Python中两个比例之差的置信区间

Question

例如，在 AB 测试中，A 群体可能有 1000 个数据点，其中 100 个是成功的。而 B 可能有 2000 个数据点和 220 个成功。这使 A 的成功比例为 0.1，B 为 0.11，其 delta 为 0.01。如何在python中围绕这个delta计算这个置信区间？

统计模型可以对一个样本执行此操作，但似乎没有一个包来处理 AB 测试所必需的两个样本之间的差异。( http://www.statsmodels.org/dev/generated/statsmodels.stats.proportion.proportion_confint.html )

Answer 1

我无法从 Statsmodels 中找到此功能。但是，该网站详细介绍了用于生成置信区间的数学方法以及以下函数的来源：

def two_proprotions_confint(success_a, size_a, success_b, size_b, significance = 0.05):
    """
    A/B test for two proportions;
    given a success a trial size of group A and B compute
    its confidence interval;
    resulting confidence interval matches R's prop.test function

    Parameters
    ----------
    success_a, success_b : int
        Number of successes in each group

    size_a, size_b : int
        Size, or number of observations in each group

    significance : float, default 0.05
        Often denoted as alpha. Governs the chance of a false positive.
        A significance level of 0.05 means that there is a 5% chance of
        a false positive. In other words, our confidence level is
        1 - 0.05 = 0.95

    Returns
    -------
    prop_diff : float
        Difference between the two proportion

    confint : 1d ndarray
        Confidence interval of the two proportion test
    """
    prop_a = success_a / size_a
    prop_b = success_b / size_b
    var = prop_a * (1 - prop_a) / size_a + prop_b * (1 - prop_b) / size_b
    se = np.sqrt(var)

    # z critical value
    confidence = 1 - significance
    z = stats.norm(loc = 0, scale = 1).ppf(confidence + significance / 2)

    # standard formula for the confidence interval
    # point-estimtate +- z * standard-error
    prop_diff = prop_b - prop_a
    confint = prop_diff + np.array([-1, 1]) * z * se
    return prop_diff, confint

Answer 2

statsmodels 包现在有confint_proportions_2indep，它获取比较两个比例的置信区间，您可以在文档中查看详细信息https://www.statsmodels.org/stable/ generated/statsmodels.stats.proportion.confint_proportions_2indep.html