Editing Effect size (section)

== Confidence intervals by means of noncentrality parameters ==

Confidence intervals of standardized effect sizes, especially Cohen's <math>{d}</math> and <math>f^2</math>, rely on the calculation of confidence intervals of [[noncentrality parameter]]s (''ncp''). A common approach to construct the confidence interval of ''ncp'' is to find the critical ''ncp'' values to fit the observed statistic to tail [[quantile]]s ''α''/2 and (1&nbsp;−&nbsp;''α''/2). The SAS and R-package MBESS provides functions to find critical values of ''ncp''.

=== ''t''-test for mean difference of single group or two related groups ===
For a single group, ''M'' denotes the sample mean, ''μ'' the population mean, ''SD'' the sample's standard deviation, ''σ'' the population's standard deviation, and ''n'' is the sample size of the group. The ''t'' value is used to test the hypothesis on the difference between the mean and a baseline&nbsp;''μ''<sub>baseline</sub>. Usually, ''μ''<sub>baseline</sub> is zero. In the case of two related groups, the single group is constructed by the differences in pair of samples, while ''SD'' and ''σ'' denote the sample's and population's standard deviations of differences rather than within original two groups.
<math display="block">t := \frac{M - \mu_{\text{baseline}}}{\text{SE}} = \frac{M- \mu_{\text{baseline}}}{\text{SD}/\sqrt{n}}=\frac{\sqrt{n} \left( \frac{M-\mu}{\sigma} \right) + \sqrt{n} \left( \frac{\mu-\mu_\text{baseline}}{\sigma}\right) }{\frac{\text{SD}} \sigma}</math>
<math display="block">ncp=\sqrt{n} \left( \frac{\mu-\mu_\text{baseline}}{\sigma} \right) </math>
and Cohen's
<math display="block">d := \frac{M-\mu_\text{baseline}}{\text{SD}}</math>

is the point estimate of
<math display="block">\frac{\mu-\mu_\text{baseline}} \sigma.</math>

So,
:<math>\tilde{d}=\frac{ncp}{\sqrt n}.</math>

=== ''t''-test for mean difference between two independent groups ===
''n''<sub>1</sub> or ''n''<sub>2</sub> are the respective sample sizes.
<math display="block">t:=\frac{M_1-M_2}{\text{SD}_\text{within}/\sqrt{\frac{n_1 n_2}{n_1+n_2}}},</math>

wherein
<math display="block">\text{SD}_\text{within}:=\sqrt{\frac{\text{SS}_\text{within}}{\text{df}_\text{within}}} = \sqrt{\frac{(n_1-1)\text{SD}_1^2+(n_2-1) \text{SD}_2^2}{n_1+n_2-2}}.</math>
<math display="block">ncp=\sqrt{\frac{n_1 n_2}{n_1+n_2}}\frac{\mu_1-\mu_2} \sigma </math>

and Cohen's
<math display="block">d:=\frac{M_1-M_2}{SD_\text{within}}</math> is the point estimate of <math>\frac{\mu_1-\mu_2} \sigma.</math>

So,
<math display="block">\tilde{d}=\frac{ncp}{\sqrt{\frac{n_1 n_2}{n_1+n_2}}}.</math>

=== 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>.