例如，在 AB 测试中，A 总体可能有 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 次成功。这使 A 的成功比例为 0.1，B 的成功比例为 0.11，其增量为 0.01。我如何在 Python 中计算此增量周围的置信区间？
Stats 模型可以针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 )
最佳答案
我无法从 Statsmodels 中找到与此相关的函数。然而，this website复习数学以生成置信区间以及作为以下函数的来源:
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
关于python - Python中两个比例之间差异的置信区间，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47570903/