Chi Square Goodness of Fit Calculator

Hypothesis Test

Chi-Square Goodness of Fit Calculator

Enter your observed counts for each category, set the hypothesized distribution, then calculate.

Assume equal expected proportions across all categories

What Is a Chi-Square Goodness-of-Fit Test?

The chi-square goodness-of-fit test determines whether the distribution of a single categorical variable in your sample matches a hypothesized or claimed distribution across its categories. It compares the counts you actually observed in each category to the counts you would expect if the hypothesized proportions were exactly true. A large gap between observed and expected counts is evidence against the hypothesized distribution.

When to Use It

You have one categorical variable with two or more categories (e.g., die faces, product color, survey response option).
You want to compare your sample's distribution to a known, claimed, or theoretical distribution — equal proportions or specific percentages.
Your data is counts, not percentages or continuous measurements.
Each observation falls into exactly one category.
Expected counts in each category are large enough for the approximation to hold — generally at least 5.

This test examines a single categorical variable against a hypothesized distribution. If you're instead asking whether two categorical variables are related to each other, use our Chi-Square Test of Independence Calculator.

The Formula

\[ \chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i} \] \[ E_i = n \times p_i \] \( df = k - 1 \)

\(O_i\) is the observed count in category \(i\); \(E_i\) is the count expected under the hypothesized distribution; \(p_i\) is the hypothesized proportion for category \(i\); \(n\) is the total number of observations; \(k\) is the number of categories.

Worked Example

A retailer claims their scented candles sell in the ratio Red 30%, Blue 25%, Green 25%, Yellow 20%. In a sample of 150 candles sold, they recorded: Red 60, Blue 45, Green 25, Yellow 20.

	Red	Blue	Green	Yellow
Observed	60	45	25	20
Expected %	30%	25%	25%	20%

Total observed: n = 150.
Expected counts: \(E = n \times p\), e.g. Red: \(150 \times 0.30 = 45\); computed the same way for all 4 categories (45, 37.5, 37.5, 30).
Chi-square contributions: 5.0 + 1.5 + 4.1667 + 3.3333 = 14.0.
Degrees of freedom: \(df = k - 1 = 4 - 1 = 3\).
Critical value at \(\alpha = 0.05\), df = 3, is 7.8147. Since 14.0 > 7.8147 (p = 0.0029), we reject the null hypothesis — the observed sales distribution does not match the claimed 30/25/25/20 ratio.

Your Results

Chi-Square (χ²)

p-value

Critical Value

Total N

Cohen's w

Decision

Step 1 — Total Observations

Step 2 — Expected Counts

Step 3 — Chi-Square Contribution per Category

Step 4 — Sum the Contributions

Step 5 — Degrees of Freedom

Step 6 — Compare to the Critical Value

Chi-square distribution Rejection region Test statistic Critical value

Category	Observed (O)	Expected %

Chi Square Goodness of Fit Calculator