Question
Calculate Karl Pearson's coefficient of correlation by actual mean method:
✓
Answer
Let X denotes price and Y denotes quantity.
$\overline{\text{X}}=\frac{\Sigma\text{X}}{\text{N}}=\frac{70}{5}=14$ $\overline{\text{Y}}=\frac{\Sigma\text{Y}}{\text{N}}=\frac{120}{5}=24$
$\text{r}=\frac{\Sigma\text{(X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})}{\sqrt{\Sigma\text{(X}-\overline{\text{X}}) ^2}\sqrt{\Sigma(\text{Y}-\overline{\text{Y}})^2}}$
$\frac{180}{\sqrt{40}\sqrt{70}}=\frac{18}{6.32\times8.36}=\frac{18}{52.83}=0.34$
| X | $\text{X}-\overline{\text{X}}$ | $(\text{X}-\overline{\text{X}})^2$ | Y | $\text{Y}-\overline{\text{Y}}$ | $(\text{Y}-\overline{\text{Y}})^2$ | $(\text{X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})$ |
| 10 | -4 | 16 | 20 | -4 | 16 | 16 |
| 12 | -2 | 4 | 29 | 5 | 25 | -10 |
| 14 | 0 | 0 | 21 | -3 | 9 | 0 |
| 16 | 2 | 1 | 22 | -2 | 4 | -4 |
| 18 | 4 | 16 | 28 | 4 | 16 | 16 |
| $\Sigma\text{X}=70\ \text{N}=5$ | $\Sigma(\text{X}-\overline{\text{X}})^2=40$ | $\Sigma\text{Y}=120$ | $\Sigma(\text{Y}-\overline{\text{Y}})^2=70$ | $\Sigma(\text{X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})=18$ |
$\text{r}=\frac{\Sigma\text{(X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})}{\sqrt{\Sigma\text{(X}-\overline{\text{X}}) ^2}\sqrt{\Sigma(\text{Y}-\overline{\text{Y}})^2}}$
$\frac{180}{\sqrt{40}\sqrt{70}}=\frac{18}{6.32\times8.36}=\frac{18}{52.83}=0.34$
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