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31. Correlazione per caratteri qualitativi /4

Denominatore

Ricordiamo che: 

\[ \color{brown}{ \boxed{\sum_{k=1}^{n}(r_k - \bar{r})^2 = \sum_{k=1}^{n}(s_k - \bar{s})^2 = \sum_{k=1}^{n}{r_k^2} - \frac{n(n+1)^2}{4} } } \]

e che:

\[ \color{brown}{ \boxed{\sum_{k=1}^{n}{r_k^2}  = \frac{n(n+1)(2n+1)}{6} } } \]

è una somma telescopica.

\[ \begin{eqnarray}\nonumber \sum_{k=1}^{n}(r_k - \bar{r})^2 &=& \frac{n(n+1)(2n+1)}{6}  - \frac{n(n+1)^2}{4} \\ &=& n(n+1) \frac{n-1}{12} \\[8pt] &=& \frac{n(n^2-1)}{12} \end{eqnarray} \]

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