Editing Effect size (section)

=== One-way ANOVA test for mean difference across multiple independent groups ===
One-way ANOVA test applies [[noncentral F distribution]]. While with a given population standard deviation <math>\sigma</math>, the same test question applies [[noncentral chi-squared distribution]].
<math display="block">F := \frac{\frac{\text{SS}_\text{between}}{\sigma^2}/\text{df}_\text{between}}{\frac{\text{SS}_\text{within}}{\sigma^2}/\text{df}_\text{within}}</math>

For each ''j''-th sample within ''i''-th group ''X''<sub>''i'',''j''</sub>, denote
<math display="block">M_i (X_{i,j}) := \frac{\sum_{w=1}^{n_i} X_{i,w}}{n_i};\; \mu_i (X_{i,j}) := \mu_i.</math>

While,
<math display="block">\begin{align}
\text{SS}_\text{between}/\sigma^2
& = \frac{\text{SS}\left(M_i (X_{i,j}); i=1,2,\dots,K,\; j=1,2,\dots,n_{i}\right)}{\sigma^2}\\
& = \text{SS}\left(\frac{M_i(X_{i,j}-\mu_i)}{\sigma}+\frac{\mu_i}{\sigma};i=1,2,\dots,K,\; j=1,2,\dots,n_i \right)\\
& \sim \chi^2\left(\text{df}=K-1,\; ncp=SS\left(\frac{\mu_i(X_{i,j})}{\sigma};i=1,2,\dots,K,\; j=1,2,\dots,n_i\right)\right)
\end{align}</math>

So, both ''ncp''(''s'') of ''F'' and <math>\chi^2</math> equate
<math display="block">\text{SS}\left(\mu_i(X_{i,j})/\sigma;i=1,2,\dots,K,\; j=1,2,\dots,n_i \right).</math>

In case of <math>n:=n_1=n_2=\cdots=n_K</math> for ''K'' independent groups of same size, the total sample size is ''N''&nbsp;:=&nbsp;''n''·''K''.
<math display="block">\text{Cohens }\tilde{f}^2 := \frac{\text{SS}(\mu_1,\mu_2, \dots ,\mu_K)}{K\cdot\sigma^2} = \frac{\text{SS} \left(\mu_i(X_{i,j})/\sigma; i=1,2,\dots,K,\; j=1,2,\dots,n_i \right)}{n\cdot K} = \frac{ncp}{n\cdot K}=\frac{ncp}N.</math>

The ''t''-test for a pair of independent groups is a special case of one-way ANOVA. Note that the noncentrality parameter <math>ncp_F</math> of ''F'' is not comparable to the noncentrality parameter <math>ncp_t</math> of the corresponding ''t''. Actually, <math>ncp_F = ncp_t^2</math>, and <math>\tilde{f} = \left|\frac{\tilde{d}}{2}\right|</math>.