Consider the Friedman statistic Square each term in the

Chapter 15, Problem 44E

(choose chapter or problem)

Consider the Friedman statistic

$F_{r}=\frac{12 \dot{b}}{k(k+1)} \sum_{i=1}^{k}\left(\overline{R_{i}}-\bar{R}\right)^{2}$.

Square each term in the sum, and show that an alternative form of $F_{r}$ is

$F_{r}=\frac{12}{b k(k+1)} \sum_{i=1}^{k} R_{i}^{2}-3 b(k+1)$

[Hint: Recall that $R_{i}^{-}=R_{i} / b, R^{-}=(k+1) / 2$ and note that $\Sigma_{i=1}^{k} R_{i}=$ sum of all of the ranks $=b k(k+1) / 2$].

Equation Transcription:

${F}_{r}=\frac{12b}{k(k+1)}\sum\limits_{i=1}^{k}\ {\left({\overline{R{\ }_{i}}-\overline{R}}\right)}^{2}$

${F}_{r}$

${F}_{r}=\frac{12}{bk(k+1)}\sum\limits_{i=1}^{k}\ {R}_{i}^{2}-3b(k+1)$

${{R}^{-}}_{i}={R}_{i}/b,{\ R}^{-}=(k+1)/2$

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

$=\ bk(k+1)/2$

Text Transcription:

F_r = 12b/k(k+1) the sum from i = 1 to k (R_i bar - R bar)^2

F_r

F_r = 12/bk(k+1) the sum of i=1 to k R_i^2 - 3b (k+1)

R^-_1 = R_1/b, R^- = (k+1)/2

the sum of i=1 to k = R_1 =

= bk(k+1)/2

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