Presentation Chapter 6 MTH1022

7/28/2019 Presentation Chapter 6 MTH1022

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TOPIC 6CORRELATION & REGRESSION

Module Outcomes:

[MO1] Identify basic mathematical concepts, skills and mathematical

techniques for algebra, calculus and data handling.[MO2]Apply the mathematical calculations, formulas, statistical

methods and calculus techniques for problem solving in industry.

[MO3]Analyse calculus and statistical problems in industry.


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INTRODUCTION

Positive linear relationship:

An increase in one variable cause another variable to

increase

Example: Sugar pr ice and Food pr ice

Negative linear relationship:

An increase in one variable cause another variable to

decrease

Example: Energy price and companies profits


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THESCATTERDIAGRAM

Use to determine whether a relationship exist

between two variables.

The independent variable is normally labeled on the

horizontal axis and dependent variable on the

vertical axis.


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Rectangular coordinate

Two quantitative variables

One variable is called independent (X) and

the second is called dependent (Y)

Points are not joined

No frequency table

Scatter diagram

Y

* *

*

X


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SCATTERDIAGRAMOFWEIGHTANDSYSTOLICBLOOD

PRESSURE

80

100

120

140

160

180

200

220

60 70 80 90 100 110 120

wt (kg)

SBP(mmHg)

Wt.

(kg)

67 69 85 83 74 81 97 92 114 85

SBP

mHg)

120 125 140 160 130 180 150 140 200 130


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80

100

120

140

160

180

200

220

60 70 80 90 100 110 120Wt (kg)

SBP(mmHg)

Scatter diagram of weight and systolic blood pressure


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No relationship between weight and pulse rate


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THE SCATTER PLOTS

The pattern of data is indicative of the type of

relationship between your two variables:

positive relationship

If x increase, y will increase as well. negative relationship

If x increase, y will decrease and if x decrease, y will

increase.

no relationship

If the slope is zero, so the two variable are not

related.


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POSITIVERELATIONSHIP


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0

2

4

6

8

10

12

14

16

18

0 10 20 30 40 50 60 70 80 90

Age in Weeks

HeightinCM


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NEGATIVERELATIONSHIP

Reliability

Age of Car


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NORELATION


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Ex: Draw a scatter diagram for the following data and state the type of

relationship between the variables.

P 3 4 6 8 11 17 18 22q 10 20 3 7 4 1 12 5

0

5

10

15

20

25

0 5 10 15 20 25

q

q

Based on the scatter diagram, it is determined that there is no relationship between

the two variables.


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WHATISCORRELATION???

Correlation analysis is a statistical method used to

measure the STRENGTH of the relationship

between two variables which are independent

variable (X) and dependent variable (Y).


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Pearsons correlation coefficient is used to measure

the strength of the relationship between two variablesthat are quantitative in nature

This method is not suitable for qualitative data,

especially rank data


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SIMPLE CORRELATIONCOEFFICIENT () It is also called Pearson's correlation or product

moment correlation coefficient.

It measures the nature and strength between two

variables of the quantitative type.


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The sign ofrdenotes the nature of

association

while the value ofrdenotes the

strength of association.


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If the sign is +ve this means the relation isdirect (an increase in one variable is associated

with an increase in the other variable and adecrease in one variable is associated with adecrease in the other variable).

While if the sign is -ve this means an inverse orindirect relationship (which means an increasein one variable is associated with a decrease inthe other).


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The value of r ranges between ( -1) and ( +1)

The value of r denotes the strength of theassociation as illustrated

by the following diagram.

-110

-0.25-0.75 0.750.25

strong strongintermediate intermediateweak weak

no relation

perfect

correlation

perfect

correlation

Directindirect


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Ifr= Zero this means no association orcorrelation between the two variables.

If0 < r< 0.25 = weak correlation.

If0.25 r< 0.75 = intermediate correlation.

If0.75 r< 1 = strong correlation.

Ifr= l = perfect correlation.


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Example : The table below show the interest rates for car loans and the average

number of customers who apply for the loans in a month from a finance company.

Interest rate

in %, x 6.0 6.2 6.5 6.8 7.0

Number of

applicants, y80 80 78 75 70

x y xy6.0 80 36 6400 480

6.2 80 38.44 6400 496

6.5 78 42.25 6084 507

6.8 75 46.24 5625 510

7.0 70 49 4900 490

= 3 2 . 5 = 3 8 3 =211.93 =29409 =2483


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EXAMPLE:

Weight

(Kg)

Age

(years)

serial

No

1271

862

1283

1054

1165

1396

A sample of 6 children was selected, data about their

age in years and weight in kilograms was recorded as

shown in the following table . It is required to find the

correlation between age and weight.


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These 2 variables are of the quantitative type, one variable

(Age) is called the independent and denoted as (X)

variable and the other (weight) is called the dependent and

denoted as (Y) variables to find the relation between ageand weight compute the simple correlation coefficient

using the following formula:


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Y2X2xy

Weight

(Kg)

(y)

Age

(years)

(x)

Serial

n.

14449841271

643648862

14464961283

10025501054

12136661165

169811171396

y2=

742

x2=

291

xy=

461

y=

66

x=

41

Total


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r = 0.759

strong direct correlation


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EXAMPLE: RELATIONSHIPBETWEEN ANXIETYAND

TEST SCORES

Anxiety

(X)

Test score

(Y)X2 Y2 XY

10 2 100 4 20

8 3 64 9 242 9 4 81 18

1 7 1 49 7

5 6 25 36 306 5 36 25 30

X = 32 Y = 32 X2 = 230 Y2 = 204 XY=129


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CALCULATING CORRELATION COEFFICIENT

94.0

)200)(356(

1024774

32)204(632)230(6

)32)(32()129)(6(22

r

r = - 0.94

Indirect strong correlation


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=0.09

Pearsons product moment correlation coefficient is -0.09. since this value isclose to -1.0, we conclude that there is a very strong negative linear

relationship between variable x and y.


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=0.953

Pearsons product moment correlation coefficient is 0.953. since this value

is close to 1.0, we conclude that there is a very strong positive linear

relationship between variable x and y.


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SPEARMANS RANK CORRELATION COEFFICIENT

1. Spearmans rank correlation coefficient is a measure ofassociation between two variables that are at least of

ordinal scale.

2. Suitable for qualitative data and quantitative data but

quantitative data must transformed into qualitative data.

= 1 6

( 1)To compute Spearmans rank correlation:

1. Make a list of the n subject ( or observation)

2. Enter the rank of X and rank of Y variable.

3. Find = .4. Find .


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Example:

Six models of cars were rated for style and performance by customers at two

different locations. The table below shows the average ratings by the customers.

The scale is from 1 to 100.

Car model J K L M N P

Location 1 50 78 32 90 63 42

Location 2 40 49 33 55 80 33

Is there a relationship between the two ratings. Prove by calculating the

Spearmans rank correlation coefficient.


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Car

model

Location 1

x

Location 2

y

Rank of x

Rank of y

= - ()

J 50 40 3 3 0 0

K 78 49 5 4 1 1

L 32 33 1 1.5 -0.5 0.25

M 90 55 6 5 1 1

N 63 80 4 6 -2 4

P 42 33 2 1.5 0.5 0.25

=6.5

=1- (.)()

= 1- = 0.81 = 1

6 ( 1)

Spearmans rank correlation coefficient is 0. since this value is close to 1.0, we

conclude that there is a very strong positive linear relationship between variable

x and y.


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Example :

In a singing competition, two judges ranked the seven finalists as shown in the

table below. Compute spearmans rank correlation coefficient of the two sets of

rating. What can be said about the ratings of the two judges?

Finalists A B C D E F

Judge 1 5 6 3 9 4 8

Judge 2 3 4 1 8 5 10


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Finalists Judge 1

x

Judge 2

y

Rank of x

Rank of y

()

A 5 3 3 2 1 1

B 6 4 4 3 1 1C 3 1 1 1 0 0

D 9 8 6 5 1 1

E 4 5 2 4 -2 4

F 8 10 5 6 -1 1

G 10 11 7 7 0 0

=8

=1- ()()

= 1-

= 0.86 = 1 6

( 1)Spearmans rank correlation coefficient is 0.86. since this value is close to 1.0,

we conclude that there is a very strong positive linear relationship between

variable x and y.


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= +

The method to find the equation is by using Least Squares Method.

General equation:

= ( ) =

Where:

n is the number of data


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Example:

Find the least squares regression line of y on x for the following data.

X 3 6 9 11 16 18

y 2 8 11 14 19 21

Determine the values of y when (a) x = 5 and (b) x = 14


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x y xy3 2 9 4 6

6 8 36 64 48

9 11 81 121 9911 14 121 196 154

16 19 256 361 304

18 21 324 441 378

= 6 3 = 7 5 =827 =1187 = 9 8 9

The regression line is y = a + bx and

=

( )

= 6 989 (63)(75)6 827 (63) =

= 1.218

=

= 756 1.218 636 = - 0.289

Thus, the regression line is y = - 0.289+1.218x

When x = 5, y = - 0.289+(1.218 5 ) = 5.801When x = 14, y = - 0.289+(1.218 14 ) = 16.763


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Example : The table below gives the pairs of measurements for two variables x

and y.

X 5 9 12 15 18 20

y 3 5 8 9 10 12

Find the regression line of y on x using the least squares method. Estimates thevalue of y when x = 11.


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x y xy 5 3 15 25 9

9 5 45 81 25

12 8 96 144 6415 9 135 225 81

18 10 180 324 100

20 12 240 400 144

= 79 = 47 = 711 = 1199 = 423

= ( )

= ()()

()

=

= 0.58

=

= 47

60.58 79

6

= 7.63 7.637

= 0.196

Thus, the regression line is y = 0.196+0.58x

When x = 11, y = 0.196+ (0.58 11)=6.679


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Presentation Chapter 6 MTH1022