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M P Keshari » Rank Correlation

Rank Correlation

This item was filled under [ CBSE XII, ISC ]

ISC – Maths – XII – Important Q.A.

Rank Correlation.

Q.1. Calculate Spearman’s rank correlation coefficient between the advertisement cost and sales from the following date :

Advertisement cost Rs. (in thousand) 39 65 62 90 82 75 25 98 36 78
Sales Rs. (in lakhs) 47 53 58 86 62 68 60 91 51 84

Solution : –

Advertisement cost Rs. (in thousand) X Sales Rs. (in lakhs)            Y Rx Ry D = Rx – Ry D2
39 47 8 10 – 2 4
65 53 6 8 – 2 4
62 58 7 7 0 0
90 86 2 2 0 0
82 62 3 5 – 2 4
75 68 5 4 1 1
25 60 10 6 4 16
98 91 1 1 0 0
36 51 9 9 0 0
78 84 4 3 1 1
Total → 30

Spearman’s Rank Correlation Coefficient = r = 1 – [6 ∑ D2 ]/[N(N2 – 1)]

= 1 – [6 × 30]/[10(102 – 1)]

= 1 – [180/(10 × 99)]

= 1 – (2/11) = 9/11 = 0.818

= 0.82. [Ans.]

Q.2. Internal and External assessment were conducted on a group of 10 students who were studying in a postgraduate class in a college. The following marks were obtained in the assessment :

Roll no. of students: 1 2 3 4 5 6 7 8 9 10
Internal Assessment: 45 62 67 32 12 38 47 68 42 85
External Assessment: 39 48 65 32 20 35 45 77 30 62

Find the Spearman’s Rank Correlation Coefficient and comment on the result.

Solution : –

Roll No. Internal x External y Rank of x Rank of y d = x – y d2
1 45 39 6 6 0 0
2 62 48 4 4 0 0
3 67 65 3 2 1 1
4 32 32 9 8 1 1
5 12 20 10 10 0 0
6 38 35 8 7 1 1
7 47 45 5 5 0 0
8 68 77 2 1 1 1
9 42 30 7 9 –2 4
10 85 62 1 3 –2 4
Total → 12

Here, n = 10, r = 1 – [6∑d2]/[n(n2 – 1)]

= 1 – (6×12)/[10(99)]

= 1 – 0.07 = 0.93 . [Ans.]

Comment : There is a high positive correlation between the internal and external

assessment.

Q.3.  The mathematical aptitude score (MAS) of ten computer programmers with job performance rating (JPR) is given below. Calculate Spearman’s rating of rank correlation and state whether those who have aptitude for maths are likely to be better programmers.

Person A B C D E F G H I J
MAS 2 5 0 4 3 1 6 8 7 9
JPR 8 16 8 9 5 4 3 17 8 12

Solution : –

Person MAS JPR R1 R2 d = R1 – R2 d2
A 2 8 8 6 2 4
B 5 16 5 2 3 9
C 0 8 10 6 4 16
D 4 9 6 4 2 4
E 3 5 7 8 – 1 1
F 1 4 9 9 0 0
G 6 3 4 10 – 6 36
H 8 17 2 1 1 1
I 7 8 3 6 – 3 9
J 9 12 1 3 – 2 4
Total → 84

Now,      r = 1 – 6 [∑d2 + 1/12(32 – 3)]/n(n2 – 1)

= 1 – 6 (84 + 2)/[10(100) – 1]

= 1 – (6×86)/990

= (990 – 516)/990

= 474/990 = 0.478 = 0.48. [Ans.]

Q.4. The marks obtained by nine students in Physics and Mathematics are given below :

Physics 35 23 47 17 10 43 9 6 28
Mathematics 30 33 45 23 8 49 12 4 31

Calculate Spearman’s coefficient of rank correlation and interpret the result.

Solution : –

Physics Mathematics R1 R2 R1 – R2 = d d2
35 30 3 5 –2 4
23 33 5 3 2 4
47 45 1 2 –1 1
17 23 6 6 0 0
10 8 7 8 –1 1
43 49 2 1 1 1
9 12 8 7 1 1
6 4 9 9 0 0
28 31 4 4 0 0
Total → 12

Here, n = 9,

r = 1 – [6∑d2]/[n(n2 – 1)]

= 1 – [6×12]/[9(81 – 1)]

= 1 – 72/[9×80]

= 1 – 1/10 = 9/10 = +0.9. [Ans.]

Q.5. The following table gives the two kinds of assessment of ten post-graduate students performance :

Students Marks in Internal Assessment Marks in External Assessment
1 45 39
2 62 48
3 67 65
4 32 32
5 12 20
6 38 35
7 47 45
8 67 77
9 42 30
10 85 62

Find Spearman’s coefficient of rank correlation and interpret the result.

Solution :–

Students S.No. Marks in Internal Assessment x Marks in External Assessment y Rank of x Rank of y x – y = d d2
1 45 39 6 6 0 0
2 62 48 4 4 0 0
3 67 65 2.5 2 0.5 0.25
4 32 32 9 8 1 1
5 12 20 10 10 0 0
6 38 35 8 7 1 1
7 47 45 5 5 0 0
8 67 77 2.5 1 1.5 2.25
9 42 30 7 9 – 2 4
10 85 62 1 3 – 2 4
Total → 12.50

Here n = 10, r = 1 – [6∑d2]/[∑n(n2 – 1)]

= 1 – [6×12.50]/[10×99]

= 1 – 0.008 = 0.992 [Ans.]

Q.6. A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that students in a pair have nearly equal scores in an intelligent test. In each pair one student was taught by method A and the other by method B and examined after the  course. The marks obtained by them are tabulated below :

Pair 1 2 3 4 5 6 7 8 9 10 11
Method A(marks) 24 29 19 14 30 19 27 30 20 28 11
Method B(marks) 37 35 16 26 23 27 19 20 16 11 21

Find the rank correlation coefficient.

Solution : –

Pair Method A Method B Rank  x Rank  y d = x – y d2
1 24 37 6 1 5 25
2 29 35 3 2 1 1
3 19 16 8.5 9.5 –1 1
4 14 26 10 4 6 36
5 30 23 1.5 5 –3.5 12.2
6 19 27 8.5 3 5.5 30.2
7 27 19 5 8 –3 9.2
8 30 20 1.5 7 –5.5 30.2
9 20 16 7 9.5 –2.5 6.2
10 28 11 4 11 –7 49
11 11 21 11 6 5 25
Total → 225

Correlation Coefficient r = 1 – (6∑d2)/[n(n2 – 1)]

= 1 – (6×225)/(11×120)

= 1 – 45/44 = 1 – 1.0227 = – 0.0227. [Ans.]

Q.7. An examination of 8 applicants for a clerical post was taken by a firm. The marks obtained by the applicants in the Reasoning and Aptitude tests are given below :

Applicants A B C D E F G H
Reasoning Test 20 28 15 60 40 80 20 12
Aptitude Test 30 50 40 20 10 60 30 30

Calculate the Spearman’s coefficient of rank correlation from the data given above.

Solution : –

Reasoning Test Aptitude Test R1 R2 D = R1 – R2 D2
20 30 5.5 5 0.5 0.25
28 50 4 2 2 4
15 40 7 3 4 16
60 20 2 7 – 5 25
40 10 3 8 – 5 25
80 60 1 1 0 0
20 30 5.5 5 0.5 0.25
12 30 8 5 3 9
ΣD2 = 79.5

Using ,    r = 1 – 6[ΣD2 + 1/12(m3 – m) + 1/12(m3 – m)]/[n(n2 – 1)]

= 1 – 6[79.5 + 6/12 + 24/12]/(8 × 63)

= 1 – 6[82]/(8 × 63)

= 1 – 3/126 = 0.02. [Ans.]

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