パチンコで使える９５％信頼区間

\[ \begin{align} R-1.96\sqrt{\frac{R(1-R)}{u}}\le p\le R+1.96\sqrt{\frac{R(1-R)}{u}}\\ \text{ここで標本比率}R=\frac{s}{u}\\ -1.96\frac{\sqrt{R(1-R)}}{\sqrt{u}}\le p-R\le1.96\frac{\sqrt{R(1-R)}}{\sqrt{u}}\\ -1.96\le(p-R)\frac{\sqrt{u}}{\sqrt{R(1-R)}}\le1.96\\ -1.96\le\frac{(p-R)\sqrt{u}}{\sqrt{R(1-R)}}\le1.96\\ -1.96\le\frac{(p-R)u}{\sqrt{R(1-R)}\sqrt{u}}\le1.96\\ -1.96\le\frac{pu-Ru}{\sqrt{uR(1-R)}}\le1.96\\ -1.96\le\frac{pu-\frac{s}{u}u}{\sqrt{u\frac{s}{u}(1-\frac{s}{u})}}\le1.96\\ -1.96\le\frac{pu-s}{\sqrt{s(1-\frac{s}{u})}}\le1.96\\ -1.96\le\frac{up-s}{\sqrt{s-\frac{s^2}{u}}}\le1.96\\ \color{red}{\text{ここにかなりの飛躍があります}}\\ -1.96\le\frac{s-up}{\sqrt{up(1-p)}}\le1.96\\ \left|\frac{s-up}{\sqrt{up(1-p)}}\right|\le1.96\\ \frac{(s-up)^2}{up(1-p)}\le1.96^2\\ (s-up)^2\le1.96^2up(1-p)\\ (s-up)^2-1.96^2up(1-p)\le0\\ s^2-2sup+u^2p^2-1.96^2up+1.96^2up^2\le0\\ (u^2+1.96^2u)p^2-(2su+1.96^2u)p+s^2\le0\\ \text{区間をやめて境界を求める(等号にする)}\\ (u^2+1.96^2u)p^2-(2su+1.96^2u)p+s^2=0\\ \text{係数をa、b、cにする}\\ ap^2+bp+c=0\\ a=u^2+1.96^2u\\ b=-(2su+1.96^2u)\\ c=s^2\\ p=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\ CI_{low}=\frac{-b-\sqrt{b^2-4ac}}{2a}\times250\\ CI_{high}=\frac{-b+\sqrt{b^2-4ac}}{2a}\times250\\ \end{align} \]