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第866行： − + + − \widehat{p\,} + \frac{z^2}{2n} + z}</math>+ − : A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}+ − \sqrt{ + − \frac{\widehat{p\,}(1 - \widehat{p\,})}{n} ++ − ==== Agresti–Coull method ====+ − − \frac{z^2}{4 n^2} − + − } − − − }{ 第918行：
第913行： − + 第946行：
第941行： − + − + 第960行：
第955行： − + + − \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt] + − − \widehat{p\,} + \frac{z^2}{2n} + z − &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}+ − + − \end{align}</math>+ 第980行：
第974行： − + − \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>+ 第990行：
第984行： − + − 1 + \frac{z^2}{n}+ − \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]+ − }</math><ref>{{cite book}}+ − − &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right) − \end{align}</math> − − 在上面的表达式中用 i = k-m 代替后，我们得到了 第1,017行：
第1,006行： − − 注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为 − − <math>\begin{align} 第1,028行：
第1,013行： − \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]+ − &= \binom{n}{m} (pq)^m (1-pq)^{n-m}+ − − − − \end{align}</math> 第1,079行：
第1,060行： − + − − \operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\ 第1,087行：
第1,066行： − &= \binom{n+m}k p^k (1-p)^{n+m-k}+ − \end{align}</math> 第1,173行：
第1,151行： − + − − \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt] − + − + − + 第1,187行：
第1,163行： − \end{align}</math> 第1,203行：
第1,178行： − + − − \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt] 第1,259行：
第1,232行： − − 二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref> − − 第1,290行：
第1,259行： + − +
无编辑摘要
f(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.
f(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.
''f''(''k'', ''n'', ''p'')对''k'' < ''M'' 是单调递增的，对''k'' > ''M'' 是单调递减的，但(''n'' + 1)''p''是整数的情况除外。在这种情况下，有(''n'' + 1)''p'' 和 (''n'' + 1)''p'' −1 两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。
''f''(''k'', ''n'', ''p'')对''k'' < ''M'' 是单调递增的，对''k'' > ''M'' 是单调递减的，但(''n'' + 1)''p''是整数的情况除外。在这种情况下，有(''n'' + 1)''p'' 和 (''n'' + 1)''p'' −1 两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。
It can also be represented in terms of the regularized incomplete beta function, as follows:
It can also be represented in terms of the regularized incomplete beta function, as follows:
在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
* The median is unique and equal to ''m'' = [[Rounding|round]](''np'') when |''m'' − ''np''| ≤ min{''p'', 1 − ''p''} (except for the case when ''p'' = {{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/>
* The median is unique and equal to ''m'' = [[Rounding|round]](''np'') when |''m'' − ''np''| ≤ min{''p'', 1 − ''p''} (except for the case when ''p'' = {{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/>
*中位数是唯一的并且等于''m'' = [[Rounding|round]](''np'')，此时|''m'' − ''np''| ≤ min{''p'', 1 − ''p''}（''p'' = \\sfrac{1}{2} 和 ''n'' 是奇数的情况除外）
*中位数是唯一的并且等于''m'' = [[Rounding|round]](''np'')，此时|''m'' − ''np''| ≤ min{''p'', 1 − ''p''}（<math>''p'' = {{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）
which implies the simpler but looser bound
which implies the simpler but looser bound
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:
对于''p'' = 1/2且''n''是奇数，任意''m''满足\\sfrac{1}{2}(''n'' − 1) ≤ ''m'' ≤ \\sfrac{1}{2} (''n'' + 1)是一个二项分布的中位数。如果''p'' = 1/2且''n'' 是偶数，那么''m'' = ''n''/2是唯一的中位数:
对于''p'' = 1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}} (''n'' − 1) ≤ ''m'' ≤ {{sfrac|1|2}} (''n'' + 1)是一个二项分布的中位数。如果''p'' = 1/2且''n'' 是偶数，那么''m'' = ''n''/2是唯一的中位数:
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math>
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math>
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
<math>\frac{
<math>\frac{}\widehat{p\,} + \frac{z^2}{2n} + z</math>
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};
可以添加一个0.5/''n''连续调整。（2012年7月更新）
<math> \sqrt{\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
==== Agresti–Coull method ====
阿格里斯蒂-库尔方法 Agresti–Coull method
<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>
<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>
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math>
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1) − 1)/n + ((1/p2) − 1)/m。
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1) − 1)/n + ((1/p2) − 1)/m</math>。
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1 − ''x'')-th quantile'.
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1 − ''x'')-th quantile'.
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1 − ''x'')-th分位数的简写。
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1 − ''x'')-''th''分位数的简写。
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math> = 0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math> = 0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math> = 0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math> = 0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。
<math>\begin{align}
<math>\begin{align}</math>
:: <math>\frac{}
:: <math>\frac{}
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
<math> \widehat{p\,} + \frac{z^2}{2n} + z</math>
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>
\sqrt{
\sqrt{}
<math> \end{align}</math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
<math> \frac{z^2}{4 n^2}
<math> \frac{z^2}{4 n^2}</math>
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields
}{
}{
<math>\begin{align}
<math>\begin{align}</math>
<math> 1 + \frac{z^2}{n}</math>
<math> \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
| chapter = Confidence intervals
| chapter = Confidence intervals
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
After substituting i = k - m in the expression above, we get
After substituting i = k - m in the expression above, we get
| title = Engineering Statistics Handbook
| title = Engineering Statistics Handbook
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields
| year = 2012
| year = 2012
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3
| access-date = 2017-07-23
| access-date = 2017-07-23
<math> \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]</math>
}}</ref>
}}</ref>
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math>
==== Comparison ====
==== Comparison ====
:<math>\begin{align}
:<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math>
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
<math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math>
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) is given by the normal distribution
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) is given by the normal distribution
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，
:<math>\begin{align}
:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.
<math>P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math>
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.
<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
:<math>\begin{align}
:<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math>
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref>
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref>
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二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref>