Chi-Square Tests

The z-tests and t-tests you have learned so far work well for comparing means and proportions — but what about categorical data? When your variable is not a number but a category (eye color, political party, treatment outcome, product preference), you cannot compute a mean or a standard deviation. Instead, you work with counts — how many observations fall into each category. The chi-square test (pronounced “ky-square” and written $\chi^2$) is the standard tool for testing hypotheses about categorical data. In this lesson you will learn two versions: the goodness-of-fit test for one categorical variable, and the test of independence for two categorical variables.

The Chi-Square Statistic

Both chi-square tests use the same core statistic:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

where $O$ is the observed count (what you actually counted in the data) and $E$ is the expected count (what you would expect if the null hypothesis were true). The sum is taken over all categories or cells.

The logic is straightforward:

• For each category, compute how far the observed count is from the expected count
• Square the difference (so negative and positive deviations both contribute positively)
• Divide by the expected count (a difference of 10 matters more when the expected count is 20 than when it is 2000)
• Add up all the terms

A large $\chi^2$ value means the observed data is far from what the null hypothesis predicts — evidence against $H_0$. A small $\chi^2$ value means the data fits the expected pattern well — no reason to reject $H_0$. The chi-square statistic is always non-negative (zero or positive), and the test is always right-tailed: you reject $H_0$ only when $\chi^2$ is large enough.

The degrees of freedom depend on which test you are performing, as described below.

Chi-Square Goodness-of-Fit Test

The goodness-of-fit test asks: does a single categorical variable follow a specified distribution? You have observed counts for each category, and you compare them to what you would expect under a hypothesized distribution.

• $H_0$: the distribution of the variable matches the expected proportions
• $H_a$: the distribution does not match the expected proportions
• Degrees of freedom: $df = (\text{number of categories}) - 1$

Example 1: Testing a Die for Fairness

You roll a six-sided die 120 times and record the results:

If the die is fair, each face should appear $120 / 6 = 20$ times. Is there evidence the die is unfair? Test at $\alpha = 0.05$.

Step 1: State the hypotheses.

$H_0: \text{the die is fair (each face has probability } 1/6\text{)}$

$H_a: \text{the die is not fair (at least one face differs from } 1/6\text{)}$

Step 2: Calculate the chi-square statistic.

$\chi^2 = \frac{(15-20)^2}{20} + \frac{(22-20)^2}{20} + \frac{(25-20)^2}{20} + \frac{(18-20)^2}{20} + \frac{(17-20)^2}{20} + \frac{(23-20)^2}{20}$

$= \frac{25}{20} + \frac{4}{20} + \frac{25}{20} + \frac{4}{20} + \frac{9}{20} + \frac{9}{20}$

$= 1.25 + 0.20 + 1.25 + 0.20 + 0.45 + 0.45 = 3.80$

Step 3: Find degrees of freedom and the critical value.

$df = 6 - 1 = 5$

At $\alpha = 0.05$ with $df = 5$, the critical value is $\chi^2_{0.05} = 11.070$.

Step 4: Make a decision. Since $3.80$ is less than $11.070$, we fail to reject $H_0$.

Step 5: Conclusion in context. There is no statistically significant evidence that the die is unfair. The observed variation (some faces appearing slightly more or less than 20 times) is well within what random chance would produce with a fair die.

Example 2: Customer Preference

A store manager believes customers prefer four product flavors equally. She surveys 200 customers:

$\chi^2 = \frac{(62-50)^2}{50} + \frac{(48-50)^2}{50} + \frac{(55-50)^2}{50} + \frac{(35-50)^2}{50}$

$= \frac{144}{50} + \frac{4}{50} + \frac{25}{50} + \frac{225}{50} = 2.88 + 0.08 + 0.50 + 4.50 = 7.96$

With $df = 4 - 1 = 3$ and $\alpha = 0.05$, the critical value is $\chi^2_{0.05} = 7.815$. Since $7.96 > 7.815$, we reject $H_0$. There is evidence that customer preferences are not equally distributed. Mint appears to be the least popular flavor, while vanilla is the most popular.

Chi-Square Test of Independence

The test of independence asks: are two categorical variables related (associated), or are they independent? The data is organized in a contingency table (also called a two-way table) with rows for one variable and columns for the other.

• $H_0$: the two variables are independent (no association)
• $H_a$: the two variables are not independent (there is an association)
• Expected count for each cell: $E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$
• Degrees of freedom: $df = (r - 1)(c - 1)$, where $r$ is the number of rows and $c$ is the number of columns

Example 3: Smoking and Lung Disease

A health researcher collects data on 1000 adults:

Is there an association between smoking status and lung disease? Test at $\alpha = 0.05$.

Step 1: State the hypotheses.

$H_0: \text{smoking status and lung disease are independent}$

$H_a: \text{smoking status and lung disease are associated}$

Step 2: Calculate expected counts under independence.

$E(\text{Smoker, Disease}) = \frac{300 \times 150}{1000} = 45$

$E(\text{Smoker, No Disease}) = \frac{300 \times 850}{1000} = 255$

$E(\text{Non-smoker, Disease}) = \frac{700 \times 150}{1000} = 105$

$E(\text{Non-smoker, No Disease}) = \frac{700 \times 850}{1000} = 595$

Verify: $45 + 255 = 300$ ✓, $105 + 595 = 700$ ✓, $45 + 105 = 150$ ✓, $255 + 595 = 850$ ✓.

Step 3: Calculate the chi-square statistic.

$\chi^2 = \frac{(90-45)^2}{45} + \frac{(210-255)^2}{255} + \frac{(60-105)^2}{105} + \frac{(640-595)^2}{595}$

$= \frac{2025}{45} + \frac{2025}{255} + \frac{2025}{105} + \frac{2025}{595}$

$= 45.000 + 7.941 + 19.286 + 3.403 = 75.630$

Notice that every cell has the same squared difference ($45^2 = 2025$). This happens in a 2-by-2 table when the row and column totals are fixed.

Step 4: Find degrees of freedom and the critical value.

$df = (2-1)(2-1) = 1$

At $\alpha = 0.05$ with $df = 1$, the critical value is $\chi^2_{0.05} = 3.841$.

Step 5: Make a decision. Since $75.630$ far exceeds $3.841$, we reject $H_0$.

Step 6: Conclusion in context. There is overwhelming evidence of an association between smoking status and lung disease. Smokers had a lung disease rate of $90/300 = 30\%$, compared to $60/700 = 8.6\%$ for non-smokers. While this test does not prove causation, the strong association is consistent with the well-established medical understanding that smoking increases lung disease risk.

Conditions for Chi-Square Tests

Both chi-square tests require the following conditions:

• Random sample or random assignment — the data must come from a properly designed study
• Independence — each observation must be independent of the others; one person’s response cannot influence another’s
• Expected counts are at least 5 — every cell in the table must have an expected count of 5 or more. This ensures the chi-square distribution is a good approximation for the test statistic. If any expected count is below 5, consider combining categories or using Fisher’s exact test (for 2-by-2 tables)

Note that the “at least 5” rule applies to expected counts, not observed counts. An observed count of 0 or 2 is fine as long as the expected count for that cell is at least 5.

Chi-Square Critical Values Reference Table

To use the table: find your degrees of freedom in the left column and your chosen $\alpha$ across the top. Reject $H_0$ if your calculated $\chi^2$ exceeds the table value.

Real-World Application: Nursing — Treatment Effectiveness Across Treatment Types

A nurse researcher wants to know if the effectiveness of three wound-care treatments differs among three different treatments. She classifies 360 patients by treatment outcome (healed vs. not healed) and treatment type:

Expected counts (each row total is 120, each column total is 245 or 115):

$E(\text{A, Healed}) = \frac{120 \times 245}{360} = 81.67 \qquad E(\text{A, Not Healed}) = \frac{120 \times 115}{360} = 38.33$

By symmetry (all row totals are 120), every row has the same expected counts: 81.67 healed and 38.33 not healed.

$\chi^2 = \frac{(85-81.67)^2}{81.67} + \frac{(35-38.33)^2}{38.33} + \frac{(70-81.67)^2}{81.67} + \frac{(50-38.33)^2}{38.33} + \frac{(90-81.67)^2}{81.67} + \frac{(30-38.33)^2}{38.33}$

$= \frac{11.09}{81.67} + \frac{11.09}{38.33} + \frac{136.19}{81.67} + \frac{136.19}{38.33} + \frac{69.39}{81.67} + \frac{69.39}{38.33}$

$= 0.136 + 0.289 + 1.668 + 3.553 + 0.850 + 1.810 = 8.306$

With $df = (3-1)(2-1) = 2$ and $\alpha = 0.05$, the critical value is $5.991$. Since $8.306 > 5.991$, we reject $H_0$. There is significant evidence that healing rates differ among the three treatments. Treatment C has the highest healing rate (75%), Treatment A is close behind (70.8%), and Treatment B is notably lower (58.3%). This information helps the nursing team prioritize Treatment C for patients who need the best chance of healing and investigate why Treatment B underperforms.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The chi-square statistic $\chi^2 = \sum \frac{(O-E)^2}{E}$ measures how far observed counts are from expected counts — it works exclusively with counts, not proportions or means
• The goodness-of-fit test checks whether a single categorical variable follows a specified distribution, with $df = k - 1$ where $k$ is the number of categories
• The test of independence checks whether two categorical variables are associated, with $df = (r-1)(c-1)$ and expected counts computed as $E = \frac{\text{row total} \times \text{col total}}{\text{grand total}}$
• Both tests are always right-tailed — you only reject $H_0$ when $\chi^2$ is large
• The key condition is that all expected counts must be at least 5
• A significant chi-square test tells you that the variables are associated, but it does not tell you the direction or strength — examine the observed percentages to interpret the nature of the relationship
• Chi-square tests are essential in medical research (treatment outcomes), quality control (defect patterns), survey analysis (opinion by demographic), and genetics (phenotype ratios)

Return to Statistics for more topics in this section.