Question

Calculate coefficient of correlation of the following data with rank difference and Karl Pearson's methods:

Economics (Marks)	77	54	27	52	14	35	90	25	56	60
Hindi (Marks)	35	58	60	46	50	40	35	56	44	42

Answer 1

Karl Pearson's Method

Economics (X)	$d_x=X-A$	dx2	History (Y)	$d_y=Y-B$	dy2	dxdy
77 54 27 52 14 $\boxed{\mathrm{A}=35}$ 90 25 56 60	42 19 –8 17 –21 0 55 –10 21 25	1764 361 64 289 441 0 3025 100 441 625	35 58 60 46 $\boxed{\mathrm{B}=50}$ 40 35 56 44 42	–15 8 10 –4 0 –10 –15 6 –6 –8	225 64 100 16 0 100 225 36 36 64	–630 152 –80 –68 0 0 –825 –60 –126 –200
N = 10	∑dx= 140	∑dx2 = 7110	N = 10	∑dy = –34	∑dy2 = 860	∑dxdy=–1837

$$\begin{gathered} r=\frac{{\sum }_{}{d}_x{d}_y-{\displaystyle \frac{\left({\sum }_{}^{}{d}_x\right)\times \left({\sum }_{}^{}{d}_y\right)}{N}}}{\sqrt{{\sum }_{}^{}{d}_x^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_x\right)}^2}{N}}}\times \sqrt{{\sum }_{}^{}{d}_y^{\mathit{2}}-{\displaystyle \frac{{\left({\sum }_{}^{}{d}_y\right)}^2}{N}}}} \\ \mathrm{or},r=\frac{-1831-{\displaystyle \frac{140\times (-34)}{10}}}{\sqrt{7110-{\displaystyle \frac{(140{)}^2}{10}}}\sqrt{860-{\displaystyle \frac{(-34{)}^2}{10}}}}=\frac{-1355}{71.76\times 27.39} \\ \therefore r=-0.689 \end{gathered}$$

Rank Difference Method

Economics	R1	History	R2	D = R1 – R2	D2
77 54 27 52 14 35 90 25 56 60	2 5 8 6 10 7 1 9 4 3	35 58 60 46 50 40 35 56 44 42	9.5 2 1 5 4 8 9.5 3 6 7	–7.5 3 7 1 6 –1 –8.5 6 –2 –4	56.25 9 49 1 36 1 72.25 36 4 16
N =10					${\sum }_{}{D}^2=280.5$

$$\begin{gathered} r_k=1-\frac{6\left[\sum D^2+{\displaystyle \frac{1}{12}}\left(M_1^3-M_1\right)\right]}{N^3-N} \\ \mathrm{Here,} \\ {M}_1=\mathrm{2,}\mathrm{as}\mathrm{two}\mathrm{students}\mathrm{have}\mathrm{scored}35\mathrm{marks}\mathrm{in}\mathrm{History}. \\ \mathrm{or},r_k=1-\frac{6\left[280.50+{\displaystyle \frac{1}{12}}(2^3-2)\right]}{1000-10} \\ \therefore ,r_k=\frac{990-1686}{990}=-0.703 \end{gathered}$$