如何使用 Python 计算 Difference in Differences 方法的置信区间？

Question

我正在尝试分析实验前后每位用户的总活跃分钟数。在这里，我包括了相关的 user data before and after the experiment - variant_number = 0 表示对照组，而 1 表示治疗组。具体来说，我对平均值（每位用户的平均总活跃分钟数）感兴趣。

首先，我计算了治疗结果的前后差异和对照结果的前后差异（分别为-183.7 和19.4）。本例中差值之差=203.1。

我想知道如何使用 Python 构建差异差异的 95% 置信区间？（如果需要我可以提供更多code/context）

Answer 1

您可以使用线性模型并测量交互作用（下图group[T.1]:period[T.pre]）。这些模拟数据的平均差异为 -223.1779，交互作用的 p 值是 p < 5e-4 非常显着，95% 置信区间是 [-276.360, -169.995].

import statsmodels.api as sm
import statsmodels.formula.api as smf
import pandas as pd
import numpy as np
np.random.seed(14)

minutes_0_pre = np.random.normal(loc=478, scale=1821, size=39776)
minutes_1_pre = np.random.normal(loc=275, scale=1078, size=9921)

minutes_0_post = np.random.normal(loc=458, scale=1653, size=37425)
minutes_1_post = np.random.normal(loc=458, scale=1681, size=9208)

df = pd.DataFrame({'minutes': np.concatenate((minutes_0_pre, minutes_1_pre, minutes_0_post, minutes_1_post)),
                   'group': np.concatenate((np.repeat(a='0', repeats=minutes_0_pre.size),
                                            np.repeat(a='1', repeats=minutes_1_pre.size),
                                            np.repeat(a='0', repeats=minutes_0_post.size),
                                            np.repeat(a='1', repeats=minutes_1_post.size))),
                   'period': np.concatenate((np.repeat(a='pre', repeats=minutes_0_pre.size + minutes_1_pre.size),
                                            np.repeat(a='post', repeats=minutes_0_post.size + minutes_1_post.size)))
                   })
model = smf.glm('minutes ~ group * period', df, family=sm.families.Gaussian()).fit()
print(model.summary())

输出：

Generalized Linear Model Regression Results                  
==============================================================================
Dep. Variable:                minutes   No. Observations:                96330
Model:                            GLM   Df Residuals:                    96326
Model Family:                Gaussian   Df Model:                            3
Link Function:               identity   Scale:                      2.8182e+06
Method:                          IRLS   Log-Likelihood:            -8.5201e+05
Date:                Mon, 18 Jan 2021   Deviance:                   2.7147e+11
Time:                        23:05:53   Pearson chi2:                 2.71e+11
No. Iterations:                     3                                         
Covariance Type:            nonrobust                                         
============================================================================================
                               coef    std err          z      P>|z|      [0.025      0.975]
--------------------------------------------------------------------------------------------
Intercept                  456.2792      8.678     52.581      0.000     439.271     473.287
group[T.1]                  14.9314     19.529      0.765      0.445     -23.344      53.207
period[T.pre]               21.7417     12.089      1.798      0.072      -1.953      45.437
group[T.1]:period[T.pre]  -223.1779     27.134     -8.225      0.000    -276.360    -169.995
============================================================================================

编辑：

由于您的汇总统计数据显示您的分布严重偏斜，自举法实际上是一种更可靠的估计置信区间的方法：

r = 1000
bootstrap = np.zeros(r)

for i in range(0, r):
    sample_index = np.random.choice(a=range(0, df.shape[0]), size=df.shape[0], replace=True)
    df_sample = df.iloc[sample_index]
    model = smf.glm('minutes ~ group * period', df_sample, family=sm.families.Gaussian()).fit()
    bootstrap[i] = model.params.iloc[3]  # interaction

bootstrap = pd.DataFrame(bootstrap, columns=['interaction'])
print(bootstrap.quantile([0.025, 0.975]).T)

输出：

                  0.025       0.975
interaction -273.524899 -175.373177

如何使用 Python 计算 Difference in Differences 方法的置信区间？

How to compute the confidence interval of the Difference in Differences method using Python?

python

ab-testing

experimental-design

hypothesis-test

t-test

编辑：