Formula 1 AD=[∑i=1n1−2in[ln(Fo(Z(i)))+ln(1−Fo(Z(n+1−i)))]−n]\mathrm{AD}=\left[\sum_{\mathrm{i}=1}^{\mathrm{n}} \frac{1-2 \mathrm{i}}{\mathrm{n}}\left[\ln \left(\mathrm{F}_{\mathrm{o}}\left(\mathrm{Z}_{(\mathrm{i})}\right)\right)+\ln \left(1-\mathrm{F}_{\mathrm{o}}\left(\mathrm{Z}_{(\mathrm{n}+1-\mathrm{i})}\right)\right)\right]-\mathrm{n}\right]AD=[i=1∑nn1−2i[ln(Fo(Z(i)))+ln(1−Fo(Z(n+1−i)))]−n] where Fo\mathrm{F}_{\mathrm{o}}Fo is the standard normal distribution function*. If AD>0.752/(1+0.75/n+2.25/n2)\mathrm{AD}>0.752 /\left(1+0.75 / \mathrm{n}+2.25 / \mathrm{n}^{2}\right)AD>0.752/(1+0.75/n+2.25/n2) Formula 2 AD>0.752/(1+0.75/n+2.25/n2)\mathrm{AD}>0.752 /\left(1+0.75 / \mathrm{n}+2.25 / \mathrm{n}^{2}\right)AD>0.752/(1+0.75/n+2.25/n2)