• 二项分布中的\(p\) 服从Beta分布 $ {\rm beta}(\alpha, \beta)$, 密度函数 \(\frac1{B(\alpha, \beta)} x^{\alpha-1} (1-x)^{\beta -1}\)

• 均值 \(\frac \alpha {\alpha + \beta}\)

• 方差 \(\frac {\alpha \beta} {(\alpha+\beta)^2 (\alpha+ \beta + 1) } ​\)

  from scipy.stats import beta

  def confidence(n_bad, n_good, tol=2):
  ''' 返回估计的坏率p, 以及在tol倍标准差下的可信度'''
  a, b = n_bad+1, n_good+1
  p = a / (a+b)
  v = beta.std(a, b)
  up, low = min(1, p + vtol), max(0, p - vtol)
  d = beta.cdf(up, a,b) - beta.cdf(low, a,b)
  return p, v, d

  test_set = [
  (500, 20000, 2),
  (1000, 200000, 2),
  (2000, 200000, 2),
  (5000, 200000, 2),
  (500, 100000, 2),
  (1000, 100000, 2),
  (2000, 100000, 2),
  (5000, 100000, 2),
  (2000, 10000, 2),
  ]

  print(" bad; total; 均值p; 标准差v; 均值的相对误差e; 置信度")
  for (n_bad, n_good, tol) in test_set:
  p,v,d = confidence(n_bad, n_good, tol)

  ss = ('{:5d};{:7d}; p={p:0.4f}; v={v:0.6f}; e={e:0.3f}; '
       + '均值在[p - {t}v, p + {t}v]的概率 {d:2.2f}%'
       ).format(n_bad, n_bad+n_good, p=p,v=v, c=v/p, d =d*100,t=tol, e=tol*v/p)
  print(ss)

  bad; total; 均值p; 标准差v; 均值的相对误差e; 置信度
  500; 20500; p=0.0244; v=0.001078; e=0.088; 均值在[p - 2v, p + 2v]的概率 95.46%
  1000; 201000; p=0.0050; v=0.000157; e=0.063; 均值在[p - 2v, p + 2v]的概率 95.46%
  2000; 202000; p=0.0099; v=0.000220; e=0.044; 均值在[p - 2v, p + 2v]的概率 95.45%
  5000; 205000; p=0.0244; v=0.000341; e=0.028; 均值在[p - 2v, p + 2v]的概率 95.45%
  500; 100500; p=0.0050; v=0.000222; e=0.089; 均值在[p - 2v, p + 2v]的概率 95.46%
  1000; 101000; p=0.0099; v=0.000312; e=0.063; 均值在[p - 2v, p + 2v]的概率 95.46%
  2000; 102000; p=0.0196; v=0.000434; e=0.044; 均值在[p - 2v, p + 2v]的概率 95.45%
  5000; 105000; p=0.0476; v=0.000657; e=0.028; 均值在[p - 2v, p + 2v]的概率 95.45%
  2000; 12000; p=0.1667; v=0.003402; e=0.041; 均值在[p - 2v, p + 2v]的概率 95.45%

结论: 坏样本大于2000以上, 在95%置信度下, 坏率的相对误差<5%