Answer to Question #256369 in Statistics and Probability for Trevz

Question

 Marks of 12 students in Arithmetic and Algebra are given below:

Arithmetic 60 34 40 50 45 40 22 43 42 66 64 46

Algebra 75 32 33 40 45 33 12 30 34 72 41 57

Calculate the rank correlation coefficient

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & x & Rank(x)& y & Rank(y) \\ \hline & 60 & 10 & 75 & 12 \\ \hdashline & 34 & 2 & 32 & 3 \\ \hdashline & 40 & 3.5 & 33 & 4.5 \\ \hdashline & 50 & 9 & 40 & 7 \\ \hdashline & 45 & 7 & 45 & 9 \\ \hdashline & 40 & 3.5 & 33 & 4.5 \\ \hdashline & 22 & 1 & 12 & 1 \\ \hdashline & 43 & 6 & 30 & 2 \\ \hdashline & 42 & 5 & 34 & 6 \\ \hdashline & 66 & 12 & 72 & 11 \\ \hdashline & 64 & 11 & 41& 8 \\ \hdashline & 46 & 8 & 57 & 10 \\ \hdashline Sum= & & 78 & & 78 \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Rank(x) Rank(x) & ( Rank(x))^2 & ( Rank(y))^2 \\ \hline & 120 & 100 & 144 \\ \hdashline & 6 & 4 & 9 \\ \hdashline & 15.75 & 12.25 & 20.25 \\ \hdashline & 63 & 81 & 49 \\ \hdashline & 63 & 49 & 81 \\ \hdashline & 15.75 & 12.25 & 20.25 \\ \hdashline & 1 & 1 & 1 \\ \hdashline & 12 & 36 & 4 \\ \hdashline & 30 & 25 & 36 \\ \hdashline & 132 & 144 & 121 \\ \hdashline & 88 & 121& 64 \\ \hdashline & 80 & 64 & 100 \\ \hdashline Sum= & 626.5& 649.5 & 649.5 \\ \end{array}$$

$$SS_{xx}=\displaystyle\sum_{i=1}^n(Rank(x_i))^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(x_i))^2$$

$$=649.5-\dfrac{1}{12}(78)^2=142.5$$

$$SS_{yy}=\displaystyle\sum_{i=1}^n(Rank(y_i))^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(y_i))^2$$

$$=649.5-\dfrac{1}{12}(78)^2=142.5$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nRank(x_i)Rank(y_i)$$

$$-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(x_i))(\displaystyle\sum_{i=1}^nRank(y_i))$$

$$=626.5-\dfrac{1}{12}(78)(78)=119.5$$

Therefore, based on this information, the sample Spearman correlation coefficient is computed as follows

$$r_S=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}=\dfrac{119.5}{\sqrt{142.5}\sqrt{142.5}}\approx0.8386$$