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If the rank correlation coefficient between marks in management and mathematics for a group of student in 0.6 and the sum of squares of the differences in ranks in 66, what is the number of students in the group?
- a)10
- b)9
- c)8
- d)11
Correct answer is option 'A'. Can you explain this answer?
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If the rank correlation coefficient between marks in management and ma...
Given:
- Rank correlation coefficient = 0.6
- Sum of squares of differences in ranks = 66
To find:
- Number of students in the group
Solution:
Let there be 'n' students in the group.
Finding the value of n:
- The sum of the first 'n' natural numbers is n(n+1)/2.
- Therefore, the sum of the ranks of the students in management and mathematics combined is n(n+1).
- Since each student has a rank in both subjects, the sum of their ranks should be counted twice, so we need to divide by 2.
- Therefore, the sum of the ranks in each subject is n(n+1)/2.
Finding the value of the rank correlation coefficient:
- The formula for rank correlation coefficient is given as:
r = 1 - (6∑d^2)/(n(n^2-1))
where,
d = difference in ranks
n = number of pairs of observations
- Here, the sum of squares of differences in ranks is given as 66.
- So, 6∑d^2 = 6(66) = 396.
Substituting the given values in the formula for r, we get:
0.6 = 1 - (396)/(n(n^2-1))
Simplifying this expression, we get:
0.6n^3 - 0.6n - 396.4 = 0
Using trial and error method, we can find that n=10 satisfies the above equation.
Therefore, the number of students in the group is 10.
Answer: Option (a) 10.
- Rank correlation coefficient = 0.6
- Sum of squares of differences in ranks = 66
To find:
- Number of students in the group
Solution:
Let there be 'n' students in the group.
Finding the value of n:
- The sum of the first 'n' natural numbers is n(n+1)/2.
- Therefore, the sum of the ranks of the students in management and mathematics combined is n(n+1).
- Since each student has a rank in both subjects, the sum of their ranks should be counted twice, so we need to divide by 2.
- Therefore, the sum of the ranks in each subject is n(n+1)/2.
Finding the value of the rank correlation coefficient:
- The formula for rank correlation coefficient is given as:
r = 1 - (6∑d^2)/(n(n^2-1))
where,
d = difference in ranks
n = number of pairs of observations
- Here, the sum of squares of differences in ranks is given as 66.
- So, 6∑d^2 = 6(66) = 396.
Substituting the given values in the formula for r, we get:
0.6 = 1 - (396)/(n(n^2-1))
Simplifying this expression, we get:
0.6n^3 - 0.6n - 396.4 = 0
Using trial and error method, we can find that n=10 satisfies the above equation.
Therefore, the number of students in the group is 10.
Answer: Option (a) 10.
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If the rank correlation coefficient between marks in management and mathematics for a group of student in 0.6 and the sum of squares of the differences in ranks in 66, what is the number of students in the group?a)10b)9c)8d)11Correct answer is option 'A'. Can you explain this answer?
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