Tables for the Calculation of $\mathrm B$- and $Z$-Distribution Functions

L. N. Bol'shev ; B. V. Gladkov ; M. V. Ščeglova

Teoriâ veroâtnostej i ee primeneniâ, Tome 6 (1961) no. 4, pp. 446-455

Résumé

Tables for the calculation of $\mathrm B$ and $Z$-distribution functions $I_x(p,q)$ and $F_{2p,2q}(z)$ are given. The published tables can be used as a certain supplement to Karl Pearson's “Tables of the Incomplete Beta-function”, Biometric Laboratory, London (1934), and permit the calculation of $I_x(p,q)$ and $F_{2p,2q}(z)$ for $q\geq50$ and $p\leq q$. The error of calculation is not higher than $5\cdot10^{-5}$ (for the case $q \geq160$ the error does not exceed $5\cdot10^{-6}$). The paper gives regions of the parameters $p$ and $q$, where Tables I, II and [3] can be used for the calculation of $I_x(p,q)$. For example let us obtain $I_{0.3}(28;73)$ and $I_{0.3}(16;85)$. The point $(p,q)$ in the first case belongs to region I and for this reason we shall use Table I. With the help of (2) and (3) $$w^2=\frac1{p}+\frac1{q}=0.035714+0.13699=0.049413,\,w = 0.222290,\\v=\frac1{w}\left(\frac1{p}-\frac1{q}\right)=\frac{0.035714- 0.13699}{0.222290}=0.09904,\\u=\frac1{w}\lg\frac{qx}{p(1 - x)}=\frac{\lg219-\lg196}{0.43429\cdot 0.22229}=\frac{2.34044-2.29226}{0.09654}=0.49917.\\$$ Then: from tables [1] we have $\Phi(u)=0.69117$, and from Table I: $\varphi _1(u,v)=0.01304,\varphi _2(u,v)=-0.01140$. Using formula (1) we finally obtain $$I_{0.3}(28;73)=\Phi(u)+\varphi _1(u,v)+w^2(u,v)=\\=0.69117+0.01304-0.04941\cdot0.01140=0.70365.$$ The exact value of $I_{0.3}(28;73)$ to five decimal places is 0.70364 (f. [6]). In the second example the parametric point $(p,q)$ is in the region II. That is why we shall use Table II for obtaining $I_{0.3}(16;85)$. With the help of (7) $$2q+p-1=185\,{\text{ and }}y=\frac{x(2q+p-1)}{2-x}=\frac{0.3\cdot185}{1.7}=32.647.$$ Then by virtue of [4] and II, $I(y,p)=0.99954$ and $\gamma(y,p)=11$. Formula (6) gives the final value $$I_{0.3}(16;85)=I(y,p)+\frac{\gamma (y,p)}{6(2q+p-1)^2}=0.99954+\frac{11}{6(185)^2}=0.99959,$$ the exact value of $I_{0.3}(16;85)$ to five decimal places is equal to 0.99959.