# Calculate rank correlation

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The ranks of same 15 students in two subjects mathematics and statistics are given below:

(1,10) (2,7) (3,2)(4,6) (5,4) (6,8) (7,3) (8,1) (9,11) (10,15) (11,9) (12,5) (13,14) (14,12) (15,13)

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Let xi denotes the ranks of students in math & yi denotes the ranks of students in statistics.

 xi yi xi - $\bar {\text x}$ yi - $\bar y$ (xi - $\bar {\text x}$)2 (yi - $\bar y$)2 (xi - $\bar {\text x}$ )(yi - $\bar{\text y}$) 1 10 -7 2 49 7 -14 2 7 -6 -1 36 1 6 3 2 -5 -6 25 36 30 4 6 -4 -2 16 4 8 5 4 -3 -4 9 16 12 6 8 -2 0 4 0 0 7 3 -1 -5 1 25 5 8 1 0 -7 0 49 0 9 11 1 3 1 9 3 10 15 2 7 4 49 14 11 9 3 1 9 1 3 12 5 4 -3 16 9 -12 13 14 6 4 36 16 24 14 12 6 4 36 16 24 15 13 7 5 49 25 35 Σxi = 120 Σyi = 120 Σ(xi - $\bar{\text x}$)2  = 280 Σ(yi - $\bar{\text y}$)2 = 280 Σ(xi - $\bar {\text x}$) (yi - $\bar {\text y}$) = 144

Σxi = 1 + 2 + 3....+15 = $\frac{15(15+1)}2$ = $\frac{15\times16}2$ = 15 x 8 = 120

Σyi = 1 + 2 + 3 + ....15 = 120

$\therefore$ $\bar {\text x} = \frac{\sum\text x_i}{n}=\frac{120}{15}$ = 8

$\bar y$ = $\frac{\sum\text y_i}{n}=\frac{120}{15}$ = 8

$\therefore$ Rank correlation = $\frac{\sum(X_i-\bar x)(y_i-\bar y)}{\sqrt{\sum(x_i-\bar x)^2\sum(y_i-\bar y)^2}}$

$=\frac{144}{\sqrt{280\times280}}=\frac{144}{280}$ = $\frac{18}{35}$ = 0.5143