The Kruskal–Wallis statistic is Perform the indicated

Chapter 15, Problem 35E

(choose chapter or problem)

The Kruskal-Wallis statistic is

$H=\frac{12}{n(n+1)} \sum_{i=1}^{k} n_{i}\left(R_{i}^{-}-\frac{n+1}{2}\right)^{2}$

Perform the indicated squaring of each term in the sum and add the resulting values to show that

$H=\frac{12}{n(n+1)} \sum_{i=1}^{k} \frac{R_{i}^{2}}{n_{i}}-3(n+1)$.

[Hint: Recall that $R_{i}^{-}=R_{i} / n_{i}$ and that $\Sigma_{i=1}^{k} R_{i}=$ sum of the first n integers $=n(n+1) / 2$. ]

Equation Transcription:

$H=\frac{12}{n(n+1)}\sum\limits_{i=1}^{k}\ {n}_{i}{\left({{{R}^{-}}_{i}-\frac{n+1}{2}}\right)}^{2}$

$H=\frac{12}{n(n+1)}\sum\limits_{i=1}^{k}\ \frac{{R}_{i}^{2}}{{n}_{i}}-3(n+1)$

${{\ R}^{-}}_{i}={R}_{i}/{n}_{i}$

${\Sigma{}}_{i=1}^{k}\ {R}_{i}=$

$=n(n+1)/2$

Text Transcription:

H = 12/n(n+1) the sum of i=1 to k n_i (R^-_i - n+1/2)^2

H = 12/n(n+1) the sum of i=1 to k R_i^2/n_i - 3(n+1)

R^-_i = R_i/n_i

the sum of i=1 to k R_i =

= n(n+1)/2

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