sim_art.md

sim_art.md

Dietmar G. Schrausser

Abstract

Introduction

Dispersion measures

Sum score

$$\Sigma x=\sum_{i=1}^n x_i$$

Range, values $x$ sorted in ascending order, distribution

$$R_x=x_n-x_0$$

Varianz

$$\sigma^2=\sum_{i=1}^n \frac{(x_i-\overline x)^2}{n}$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}$$

Population estimator

$$\hat\sigma=\sigma\cdot\sqrt{\frac{n}{n-1}}$$

$$=\sqrt{\sigma^2\cdot\frac{n}{n-1}}$$

Coefficient of variation $\omega$

$$\omega=\frac{\sigma}{\overline x}$$

Method

Mean dispersion $theta=\overline d$

Schrausser (2022, p. 33).

$$\overline d=\frac{\sum_{i=1}^n{|x_i-\overline x|}}{\sum_{i=1}^n1};x_i\ne \overline x,$$

$$\hat{\overline \delta}=\sigma\cdot\zeta\cdot\sqrt{\frac{n}{n-1}}=\hat\sigma\cdot\zeta$$

With n>=10, a correction factor is suggested by results of the simulation

$$\hat{\overline \delta}=\hat\sigma\cdot\zeta\cdot (1-\frac{3}{800\cdot\zeta^{-1}})^{-1}$$

Simulation theta d, sd

sim0.pas [MDN.pas] ->sim01.pas [simp.pas]

$$\hat{\overline d}=\sigma_0\cdot\zeta\cdot\sqrt{\frac{n-1}{n}}$$

$$\hat{\overline d}=\sigma_0\cdot\zeta\cdot\sqrt{\frac{n-1}{n}}\cdot(1-\frac{3}{1000})$$

$$=\sigma_0\cdot\zeta\cdot\sqrt{\frac{n-1}{n}}\cdot(1-\frac{3}{10^3})$$

$$=\sigma_0\cdot\zeta\cdot\sqrt{\frac{n-1}{n}}\cdot (1-\frac{3}{800\cdot\zeta^{-1}})$$

$$=\sigma_0\cdot\zeta\cdot\sqrt{\frac{n-1}{n}}\cdot0.997$$

where

$$\zeta_{\overline d}=\overline\delta=\frac{4}{5}$$

Standard error $\hat\sigma_{\overline d}$

$$\hat\sigma_{\overline d}=\sqrt{\frac{\hat\sigma^2\cdot\frac{n}{n-1}}{n}}\cdot\frac{1}{2\cdot\zeta\cdot\frac{n}{n-1}}$$

$$=\hat\sigma\cdot\sqrt{\frac{\frac{n}{n-1}}{n}}\cdot\frac{1}{2\cdot\zeta\cdot\frac{n}{n-1}}$$

$$=\hat\sigma\cdot\frac{\sqrt{\frac{n}{n-1}}}{2\cdot\zeta\cdot\frac{n}{n-1}\cdot\sqrt{n}}$$

$$=\hat\sigma\cdot\frac{\sqrt{\frac{n-1}{n}}}{2\cdot\zeta\cdot\sqrt{n}}$$

$$=\sigma\cdot\sqrt{\frac{n}{n-1}}\cdot\sqrt{\frac{\frac{n}{n-1}}{n}}\cdot\frac{1}{2\cdot\zeta\cdot\frac{n}{n-1}}$$

$$=\sigma\cdot\frac{1}{2\cdot\zeta\cdot\sqrt{n}}$$

With n>=20, results of the simulation suggest a correction factor

$$\hat\sigma_{\overline d}=\sigma\cdot\frac{1}{2\cdot\zeta\cdot\sqrt{n}}\cdot(1-\frac{1}{50\cdot\zeta})$$

$$=\sigma\cdot\frac{1}{2\cdot\zeta\cdot\sqrt{n}}\cdot(1-\frac{1}{40})$$

$$CI_p=\hat{\overline{\delta}}±z_{(1-\frac{1-p}{2})}\cdot \hat\sigma_{\overline d}$$

Note: $$\sqrt{\frac{a}{b}}=\frac{a}{b}\cdot\sqrt{\frac{b}{a}}$$

Simulation theta diff dd, sdd

Paired

Mean dispersion of differences

sim0b.pas [MDN.pas] ->sim03.pas [simp.pas]

$$\hat\sigma_{\overline d_d}=\sigma_0\cdot\frac{\sqrt{\frac{\frac{n}{n-1}}{n}}\cdot\frac{1}{2\cdot\zeta\cdot\frac{n}{n-1}}}{\zeta}\cdot\frac{1}{\sqrt\zeta}$$

Unpaired

Difference of Mean dispersion

sim0a.pas [MDN.pas] ->sim02.pas [simp.pas]

$$\hat\sigma_{\overline d_d}=$$

Difference of $\overline d$

Paired

$$\theta=\overline d_{d}=\frac{\sum_{i=1}^n{|x_{d,i}-\overline x_d}|}{n}$$

$$x_d={x_{1}-x_{2}}$$

$$\hat\theta=\frac{\sigma_0}{\sqrt{\zeta}}$$

with

$$z=\frac{\overline d_{d}}{\hat\sigma_{\overline d_d}}$$

$$\hat\sigma_{\overline d_d}=$$

Unpaired

$$\theta=d_{\overline d}=\overline d_1-\overline d_2$$

$$\hat\theta=0$$

with

$$z=\frac{d_{\overline d}}{\hat\sigma_{d_\overline d}}$$

$$\hat\sigma_{d_\overline d}=\sigma\cdot\frac{\sqrt{\frac{n-2}{n}}}{2\cdot\zeta\cdot\sqrt{n}}\cdot\frac{n}{\zeta\cdot(n-1)}\cdot \frac{n}{n-1}$$

$$=\sigma\cdot\frac{\sqrt{\frac{n-2}{n}}\cdot n}{2\cdot\zeta^2\cdot\sqrt{n}\cdot (n-1)}\cdot \frac{n}{n-1}$$

$$=\sigma\cdot\frac{\sqrt{n-2}\cdot n}{2\cdot\zeta^2\cdot n\cdot (n-1)}\cdot \frac{n}{n-1}$$

$$=\sigma\cdot\frac{\sqrt{n-2}}{2\cdot\zeta^2\cdot(n-1)}\cdot \frac{n}{n-1}$$

$$=\sigma\cdot\frac{n\cdot\sqrt{n-2}}{2\cdot\zeta^2\cdot(n-1)^2}$$

$$\hat\sigma_{d_\overline d}=\sigma\cdot\frac{n\cdot\sqrt{n-2}}{2\cdot\zeta^2\cdot(n-1)^2}\cdot c$$

Nonlinear regression revealed a inverse relationship between $n$ and inaccuracy quotient $\hat\iota=\frac{\hat\sigma_{d_\overline d}}{\sigma_{d_\overline d}}$ ($r=0.999$, $det=99.81%$) and therefore a correction factor $\hat c$ where

$$\hat c=\frac{0.993}{n}+0.924$$

hence

$$\hat\sigma_{d_\overline d}=\sigma\cdot\frac{n\cdot\sqrt{n-2}}{2\cdot\zeta^2\cdot(n-1)^2}\cdot\frac{1}{\hat c}$$

or

$$\hat\sigma_{d_\overline d}=\sigma\cdot\frac{n\cdot\sqrt{n-2}}{2\cdot\zeta^2\cdot(n-1)^2}\cdot(\frac{1}{n}+\frac{231}{200}\cdot\zeta)^{-1}$$

Calculation of estimated levels $\alpha$

$$\alpha^{-'}_1=[\alpha^-_{1_{crit}}+(\alpha_{1_{sim}}^{-}-\alpha_{1_{asympt}}^{-})]\cdot100$$

$$\alpha^{-'}_2=[\frac{\alpha^-_{2_{crit}}}{2}+(\alpha_{2_{sim}}^{-}-\alpha_{2_{asympt}}^{-})]\cdot200$$

$$\alpha^{+'}_2 =\{1-[1-\frac{\alpha^+_{2_{crit}}}{2}+(\alpha_{2_{sim}}^{+}-\alpha_{2_{asympt}}^{+})]\}\cdot200$$

$$\alpha^{+'}_1 =\{1-[1-\alpha^+_{1_{crit}}+(\alpha_{1_{sim}}^{+}-\alpha_{1_{asympt}}^{+})]\}\cdot100$$

Discussion

Scale

References

Schrausser, D. G. (2022). Mathematical-Statistical Algorithm Interpreter, SCHRAUSSER-MAT: Function Index, Manual. Handbooks. Academia. https://doi.org/10.13140/RG.2.2.28314.52164