Question

Using assumed average in Karl Pearson's formula, calculate coefficient of correlation, given the following data:

X	78	89	97	69	59	79	68	61
Y	125	137	156	112	107	106	123	138

Answer 1

X	$d_x=X-A$	dx2	Y	$d_y=Y-B$	dy2	dxdy
78 89 97 $\boxed{\mathrm{A}=69}$ 59 79 68 61	9 20 28 0 –10 10 –1 –8	81 400 784 0 100 100 1 64	$\boxed{\mathrm{B}=125}$ 137 156 112 107 106 123 138	0 12 31 –13 –18 –19 –2 13	0 144 961 169 324 361 4 169	0 240 868 0 180 –190 2 –104
N = 8	∑dx= 48	∑dx2 =1530	N = 8	∑dy = 4	∑dy2 = 2132	∑dxdy= 996

Suppose the assumed mean is 69 and 125 for series A and series B, respectively.
$$\begin{gathered} r=\frac{{\sum }_{}{d}_x{d}_y-{\displaystyle \frac{\left({\sum }_{}^{}{d}_x\right)\times \left({\sum }_{}^{}{d}_y\right)}{N}}}{\sqrt{{\sum }_{}^{}{d}_x^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_x\right)}^2}{N}}}\times \sqrt{{\sum }_{}^{}{d}_y^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_y\right)}^2}{N}}}} \\ \mathrm{or},r=\frac{996-{\displaystyle \frac{48\times 4}{8}}}{\sqrt{1530-{\displaystyle \frac{(48{)}^2}{8}}}\sqrt{2132-{\displaystyle \frac{(4{)}^2}{8}}}}=\frac{972}{35.24\times 46.15} \\ \therefore r=+0.597 \end{gathered}$$