更改

====阿格里斯蒂-库尔方法====

====阿格里斯蒂-库尔方法====

<math>       \frac{z^2}{4 n^2}</math>

:<math>\frac{z^2}{4 n^2}</math><ref name="Agresti1988">{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref>

 

:<math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math>

<ref name="Agresti1988">{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref>

:<math>1 + \frac{z^2}{n}</math>

 

::<math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math>

 

<math>1 + \frac{z^2}{n}</math>

这里''p''的估计量被修改为

这里''p''的估计量被修改为

 

:<math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>

 

::<math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>

确切的(克洛佩尔-皮尔森)方法是最保守的。

确切的(克洛佩尔-皮尔森)方法是最保守的。