11.4286
1
80
40
40
45
35
22.5
17.5
22.5
17.5
17.5
0.378
11.4286
1
80
40
40
45
35
22.5
17.5
22.5
17.5
17.5
0.378
The Chi-Square Test of Independence Calculator determines whether two categorical variables are statistically associated. This non-parametric test compares observed frequencies in a contingency table to the frequencies expected under the null hypothesis of independence.
Enter the four cell counts of your 2×2 contingency table to compute the chi-square statistic, expected frequencies, degrees of freedom, and Cramér's V effect size measure.
The chi-square test of independence evaluates whether the distribution of one categorical variable differs across levels of another. The procedure compares observed cell frequencies to expected frequencies calculated under the assumption of independence.
For each cell, the expected frequency is:
$$E_{ij} = \frac{R_i \times C_j}{N}$$
Where \(R_i\) is the row total, \(C_j\) is the column total, and \(N\) is the grand total. The chi-square statistic is then:
$$\chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$$
For a 2×2 table, df = (r−1)(c−1) = 1. The statistic follows a chi-square distribution under the null hypothesis. A shortcut formula for 2×2 tables exists:
$$\chi^2 = \frac{N(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$$
Cramér's V measures the strength of association: \(V = \sqrt{\chi^2 / N}\) for a 2×2 table. V ranges from 0 (no association) to 1 (perfect association).
Interpreting the chi-square test involves:
Inputs
Results
Testing association between smoking status and lung disease. χ² = 10.42 >> 3.841 (critical at α=0.05), indicating a significant association.
Inputs
Results
Testing whether gender is related to product preference. χ² = 0.25 < 3.841, so we cannot reject independence — no significant association.
Use this test when you have two categorical variables and want to determine if they are associated. Both variables should have two or more levels, and the data should consist of frequency counts (not percentages or means). Examples include testing whether gender is associated with voting preference, or whether treatment group is related to outcome category.
A common rule of thumb is that all expected frequencies should be at least 5. When this condition is violated, the chi-square approximation may be inaccurate. For 2×2 tables with small expected frequencies, Fisher's exact test is the preferred alternative. Some statisticians use a threshold of 1 with Yates' correction applied.
Yates' correction adjusts for the fact that the chi-square distribution is continuous while the test statistic is discrete. The corrected formula subtracts 0.5 from each |O − E| before squaring: χ² = Σ(|O−E| − 0.5)²/E. It is conservative and primarily used for 2×2 tables with small sample sizes. Many modern statisticians prefer Fisher's exact test over Yates' correction.
The test of independence examines whether two categorical variables are associated (using a contingency table). The goodness of fit test examines whether a single categorical variable follows a specified distribution. Independence uses a two-way table; goodness of fit uses a one-way table comparing observed to hypothesized frequencies.
No. The standard chi-square test of independence requires independent observations. For paired or matched categorical data (e.g., before/after measurements on the same subjects), use McNemar's test instead. McNemar's test specifically analyzes the discordant pairs in a 2×2 table of matched data.
Chi-square is influenced by sample size — a larger sample will produce a larger χ² even if the association strength is the same. Cramér's V normalizes for sample size and table dimensions, providing a standardized measure of association strength from 0 to 1. This makes it useful for comparing association strength across studies with different sample sizes.
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