spearman's cooeficient of rank correlation

8/7/2019 spearman's cooeficient of rank correlation

1/17

42510011 0010 1010 1101 0001 0100 1011

Spearmans Coefficient of Rank

Correlation, r s

By :

Dr.Wan Azlinda Binti Wan Mohamed


2/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Spearmans rank-order correlationcoefficient

The correlation coefficient is used when oneor more variables is measured on an ordinal(ranking) scale Describes the linear relationship betweentwo variables measured using ranked scores

Symbol used r s (The subscript s stands for Spearman; Charles Spearman invented thisone)


3/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Computational Formula for the

Spearman Rank-Order CorrelationCoefficient is:

R s = 1 6( D 2)-----------N (N 2 -1)

N is the number of pair ranksD is the difference between the two ranks in

each pair


4/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Running the Spearman Rank-Order Correlation Test

1. Determine the difference between theranks for each subjects

2. Square each difference and sum them3. Calculate the rho statistics.4. Compare the obtained rho value with the

critical value


5/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Summary of the Spearman Rank-Order Correlation Test

Hypotheses:H0 : Rho = 0Ha : Rho 0, or Rho < 0, or Rho > 0Assumptiojns:Subjects are randomly selectedObservations are ranked order Decision Rules:n = number of pairs of ranksIf rho obt rho crit , reject H 0If rho obt < rho crit , do not reject H 0Formula

rho = 1 6( D 2)

n (n 2 -1)


6/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 Sample dataParticipant Observer A: X Observer B: Y

1 4 3

2 1 23 9 8

4 8 6

5 3 5

6 5 4

7 6 7

8 2 1

9 7 9


7/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 SolutionParticipant Observer A: X Observer B: Y D D 21 4 3 1 1

2 1 2 -1 1

3 9 8 1 14 8 6 2 4

5 3 5 -2 4

6 5 4 1 1

7 6 7 -1 1

8 2 1 1 1

9 7 9 -2 4

D 2=18


8/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 SolutionRs = 1 6( D 2)

-----------

N (N2

-1)

= 1 (6(18))----------9 (9 2 -1)

= 1 - ((108)/720)= 1 0.15

= + .85


9/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

What does the value of r s tell you?

Spearmans rank correlation coefficient is actually derivedfrom the product-moment correlation coefficient , suchthat:

-1 r s 1 r s = 0.85 Means that a child receiving a particular rankingfrom one observer tended to receive very close to the sameranking from other observer

r s = +1 means the ranking is in complete agreement r s = 0 means that there is no correlation between therankings r s = -1 means that the ranking are in complete

disagreement. In fact they are in exact reverse order.


10/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Exercise:

The marks of eight candidates in E nglish andMathematics are:

Candidate 1 2 3 4 5 6 7 8E nglish (x) 50 58 35 86 76 43 40 60

Maths (y) 65 72 54 82 32 74 40 53

Rank the results and hence find Spearmans rank correlation coefficient between the two sets of marks.Comment on the value obtained,


11/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 SolutionE nglish(x)

50 58 35 86 76 43 40 60

Maths(y)

65 72 54 82 32 74 40 53

Rank x 4 5 1 8 7 3 2 6

Rank y 5 6 4 8 1 7 2 3

D -1 -1 -3 0 6 -4 0 3

D2 1 1 9 0 36 16 0 9 D 2=72


12/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 SolutionR s = 1 6( D 2)

-----------

N (N2

-1)

= 1 (6(72))----------8 (8 2 -1)

= 1 - ((432)/504)= 1 0.857

= .142

Spearmans coefficient of rank correlation is 0.142

This appears to show avery weak positivecorrelation between theE nglish and Mathematicsranking


13/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 Tied Ranks A tied rank occurs when two participants receive

the same rank on the same variable (e.g two

person are tied for first on variable x) Tied ranks result in an incorrect value of r s Resolve (correct) any tied ranks before computing

r s

Therefore, for each participant at a tied rank,assign the mean of the ranks that would have beenused had there not been a tie


14/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Example

Runner Race X Race Y To resolve ties New Y

A 4 1 Tie uses ranks 1and 2, becomes 1.5

1.5

B 3 1 Tie uses ranks 1and 2, becomes 1.5

1.5

C 2 2 Becomes 3rd 3

D 1 3 Becomes 4th 4


15/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Example

Runner Race X New Y D D 2

A 4 1.5 2.5 6.25

B 3 1.5 1.5 2.25

C 2 3 -1 1

D 1 4 -3 9

D2= 18.5


16/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11 SolutionR s = 1 6( D 2)

-----------

N (N2

-1)

= 1 (6(18.5))----------4 (4 2 -1)

= 1 - ((111)/60)= 1 1.85

= - .85


17/17

4251

0 0 11 0 0 1 0 1 0 1 0 11 0 1 0 0 0 1 0 1 0 0 1 0 11

Thank You

spearman's cooeficient of rank correlation

Documents