Non-parametric test.

2.  INTRODUCTION
 DEFINITION
 TEST STATISTICS
 KRC TABLE
 EXAMPLES
 PROPERTIES
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3. Sir Maurice George Kendall
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 Sir Maurice George Kendall, FBA (A british
Academy) (6 September 1907 – 29 March 1983) was
a British statistician, widely known for his
contribution to statistics. The Kendall tau rank
correlationis named after him

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Prof. Maurice George Kendall
Sir Maurice Kendall (1907-1983)
President of the IASC (International Accounting
standard committee) (1979-1981)
London, UK

5. IN STATISTICS, THE KENDALL RANK
CORRELATION COEFFICIENT,
COMMONLY REFERRED TO
AS KENDALL'S TAU
COEFFICIENT (AFTER THE GREEK
LETTER Τ), IS A STATISTIC USED TO
MEASURE THE ORDINAL
ASSOCIATION BETWEEN TWO
MEASURED QUANTITIES
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6. Hypothesis
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 H0: There is no association between two variables
H1: There is an association between two variables

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Test Statistics

8. Continued Test Statistics
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 Where nc is the number of concordant pairs and nd is
the number of discordant pairs
 And
𝑛 𝑛−1
2
no. of possible pairs
 i.e : a,b,c,d is an arrange data so the no. of
possible pairs are
4 4−1
2
=6
as shown below
(a,b),(a,c),(a,d),(b,c), (b,d), (c,d)
For these pairs we subtract 2nd value from 1st, if the
value become positive so it will be concordant and if
negative then the pair is said to be discordant

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Critical values of τ for
α= 0.05 and α =0 .01.
N 4 5 6 7 8 9 10
.05 1 .8000 .7333 .6190 .5714 .5000 .4667
.01 - 1 .8667 .8095 .7143 .6667 .6000
Decision rule: if 𝜏 cal is greater than 𝜏tab then we reject Ho,
OR)
If 𝜏cal > 𝜏tab (Reject Ho)

10. 𝑍 = 𝜏/𝛿𝜏
WHERE
𝜹𝝉 =
𝟐 𝟐𝒏+𝟓
𝟗𝒏 𝒏−𝟏
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For n>10 we use normal
approximation

11. SPEARMAN'S RANK CORRELATION IS
SATISFACTORY FOR TESTING A NULL
HYPOTHESIS OF INDEPENDENCE BETWEEN
TWO VARIABLES BUT IT IS DIFFICULT TO
INTERPRET WHEN THE NULL HYPOTHESIS
IS REJECTED. KENDALL'S RANK
CORRELATION IMPROVES UPON THIS BY
REFLECTING THE STRENGTH OF THE
DEPENDENCE BETWEEN THE VARIABLES
BEING COMPARED.
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12. Example
Assume that we are ranking the grades and IQ
levels for a group of students
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Students Grades IQ
a 1 1
b 2 4
e 5 2
c 3 3
d 4 5

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1. Testing of hypothesis:
H0: The IQ is independent of Grades
H1: The IQ is not independent of Grades
2. Level of significance α=0.05
3. Test statistics:
Where nc is the number of concordant pairs and nd is the number of
discordant pairs

14. Calculations
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 4.Calculations: Sort these in an order such that
one of the variables is in the proper order, let do
this for grades we will then examine the
variable IQ for the value of Tau ( 𝜏),
Students Grades IQ
a 1 1
b 2 4
c 3 3
d 4 5
e 5 2

15. Calculations Continued
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Pairs (1,4), (1,3), (1,5) ,(1,2), (4,3), (4,5), (4,2),
(3,5), (3,2), (5,2)
Where Nc=6 and Nd=4 and
𝒏 𝒏−𝟏
𝟐
=10
So 𝜏 =6-4/10
=0.2

16. Critical Region And Conclusion :
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Critical Region:
𝜏 Tab(5,0.05)=0.8000
Where 𝜏cal= 0.2
Conclusion : Since 𝜏cal is not greater than 𝜏tab value so
we donot reject our Ho and conclude that the IQ is
independent of grades

17. 1 . T H E D E N O M I N A T O R I S T H E T O T A L N U M B E R O F
P A I R C O M B I N A T I O N S , S O T H E C O E F F I C I E N T
M U S T B E I N T H E R A N G E − 1 ≤ Τ ≤ 1 .
2 . I F T H E A G R E E M E N T B E T W E E N T H E T W O
R A N K I N G S I S P E R F E C T ( I . E . , T H E T W O R A N K I N G S
A R E T H E S A M E ) T H E C O E F F I C I E N T H A S V A L U E 1 .
3 . I F T H E D I S A G R E E M E N T B E T W E E N T H E T W O
R A N K I N G S I S P E R F E C T ( I . E . , O N E R A N K I N G I S T H E
R E V E R S E O F T H E O T H E R ) T H E C O E F F I C I E N T H A S
V A L U E − 1 .
4 . I F X A N D Y A R E I N D E P E N D E N T , T H E N W E W O U L D
E X P E C T T H E C O E F F I C I E N T T O B E A P P R O X I M A T E L Y
Z E R O .
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