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Exercises · Q9
Q.

Calculate Karl Pearson's coefficient of correlation for the data:

xx246810
yy579811
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✓ Free question

Here n=5n = 5.

xˉ=2+4+6+8+105=305=6,yˉ=5+7+9+8+115=405=8.\bar x = \frac{2+4+6+8+10}{5} = \frac{30}{5} = 6, \qquad \bar y = \frac{5+7+9+8+11}{5} = \frac{40}{5} = 8.

Take dx=x−6d_x = x - 6 and dy=y−8d_y = y - 8:

xxyydxd_xdyd_ydxdyd_x d_ydx2d_x^2dy2d_y^2
25−4-4−3-312169
47−2-2−1-1241
6901001
8820040
10114312169
Total00264020

r=∑dxdy∑dx2 ∑dy2=2640 20=26800=2628.2843=0.9192.r = \frac{\sum d_x d_y}{\sqrt{\sum d_x^2}\,\sqrt{\sum d_y^2}} = \frac{26}{\sqrt{40}\,\sqrt{20}} = \frac{26}{\sqrt{800}} = \frac{26}{28.2843} = 0.9192.

Independent check. 40×20=800=28.2843\sqrt{40 \times 20} = \sqrt{800} = 28.2843 and 26/28.2843=0.9192426/28.2843 = 0.91924 — confirmed.

✓Final answer

r≈+0.919r \approx +0.919 — a high degree of positive correlation.

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