blocks|key|684286|text|测试统计和p值是正确的(也许也是可以理解的)。让我逐步解释一下结果。在维基百科(https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data)上，题为“维基百科”(https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data)中的分类数据的“类数据”的样本检验的章节(https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data)也可能会有所帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|684287|，|684288|，期望计数是如果我们假设独立的话，最终会出现在表中的一个给定单元格中的观测数。黑和黑的分数分别为0.15468974和0.12061524。在独立的情况下，我们期望样本中的observations).|unordered-list-item|684289|All观测值被标记为Black和No+(注:+275015是0.15468974x0.12061524x275015=5131.21350472的总数，其他期望计数是相似的)。请注意，期望计数和观察计数(即表中的数字)之间的差异相当大。这应该是独立的空假设可能是错误的第一个指示。|684290|计算测试统计量，对单元格中的每个元素计算(Obs)%5E2/Exp，并对表中的所有元素进行求和。其结果确实是5207.162302393083+(参见下面的代码)。在空假设下，这个检验统计量是由2df分布的chi2+(正如您已经提到的)。与此分布相比，值5207.162302393083确实处于分布的尾部，因此很难在独立性为零的情况下观察到此结果。因此p值等于零。|684291|。|684292|下面发布的代码复制了所有的数字，并以2自由度绘制了chi2发行版的PDF。我希望这能帮到你。|684293|import+numpy+as+np
import+matplotlib.pyplot+as+plt
from+scipy.stats+import+chi2

#+Data+and+properties
TrueCounts+=+np.array(+[+[9170,33372],[15137,152307],[8864,56165]+])
Datadimension+=+TrueCounts.shape
TotalCounts+=+np.sum(TrueCounts)
print(TotalCounts)

#+Fractions
fracAnswer+=+np.sum(TrueCounts,+axis=0)/TotalCounts
fracRace+=+np.sum(TrueCounts,+axis=1)/TotalCounts

#+Caculate+expected+counts
ExpCounts+=+np.zeros(np.shape(TrueCounts))
for+iter1+in+range(Datadimension[0]):
++++for+iter2+in+range(Datadimension[1]):
++++++++ExpCounts[iter1,+iter2]+=+fracRace[iter1]*fracAnswer[iter2]*TotalCounts


print('===+True+and+expected+counts+===')
print(fracAnswer)
print(fracRace)

print('===+True+and+expected+counts+===')
print(TrueCounts)
print(ExpCounts)

print('===+Test+summary+===')
TestStat+=+np.sum(+(TrueCounts-ExpCounts)**2/ExpCounts+)
print(TestStat)

#+Make+ch2+plot+for+comparison
x+=+np.arange(0,+20,+0.05)
plt.plot(x,+chi2.pdf(x,+df=2))
plt.show()|code-block|syntax|javascript|684294|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data|1|2^0|14|2K|0|40|2K|1|76|2K|2|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]|$A|19|B|1A|1|1B]|$A|1C|B|1D|1|1E]]|C|$]]|$1|D|3|E|5|6|7|1F|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|1G|8|@]|9|@]|C|$]]|$1|I|3|J|5|H|7|1H|8|@]|9|@]|C|$]]|$1|K|3|L|5|H|7|1I|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|1J|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1K|8|@]|9|@]|C|$]]|$1|Q|3|R|5|S|7|1L|8|@]|9|@]|C|$T|U]]|$1|V|3|-4|5|6|7|1M|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|12]]|14|$5|Y|Z|10|C|$11|12]]]]

The test statistic and p-value are correct (and perhaps also understandable). Let me stepwise explain the outcome. The section entitled ``Example chi-squared test for categorical data'' on wikipedia (<a href="https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Chi-squared_test#Example_chi-squared_test_for_categorical_data</a>) might help as well.
<ul>
<li>The expected count is the number of observations that would end up in a given cell of the table if we would assume independence. The fractions of Black and No are 0.15468974 and 0.12061524, respectively. Under independence, we expect 0.15468974x0.12061524x275015=5131.21350472 observations in the sample to be labeled as Black and No (Note: 275015 is the total number of observations).</li>
<li>All other expected counts are calculated similar. Note that the differences between the expected and observed counts (i.e. the numbers in your table) are rather large. This should be a first indication that the null hypothesis of independence might be false.</li>
<li>The test statistic is calculated be computing (Obs-Exp)^2/Exp for each element in the cell and summing all the elements in the table. The result is indeed 5207.162302393083 (see code below). Under the null hypothesis, this test statistic is chi2 distributed with 2 df (as you already mentioned). Compared to this distribution, the value 5207.162302393083 is truly far in the tail of the distribution making it very very unlikely to observe this outcome under the null of independence. The p-value is therefore equal to zero.</li>
</ul>
The code posted below replicates all the numbers and plots the PDF of the chi2 distribution with 2 degrees of freedom. I hope this helps.
<pre><code>import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import chi2

# Data and properties
TrueCounts = np.array( [ [9170,33372],[15137,152307],[8864,56165] ])
Datadimension = TrueCounts.shape
TotalCounts = np.sum(TrueCounts)
print(TotalCounts)

# Fractions
fracAnswer = np.sum(TrueCounts, axis=0)/TotalCounts
fracRace = np.sum(TrueCounts, axis=1)/TotalCounts

# Caculate expected counts
ExpCounts = np.zeros(np.shape(TrueCounts))
for iter1 in range(Datadimension[0]):
 for iter2 in range(Datadimension[1]):
 ExpCounts[iter1, iter2] = fracRace[iter1]*fracAnswer[iter2]*TotalCounts


print('=== True and expected counts ===')
print(fracAnswer)
print(fracRace)

print('=== True and expected counts ===')
print(TrueCounts)
print(ExpCounts)

print('=== Test summary ===')
TestStat = np.sum( (TrueCounts-ExpCounts)**2/ExpCounts )
print(TestStat)

# Make ch2 plot for comparison
x = np.arange(0, 20, 0.05)
plt.plot(x, chi2.pdf(x, df=2))
plt.show()
</code></pre>

I'm running a chi square test on some categorical values pertaining to race, and whether different racial groups participated in a clinic. As there's about a dozen different races in this data, I bucketed them down to 'White', 'Black' and 'Other', just for the purposes of testing (as the correlations indicated most of the activity occurring between 'White and 'Black'. However, using Python's .chi2_contingency() method, I'm getting results back that seem unusual. The table is below:
<pre><code> Appointment Status No Yes

Black 9170 33372
White 15137 152307
Other 8864 56165
</code></pre>
The Python method returns the following:
<pre><code>X^2: 5207.16
p-value: 0.0
df: 2
expected values array: array([[ 5131.21350472, 37410.78649528],
 [ 7843.48838791, 57185.51161209],
 [ 20196.29810738, 147247.70189262]]))
</code></pre>
The df is good, but the chi square value and p-value both don't seem right. Is there something anyone can see that I might be doing methodologically that might be producing these values, or might there be something going on behind the scenes in Python that's doing this? Thanks!

Unusual results from running chi square test with stats.chi2_contingency()

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在对一些与种族有关的分类值进行一次卡方检验，以及不同种族群体是否参加了一个诊所。由于这个数据中有十几个不同的种族，我把它们分成“白色”、“黑色”和“其他”，只是为了测试(因为相关性表明大多数活动发生在“白人”和“黑人”之间)。然而，使用Python的.chi2_contingency()方法，我得到的结果看起来很不...

用stats.chi2_contingency()运行卡方检验的异常结果-腾讯云开发者社区-腾讯云

问用stats.chi2_contingency()运行卡方检验的异常结果
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