我需要找到 95% 置信区间为 2.5 和 97.5 分位数
boot_mean_diff = []
for i in range(3000):
boot_before = before_proportion
boot_after = after_proportion
boot_mean_diff.append(boot_after.mean()-boot_before.mean())
# Calculating a 95% confidence interval from boot_mean_diff
boot_mean_diff=pd.Series(boot_mean_diff)
#boot_mean_diff1=boot.boot_mean_diff(frac=1,replace=True)
confidence_interval = pd.Series(boot_mean_diff).quantile([0.025,0.975])
confidence_interval
但是，我收到以下错误----
AssertionError: confidence_interval should be calculated as the [0.025, 0.975] quantiles of boot_mean_diff.
最佳答案
我也面临同样的错误。尝试下面的代码
# A bootstrap analysis of the reduction of deaths due to handwashing
boot_mean_diff = []
for i in range(3000):
boot_before = before_proportion.sample(frac=1, replace=True)
boot_after = after_proportion.sample(frac=1, replace=True)
boot_mean_diff.append(np.mean(boot_after) - np.mean(boot_before))
# Calculating a 95% confidence interval from boot_mean_diff
confidence_interval = pd.Series(boot_mean_diff).quantile([0.025, 0.975])
print(confidence_interval)
我使用了 import numpy as np，因此使用了 np.mean() 你可以直接尝试使用 boot_after.mean()
关于python - 将置信区间计算为分位数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56473741/