Independence test

Chi-square test of independence

The chi-square test of independence helps you decide whether two categorical variables are independent (unrelated) or whether there’s a statistically significant association between them. You use this test when your data are frequency counts (how many observations fall into each category combination) for two categorical variables.

Definitions

Chi-square test of independence

A hypothesis test used to determine whether two categorical variables, organized in a contingency table, are independent or associated.

Observed frequency ($O_{ij}$)

The actual count in the cell at the intersection of row $i$ and column $j$.

Expected frequency ($E_{ij}$)

The count expected in cell $(i,j)$ if the two variables were independent.

Hypotheses:

Null hypothesis ($H_0$): The two variables are independent and there is no association between them. Knowing the value of one variable provides no information about the value of the other.

Alternative hypothesis ($H_a$): The two variables are not independent and there is an association between them. The distribution of one variable depends on the distribution of the other.

When to use this test

Some situations where the chi-square test of independence is appropriate:

1. Is a customer’s preference for a product independent of the state that the customer lives in?
2. Is level of education independent of opinion on a policy?
3. Is the number of defective products independent of the time of day they were made?

Key equations

Definitions

Chi-square test statistic

Measures the total discrepancy between observed and expected frequencies. A large ${\chi }^2$ value indicates a large discrepancy, providing evidence against the null hypothesis of independence.

$$\begin{array}{rl}{\chi }^2& =\sum \frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}\end{array}$$

Expected frequency

The count expected in a cell if the two variables were independent.

$$\begin{array}{rl}{E}_{ij}& =\frac{({\text{Row Total}}_i)({\text{Column Total}}_j)}{\text{Grand Total}}\end{array}$$

Degrees of freedom ($df$)

Defines the shape of the chi-square distribution used for the test, where $r$ is the number of rows and $c$ is the number of columns.

$$\begin{array}{rl}df& =(r-1)(c-1)\end{array}$$

Cramér’s V

A measure of the strength of the association (effect size).

$$\begin{array}{rl}V& =\sqrt{\frac{{\chi }^2}{n\cdot (\min (r,c)-1)}}\end{array}$$

Decision rule

• If ${\chi }_{\text{calculated}}^2>{\chi }_{\text{critical}}^2$: Reject the null hypothesis $H_0$ and conclude there is a statistically significant association between the variables.
• If ${\chi }_{\text{calculated}}^2\le {\chi }_{\text{critical}}^2$: Fail to reject $H_0$. There is not enough evidence to conclude an association exists.

For the test to be valid, the following assumptions must be met:

1. Both variables must be categorical.
2. The observations must be independent of one another (one person’s response doesn’t influence another person’s response).
3. No more than $20\%$ of the expected cells should have a frequency less than $5$, and no expected cell count should be less than $1$. Otherwise, the test may not be reliable.

Practice problem

Example:

A researcher wants to know if there is an association between a person’s age group and their preferred social media platform. They survey $300$ people and obtain the following results:

	Instagram	TikTok	Facebook	Total
Teens	$60$	$40$	$10$	$110$
Adults	$40$	$30$	$120$	$190$
Total	$100$	$70$	$130$	$300$

Part a.

State the hypotheses.

Solution:

Part b.

Calculate the expected frequencies $E_{ij}$.

Use the formula:

$$\begin{array}{rl}{E}_{ij}& =\frac{({\text{Row Total}}_i)({\text{Column Total}}_j)}{\text{Grand Total}}\end{array}$$

Compute for each cell:

$$\begin{array}{rl}{E}_{T,IG}& =\frac{(110)(100)}{300}\\ & =\frac{11,000}{300}\\ & \approx 36.67\end{array}$$

$$\begin{array}{rl}{E}_{T,TT}& =\frac{(110)(70)}{300}\\ & =\frac{7,700}{300}\\ & \approx 25.67\end{array}$$

$$\begin{array}{rl}{E}_{T,FB}& =\frac{(110)(130)}{300}\\ & =\frac{14,300}{300}\\ & \approx 47.67\end{array}$$

$$\begin{array}{rl}{E}_{A,IG}& =\frac{(190)(100)}{300}\\ & =\frac{19,000}{300}\\ & \approx 63.33\end{array}$$

$$\begin{array}{rl}{E}_{A,TT}& =\frac{(190)(70)}{300}\\ & =\frac{13,300}{300}\\ & \approx 44.33\end{array}$$

$$\begin{array}{rl}{E}_{A,FB}& =\frac{(190)(130)}{300}\\ & =\frac{24,700}{300}\\ & \approx 82.33\end{array}$$

Solution:

Part c.

Calculate the test statistic ${\chi }^2$.

Use the formula:

$$\begin{array}{rl}{\chi }^2& =\sum \frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}\end{array}$$

Compute each term:

$$\begin{array}{rl}\frac{(60-36.67{)}^2}{36.67}& =\frac{(23.33{)}^2}{36.67}\\ & =\frac{544.29}{36.67}\\ & \approx 14.84\end{array}$$

$$\begin{array}{rl}\frac{(40-25.67{)}^2}{25.67}& =\frac{(14.33{)}^2}{25.67}\\ & =\frac{205.35}{25.67}\\ & \approx 8.00\end{array}$$

$$\begin{array}{rl}\frac{(10-47.67{)}^2}{47.67}& =\frac{(-37.67{)}^2}{47.67}\\ & =\frac{1,419.03}{47.67}\\ & \approx 29.78\end{array}$$

$$\begin{array}{rl}\frac{(40-63.33{)}^2}{63.33}& =\frac{(-23.33{)}^2}{63.33}\\ & =\frac{544.29}{63.33}\\ & \approx 8.59\end{array}$$

$$\begin{array}{rl}\frac{(30-44.33{)}^2}{44.33}& =\frac{(-14.33{)}^2}{44.33}\\ & =\frac{205.35}{44.33}\\ & \approx 4.63\end{array}$$

$$\begin{array}{rl}\frac{(120-82.33{)}^2}{82.33}& =\frac{(37.67{)}^2}{82.33}\\ & =\frac{1,419.03}{82.33}\\ & \approx 17.24\end{array}$$

Sum the terms:

$$\begin{array}{rl}{\chi }^2& \approx 14.84+8.00+29.78+8.59+4.63+17.24\end{array}$$

Solution:

Part d.

Determine the degrees of freedom $df$.

Use the formula:

$$\begin{array}{rl}df& =(r-1)(c-1)\end{array}$$

There are $r=2$ rows (Teens, Adults) and $c=3$ columns (Instagram, TikTok, Facebook).

$$\begin{array}{rl}df& =(2-1)(3-1)\\ & =(1)(2)\end{array}$$

Solution:

Part e.

Make a decision using the decision rule.

Compare the test statistic to the critical value from a chi-square distribution table with $df=2$ and a significance level of $\alpha =0.05$:

$$\begin{array}{rl}{\chi }_{\text{critical}}^2& =5.991\end{array}$$

Compare the values:

$$\begin{array}{rl}83.08& >5.991\end{array}$$

Solution:

Part f.

Interpret the results in context.

Solution: