我正在寻找Python中的测试:
> survivors <- matrix(c(1781,1443,135,47), ncol=2)
> colnames(survivors) <- c('survived','died')
> rownames(survivors) <- c('no seat belt','seat belt')
> survivors
survived died
no seat belt 1781 135
seat belt 1443 47
> prop.test(survivors)
2-sample test for equality of proportions with continuity correction
data: survivors
X-squared = 24.3328, df = 1, p-value = 8.105e-07
alternative hypothesis: two.sided
95 percent confidence interval:
-0.05400606 -0.02382527
sample estimates:
prop 1 prop 2
0.9295407 0.9684564
我最感兴趣的是p-value
计算。
示例来自here
答案 0 :(得分:18)
我想我明白了:
In [11]: from scipy import stats
In [12]: import numpy as np
In [13]: survivors = np.array([[1781,135], [1443, 47]])
In [14]: stats.chi2_contingency(survivors)
Out[14]:
(24.332761232771361, # x-squared
8.1048817984512269e-07, # p-value
1,
array([[ 1813.61832061, 102.38167939],
[ 1410.38167939, 79.61832061]]))
答案 1 :(得分:1)
添加到@Akavall的答案:如果您没有明确拥有“失败”计数(示例中的死亡人数),则R的prop.test
可让您仅指定试验的总数,例如prop.test(c(1781, 1443), c(1781+135, 1443+47))
将为您提供与您生成的列联表相同的结果。
Scipy的chi2_contingency
明确要求失败计数和完整的应急表。如果您没有明确的失败计数,而只想检查两个样本中成功总数所占的比例是否相等,则可以使用来破解scipy函数
survivors = np.array([[1781, total1 - 1781], [1443, total2 - 47]])
chi2_contingency(survivors)
# Result:
(24.332761232771361, 8.1048817984512269e-07, 1,
array([[ 1813.61832061, 102.38167939],
[ 1410.38167939, 79.61832061]]))
花些时间弄清楚这个问题。希望对别人有帮助。