Subject: Applied Mathematics 4

Topic: Correlation

Difficulty: Medium

x: 32, 55, 49, 60, 43, 37, 43, 49 , 10, 20

y: 40, 30, 70, 20, 30, 50, 72, 60, 45, 25

Solution: $$ \begin{array}{|c|c|c|c|c|c|} \hline x & y & R_{x} & R_{y} & d=R_{x}-R_{y} & d^{2} \\ \hline 32 & 40 & 3 & 5 & -2 & 4 \\ \hline 55 & 30 & 9 & 3.5 & 5.5 & 30.25 \\ \hline 49 & 70 & 7.5 & 9 & -1.5 & 2.25 \\ \hline 60 & 20 & 10 & 1 & 9 & 81 \\ \hline 43 & 30 & 5.5 & 3.5 & 2 & 4 \\ \hline 37 & 50 & 4 & 7 & -3 & 9 \\ \hline 43 & 72 & 5.5 & 10 & -4.5 & 20.25 \\ \hline 49 & 60 & 7.5 & 8 & -0.5 & 0.25 \\ \hline 10 & 45 & 1 & 6 & -5 & 25 \\ \hline 20 & 25 & 2 & 2 & 0 & 0 \\ \hline & & & & & \Sigma d^{2}=176 \\ \hline \end{array} $$

Correction factor (C.F)

When $x=49 \Leftrightarrow C_{1}=\frac{2^{3}-2}{12}=0.5$

When $x=43 \Longleftrightarrow C_{2}=\frac{2^{3}-2}{12}=0.5$

when $y=30 \Rightarrow C_{3}=\frac{2^{3}- 2}{12}=0.5$

By Spearman's rank corelation formula with equal ranks, $$ \begin{aligned} R &=1-\frac{6\left[\Sigma d^{2}+C \cdot F\right]}{N^{3}-N} \\ &=1-\frac{6\left[\Sigma d^{2}+C_{1}+C_{2}+C_{3}\right]}{N^{3}-N} \\ &=1-\frac{6[176+0.5+0.5+0.5]}{10^{3}-10} \\ &=-0.07576 \end{aligned} $$

Thus , it is negative corelation between x and y.