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Chi-Squared Test Guide

What should I know about Chi-Squared Formula and Steps?

χ² = Σ (O − E)² / E. Where O = observed frequency, E = expected frequency. Steps: state the null hypothesis (H₀: no significant difference / variables are independent). Calculate expected frequencies. Calculate (O−E)²/E for each category. Sum all values to get χ². Find degrees of freedom (df = number of categories − 1 for goodness-of-fit). Compare to critical value at p = 0.05 (5% significance level). If χ² > critical value: reject H₀ (result is statistically significant). If χ² is less than the critical value: fail to reject H₀ — there is not enough evidence to conclude the observed and expected frequencies differ significantly.

What do I need to know about Degrees of Freedom and Critical Values?

For goodness-of-fit: df = number of categories − 1. Critical values at p = 0.05: df 1: 3.841, df 2: 5.991, df 3: 7.815, df 4: 9.488, df 5: 11.070. Example: genetics cross expecting 3:1 ratio (2 categories) → df = 1 → critical value = 3.841. If calculated χ² = 2.1: 2.1 < 3.841, do not reject H₀ — data consistent with expected ratio. If χ² = 5.2: 5.2 > 3.841, reject H₀ — data significantly differs from expected 3:1 ratio.

What should I know about Chi-Squared in Biology (Genetics)?

The chi-squared test is used to test whether observed genetic ratios match expected Mendelian ratios. Example: cross expected to give 3:1 phenotype ratio. Observed: 72 dominant, 28 recessive (n=100). Expected: 75 dominant, 25 recessive. χ² = (72−75)²/75 + (28−25)²/25 = 9/75 + 9/25 = 0.12 + 0.36 = 0.48. df = 1, critical value = 3.841. 0.48 < 3.841 — do not reject H₀. The observed results are consistent with the expected 3:1 Mendelian ratio. Always state what the null hypothesis is before calculating chi-squared, since the whole test is a comparison against that specific expected ratio.

What's the key thing to understand about Assumptions and Limitations?

Chi-squared assumptions: data must be frequencies (counts), not proportions or percentages. Expected frequency in each category must be ≥ 5 (small expected frequencies inflate χ² artificially). Observations must be independent. The test does not work for continuous data (use t-test or ANOVA instead). For a 2×2 contingency table with small samples (any expected count < 5), use Fisher's exact test instead. Chi-squared only detects whether differences exist — it does not indicate the direction or magnitude of any relationship, which is why it's often paired with other descriptive statistics for a fuller picture.

What should I know about Choosing Between Chi-Squared Tests?

Chi-squared has two common uses that are easy to confuse. A goodness-of-fit test checks whether one categorical dataset matches an expected distribution (e.g. are dice rolls uniformly distributed across six faces?) — degrees of freedom = categories−1. A test of independence checks whether two categorical variables are related (e.g. is smoking status independent of lung disease?), using a contingency table where degrees of freedom = (rows−1)×(columns−1). Both use the same core formula, Σ(O−E)²/E, but the expected values E are calculated differently: goodness-of-fit uses a theoretical proportion, while independence tests use row-total×column-total/grand-total for each cell. Picking the wrong test for your data structure is a common statistics error — if you have one categorical variable compared to a theoretical model, use goodness-of-fit; if you have two categorical variables measured on the same subjects, use the independence test.

What do I need to know about Expected Frequency Requirements?

The chi-squared test's results are only reliable when expected frequencies are large enough — the standard rule of thumb is that no expected cell value should fall below 5. If this condition is violated (common with small sample sizes or rare categories), the chi-squared approximation to the true sampling distribution becomes inaccurate, and Fisher's exact test should be used instead, especially for 2×2 tables. This is a frequently overlooked assumption: a chi-squared value can be calculated and will produce a p-value regardless of sample size, but that p-value is only trustworthy when the expected-frequency condition holds. In practice this means chi-squared tests need a reasonably large sample — a rough guide is a total sample size of at least 20, and ideally spread so no category is too thin — before the result should be treated as statistically valid rather than a rough guide.

What's the difference between Goodness-of-Fit and Test of Independence?

The chi-squared statistic is used for two related but distinct kinds of tests. A goodness-of-fit test checks whether a single categorical variable's observed frequencies match an expected theoretical distribution — for example, checking whether a die's roll outcomes match the expected 1/6 probability for each face. A test of independence checks whether two categorical variables are related to each other, using a contingency table — for example, testing whether smoking status and lung disease diagnosis are associated in a sample of patients. Both use the same underlying formula, Σ(observed - expected)²/expected, but differ in how "expected" values are calculated: goodness-of-fit uses a specified theoretical proportion, while a test of independence calculates expected values from the table's own row and column totals, assuming the two variables are unrelated, then checks how far the actual data strays from that assumption.

What should I know about Chi-Squared Critical Values at Different Significance Levels?

The critical value a calculated chi-squared statistic must exceed to be 'significant' depends on both the degrees of freedom and the chosen significance level (how strict the test is):

A stricter significance level (p=0.01 rather than p=0.05) demands a larger chi-squared value before rejecting the null hypothesis, reducing the chance of a false positive at the cost of needing stronger evidence.

What should I know about a Worked Example: Test of Independence?

As a worked example of a test of independence (rather than goodness-of-fit), suppose a survey of 200 people records whether they exercise regularly and whether they report good sleep quality: 60 exercise and sleep well, 40 exercise but sleep poorly, 50 don't exercise but sleep well, and 50 don't exercise and sleep poorly. Row totals: 100 exercisers, 100 non-exercisers. Column totals: 110 good sleepers, 90 poor sleepers. Grand total: 200. Expected count for each cell = (row total × column total)/grand total. Exercise & good sleep: (100×110)/200 = 55. Exercise & poor sleep: (100×90)/200 = 45. No exercise & good sleep: 55. No exercise & poor sleep: 45. χ² = Σ(O−E)²/E = (60−55)²/55 + (40−45)²/45 + (50−55)²/55 + (50−45)²/45 = 25/55+25/45+25/55+25/45 ≈ 0.4545+0.5556+0.4545+0.5556 = 2.02. Degrees of freedom for a 2×2 independence test = (rows−1)×(columns−1) = 1, so the critical value at p=0.05 is 3.841. Since 2.02 < 3.841, there is not enough evidence to reject the null hypothesis that exercise and sleep quality are independent in this sample.

What should I know about p-values and Chi-Squared?

A chi-squared statistic on its own is just a number — its meaning comes from converting it into a p-value, the probability of observing a chi-squared value at least that large purely by chance if the null hypothesis were actually true. Comparing a calculated χ² directly against a critical value is mathematically equivalent to comparing a p-value against a significance level such as 0.05, but modern statistical software usually reports the p-value directly rather than requiring a lookup table. A smaller p-value means the observed data would be rarer under the null hypothesis, so a p-value below the chosen significance level leads to rejecting the null hypothesis. It's worth remembering that 'statistically significant' is not the same as 'practically important' — with a very large sample size, even a tiny, practically meaningless deviation from the expected frequencies can produce a significant chi-squared result, which is why effect size measures (such as Cramér's V for contingency tables) are often reported alongside the raw chi-squared statistic and its p-value.

What should I know about Yates' Continuity Correction?

For a 2×2 contingency table, especially with smaller sample sizes, the standard chi-squared formula can slightly overstate significance because it approximates a discrete (count-based) distribution with a continuous one. Yates' continuity correction adjusts for this by subtracting 0.5 from each |O−E| difference before squaring: χ²_Yates = Σ(|O−E|−0.5)²/E. This correction makes the test slightly more conservative (less likely to find significance), and is most relevant for small 2×2 tables — for larger tables or larger sample sizes, the correction's effect becomes negligible and is often omitted in modern statistical practice.

What should I know about One-Tailed vs Two-Tailed Testing?

The chi-squared test is inherently one-tailed: because every term in the formula is squared, both an observed frequency that's too high and one that's too low contribute equally to a larger χ², so there's no separate 'two-tailed' version to choose between, unlike a z-test or t-test where the direction of a difference matters.