Ito T, Kato T, Takagishi K, Okabe S, Sano D
Division of Environmental Engineering, Faculty of Engineering, Hokkaido University, North 13, West 8, Kita-ku, Sapporo, Hokkaido, 060-8628, Japan E-mail:
Department of Computer Science, Graduate School of Engineering, Gunma University, Tenjinmachi 1-5-1, Kiryu, Gunma, 376-8515, Japan.
Water Sci Technol. 2015;72(10):1789-95. doi: 10.2166/wst.2015.402.
Left-censored datasets of virus density in wastewater samples make it difficult to evaluate the virus removal efficiency in wastewater treatment processes. In the present study, we modeled the probabilistic distribution of virus removal efficiency in a wastewater treatment process with a Bayesian approach, and investigated how many detect samples in influent and effluent are necessary for accurate estimation. One hundred left-censored data of virus density in wastewater (influent and effluent) were artificially generated based on assumed log-normal distributions and the posterior predictive distribution of virus density, and the log-ratio distribution were estimated. The estimation accuracy of distributions was quantified by Bhattacharyya coefficient. When it is assumed that the accurate estimation of posterior predictive distributions is possible when a 100% positive rate is obtained for 12 pairs of influent and effluent, 11 out of 144, 60 out of 324, and 201 out of 576 combinations of detect samples gave an accurate estimation at the significant level of 0.01 in a Kruskal-Wallis test when the total sample number was 12, 18, and 24, respectively. The combinations with the minimum number of detect samples were (12, 9), (16, 10), and (21, 8) when the total sample number was 12, 18, and 24, respectively.
污水样本中病毒密度的左删失数据集使得评估污水处理过程中的病毒去除效率变得困难。在本研究中，我们采用贝叶斯方法对污水处理过程中病毒去除效率的概率分布进行建模，并研究进水和出水需要检测多少样本才能进行准确估计。基于假设的对数正态分布和病毒密度的后验预测分布，人工生成了100个污水（进水和出水）中病毒密度的左删失数据，并估计了对数比分布。分布的估计精度通过 Bhattacharyya 系数进行量化。当假设当12对进水和出水的阳性率达到100%时能够准确估计后验预测分布时，在总样本数分别为12、18和24时，144种检测样本组合中的11种、324种中的60种以及576种中的201种在 Kruskal-Wallis 检验中在0.01的显著水平下给出了准确估计。当总样本数分别为12、18和24时，检测样本数最少的组合分别为(12, 9)、(16, 10) 和(21, 8)。
