Homogeneity test

Chi-Square Test for Homogeneity

The chi-square test for homogeneity is used to determine whether two or more populations have the same distribution of a single categorical variable. Unlike the test of independence (which uses one sample classified by two variables), the test for homogeneity uses separate samples from different populations and compares them on the same categorical variable.

Definitions

Chi-Square Test for Homogeneity: A statistical test used to compare the distribution of a categorical variable across two or more populations or groups using separate random samples from each.

Expected Frequency: The count we would expect in each cell if the null hypothesis (that all populations have the same distribution) were true. Calculated as:

$$\text{Expected}=\frac{(\text{row total})(\text{column total})}{\text{grand total}}$$

Degrees of Freedom: For a chi-square test for homogeneity, $df=(r-1)(c-1)$, where $r$ is the number of rows and $c$ is the number of columns.

When to Use It

Use a chi-square test for homogeneity when:

• You have separate random samples from two or more populations (or treatment groups).
• You want to compare the distribution of a single categorical variable across those groups.

Hypotheses for a Chi-Square Test for Homogeneity

$H_0$: The distribution of the categorical variable is the same across all populations.

$H_a$: The distribution of the categorical variable is not the same across all populations (at least one population differs).

Conditions for Inference

1. Random: The data come from separate random samples from each population (or from a randomized experiment).
2. 10% (if sampling without replacement): Each sample size is less than 10% of its respective population.
3. Large Counts: All expected counts are at least 5.

Practice problem

A manufacturing company wants to determine whether its safety training program produces the same proficiency rates across four departments. The company takes independent random samples from each department and records whether each worker is proficient or not proficient on the post-training assessment.

Department	Sampled	Proficient	Not Proficient
Welding	90	68	22
Assembly	110	75	35
Shipping	80	50	30
Quality Control	100	85	15

Test, at the $\alpha =0.05$ level, whether the distribution of proficiency is the same across all four departments.

Part a. State the hypotheses.

Part b. Organize the observed data into a two-way table.

	Welding	Assembly	Shipping	Quality Control	Total
Proficient	68	75	50	85	278
Not Proficient	22	35	30	15	102
Total	90	110	80	100	380

Part c. Check the conditions for inference.

Random: The problem states the company took independent random samples from each department. ✓

10%: Each sample is reasonably less than 10% of the workers in its department (assuming each department has at least 10 times the sample size). ✓

Large Counts: Compute expected counts using $\text{Expected}=\frac{(\text{row total})(\text{column total})}{380}$.

$\text{Expected (Proficient, Welding)}$ $\begin{array}{rl}& =\frac{(278)(90)}{380}\\ & \approx 65.84\end{array}$

$\text{Expected (Proficient, Assembly)}$ $\begin{array}{rl}& =\frac{(278)(110)}{380}\\ & \approx 80.47\end{array}$

$\text{Expected (Proficient, Shipping)}$ $\begin{array}{rl}& =\frac{(278)(80)}{380}\\ & \approx 58.53\end{array}$

$\text{Expected (Proficient, Quality Control)}$ $\begin{array}{rl}& =\frac{(278)(100)}{380}\\ & \approx 73.16\end{array}$

$\text{Expected (Not Proficient, Welding)}$ $\begin{array}{rl}& =\frac{(102)(90)}{380}\\ & \approx 24.16\end{array}$

$\text{Expected (Not Proficient, Assembly)}$ $\begin{array}{rl}& =\frac{(102)(110)}{380}\\ & \approx 29.53\end{array}$

$\text{Expected (Not Proficient, Shipping)}$ $\begin{array}{rl}& =\frac{(102)(80)}{380}\\ & \approx 21.47\end{array}$

$\text{Expected (Not Proficient, Quality Control)}$ $\begin{array}{rl}& =\frac{(102)(100)}{380}\\ & \approx 26.84\end{array}$

Part d. Find the chi-square test statistic and p-value.

Using the TI-84:

1. Enter the observed counts into Matrix [A] (2 rows × 4 columns).
2. Press {`STAT`} → {`TESTS`} → {`χ²-Test`}.
3. Set Observed: [A], Expected: [B] (the calculator will fill [B] automatically).
4. Highlight {`Calculate`} and press {`ENTER`}.

The degrees of freedom: $df=(2-1)(4-1)=3$.

Part e. Interpret the results in context.

Compare the p-value to $\alpha =0.05$.