来自重采样方法与 scipy.stats.chi2_contigency 的卡方检验 P 值

Question

本题参考“O'Relly Practical Statistics for Data Scientist 2nd Edition”第3章，会话卡方检验。

这本书提供了一个卡方测试用例的例子，它假设一个网站有三个不同的标题，有 1000 名访问者。 结果显示了每个标题的点击次数。

观察到的数据如下：

Headline   A    B    C
Click      14   8    12
No-click   986  992  988

预期值计算如下：

Headline   A        B        C
Click      11.13    11.13    11.13
No-click   988.67   988.67   988.67

Pearson 残差定义为：

桌子现在在哪里：

Headline   A        B        C
Click      0.792    -0.990   0.198
No-click   -0.085   0.106   -0.021

卡方统计量是 Pearson 残差平方的总和： . 这是 1.666

到现在为止还挺好。 现在是重采样部分：

1. Assuming a box of 34 ones and 2966 zeros
2. Shuffle, and take three samples of 1000 and count how many ones(Clicks)
3. Find the squared differences between the shuffled counts and expected counts then sum them.
4. Repeat steps 2 to 3, a few thousand times.
5. The P-value is how often does the resampled sum of squared deviations exceed the observed.

本书提供的重采样python测试代码如下：（可从https://github.com/gedeck/practical-statistics-for-data-scientists/tree/master/python/code下载）

## Practical Statistics for Data Scientists (Python)
## Chapter 3. Statistial Experiments and Significance Testing
# > (c) 2019 Peter C. Bruce, Andrew Bruce, Peter Gedeck

# Import required Python packages.

from pathlib import Path
import random

import pandas as pd
import numpy as np

from scipy import stats
import statsmodels.api as sm
import statsmodels.formula.api as smf
from statsmodels.stats import power

import matplotlib.pylab as plt

DATA = Path('.').resolve().parents[1] / 'data'

# Define paths to data sets. If you don't keep your data in the same directory as the code, adapt the path names.

CLICK_RATE_CSV = DATA / 'click_rates.csv'

...

## Chi-Square Test
### Chi-Square Test: A Resampling Approach

# Table 3-4
click_rate = pd.read_csv(CLICK_RATE_CSV)
clicks = click_rate.pivot(index='Click', columns='Headline', values='Rate')
print(clicks)

# Table 3-5
row_average = clicks.mean(axis=1)
pd.DataFrame({
    'Headline A': row_average,
    'Headline B': row_average,
    'Headline C': row_average,
})

# Resampling approach
box = [1] * 34
box.extend([0] * 2966)
random.shuffle(box)

def chi2(observed, expected):
    pearson_residuals = []
    for row, expect in zip(observed, expected):
        pearson_residuals.append([(observe - expect) ** 2 / expect
                                  for observe in row])
    # return sum of squares
    return np.sum(pearson_residuals)

expected_clicks = 34 / 3
expected_noclicks = 1000 - expected_clicks
expected = [34 / 3, 1000 - 34 / 3]
chi2observed = chi2(clicks.values, expected)

def perm_fun(box):
    sample_clicks = [sum(random.sample(box, 1000)),
                     sum(random.sample(box, 1000)),
                     sum(random.sample(box, 1000))]
    sample_noclicks = [1000 - n for n in sample_clicks]
    return chi2([sample_clicks, sample_noclicks], expected)

perm_chi2 = [perm_fun(box) for _ in range(2000)]

resampled_p_value = sum(perm_chi2 > chi2observed) / len(perm_chi2)

print(f'Observed chi2: {chi2observed:.4f}')
print(f'Resampled p-value: {resampled_p_value:.4f}')

chisq, pvalue, df, expected = stats.chi2_contingency(clicks)
print(f'Observed chi2: {chi2observed:.4f}')
print(f'p-value: {pvalue:.4f}')

现在，我运行 perm_fun(box) 2,000 次并获得了 0.4775 的重采样 P 值。 但是，如果我运行 perm_fun(box) 10,000 次和 100,000 次，两次都能够获得 0.84 的重采样 P 值。 在我看来，P 值应该在 0.84 左右。 为什么 stats.chi2_contigency 显示的数字如此之小？

我运行 2000 次的结果是：

Observed chi2: 1.6659
Resampled p-value: 0.8300
Observed chi2: 1.6659
p-value: 0.4348

如果我运行它 10,000 次，结果是：

Observed chi2: 1.6659
Resampled p-value: 0.8386
Observed chi2: 1.6659
p-value: 0.4348

软件版本：

pandas.__version__:         0.25.1
numpy.__version__:          1.16.5
scipy.__version__:          1.3.1
statsmodels.__version__:    0.10.1
sys.version_info:           3.7.4

Answer 1

我运行了您的代码，尝试了 2000、10000 和 100000 次循环，并且所有 3 次都接近 0.47。 但是，我确实在这一行遇到了一个我必须修复的错误：

resampled_p_value = sum(perm_chi2 > chi2observed) / len(perm_chi2)

这里perm_chi2是一个列表， chi2observed是一个浮点数，所以我想知道这段代码是如何为你运行的（也许你为修复它所做的一切都是错误的根源）。 无论如何，将其更改为预期的

resampled_p_value = sum([1*(x > chi2observed) for x in perm_chi2]) / len(perm_chi2)

允许我运行它并接近 0.47。

确保在更改迭代次数时，只更改 2000，而不更改其他任何数字。