kendall’s tau

8/3/2019 Kendalls Tau

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STATISTICAL METHODS AND ADVANCED

APPLICATIONS(SMS 3123)

TITLE : KENDALLS TAU

GROUP : 16

GROUPS MEMBER :36) NOR AMIN BIN ROZZAHLIM (D20091036056)

28) NOOR IDAYU BT BADARUDDIN (D20091036044)

49) NOOR AMALINA BT OSMAN (D20091036072)

LECTURERS NAME : MR. LIM KIAN BOON


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KENDALLS

TAU


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LEARNING OUTCOMETo test the null hypothesisthat x and y are independent

(which implies ) againstone of the followingalternative:

0or0,,0

0


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ASSUMPTIONS

1. The data consists of a random sample of n pairsof numeric or nonnumeric observations.

Each pair of observations represents two

measurement taken on the same unit ofassociation.

2. The data are measured on at least an ordinal

scale, so that we can rank each X observation inrelation to all other observed Xs and each Yobservation in relation to all other observed Ys.

)( , ii YX


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KENDALLS TAU TEST :There are two types of Kendalls Tau Test :

1. Small sample ( Ties and non-ties)2. Large sample approximation


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HYPOTHESES

CASE A ()Two sided

CASE B (>)One sided

CASE C (0

H0: X and Y are

independent

H1:


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Small Samples(Non-ties)


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TEST STATISTIC

Test statistic:

where :

S = P - Q

n=number of (X, Y) observations (or rank)

2)1(

^

nn

S


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To obtain P an Q , follow the steps:

1. Arrange the observations in a column

according to the magnitude of the Xs from smallestto largest X values. Then we say Xs are in naturalorder.

2. Compare each Y value, one at a time with each Yvalue appearing below it. In making these comparison,we say that a pair of Y values is in natural orderifthe Y below is larger than the Y above. We say thata pair of Y values is in reverse natural orderif the

Y below is smaller than the Y above.

3. P : Total number of pairs in natural order andQ : Total number of pairs in reverse natural order.

4. S = P Q

)( , ii YX


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Let say we are given data as below:

Rearrange the data to get the value of P and Qso that we get value of S :

S = 2

1 = 1

X Y

3 7

1 6

2 3

X Y Y pairs in naturalorder

Y pairs in reversenatural order

1 6 1 1

2 3 1 0

3 7 0 0

P = 2 Q = 1


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If all the Y pairs are in natural order, then

and we have

indicating perfect direct correlation between theranking of X and Y

2)1(0

2)1(

,0,2

)1(

nnnnS

QPSQnn

P

1

2

)1(2

)1(

nn

nn


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If all the Y pairs in reverse natural orders we have

and

indicating a perfect inverse correlation between the

and Y rankings.Thus, cannot be greater than +1 or smallerthan -1.

2

)1(

2

)1(0

,

2

)1(,0

nnnnS

QPSnn

QP

1

2

)1(2

)1(

nn

nn


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Small Samples(Ties)


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If many ties are present, we may compute byusing the following special formula:

Q-PS

rankgivenaattiedaren thatobservatioyofnumberrankgivenaattiedaren thatobservatioxofnumber

)1(2

1

)1(2

1

)1(2

1)1(

2

1

^

y

x

yyy

xxx

yx

tt

ttT

ttT

where

TnnTnn

S


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To obtain P and Q, follow the steps:

1. List the observation in ascending(natural) order

according to the magnitude of the Xs.2. Within the tied observations of the Xs, arrange

the Y values in ascending order of magnitude.

3. Count the number of Y pairs in natural order and

the number of Y pairs in reverse natural order asdescribe before, but do not compare a Y valueaccompanying a tied X value(say, ) with anyother Y value accompanying another X value thatis tied with

aX

aX


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We are given data as below:

Rearrange data to get the value of P and Q so that we

get S:

S = 0 2 = -2 ,

X Y

1 3

1 2

2 4

X Y Y pairs innatural order

Y pairs inreverse

natural order

1 2 0 1

1 3 0 1

2 1 0 0

P = 0 Q = 2

0t,2t yx


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DECISION Refer table A.22, page 579

If is unknown, assume =0.05CASE A

()CASE B

(>)CASE C

(


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LARGE SAMPLEAPPROXIMATION


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LARGE SAMPLE

APPROXIMATION

)52(2)1(3

^

nnnz

If n>40, use this formula to compute


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EXAMPLE-NON TIESCravens and Woodruff * conducted a study to design andtest a methodology for analytically determining standards ofsales performance. They reported the data on benchmarkachievement and management rating for 25 sales territoriesshown in the Table 9.7. They computed benchmarkachievement as being sales volume divided by benchmarksales, and based management ratings on salespersonmotivation and effort.We wish to compute for these data to see whether thereis sufficient evidence to conclude that benchmarkachievement and management rating are directly related.Although the data are reported as ranks, we follow the sameprocedure in computing as we would if the data were

reported in absolute quantities .

*Cravens, David W., and Robert B. Woodruff, An Approach for Determining CriteriaofSales Perfomance,J. Appl. Psychol., 57 (1973), 242-247.

0.005use


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Table 9.7

Territory Benchmarkachievement

(X)

Managementrating (Y)


(X)


1 2 4 14 11 10

2 9 2 15 1 1

3 7 20 16 21 14

4 23 17 17 14 15

5 5 5 18 3 11

6 17 7 19 13 13

7 16 6 20 18 19

8 25 24 21 22 25

9 4 3 22 19 16

10 10 21 23 24 23

11 20 18 24 6 22

12 15 9 25 12 12

13 8 8


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SOLUTION:

1) HYPOTHESES

H0: Benchmark achievement and

management rating are independent(=0

)

H1: Benchmark achievement and management

rating are directly related (>0) (claim)


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2)TEST STATISTICS:Firstly, arrange the data as in the Table 9.7 so that the Xranks are in natural order. Then, based on the definitions of

natural order and reverse natural order of Y, find thenumber of Y pairs in natural orders and reverse naturalorders. After we complete the arrangement, the formulashown below will be applied:

sampleofnumber

ordernaturalreverseinpairsY

ordernaturalinpairsY

n

Q

P

QPS

2

)1(

nn

S^


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X Y (X, Y) rankings

Y pairs innatural order


1 1 (1, 1) 24 0

2 4 (2, 4) 21 2

3 11 (3, 11) 14 8

4 3 (4, 3) 20 1

5 5 (5, 5) 19 1

6 22 (6, 22) 3 16

7 20 (7, 20) 4 14

8 8 (8, 8) 14 3

9 2 (9, 2) 16 010 21 (10, 21) 3 12

11 10 (11, 10) 11 3

12 12 (12, 12) 10 3

13 13 (13, 13) 9 3

Arrangement of data for computing (Table 9.7)


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X Y (X, Y) rankings

Y pairs innatural order


14 15 (14,15) 7 4

15 9 (15, 9) 8 2

16 6 (16,6) 9 0

17 7 (17, 7) 8 0

18 9 (18, 9) 3 4

19 6 (19, 6) 5 1

20 18 (20, 18) 3 2

21 14 (21, 14) 4 0

22 25 (22, 25) 0 323 17 (23, 17) 2 0

24 23 (24, 23) 1 0

25 24 (25, 24) 0 0

P = 218 Q = 82


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From the calculation :P=218, Q=82, n=25

S = P Q

= 218 82= 136

The test statistic,

453.0

2)24(25

136^


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3) DECISION

Test statistic,

From table A.22 with n=25 and =0.005critical value,

Since

So, we reject H0.

453.0^

367.0* 005.0

005.0

^

*


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4) CONCLUSION

There is enough evidence to support the claim thatthere is direct relationship between benchmarkachievement and management ranking in the

population.


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EXAMPLE-TIES

Krippner* reported the data shown in Table 9.9 on30 children(26 boys,4 girls) who attended asummer reading clinic sponsored by a universitychild-study center. The data were generated as

part of an investigation to determine which of theseveral variables appear to be related to readingimprovement manifested in a remedial program. Wewish to compute from these data and test thenull-hypothesis that there is no associationbetween IQ and reading improvement,let .*Krippner, Stanley, Correlates of Reading Improvement, J.Devel. Reading, 7 (1963)

29-39.

1.0


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Data on 30 subjects enrolled in a 5-week summer reading clinic

Table 9.9

Client Improvement(X) WISC IQ fullscale(Y)

Alvin 0.6 86

Barry 0.2 107

Chester 1.6 102Dick 0.5 104

Earl 0.9 104

Floyd 0.5 89

Gregg 0.8 109Harry 0.8 109

Ivan 0.8 101

Jacob 0.4 96

Karl 1.8 113


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Continue


Lewis 0.1 85

Marvin 0.9 100

Ned 0.2 94Oscar 1.6 104

Peter 1.6 104

Quincy 0.0 98

Ralph 1.6 115

Rita 0.2 109

Simon 0.3 94

Tony 0.0 112

Uriah 1.0 96


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Continue


Victor 1.3 113

Waldo 0.6 110

Walter 0.6 97Wanda 0.5 107

Xavier 1.7 113

York 1.6 109

Yvonne 2.2 98

Zohra 1.5 106


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1) HYPOTHESES:

0)(claim)IQ(andtimprovemenreadingbetweeniprelationshinverseordirectisThere:

tindependenareIQandtimprovemenReading

1

0 :

H

H

2) TEST STATISTIC:

Firstly we arrange the data based on the natural orderof X (ascending order).Find the Y pairs in natural order and Y pairs in reversenatural order based on their definition. After we get the

arrangement, we will get the value of P and Q to beapplied in formula shown below:


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Continue

rankgivenaattiedarethatnsobservatioYofno.

rankgivenaattiedarethatnsobservatioXofno.

)1(21),1(

21

)1(2

1)1(

2

1

y

x

yyyxxx

yx

t

t

ttTttT

TnnTnn

S


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Arrangement of data for computing in Table 9.9 :

Improvement(X) IQ(Y) Y pairs innatural order Y pairs inreversenatural order

0.0 98 19 8

0.0 112 4 24

0.1 85 27 00.2 94 21 2

0.2 107 8 15

0.2 109 5 16

0.3 94 21 2

0.4 96 19 2

0.5 89 18 1

0.5 104 9 7


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Continue

Improvement(X) IQ(Y) Y pairs innatural order Y pairs inreverse naturalorder

0.5 107 8 11

0.6 86 16 0

0.6 97 15 10.6 110 4 12

0.8 101 10 3

0.8 109 4 8

0.8 109 4 8

0.9 100 9 2

0.9 104 6 3

1.0 96 10 0

1.3 113 1 6


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Continue

Improvement(X) IQ(Y) Y pairs in naturalorder

Y pairs inreverse natural

order

1.5 106 4 4

1.6 102 2 1

1.6 104 2 1

1.6 104 2 1

1.6 109 2 1

1.6 115 0 3

1.7 113 0 11.8 113 0 1

2.2 98 0 0

P=250 Q=144


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Continue

2564.0

192

)29(3024

2

)29(30

106

192

)2(3)3(4)1(2)3(4)1(2)1(2)1(2T

242

)4(5)1(2)2(3)2(3)2(3)2(3)1(2T

106144250S

y

x


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3) DECISION

4) CONCLUSION

test).tail-ce(twosignificanoflevel0.10at the

HrejectcanweA.22,in tablegiven30nfor0.218*than

greateris0.256ofvaluecomputedourSince

00.05

IQ.and

timprovemenreadingbetweeniprelationshinverseordirect

isethat therclaimesupport thtoevidenceenoughisThere


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EXERCISE(NON-TIES)

Johnson* conducted a study to determine whether, incollegiate schools of nursing, relationships bertweencertain variables could be identified. Two variables ofinterest for which indixes were constucted wereextent of agreement (between the dean and thefaculty) on the responsibilities for decision making and

faculty satisfaction. The ranks on the two variablesof the 12 institutions that participated in the study areshown in Table 9.11. The author computed a value ofrs =-0.336 from the data, which she declared notsignificant. Compute from the data and testsignificance against the alternative that < 0.

*Johnson, Betty M.,Decision Making, Faculty Satisfaction, and The Place of theSchool of Nursing in the University,Nursing Res.,22(1973),100-107.


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Table 9.11

School Rank on faculty

satisfaction

Rank on

decision-makingagreement

A 1 12B 7 11C 6 10

D 2 9E 8 8F 4 7G 10 6H 12 5I 11 4J 5 3K 9 2L 3 1


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EXERCISE(TIES)

Pierce* points out that is most investigations oflightning discharges to earth, the estimated quantity ofelectricity passing from the cloud to the ground isaround 20 to 30 coulombs. However, Pierce cites thedata of Meese and Evans*, who reported much larger

values. Their data as reported by Pierce are shown intable 9.13, along with the distance of the observing siteof the discharge. Pierce computes a Pearson product-moment correlation coefficient of r=0.877 and a P valueof 0.01. Compute and the corresponding P value forHi: > 0.*Pierce, E.T.,TheCharge Transferred to Earth by a lightning Flash, J.FranklinInst.,286 (1968), 353-354

*Meese,A.D., and W.H.Evans.Charge Transfer in the Lightning Stroke asDetermined by the Magnetograph,J.Franklin Inst.,273(1963),375-382.

Table 9.13


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Distance, kilometres Charge, coulombs

6 23

6 46

6 46

6 47

6 94

7 80

9 13310 81

10 114

10 274

11 260

12 37815 197

15 234

18 1035

23 1065


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EXERCISE(LARGE SAMPLE)

Cravens and Woodruff * conducted a study to design andtest a methodology for analytically determining standards ofsales performance. They reported the data on benchmarkachievement and management rating for 41 sales territoriesshown in the table 9.14. They computed benchmarkachievement as being sales volume divided by benchmarksales, and based management ratings on salesperson

motivation and effort.We wish to compute for these data to see whether there issufficient evidence to conclude that benchmark achievementand management rating are directly related. Although thedata are reported as ranks, we follow the same procedure incomputing as we would if the data were reported in

absolute quantities.

*Cravens, David W., and Robert B. Woodruff, An Approach for DeterminingCriteria ofSales Perfomance,J. Appl. Psychol., 57 (1973), 242-247.

T bl 9 14


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Table 9.14


(X)



(X)


1 2 4 14 11 10

2 9 2 15 1 1

3 7 20 16 21 14

4 23 17 17 14 15

5 5 5 18 3 11

6 17 7 19 13 13

7 16 6 20 18 19

8 25 24 21 22 25

9 4 3 22 19 16

10 10 21 23 24 23

11 20 18 24 6 22

12 15 9 25 12 12

13 8 8 26 28 34

Continue


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Continue


(X)



(X)


27 30 41 35 32 30

28 26 38 36 39 33

29 29 36 37 37 3530 27 32 38 36 37

31 33 29 39 41 39

32 35 31 40 38 40

33 31 26 41 40 27

34 34 28


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TABLE

A.22


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kendall’s tau

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