Test whether the observed data follow the Poisson distribution (Test of the goodness of fit of the Poisson distribution by Python)

Problem setting

In a supermarket, ** the number of customer complaints (per day) ** was aggregated for $ 365 $ a day, and the results were as follows.

Can we think of this data as ** following a Poisson distribution **? Test at the significance level $ 5 % $.

Visualization

Let's graph the "** expected frequency " and " observed frequency **" calculated by hypothesizing that they follow the Poisson distribution.

"** Expected frequency " is calculated as follows. First, multiply the "number of complaints" by the " observation frequency " and divide by the number of observation days $ 365 $ to get the ** average number of complaints per day ** $ \ mu = 2.931 . Then, find the probability mass ( k = 0.7 $) for the ** Poisson distribution ** with an average of $ 2.931 $, and multiply it by $ 365 $ for the number of observation days to obtain " expected frequency **".

Visualization of observed and expected frequencies

import numpy as np
import scipy.stats as st
import japanize_matplotlib
import matplotlib.pyplot as plt

n = np.arange(0,7+1)
f = np.array((22,49,82,86,63,39,16,8)) #frequency

mu = (n*f).sum()/f.sum() #Calculate average number of claims 2.931

Po = st.poisson(mu=mu) # μ=2.Poisson distribution of 931
xp = Po.pmf(k=n)       # k=1...Probability mass of 7
xp[-1] = 1 - st.poisson.cdf(n[-2],mu=mu) # k>=Probability mass of 7 k=Integrated into 7
fp = xp * f.sum()      #Expected frequency

plt.figure(figsize=(6,3),dpi=120,facecolor='white')
plt.bar(n,f,width=0.9, label='Observation frequency')
plt.plot(n,fp,'o--', c='tab:orange', label=f'Expected frequency$Po\,(\\lambda={mu:.2f})$')
plt.tick_params(axis='x',length=0)
plt.legend(framealpha=1, fancybox=False)
plt.gca().set_axisbelow(True)
plt.grid(axis='y')
plt.show()

The execution result is as follows.

Goodness of fit test

Is the hypothesis that "the object follows a Poisson distribution" (null hypothesis) correct? Is evaluated by ** statistical test **.

If this hypothesis is correct, then the following ** statistic ** $ t calculated from ** observation frequency ** $ \ boldsymbol {x} $ and ** hypothesis-based expected frequency ** $ \ boldsymbol {x} _p $ The test is performed using the property that $ follows (approximately) ** chi-square distribution ** (degrees of freedom $ n-2 $).

In the current problem, the number of claims is in the $ 8 $ range from $ 0 $ to $ 7 $, so ** degrees of freedom ** is $ n-2 = 8-2 = 6 $.

Let's draw a probability density function for a chi-square distribution with $ 6 $ degrees of freedom.

Chi-square distribution with 6 degrees of freedom

import numpy as np
import scipy.stats as st
import japanize_matplotlib
import matplotlib.pyplot as plt

x = np.linspace(0,25,200)
p = st.chi2.pdf(x, df=6)

x_p95 = st.chi2.ppf(0.95, df=6)

y_range = (0.00,0.16)
plt.figure(figsize=(6,3),dpi=120,facecolor='white')
plt.plot(x,p)
plt.vlines(x_p95,*y_range,lw=1,ls=':',color='tab:red')

x2 = np.linspace(0,x_p95,200)
plt.fill_between(x2,st.chi2.pdf(x2,df=6),np.zeros(len(x2)),facecolor='tab:blue',alpha=0.3)

plt.text(5,0.05,'0.95',ha='center',va='center')
plt.text(x_p95,-0.004,f'{x_p95:.2f}',c='tab:red', ha='center',va='top')

plt.text(x_p95,0.08,f'→ Rejection area',c='tab:red')

arrowprops=dict(arrowstyle='->',connectionstyle='angle,angleA=0,angleB=60, rad=1')
kw = dict(xycoords='data',textcoords='data',ha='left', arrowprops=arrowprops)
plt.gca().annotate('0.05', xy=(13.3, 0.003), xytext=(15.5,0.016), **kw)

plt.xlim(0,25)
plt.ylim(*y_range)

plt.xlabel('Statistics$t$')
plt.show()

The execution result is as follows.

In fact, find the statistic $ t $ and its corresponding $ p $ value.

Calculation of test statistic and p-value

t = ( ((f-fp)**2)/fp ).sum()        #Statistic t
p = 1 - st.chi2.cdf(t,df=len(n)-2)  #6 degrees of freedom(8-2)P-value corresponding to t in the chi-square distribution of

print(f'Statistic t= {t:.2f}, P-value= {p:.2f} ',end='')
if p >= 0.05 :
  print('（ >= 0.05 ) \n Null hypothesis (following Poisson distribution) is not rejected')
else :
  print('（ < 0.05 ) \n Null hypothesis (following Poisson distribution) is rejected')

The execution result is as follows.

Statistic t= 3.22, p-value= 0.78 （ >= 0.05 ) 
Null hypothesis (following Poisson distribution) is not rejected