The chi square table of critical values is a fundamental tool in statistical hypothesis testing, especially when evaluating categorical data. Worth adding: researchers, analysts, and students rely on this table to decide whether an observed distribution differs significantly from an expected one. By comparing a calculated chi‑square statistic to the appropriate critical value, you can determine if the null hypothesis should be rejected at a chosen significance level. This article explains what the table represents, how to interpret its entries, and walks through practical examples so you can apply the chi‑square test confidently in your own work.
The official docs gloss over this. That's a mistake And that's really what it comes down to..
What Is the Chi‑Square Test?
The chi‑square test assesses the goodness‑of‑fit between observed frequencies and expected frequencies under a specific hypothesis. It is non‑parametric, meaning it does not assume a normal distribution of the underlying data. That said, the test statistic follows a chi‑square distribution, which is defined by its degrees of freedom (df). The shape of this distribution changes with df, which is why critical values differ across rows of the table.
Key Concepts
- Observed frequency (O): The actual count recorded in each category.
- Expected frequency (E): The count you would anticipate if the null hypothesis were true.
- Chi‑square statistic (χ²): Calculated as Σ[(O − E)² / E] across all categories.
- Degrees of freedom (df): For a goodness‑of‑fit test, df = (number of categories − 1). For a test of independence in a contingency table, df = (rows − 1) × (columns − 1).
- Significance level (α): The probability of rejecting the null hypothesis when it is true; common choices are 0.05, 0.01, and 0.001.
- Critical value: The threshold from the chi‑square table that the test statistic must exceed to reject H₀ at the given α and df.
How the Chi‑Square Table Is Structured
A typical chi‑square table of critical values lists degrees of freedom down the leftmost column and significance levels (or corresponding tail probabilities) across the top row. Each cell contains the χ² value that separates the acceptance region from the rejection region for that combination of df and α Most people skip this — try not to..
Example Layout (illustrative only)
| df \ α | 0.Consider this: 10 | 0. Think about it: 05 | 0. Day to day, 025 | 0. 01 | 0.Here's the thing — 005 |
|---|---|---|---|---|---|
| 1 | 2. 71 | 3.84 | 5.02 | 6.63 | 7.88 |
| 2 | 4.61 | 5.99 | 7.38 | 9.21 | 10.60 |
| 3 | 6.And 25 | 7. 81 | 9.35 | 11.34 | 12. |
And yeah — that's actually more nuanced than it sounds.
Note: The numbers above are rounded for readability; actual tables provide more decimal places.
When you look up a critical value, you first locate the row matching your degrees of freedom, then move horizontally to the column representing your chosen α. The intersecting figure is the critical χ².
Step‑by‑Step Guide to Using the Table
-
State the hypotheses
- H₀: Observed frequencies follow the expected distribution (or variables are independent).
- H₁: At least one category deviates from expectation (or variables are associated).
-
Calculate the chi‑square statistic
Compute χ² = Σ[(O − E)² / E] for all cells. -
Determine degrees of freedom
- Goodness‑of‑fit: df = k − 1 (k = number of categories).
- Test of independence: df = (r − 1)(c − 1) (r = rows, c = columns).
-
Choose a significance level (α)
Commonly 0.05; adjust based on the cost of Type I error. -
Locate the critical value
Find the row for your df, then the column for your α. Record the χ²₍crit₎. -
Make a decision
- If χ² > χ²₍crit₎ → reject H₀ (significant difference/association).
- If χ² ≤ χ²₍crit₎ → fail to reject H₀ (no sufficient evidence).
-
Report results
Include χ² value, df, p‑value (if available), and the conclusion in plain language Small thing, real impact..
Worked Example: Goodness‑of‑Fit
A genetics researcher expects a 3:1 phenotypic ratio (dominant:recessive) in a pea plant cross. Out of 200 plants, 150 show the dominant trait and 50 show the recessive trait.
-
Expected counts
- Dominant: 200 × ¾ = 150
- Recessive: 200 × ¼ = 50
-
Chi‑square calculation
- Dominant: (150 − 150)² / 150 = 0
- Recessive: (50 − 50)² / 50 = 0
- χ² = 0 + 0 = 0
-
Degrees of freedom
df = 2 − 1 = 1 -
Choose α = 0.05
-
Critical value (df = 1, α = 0.05) → 3.84
-
Decision
χ² = 0 ≤ 3.84 → fail to reject H₀. The observed ratio matches the expected 3:1 ratio.
Worked Example: Test of Independence
A survey examines whether preference for a new product (Like/Dislike) is related to age group (Under 30,
