(X) Productivity in

units(Y)

XY X2

3.6 9.3 33.48 12.96

4.8 10.2 48.96 23.04

2.4 9.7 23.28 5.76

7.2 11.5 82.8 51.84

6.9 12 82.8 47.61

8.4 14.2 119.28 70.56

10.7 18.6 199.02 114.49

11.2 28.4 318.08 125.44

6.1 13.2 80.52 37.21

7.9 10.8 85.32 62.41

9.5 22.7 215.65 90.25

5.4 12.3 66.42 29.16

X=84.1 Y=172.9 XY=1355.61 X2

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b=1.768

a=2.01

Y=2.01+1.768X

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The Strength of Association R2

R2 = ( Explained Variance) / ( TotalVariance)

Total Variance = (ExplainedVariance)+

(UnexplainedVariance)

Explained Variance=(TotalVariance )

(Unexplainedi