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

Question

此问题参考书籍“O'Relly Practical Statistics for Data Scientists 2nd Edition”第 3 章，session Chi-Square Test。

这本书提供了一个卡方测试用例的示例，其中假设一个网站具有三个不同的标题，由 1000 名访问者运行。结果显示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

期望值计算如下：

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：

现在桌子在哪里：

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

卡方统计量是平方 Pearson 残差的总和：。这是 1.666

到目前为止一切顺利。 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.

书中提供的重采样python测试代码如下： （可以从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，我运行了 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，我两次都能获得 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 显示的数字这么小？

我运行2000次得到的结果是：

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

我运行你的代码尝试了 2000、10000 和 100000 次循环，所有这三次我都接近 .47。然而，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 .47.

确保当你改变迭代次数时，你只通过改变 2000 来做到这一点，没有其他数字。

Answer 2

排列函数：

    def perm_fun(box):
        sample_clicks = [sum(random.sample(box, 1000)),
                         sum(random.sample(box, 1000)),
                         sum(random.sample(box, 1000))]

不应该吗

    def perm_fun(box):
        random.shuffle(box)
        sample_clicks = [sum(box[0:1000]),
                         sum(box[1000:2000]),
                         sum(box[2000:3000])]

确保点击总数始终为34？