# The $\chi^2$-Test¶

Recall that we have recently been discussing relationships between variables. For example, linear regression examines the relationship between two numerical variables. Similarly, the $\chi^2$-test examines the relationship between categorical variables.

## The two types of $\chi^2$ tests¶

As we'll see, there are two somewhat different types of $\chi^2$-tests. Specifically, there's

• the $\chi^2$-test for homogeneity, which tests whether frequency counts for a single categorical variable are distributed similarly across different populations and
• the $\chi^2$-test for independence, which tests whether there is a significant association between two categorical variables from a single population.

### Imports from scipy.stats¶

The two types of $\chi^2$ tests each has its own function in the scipy.stats module. There's:

• chisquare for homogeneity and
• chi2_contingency for independence.

Let's go ahead and import those now:

In [1]:
from scipy.stats import chisquare, chi2_contingency


## A basic example for homogeneity¶

Here's a potentially important question: Is a given pool of potential jurors in a county racially representative of that county?

### Specific data¶

Here's some specific data representing 275 jurors in a small county. Jurors identified their racial group, as shown in the table below. We would like to determine if these jurors are racially representative of the population.

Race White Black Hispanic Other Total
Representation in juries 205 26 25 19 275
Percentages for registered voters 0.72 0.07 0.12 0.09 1.00
Expected count 198 19.25 33 24.75 275

### Using chisquare¶

The chisquare function built for exactly this situation and it's pretty easy to use:

In [2]:
chisquare([205, 26, 25, 19], f_exp = [198.0, 19.25, 33.0, 24.75])

Out[2]:
Power_divergenceResult(statistic=5.8896103896103895, pvalue=0.11710619130850619)

There's a lot going on in the background here but, ultimately, we are interested in that $p$-value. If we are looking for a 95% confidence level, then we are unable to reject the null hypothesis here, in spite of the deviation from expected counts that we see in the data.

### Some formulae¶

The $p$-value is computed using the $\chi^2$ statistic, which we find as follows:

We suppose that we are to evaluate whether there is convincing evidence that a set of observed counts $O_1$, $O_2$, ..., $O_k$ in $k$ categories are unusually different from what might be expected under a null hypothesis. Call the \emph{expected counts} that are based on the null hypothesis $E_1$, $E_2$, ..., $E_k$. If each expected count is at least 5 and the null hypothesis is true, then the test statistic below follows a chi-square distribution with $k-1$ degrees of freedom: $$\chi^2 = \frac{(O_1 - E_1)^2}{E_1} + \frac{(O_2 - E_2)^2}{E_2} + \cdots + \frac{(O_k - E_k)^2}{E_k}$$ We then evaluate the area under the tail of the $\chi^2$-distribution with $k-1$ degrees of freedom. In the example above, the $\chi^2$-statistic is

In [3]:
ch_sq = ((205-198)**2/198 + (26-19.25)**2/19.25 +(25-33)**2/33 + (19-24.75)**2/24.75)
ch_sq

Out[3]:
5.8896103896103895

Geometrically, this represents the area under the curve below and to the right of $5.88$:

Note that we can use a table like the one in the back of our text to assess the null-hypothesis.

## An example for independence¶

Sometimes, we have two categorical variables and we want to know if they are independent or not. This is also called the Chi-Square test for homogeneity. One example from the R-Tutorial examines whether exercise and smoking are independent of one another using the following data:

Smokes/Exercises Frequently Some None
Never 435 420 90
Occasionally 60 20 15
Regularly 45 35 5
Heavily 35 15 5

The rows indicate how much the participant smokes and the columns indicate how much they exercise. Our null hypothesis is that these are independent; our alternative hypothesis is contrary.

We can enter a small table like this into Python and get the $p$-value with chi2_contingency as follows:

In [4]:
A = [
[435,420,90],
[60,20,15],
[45,35,5],
[35,15,5]
]
chi2_contingency(A)[1]

Out[4]:
0.00011960587467155845

It looks like we reject the null hypothesis of independence.