Question #167469

Ten competitors in a music competition are ranked by three adjudicators in the following order.

Competitor

  Competitor

A

B

C

D

E

F

G

H

I

J

Adjudicator 1

 

Adjudicator 2

 

Adjudicator 3

1

 

3

 

6

6

 

5

 

4

5

 

8

 

9

10

 

4

 

7

3

 

7

 

1

2

 

10

 

2

4

 

2

 

3

9

 

1

 

10

7

 

6

 

5

8

 

9

 

8

 Use Spearman’s rank order correlation technique  to compute:

i)       The pair of adjudicators who have the nearest approach to common tastes in music festival and therefore give the higher correlation between them.                         (3 marks)

ii)     The pair of adjudicators who have the worst taste of music and post the worst correlation.                                                                                                                                 (3 marks)

iii)   The average correlation amongst the three adjudicators.                              (2 marks)



Expert's answer


Let us calculate the spearman's coefficients-

r=1(6d2n3n)r=1-(\dfrac{6\sum d^2}{n^3-n})

d122=200d223=204d132=64\sum d^2_{12}=200\\\sum d^2{23}=204\\\sum d^2_{13}=64


Then the coefficient of adjudicator 1 w.r.t 2


Cofficient r12=1(6×200(10)310)r_{12}=1-(\dfrac{6\times 200}{(10)^3-10})


=11200990=1-\dfrac{1200}{990}


=11.21=0.21=1-1.21=-0.21


Then the coefficient of adjudicator 2 w.r.t 3


Cofficient r23=1(6×204(10)310)r_{23}=1-(\dfrac{6\times 204}{(10)^3-10})


=11224990=1-\dfrac{1224}{990}


=11.236=0.236=1-1.236=-0.236


Then the coefficient of adjudicator 1 w.r.t 3


Cofficient r13=1(6×64(10)310)r_{13}=1-(\dfrac{6\times 64}{(10)^3-10})


=1384990=1-\dfrac{384}{990}


=10.837=0.163=1-0.837=0.163


(i) As coefficient r13r_{13} has positive value. So adjudicators 1 and 3 have the nearest approach to common taste in music festival and have highest correlation between them. 


(ii) Adjudicators 2 has worst taste of music.

and coefficient value= -0.236


(iii) Average correlation coefficient =r12+r23+r133=\dfrac{r_{12}+r_{23}+r_{13}}{3}


=0.210.236+0.1633=0.2833=0.0943=\dfrac{-0.21-0.236+0.163}{3}=\dfrac{-0.283}{3}=-0.0943



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