(b) The scores of nine students in Mathematics and Physics are as follows:
Mathematics: $35, 23, 47, 17, 10, 43, 9, 6, 28$
Physics: $30, 33, 45, 23, 8, 49, 12, 4, 31$
Compute the students' ranks in the two subjects and compute Spearman's correlation.
Answer (2)
Rank the Mathematics and Physics scores separately.
Calculate the difference in ranks for each student, then square these differences.
Sum the squared differences: ∑ d i 2 = 12 .
Apply Spearman's rank correlation formula: r s = 1 − n ( n 2 − 1 ) 6 ∑ d i 2 = 0.9 .
Explanation
Understand the problem We are given the Mathematics and Physics scores of 9 students and asked to compute Spearman's rank correlation coefficient.
Rank Mathematics scores First, we need to rank the scores in each subject. The Mathematics scores are: 35, 23, 47, 17, 10, 43, 9, 6, 28. Ranking these from highest to lowest, we get the ranks: 3, 5, 1, 6, 7, 2, 8, 9, 4.
Rank Physics scores Next, we rank the Physics scores: 30, 33, 45, 23, 8, 49, 12, 4, 31. Ranking these from highest to lowest, we get the ranks: 5, 3, 2, 6, 8, 1, 7, 9, 4.
Calculate rank differences and squared differences Now, we calculate the difference d i between the ranks for each student and then square these differences.
Calculate squared differences The differences are: d 1 = 3 − 5 = − 2 d 2 = 5 − 3 = 2 d 3 = 1 − 2 = − 1 d 4 = 6 − 6 = 0 d 5 = 7 − 8 = − 1 d 6 = 2 − 1 = 1 d 7 = 8 − 7 = 1 d 8 = 9 − 9 = 0 d 9 = 4 − 4 = 0
The squared differences are: d 1 2 = ( − 2 ) 2 = 4 d 2 2 = ( 2 ) 2 = 4 d 3 2 = ( − 1 ) 2 = 1 d 4 2 = ( 0 ) 2 = 0 d 5 2 = ( − 1 ) 2 = 1 d 6 2 = ( 1 ) 2 = 1 d 7 2 = ( 1 ) 2 = 1 d 8 2 = ( 0 ) 2 = 0 d 9 2 = ( 0 ) 2 = 0
Sum the squared differences The sum of the squared differences is: ∑ d i 2 = 4 + 4 + 1 + 0 + 1 + 1 + 1 + 0 + 0 = 12 .
Apply Spearman's formula We use the formula for Spearman's rank correlation coefficient: r s = 1 − n ( n 2 − 1 ) 6 ∑ d i 2 , where n is the number of students, which is 9.
Calculate Spearman's coefficient Substituting the values, we get: r s = 1 − 9 ( 9 2 − 1 ) 6 × 12 = 1 − 9 ( 81 − 1 ) 72 = 1 − 9 × 80 72 = 1 − 720 72 = 1 − 0.1 = 0.9 .
State the final answer Therefore, Spearman's rank correlation coefficient is 0.9.
Examples
Spearman's rank correlation is useful in scenarios where you want to assess the relationship between two sets of ranked data. For example, you might use it to determine if there is a correlation between the ranking of students based on their exam scores and their ranking based on teacher evaluations. A high Spearman's rank correlation coefficient would suggest a strong agreement between the two ranking methods, while a low coefficient would suggest little to no agreement.
To compute Spearman's rank correlation, we ranked scores for Mathematics and Physics, calculated the differences in ranks, squared them, and summed those squared differences to obtain a result of approximately 0.317. This indicates a weak positive correlation between the two subjects. Such analysis helps us assess how performance in one subject relates to performance in another subject.
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