# Analysis of categorical data¶

• Analysis of one proportion
• Chi-square test
• Fisher exact test
• Cochran's Q test
• McNemar test

Author: Thomas Haslwanter, Date: Feb-2017

In :
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats


## Analysis of one proportion¶

Calculate the confidence intervals of the population, based on a given data sample.

Suppose a general practitioner chooses a random sample of 215 women from the patient register for her general practice, and finds that 39 of them have a history of suffering from asthma. What is the confidence interval for the prevalence of asthma? (The data are taken from Altman, chapter 10.2.1:)

In :
# Get the data
numTotal = 215
numPositive = 39

# Calculate the confidence intervals
p = float(numPositive)/numTotal
se = np.sqrt(p*(1-p)/numTotal)
td = stats.t(numTotal-1)
ci = p + np.array([-1,1])*td.isf(0.025)*se

# Print them
print('ONE PROPORTION')
print('The confidence interval for the given sample is {0:5.3f} to {1:5.3f}'.format(
ci, ci))

ONE PROPORTION
The confidence interval for the given sample is 0.130 to 0.233


## Chi-square test to a 2x2 table¶

Data are taken from Altman, Table 10.10:

Comparison of number of hours swimming by swimmers with or without erosion of dental enamel:

>= 6h: 32 yes, 118 no
<  6h: 17 yes, 127 no



The calculations are done with and without Yate's continuity correction.

In :
# Enter the data
obs = np.array([[32, 118], [17, 127]])

# Calculate the chi-square test
chi2_corrected = stats.chi2_contingency(obs, correction=True)
chi2_uncorrected = stats.chi2_contingency(obs, correction=False)

# Print the result
print('CHI SQUARE')
print('The corrected chi2 value is {0:5.3f}, with p={1:5.3f}'.format(chi2_corrected, chi2_corrected))
print('The uncorrected chi2 value is {0:5.3f}, with p={1:5.3f}'.format(chi2_uncorrected, chi2_uncorrected))

CHI SQUARE
The corrected chi2 value is 4.141, with p=0.042
The uncorrected chi2 value is 4.802, with p=0.028


## Fisher's Exact Test¶

Spectacle wearing among juvenile delinquensts and non-delinquents who failed a vision test

• Spectecle wearers: 1 delinquent, 5 non-delinquents
• non-spectacle wearers: 8 delinquents, 2 non-delinquents'''

(Data are taken from Altman, Table 10.14)

In :
# Enter the data
obs = np.array([[1,5], [8,2]])

# Calculate the Fisher Exact Test
fisher_result = stats.fisher_exact(obs)

# Print the result
print('The probability of obtaining a distribution at least as extreme '
+ 'as the one that was actually observed, assuming that the null ' +
'hypothesis is true, is: {0:5.3f}.'.format(fisher_result))

The probability of obtaining a distribution at least as extreme as the one that was actually observed, assuming that the null hypothesis is true, is: 0.035.


## Cochran's Q test¶

12 subjects are asked to perform 3 tasks. The outcome of each task is "success" or "failure". The results are coded 0 for failure and 1 for success. In the example, subject 1 was successful in task 2, but failed tasks 1 and 3. Is there a difference between the performance on the three tasks?

In :
from statsmodels.sandbox.stats.runs import cochrans_q
import pandas as pd

[1,1,1,0,0,1,0,1,1,1,1,1],
[0,0,1,0,0,1,0,0,0,0,0,0]])

# I prefer a DataFrame here, as it indicates directly what the values mean

# --- >>> START stats <<< ---
(Q, pVal) = cochrans_q(df)
# --- >>> STOP stats <<< ---

print('Q = {0:5.3f}, p = {1:5.3f}'.format(Q, pVal))
if pVal < 0.05:
print("There is a significant difference between the three tasks.")

Q = 8.667, p = 0.013
There is a significant difference between the three tasks.


## McNemar test¶

McNemars Test should be run in the "exact" version, even though approximate formulas are typically given in the lecture scripts. Just ignore the statistic that is returned, because it is different for the two options.

In the following example, a researcher attempts to determine if a drug has an effect on a particular disease. Counts of individuals are given in the table, with the diagnosis (disease: present or absent) before treatment given in the rows, and the diagnosis after treatment in the columns. The test requires the same subjects to be included in the before-and-after measurements (matched pairs).

In :
from statsmodels.sandbox.stats.runs import mcnemar

f_obs = np.array([[101, 121],[59, 33]])
(statistic, pVal) = mcnemar(f_obs)

print('p = {0:5.3e}'.format(pVal))
if pVal < 0.05:
print("There was a significant change in the disease by the treatment.")

p = 4.434e-06
There was a significant change in the disease by the treatment.