Hughes and Hase Problem 8.4: Is this British national lottery fair?¶
Test the hypothesis that the distribution of numbered balls in 106 drawings of the British Lottery is uniform.
Use the procedure outlined at top of p.112 in Hughes and Hase, 2and calculate the $\chi^2$ statistic:
$$ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} $$where $O_i$ represents the number of occurrences of something in the sample data, and $E_i$ gives the expected number of occurrences based on the null hypothesis.
import numpy as np
from scipy import stats
import matplotlib as mpl
import matplotlib.pyplot as plt
# Following is an Ipython magic command that puts figures in notebook.
%matplotlib notebook
# M.L. modification of matplotlib defaults
# Changes can also be put in matplotlibrc file,
# or effected using mpl.rcParams[]
plt.style.use('classic')
plt.rc('figure', figsize = (6, 4.5)) # Reduces overall size of figures
plt.rc('axes', labelsize=16, titlesize=14)
plt.rc('figure', autolayout = True) # Adjusts supblot params for new size
Data set¶
data = np.array([11,11,13,14,11,22,15,9,9,16,\
17,12,8,13,8,15,9,13,19,9,\
12,10,17,13,10,9,10,15,9,14,\
16,17,11,13,14,11,13,21,14,13,\
12,11,16,13,10,18,16,16,8])
np.mean(data), np.sum(data), 106*6, len(data)
Formulate the null hypothesis¶
The distribution is uniform.
Calculate expected number of occurrences based on null hypothesis¶
Let's consider the probability of getting a 9 in a single draw of 6 balls. The 9 could the first ball drawn, and the balls 2-6 would be anything else (not 9). The probability for this is $P_1 = 1/49$. Or the 9 could be the second ball drawn, or it could be the third ball drawn, etc. The probability of a 9 being drawn in one of the slots is
$$ P = P_1 + P_2 + P_3 + P_4 + P_5 + P_6 = 6P_1 = \frac{6}{49}. $$The expected number 9's in 106 draws is thus
$$ 106\times P = 106 \times\frac{6}{49} = 12.98. $$Nines aren't special if the numbers are distributed uniformly, the expected values ($E_i$) are all the same.
e = 106*6/49*np.ones(49)
Obeserved values ($O_i$)¶
This is the data given by Hughes and Hase in the problem statement.
o = data
Make a "histogram" of the observed data along with expectations¶
plt.figure()
x = np.linspace(1,49,49)
plt.bar(x, o, label='observed', width=0.5, alpha = 0.5)
plt.scatter(x, e, color='red', label='expected')
plt.xlabel('value')
plt.ylabel('occurrences')
plt.title('1000 Rolls')
plt.xlim(1,49)
plt.legend(loc='upper left');
Calculate $\chi^2$ statistic¶
chi2_data = np.sum((o-e)**2/e)
print('chi2 = ',chi2_data)
print('reduced chi2 = ',chi2_data/49)
Calculate the probability of getting a value of $\chi^2$ greater than value determined from data¶
p = 1 - stats.chi2.cdf(chi2_data, 49)
print('probability of getting value of chisq greater than ',chi2_data,'is ',p)
Our value of $\chi^2$ is very reasonable. Don't reject null hypothesis based on this data.¶
Version information¶
version_information is from J.R. Johansson (jrjohansson at gmail.com); see Introduction to scientific computing with Python for more information and instructions for package installation.
version_information is installed on the linux network at Bucknell
%load_ext version_information
%version_information numpy, scipy, matplotlib