Chi-Square Tests

Overview

Chi-square tests are used for categorical data to test goodness-of-fit (single variable) and independence (two variables).

Test Statistic

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

Where:

• $O$ = observed frequency
• $E$ = expected frequency

Goodness-of-Fit Test

Tests whether observed frequencies match expected frequencies.

Hypotheses

• $H_0$: The data follows the specified distribution
• $H_1$: The data does not follow the specified distribution

Expected Frequencies

$$E = n \times p$$

Where $n$ = total observations, $p$ = hypothesized proportion

Degrees of Freedom

$$df = k - 1$$

Where $k$ = number of categories

Test of Independence

Tests whether two categorical variables are associated.

Hypotheses

• $H_0$: The variables are independent
• $H_1$: The variables are associated (dependent)

Expected Frequencies

$$E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$$

Degrees of Freedom

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

Where $r$ = rows, $c$ = columns

Conditions

1. Random sampling
2. Independent observations
3. Expected frequencies $\geq 5$ (all cells)

Examples

Example 1: Goodness-of-Fit

Testing if a die is fair (600 rolls):

Face	1	2	3	4	5	6
$O$	92	108	97	103	88	112
$E$	100	100	100	100	100	100

$$\chi^2 = \frac{(92-100)^2}{100} + \frac{(108-100)^2}{100} + \frac{(97-100)^2}{100} + \frac{(103-100)^2}{100} + \frac{(88-100)^2}{100} + \frac{(112-100)^2}{100}$$ $$= 0.64 + 0.64 + 0.09 + 0.09 + 1.44 + 1.44 = 4.34$$ $$df = 6 - 1 = 5, \quad \chi^2_{\text{crit}} (\alpha=0.05) = 11.07$$ $$4.34 < 11.07 \Rightarrow \text{Fail to reject } H_0$$

The die appears fair.

Example 2: Test of Independence

Survey: Gender vs Product Preference (300 people)

	Product A	Product B	Product C	Total
Male	50	40	60	150
Female	30	50	70	150
Total	80	90	130	300

Expected values:

$$E(\text{Male, A}) = \frac{150 \times 80}{300} = 40$$ $$E(\text{Male, B}) = \frac{150 \times 90}{300} = 45$$ $$E(\text{Male, C}) = \frac{150 \times 130}{300} = 65$$

And similarly for Female row.

$$\chi^2 = \frac{(50-40)^2}{40} + \frac{(40-45)^2}{45} + \frac{(60-65)^2}{65} + \frac{(30-40)^2}{40} + \frac{(50-45)^2}{45} + \frac{(70-65)^2}{65}$$ $$= 2.5 + 0.56 + 0.38 + 2.5 + 0.56 + 0.38 = 6.88$$ $$df = (2-1)(3-1) = 2, \quad \chi^2_{\text{crit}} (\alpha=0.05) = 5.99$$ $$6.88 > 5.99 \Rightarrow \text{Reject } H_0$$

Gender and product preference are associated.

Example 3: Homogeneity Test

Same calculation as independence, but tests if distributions are the same across groups.

Interpretation

For Goodness-of-Fit

• Large $\chi^2$ → Poor fit → Reject $H_0$
• Small $\chi^2$ → Good fit → Fail to reject $H_0$

For Independence

• Large $\chi^2$ → Variables are associated
• Small $\chi^2$ → Variables appear independent

Effect Size: Cramér's V

For test of independence:

$$V = \sqrt{\frac{\chi^2}{n \times \min(r-1, c-1)}}$$

$V$	Interpretation
0.1	Small
0.3	Medium
0.5	Large

Yates' Correction

For 2×2 tables:

$$\chi^2 = \sum \frac{(\lvert O - E \rvert - 0.5)^2}{E}$$

Reduces Type I error for small samples.