14.6 ANOVA (Analysis of Variance)

One-Way ANOVA

ANOVA (analysis of variance) can be used to test whether the means of more than two groups are statistically different from each other.

Example: The data set below (URL) contains the students test1_score, gender, level, section etc.

Write a program to read the data set and display the first and last five rows.
Write a program to find test1_score means, std and size for each level

my_csv_file.csv

import pandas as pd

url='my_csv_file.csv'

x_df = pd.read_csv(url)

x_df  # print the first and last five rows

	id	gender	section	time	level	test1_score	test2_score	test3_score
0	1	female	C	17.3	low	65.4	84.3	51.9
1	2	female	D	5.1	low	72.4	86.1	52.0
2	3	female	A	2.4	low	73.8	84.2	53.1
3	4	female	A	8.8	low	70.2	87.9	52.0
4	5	male	B	15.0	low	60.3	81.4	55.5
...	...	...	...	...	...	...	...	...
175	176	male	B	13.2	high	78.8	89.3	52.5
176	177	female	B	12.6	high	76.5	83.4	50.0
177	178	male	D	8.4	high	78.1	86.3	51.0
178	179	male	C	9.5	high	76.6	86.6	52.4
179	180	male	C	16.2	high	83.1	89.4	54.1

x_df.groupby(['level']).agg({'test1_score': ['mean', 'std','size']})

	test1_score
	mean	std	size
level
high	79.613333	3.617306	60
low	68.775000	4.350282	60
medium	74.841667	3.638881	60

Example: The data set below (URL) contains the students test1_score, gender, level, section etc.

Write a program to perform ANOVA to test if there is sufficient evidence to indicate a difference in the averages of test1_scores among the three levels (low, medium, high). Use 10% level of significance.

my_csv_file.csv

import statsmodels.api as sm
from statsmodels.formula.api import ols
model = ols('test1_score ~ C(level)',data = x_df).fit()
table = sm.stats.anova_lm(model,type=2) # type 2 for equal or non-equal sample sizes
print(table)

Show expected output

             df       sum_sq      mean_sq           F        PR(>F)
C(level)    2.0  3540.854333  1770.427167  117.372972  3.555808e-33
Residual  177.0  2669.827667    15.083772         NaN           NaN

Interpretation: Since the p-value 3.555808e-33 < $α$ = 0.10, reject the null hypothesis that the averages for all three levels are equal. That is, there is a significant difference between the averages of test1_score of the three levels. The sum of squares residual is 2669.827667. The sum of squares treatment (level) is 3540.854333.

Adapted from Python for Introductory Statistics, by Simon Aman (Truman College, City Colleges of Chicago), licensed under CC BY 4.0. Changes were made: reformatted as an accessible XYZ web edition with live in-browser code cells. License: CC-BY-4.0.