Question

Calculate coefficient of correlation of the following data:

X	10	6	9	10	12	13	11	9
Y	9	4	6	9	11	13	8	4

Answer 1

X	$d_x=X-A$	dx2	Y	$d_y=Y-B$	dy2	dxdy
10 6 9 $\boxed{\mathrm{A}=10}$ 12 13 11 9	0 –4 –1 0 2 3 1 –1	0 16 1 0 9 4 1 1	9 4 6 9 $\boxed{\mathrm{B}=11}$ 13 8 4	–2 –7 –5 –2 0 2 –3 –7	4 49 25 4 0 4 9 49	0 28 5 0 0 6 –3 7
N = 8	∑dx= 0	∑dx2 = 32	N = 8	∑dy= –24	∑dy2 = 144	∑dxdy= 43

$$\begin{gathered} r=\frac{{\sum }_{}{d}_x{d}_y-{\displaystyle \frac{\left({\sum }_{}^{}{d}_x\right)\times \left({\sum }_{}^{}{d}_y\right)}{N}}}{\sqrt{{\sum }_{}^{}{d}_x^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_x\right)}^2}{N}}}\times \sqrt{{\sum }_{}^{}{d}_y^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_y\right)}^2}{N}}}} \\ \mathrm{or},r=\frac{43-{\displaystyle \frac{0\times \left(-24\right)}{8}}}{\sqrt{32-{\displaystyle \frac{(0{)}^2}{8}}}\times \sqrt{144-{\displaystyle \frac{(-24{)}^2}{8}}}}=\frac{43}{\sqrt{32}\times \sqrt{72}}=\frac{43}{5.65\times 8.48} \\ \therefore r=+0.896 \end{gathered}$$