Analysis of variance

 4.4 variance analysis

 4.4.1 One way ANOVA and python Realization

* One way ANOVA ： Consider only one factor in an experiment A The change of

example 4.22 use 4 Technology A1, A2, A3,A4 produce lamps , Several bulbs made by various processes were taken out to measure their service life , Results such as table 4.10 Shown , Try to infer whether there is a significant difference in the service life of bulbs made by these processes .

See you for the explanation P151——P155

import numpy as np
import statsmodels.api as sm
y=np.array([1620, 1670, 1700, 1750, 1800, 1580, 1600, 1640, 1720,
            1460, 1540, 1620, 1680, 1500, 1550, 1610])
x=np.hstack([np.ones(5), np.full(4,2), np.full(4,3), np.full(3,4)])// Modify this row when adding data , This row is to build the table 
d= {'x':x,'y':y}   # Construct a dictionary 
model = sm.formula.ols("y~C(x)",d).fit()   # Build the model 
anovat = sm.stats.anova_lm(model)  # One way ANOVA 
print(anovat)

 

PR=0.042004<0.05, Therefore, there is a significant difference in the service life of bulbs made by these processes

 

example 4.23（ Balanced data ） To investigate 5 Whether the labor productivity of workers is the same , Recorded each person 4 Days of production , Determine whether there is a significant difference in their productivity . 

import numpy as np
import pandas as pd
import statsmodels.api as sm
df = pd.read_excel("Pdata4_23.xlsx", header=None)
a=df.values.T.flatten()
print(a)
b=np.arange(1,6)// Modify this line directly 
x=np.tile(b,(4,1)).T.flatten()
print(x)
d={'x':x,'y':a} # Construct the dictionary needed for solving 
model = sm.formula.ols("y~C(x)",d).fit()  # Build the model 
anovat = sm.stats.anova_lm(model)  # One way ANOVA 
print(anovat)

 

PR=0.110913>0.05, So there is no significant difference

4.4.2 Two factor analysis of variance and python Realization

 1. mathematical model

It is divided into ： Two factor ANOVA with interaction effect and two factor ANOVA without interaction effect

（ Combined with the F Test method ）

 2. Realization

# Program files Pex4_24.py
import numpy as np
import statsmodels.api as sm
y=np.array([[11, 11, 13, 10], [10, 11, 9, 12],
         [9, 10, 7, 6], [7, 8, 11, 10],
         [5, 13, 12, 14], [11, 14, 13, 10]]).flatten()
A=np.tile(np.arange(1,5),(6,1)).flatten()# Temperature minute 4 Group 
B=np.tile(np.arange(1,4).reshape(3,1),(1,8)).flatten()# Concentration points 3 Group 
# Be careful , When modifying data A,B All have to be revised. , Such as 
#y=np.array([[11, 11, 13, 10,1], [10, 11, 9, 12,1],
#         [9, 10, 7, 6,1], [7, 8, 11, 10,1],
#         [5, 13, 12, 14,1], [11, 14, 13, 10,1]]).flatten()
#A=np.tile(np.arange(1,6),(6,1)).flatten()# Temperature minute 4 Group 
#B=np.tile(np.arange(1,4).reshape(3,1),(1,10)).flatten()# Concentration points 3 Group 
d={'x1':A,'x2':B,'y':y}
model = sm.formula.ols("y~C(x1)+C(x2)+C(x1):C(x2)",d).fit()  # Notice how the interaction formula is written 
anovat = sm.stats.anova_lm(model)  # Two way ANOVA was performed 
print(anovat)