Anova in Python/v3

Learn how to perform a one and two way ANOVA test using Python.

Note: this page is part of the documentation for version 3 of Plotly.py, which is not the most recent version.
See our Version 4 Migration Guide for information about how to upgrade.

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Imports¶

The tutorial below imports NumPy, Pandas, SciPy, and Statsmodels.

In [1]:

import plotly.plotly as py
import plotly.graph_objs as go
from plotly.tools import FigureFactory as FF

import numpy as np
import pandas as pd
import scipy

import statsmodels
import statsmodels.api as sm
from statsmodels.formula.api import ols

One-Way ANOVA¶

An Analysis of Variance Test or an ANOVA is a generalization of the t-tests to more than 2 groups. Our null hypothesis states that there are equal means in the populations from which the groups of data were sampled. More succinctly:

\begin{matrix} μ_{1} = μ_{2} = . . . = μ_{n} \end{matrix}

$\begin{align*} \mu_1 = \mu_2 = ... = \mu_n \end{align*}$

for $n$ groups of data. Our alternative hypothesis would be that any one of the equivalences in the above equation fail to be met.

In [2]:

moore = sm.datasets.get_rdataset("Moore", "car", cache=True)

data = moore.data
data = data.rename(columns={"partner.status" :"partner_status"})  # make name pythonic

moore_lm = ols('conformity ~ C(fcategory, Sum)*C(partner_status, Sum)', data=data).fit()
table = sm.stats.anova_lm(moore_lm, typ=2) # Type 2 ANOVA DataFrame

print(table)

                                              sum_sq    df          F  \
C(fcategory, Sum)                          11.614700   2.0   0.276958
C(partner_status, Sum)                    212.213778   1.0  10.120692
C(fcategory, Sum):C(partner_status, Sum)  175.488928   2.0   4.184623
Residual                                  817.763961  39.0        NaN

                                            PR(>F)
C(fcategory, Sum)                         0.759564
C(partner_status, Sum)                    0.002874
C(fcategory, Sum):C(partner_status, Sum)  0.022572
Residual                                       NaN

In this ANOVA test, we are dealing with an F-Statistic and not a p-value. Their connection is integral as they are two ways of expressing the same thing. When we set a significance level at the start of our statistical tests (usually 0.05), we are saying that if our variable in question takes on the 5% ends of our distribution, then we can start to make the case that there is evidence against the null, which states that the data belongs to this particular distribution.

The F value is the point such that the area of the curve past that point to the tail is just the p-value. Therefore:

\begin{matrix} P r (> F) = p \end{matrix}

$\begin{align*} Pr(>F) = p \end{align*}$

For more information on the choice of 0.05 for a significance level, check out this page.

Let us import some data for our next analysis. This time some data on tooth growth:

In [3]:

data = pd.read_csv('https://raw.githubusercontent.com/plotly/datasets/master/tooth_growth_csv')
df = data[0:10]

table = FF.create_table(df)
py.iplot(table, filename='tooth-data-sample')

Out[3]:

Two-Way ANOVA¶

In a Two-Way ANOVA, there are two variables to consider. The question is whether our variable in question (tooth length len) is related to the two other variables supp and dose by the equation:

\begin{matrix} l e n = s u p p + d o s e + s u p p \times d o s e \end{matrix}

$\begin{align*} len = supp + dose + supp \times dose \end{align*}$

In [4]:

formula = 'len ~ C(supp) + C(dose) + C(supp):C(dose)'
model = ols(formula, data).fit()
aov_table = statsmodels.stats.anova.anova_lm(model, typ=2)
print(aov_table)

                      sum_sq    df          F        PR(>F)
C(supp)           205.350000   1.0  15.571979  2.311828e-04
C(dose)          2426.434333   2.0  91.999965  4.046291e-18
C(supp):C(dose)   108.319000   2.0   4.106991  2.186027e-02
Residual          712.106000  54.0        NaN           NaN