Question

我有两个数据集（Pandas系列）-ds1和ds2-我想针对其平均值（如果是正常的）或中位数（对于非正常的）的差异计算95％置信区间。

对于均值差异，我这样计算t检验统计量和CI：

import statsmodels.api as sm
tstat, p_value, dof = sm.stats.ttest_ind(ds1, ds2)
CI = sm.stats.CompareMeans.from_data(ds1, ds2).tconfint_diff()

对于中位数，我这样做：

from scipy.stats import mannwhitneyu
U_stat, p_value = mannwhitneyu(ds1, ds2, True, "two-sided")

如何计算中位数差异的CI？

Answer 1

我碰到了一篇论文（计算一些非参数的置信区间 MICHAEL J CAMPBELL和MARTIN J GARDNER的分析得出CI公式。

基于此：

from scipy.stats import norm

ct1 = ds1.count()  #items in dataset 1
ct2 = ds2.count()  #items in dataset 2
alpha = 0.05       #95% confidence interval
N = norm.ppf(1 - alpha/2) # percent point function - inverse of cdf

# The confidence interval for the difference between the two population
# medians is derived through these nxm differences.
diffs = sorted([i-j for i in ds1 for j in ds2])

# For an approximate 100(1-a)% confidence interval first calculate K:
k = int(round(ct1*ct2/2 - (N * (ct1*ct2*(ct1+ct2+1)/12)**0.5)))

# The Kth smallest to the Kth largest of the n x m differences 
# ct1 and ct2 should be > ~20
CI = (diffs[k], diffs[len(diffs)-k])

Python Mann-Whitney置信区间

1 个答案: