I have two series of data as below. I want to create an OLS linear regression model for df1 and another OLS linear regression model for df2. And then statistically test if the y-intercepts of these two linear regression models are statistically different (p<0.05), and also test if the slopes of these two linear regression models are statistically different (p<0.05). I did the following
import numpy as np
import math
import matplotlib.pyplot as plt
import pandas as pd
import statsmodels.api as sm
np.inf == float('inf')
data1 = [1, 3, 45, 0, 25, 13, 43]
data2 = [1, 1, 1, 1, 1, 1, 1]
df1 = pd.DataFrame(data1)
df2 = pd.DataFrame(data2)
fig, ax = plt.subplots()
df1.plot(figsize=(20, 10), linewidth=5, fontsize=18, ax=ax, kind='line')
df2.plot(figsize=(20, 10), linewidth=5, fontsize=18, ax=ax, kind='line')
plt.show()
model1 = sm.OLS(df1, df1.index)
model2 = sm.OLS(df2, df2.index)
results1 = model1.fit()
results2 = model2.fit()
print(results1.summary())
print(results2.summary())
Results #1
OLS Regression Results
=======================================================================================
Dep. Variable: 0 R-squared (uncentered): 0.625
Model: OLS Adj. R-squared (uncentered): 0.563
Method: Least Squares F-statistic: 10.02
Date: Mon, 01 Mar 2021 Prob (F-statistic): 0.0194
Time: 20:34:34 Log-Likelihood: -29.262
No. Observations: 7 AIC: 60.52
Df Residuals: 6 BIC: 60.47
Df Model: 1
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
x1 5.6703 1.791 3.165 0.019 1.287 10.054
==============================================================================
Omnibus: nan Durbin-Watson: 2.956
Prob(Omnibus): nan Jarque-Bera (JB): 0.769
Skew: 0.811 Prob(JB): 0.681
Kurtosis: 2.943 Cond. No. 1.00
==============================================================================
Results #2
OLS Regression Results
=======================================================================================
Dep. Variable: 0 R-squared (uncentered): 0.692
Model: OLS Adj. R-squared (uncentered): 0.641
Method: Least Squares F-statistic: 13.50
Date: Mon, 01 Mar 2021 Prob (F-statistic): 0.0104
Time: 20:39:14 Log-Likelihood: -5.8073
No. Observations: 7 AIC: 13.61
Df Residuals: 6 BIC: 13.56
Df Model: 1
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
x1 0.2308 0.063 3.674 0.010 0.077 0.384
==============================================================================
Omnibus: nan Durbin-Watson: 0.148
Prob(Omnibus): nan Jarque-Bera (JB): 0.456
Skew: 0.000 Prob(JB): 0.796
Kurtosis: 1.750 Cond. No. 1.00
==============================================================================
This is as far I have got, but I think something is wrong. Neither of these regression outcome seems to show the y-intercept. Also, I expect the coef in results #2 to be 0 since I expect the slope to be 0 when all the values are 1, but the result shows 0.2308. Any suggestions or guiding material will be greatly appreciated.
In statsmodels an OLS model does not fit an intercept by default (see the docs).
exog array_like A nobs x k array where nobs is the number of observations and k is the number of regressors. An intercept is not included by default and should be added by the user. See statsmodels.tools.add_constant.
The documentation on the exog argument of the OLS constructor suggests using this feature of the tools module in order to add an intercept to the data.
To perform a hypothesis test on the values of the coefficients this question provides some guidance. This unfortunately only works if the variances of the residual errors is the same.
We can start by looking at whether the residuals of each distribution have the same variance (using Levine's test) and ignore coefficients of the regression model for now.
import numpy as np
import pandas as pd
from scipy.stats import levene
from statsmodels.tools import add_constant
from statsmodels.formula.api import ols ## use formula api to make the tests easier
np.inf == float('inf')
data1 = [1, 3, 45, 0, 25, 13, 43]
data2 = [1, 1, 1, 1, 1, 1, 1]
df1 = add_constant(pd.DataFrame(data1)) ## add a constant column so we fit an intercept
df1 = df1.reset_index() ## just doing this to make the index a column of the data frame
df1 = df1.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
df2 = add_constant(pd.DataFrame(data2)) ## this does nothing because the y column is already a constant
df2 = df2.reset_index()
df2 = df2.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
formula1 = 'y ~ x + const' ## define formulae
formula2 = 'y ~ x'
model1 = ols(formula1, df1).fit()
model2 = ols(formula2, df2).fit()
print(levene(model1.resid, model2.resid))
The output of the levene test looks like this:
LeveneResult(statistic=7.317386741297884, pvalue=0.019129208414097015)
So we can reject the null hypothesis that the residual distributions have the same variance at alpha=0.05.
There is no point to testing the linear regression coefficients now because the residuals don't have don't have the same distributions. It is important to remember that in a regression problem it doesn't make sense to compare the regression coefficients independent of the data they are fit on. The distribution of the regression coefficients depends on the distribution of the data.
Lets see what happens when we try the proposed test anyways. Combining the instructions above with this method from the OLS package yields the following code:
## stack the data and addd the indicator variable as described in:
## stackexchange question:
df1['c'] = 1 ## add indicator variable that tags the first groups of points
df_all = df1.append(df2, ignore_index=True).drop('const', axis=1)
df_all = df_all.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
df_all = df_all.fillna(0) ## a bunch of the values are missing in the indicator columns after stacking
df_all['int'] = df_all['x'] * df_all['c'] # construct the interaction column
print(df_all) ## look a the data
formula = 'y ~ x + c + int' ## define the linear model using the formula api
result = ols(formula, df_all).fit()
hypotheses = '(c = 0), (int = 0)'
f_test = result.f_test(hypotheses)
print(f_test)
The result of the f-test looks like this:
<F test: F=array([[4.01995453]]), p=0.05233934453138028, df_denom=10, df_num=2>
The result of the f-test means that we just barely fail to reject any of the null hypotheses specified in the hypotheses variable namely that the coefficient of the indicator variable 'c' and interaction term 'int' are zero.
From this example it is clear that the f test on the regression coefficients is not very powerful if the residuals do not have the same variance.
Note that the given example has so few points it is hard for the statistical tests to clearly distinguish the two cases even though to the human eye they are very different. This is because even though the statistical tests are designed to make few assumptions about the data but those assumption get better the more data you have. When testing statistical methods to see if they accord with your expectations it is often best to start by constructing large samples with little noise and then see how well the methods work as your data sets get smaller and noisier.
For the sake of completeness I will construct an example where the Levene test will fail to distinguish the two regression models but f test will succeed to do so. The idea is to compare the regression of a noisy data set with its reverse. The distribution of residual errors will be the same but the relationship between the variables will be very different. Note that this would not work reversing the noisy dataset given in the previous example because the data is so noisy the f test cannot distinguish between the positive and negative slope.
import numpy as np
import pandas as pd
from scipy.stats import levene
from statsmodels.tools import add_constant
from statsmodels.formula.api import ols ## use formula api to make the tests easier
n_samples = 6
noise = np.random.randn(n_samples) * 5
data1 = np.linspace(0, 30, n_samples) + noise
data2 = data1[::-1] ## reverse the time series
df1 = add_constant(pd.DataFrame(data1)) ## add a constant column so we fit an intercept
df1 = df1.reset_index() ## just doing this to make the index a column of the data frame
df1 = df1.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
df2 = add_constant(pd.DataFrame(data2)) ## this does nothing because the y column is already a constant
df2 = df2.reset_index()
df2 = df2.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
formula1 = 'y ~ x + const' ## define formulae
formula2 = 'y ~ x'
model1 = ols(formula1, df1).fit()
model2 = ols(formula2, df2).fit()
print(levene(model1.resid, model2.resid))
## stack the data and addd the indicator variable as described in:
## stackexchange question:
df1['c'] = 1 ## add indicator variable that tags the first groups of points
df_all = df1.append(df2, ignore_index=True).drop('const', axis=1)
df_all = df_all.rename(columns={'index':'x', 0:'y'}) ## the old index will now be called x and the old values are now y
df_all = df_all.fillna(0) ## a bunch of the values are missing in the indicator columns after stacking
df_all['int'] = df_all['x'] * df_all['c'] # construct the interaction column
print(df_all) ## look a the data
formula = 'y ~ x + c + int' ## define the linear model using the formula api
result = ols(formula, df_all).fit()
hypotheses = '(c = 0), (int = 0)'
f_test = result.f_test(hypotheses)
print(f_test)
The result of Levene test and the f test follow:
LeveneResult(statistic=5.451203655948632e-31, pvalue=1.0)
<F test: F=array([[10.62788052]]), p=0.005591319998324387, df_denom=8, df_num=2>
A final note since we are doing multiple comparisons on this data and stopping if we get a significant result, i.e. if the Levene test rejects the null we quit, if it doesn't then we do the f test, this is a stepwise hypothesis test and we are actually inflating our false positive error rate. We should correct our p-values for multiple comparisons before we report our results. Note that the f test is already doing this for the hypotheses we test about the regression coefficients. I am a bit fuzzy on the underlying assumptions of these testing procedures so I am not 100% sure that you are better off making the following correction but keep it in mind in case you feel you are getting false positives too often.
from statsmodels.sandbox.stats.multicomp import multipletests
print(multipletests([1, .005591], .05)) ## correct out pvalues given that we did two comparisons
The output looks like this:
(array([False, True]), array([1. , 0.01115074]), 0.025320565519103666, 0.025)
This means we rejected the second null hypothesis under the correction and that the corrected p-values looks like [1., 0.011150]. The last two values are corrections to your significance level under two different correction methods.
I hope this helps anyone trying to do this type of work. If anyone has anything to add I would welcome comments. This isn't my area of expertise so I could be making some mistakes.