如何解释statsmodels普通最小二乘法模型的结果？

Question

受Can scipy.stats identify and mask obvious outliers?的启发，我想了解statsmodel的OLS的输出。

我将代码修改为适合当今的要求，并希望了解如何解释*.outlier_test()的响应-分别是否真正地bonf(p) <0.5是用于识别某个对象的正确方法。离群值。

代码：

from random import random
import statsmodels.api as smapi
from statsmodels.formula.api import ols
import statsmodels.graphics as smgraphics
# Make data #
x = list(range(30))
y = [y*(10+random())+200 for y in x]
# Add outlier #
x.insert(6,15)
y.insert(6,220)
x.insert(6,16)
y.insert(6,295)
# Make fit #
regression = ols("data ~ x", data=dict(data=y, x=x)).fit()
# Find outliers #
test = regression.outlier_test()
print("test.columns:", test.columns)
print(test)

outliers = ((x[i],y[i]) for i,t in enumerate(test.iloc[:,2]) if t < 0.5)
print ('Outliers: ', list(outliers))
# Figure #
figure = smgraphics.regressionplots.plot_fit(regression, 1)
# Add line #
smgraphics.regressionplots.abline_plot(model_results=regression, ax=figure.axes[0])
figure.show()

输出：

test.columns: Index(['student_resid', 'unadj_p', 'bonf(p)'], dtype='object')

    student_resid       unadj_p       bonf(p)
0        0.256226  7.995850e-01  1.000000e+00
1        0.235436  8.155247e-01  1.000000e+00
2        0.266506  7.917355e-01  1.000000e+00
3        0.195602  8.462860e-01  1.000000e+00
4        0.206646  8.377301e-01  1.000000e+00
5        0.235760  8.152759e-01  1.000000e+00
6       -2.670250  1.229206e-02  3.933460e-01
7       -9.404263  2.609308e-10  8.349786e-09
8        0.160577  8.735400e-01  1.000000e+00
9        0.317017  7.535015e-01  1.000000e+00
10       0.120925  9.045843e-01  1.000000e+00
11       0.249872  8.044476e-01  1.000000e+00
12       0.250744  8.037804e-01  1.000000e+00
13       0.399508  6.924460e-01  1.000000e+00
14       0.313912  7.558347e-01  1.000000e+00
15       0.187027  8.529415e-01  1.000000e+00
16       0.019263  9.847634e-01  1.000000e+00
17       0.038839  9.692847e-01  1.000000e+00
18       0.015481  9.877546e-01  1.000000e+00
19       0.417676  6.792601e-01  1.000000e+00
20       0.153612  8.789799e-01  1.000000e+00
21       0.201890  8.414121e-01  1.000000e+00
22       0.540464  5.930042e-01  1.000000e+00
23       0.216489  8.301220e-01  1.000000e+00
24      -0.156133  8.770102e-01  1.000000e+00
25       0.477092  6.368722e-01  1.000000e+00
26       0.246855  8.067600e-01  1.000000e+00
27       0.494958  6.243592e-01  1.000000e+00
28       0.413796  6.820681e-01  1.000000e+00
29       0.067460  9.466782e-01  1.000000e+00
30       0.165854  8.694224e-01  1.000000e+00
31       0.511132  6.131286e-01  1.000000e+00

和：

Outliers:  [(16, 295), (15, 220)]

所以我的问题是：

在使用statsmodels普通最小二乘法模型时，bonf(p) <0.5真的是识别异常值的正确方法吗？