线性回归的标准差/误差

Question

所以我有:

t = [0.0, 3.0, 5.0, 7.2, 10.0, 13.0, 15.0, 20.0, 25.0, 30.0, 35.0]
U = [12.5, 10.0, 7.6, 6.0, 4.4, 3.1, 2.5, 1.5, 1.0, 0.5, 0.3]
U_0 = 12.5
y = []
for number in U:
    y.append(math.log(number/U_0, math.e))
(m, b) = np.polyfit(t, y, 1)
yp = np.polyval([m, b], t)
plt.plot(t, yp)
plt.show()

因此,通过这样做,我得到线性回归拟合m=-0.1071和b=0.0347.

如何获得m值的偏差或误差？

我想要 m = -0.1071*(1+ plus/minus error)

m是k,b在y = kx + n时是n

Answer 1

import numpy as np
import pandas as pd
import statsmodels.api as sm
import math

U = [12.5, 10.0, 7.6, 6.0, 4.4, 3.1, 2.5, 1.5, 1.0, 0.5, 0.3]
U_0 = 12.5
y = []

for number in U:
    y.append(math.log(number/U_0, math.e))

y = np.array(y)

t = np.array([0.0, 3.0, 5.0, 7.2, 10.0, 13.0, 15.0, 20.0, 25.0, 30.0, 35.0])
t = sm.add_constant(t, prepend=False)

model = sm.OLS(y,t)
result = model.fit()
result.summary()

Answer 2

您可以使用scipy.stats.linregress：

m, b, r_value, p_value, std_err = stats.linregress(t, yp)

并且标准偏差将存储在std_err...