13.2 Test of Independence

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Sometimes, categorical data can be organized by a two-way classification or contingency table. Each cell (intersection of row and column) is the observed count.

$H_{o} : The row and column variables are independent$

$H_{a} : The row and column variables are dependent$

The Chi-square test statistic is given by:

$χ^{2} = \sum_{i, j = 1}^{r c} \frac{(O_{i j} - E_{i j})^{2}}{E_{i j}}$

where

$r$ and $c$ are the number of rows and columns respectively

$E_{i j} = \frac{r_{i} c_{j}}{n}$ where $r_{i}, c_{j}$ row i, column j totals.

$df = (r - 1) (c - 1)$

Note: In order to perfrom the Chi-Square test of independence, we assume:

samples are randomly selected
sample data consist of frequency counts for each two-way table
For each cell, the expected frequency is at least 5

Use the function chi2_test_of_indp below with argument: each column from the observed frequency combined as obs = [c1, c2,...]. Where c1 is the list from column 1, c2 is the list from column 2...

def chi2_test_of_indp(obs):
    from scipy import stats
    chi2, p, dof, ex = stats.chi2_contingency(obs)
    
    print("chi-square test statitic:",chi2)
    print("p-value:", p)

Example: The contingency table below shows a random sample of white, black and Hispanic college students based on whether their family borrowed money to pay their college education. Test the hypothesis that borrowing money for college and race are independent at 1% level of significance. (adapted from Larson and Farber)

race/borrow	Yes	No
White	49	64
Black	85	123
Hispanic	85	180

$H sub 0:$ Borrowing money for college and race are independent

$H a :$ Borrowing money for college and race are dependent

c1 = [49,85,85]
c2 = [64,123,180]

obs = [c1,c2]

chi2_test_of_indp(obs)

Show expected output

chi-square test statitic: 5.993527985414218
p-value: 0.04994844064430499

Swapping the column and row still gives the same result (see below).

c1 = [49,64]
c2 = [85,123]
c3 = [85,180]

obs = [c1,c2,c3]
chi2_test_of_indp(obs)

Show expected output

chi-square test statitic: 5.993527985414217
p-value: 0.04994844064430501

At $α equals 0.05$ , the p-value is less than.05. Reject the null hypothesis that race and borrowing college money are independent.

Example: The contingency table below shows the effects of Covid19 by race in the USA. Test the hypothesis that effects of Covid19 is independent of race at 5% level of significance.

Covid19	White	Native	Asian	Black	Hispanic
Cases	100	170	70	110	190
Hospitalization	100	340	100	280	280
Death	100	240	100	200	230

Use the chi2_test_of_indp function below.

def chi2_test_of_indp(obs):
    from scipy import stats
    chi2, p, dof, ex = stats.chi2_contingency(obs)
    
    print("chi-square test statistic:",chi2)
    print("p-value:", p)

r1 = [100,170,70,110,190]
r2 = [100,340,100,280,280]
r3 = [100,240,100,200,230]

obs = [r1,r2,r3]

chi2_test_of_indp(obs)

Show expected output

chi-square test statistic: 36.38048143358525
p-value: 1.4955960363094355e-05

Interpretation: Since the p-value 1.4955960363094355e-05 is less than 5%, reject the null hypothesis. That is, there is enough evidence to reject the claim that the effects of Covid10 and race are independent.

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.