Chi-Square Test: Formula, Example and Degrees of Freedom

The chi-square test (written χ², and also called the chi-squared test) is a test on counts: how many times each category turns up in a sample. It sets those counts, the observed frequencies, against the ones you should see if a starting hypothesis were true, the expected frequencies. Karl Pearson published it in 1900. MacTutor, the history of mathematics archive at the University of St Andrews, says Pearson’s aim with the test was to end the normal distribution’s central role.

One calculation answers two questions. A test of independence asks whether two ways of classifying the same items are related; a goodness of fit test asks whether a sample behaves the way a chosen distribution predicts.

Null and alternative hypotheses in a chi-square test

Chi-square is a good place to see how probability and statistics work together: a number computed from the data is checked against a probability distribution to choose between two hypotheses. The first, the null hypothesis H₀, is the one the calculation puts on trial.

In a test of independence, the NIST handbook of statistical methods states the null hypothesis as the two classifications being independent. Put another way, the chance of landing in a given cell equals the chance of its row multiplied by the chance of its column, the same multiplication rule used for independent events when you calculate probability. The alternative hypothesis claims the product fails in at least one cell, which would mean the classifications depend on each other.

For goodness of fit, H₀ says the data follow the stated distribution and the alternative says they depart from it.

The chi-square test formula

For every cell, take the observed frequency O and the expected frequency E, square the difference and divide by E. The statistic is the sum of all those pieces:

χ² = Σ (O − E)² / E

In a test of independence the expected counts come from the table’s totals. For a single cell:

E = (row total × column total) / grand total

A large statistic means the observed counts sit far from the expected ones. Yale’s statistics course reads this, for goodness of fit, as a sign that the proposed model describes the data badly. When the null hypothesis holds, the value roughly follows a chi-square distribution, which NIST defines as the sum of the squares of ν independent standard normal variables, ν being the degrees of freedom.

Chi-square test example: wafer defects across three shifts

NIST, the US federal standards agency, works through a factory case. A plant records 309 defects on wafers, sorted by type (A, B, C, D) and by production shift (1, 2, 3), and wants to know whether the mix of defects changes from one shift to the next. The two classifications make a contingency table with 3 rows and 4 columns; expected frequencies are in brackets.

Shift	Type A	Type B	Type C	Type D	Total
1	15 (22.51)	21 (20.99)	45 (38.94)	13 (11.56)	94
2	26 (22.99)	31 (21.44)	34 (39.77)	5 (11.81)	96
3	33 (28.50)	17 (26.57)	49 (49.29)	20 (14.63)	119
Total	74	69	128	38	309

Practical example: for the cell «shift 1, type A» the row total is 94, the column total 74 and the grand total 309. The expected count is 94 × 74 / 309 = 22.51. Fifteen defects were actually observed, so this cell adds (15 − 22.51)² / 22.51 to the sum.

Repeat that for all twelve cells and add them up, and NIST arrives at χ² = 19.18. The table has (3 − 1)(4 − 1) = 6 degrees of freedom, and at a 0.05 significance level the critical value for 6 degrees is 12.5916. Since 19.18 is higher, the independence hypothesis is rejected: in NIST’s reading, the data give significant evidence that the share of each defect type changes with the shift. The example’s p-value is 0.00387.

Degrees of freedom in the chi-square test

Degrees of freedom count how many cells can still vary once the totals are fixed. In a goodness of fit test with k categories there are k − 1. Yale’s course points out that the last count has no room to move, because the counts have to add up to the total. Each parameter estimated from the data costs one more, so with d estimated parameters you end up at k − 1 − d.

NIST writes the same rule as k − c, with k counting the classes that are not empty and c equal to the estimated parameters plus one. In its normality example two parameters are estimated, so c is 3.

For a contingency table, (r − 1)(c − 1), where c now stands for columns, follows from the same reasoning. There are r·c cells, and after the constraint on the grand total rc − 1 of them are free; subtract the (r − 1) + (c − 1) parameters estimated for the row and column totals and exactly (r − 1)(c − 1) remain.

Chi-square goodness of fit test: the pea seed example

Geneticists use the goodness of fit version to check whether counts match an expected ratio. Phillip McClean’s genetics course at North Dakota State University takes a set of pea seeds whose predicted ratio is 9:3:3:1, and counts 556 of them.

Seed class	Observed	Expected
Round and yellow (9/16)	315	312.75
Round and green (3/16)	108	104.25
Wrinkled and yellow (3/16)	101	104.25
Wrinkled and green (1/16)	32	34.75
Total	556	556

Each expected count is the ratio’s fraction applied to the total, so 9/16 of 556 gives 312.75. Four classes mean 3 degrees of freedom, and the statistic comes to 0.47. At 3 degrees of freedom the table puts the 0.05 critical value at 7.82 (7.815 in the NIST table), and 0.47 corresponds to a probability above 0.90. The page concludes that the data are consistent with the 9:3:3:1 ratio.

NIST adds two cautions for this use. The test needs data grouped into classes, so loose observations first have to go into frequency tables for categorical variables, and the value of the statistic depends on how those classes are drawn. Discrete distributions, the binomial and the Poisson among them, can be tested this way too.

Chi-square independence test vs goodness of fit test

	Test of independence	Goodness of fit
Question	Are two classifications related?	Do the data follow the predicted distribution?
Data	Contingency table, r rows and c columns	One classification into k classes
Null hypothesis	The classifications are independent	The data follow the stated distribution
Expected frequencies	Row total × column total / grand total	Predicted probability × total
Degrees of freedom	(r − 1)(c − 1)	k − 1, minus estimated parameters
Example	Wafer defects by shift (NIST)	Pea seeds, 9:3:3:1 ratio

How to read the result

Pearson’s chi-square test looks at one tail only, the upper one. You find the column for 1 − α in the table and reject the null hypothesis when the statistic is larger than the value listed there. A few thresholds from the NIST table:

Degrees of freedom	α = 0.05	α = 0.01
1	3.841	6.635
2	5.991	9.210
3	7.815	11.345

McClean’s page treats 0.05 as the conventional threshold.

A significant result tells you an association is there. Bewick, Cheek and Ball, who wrote a review of tests of association in Critical Care in 2004, caution that the test by itself does not say how strong it is. To find the cells that matter most, look at the standardized residuals, which are built on the gap between observed and expected; any beyond 2 in absolute value are significant.

Practical example: in the review by Bewick and colleagues, 1706 patients with a central venous catheter are classified by catheter site and by outcome, three categories each. The expected count for the cell «internal jugular, no complications» is 1305 × 934 / 1706 = 714.5. The nine contributions add up to 51.26; with 4 degrees of freedom the critical value for P = 0.001 is 18.47, so P < 0.001. The authors write that the result suggests an association between catheter site and infectious complication.

When the chi-square test is the wrong choice

The chi-square distribution is an approximation, and it only works when expected counts are big enough. NIST says the approximation needs a large enough sample to be valid. Bewick and colleagues put a number on it, with the rule they attribute to Cochran: no cell may have an expected count under 1, and at most 20% of cells may fall under 5. Their way out is to combine categories where that makes sense. For 2×2 tables with few cases, the options are Yates’s continuity correction and Fisher’s exact test. Yates’s correction raises the smallest frequency by 0.5 and adjusts the others so the totals stay unchanged.

Practical example: a trial of intravenous nitrate in heart attack patients enrolled 95 people; 3 of the 50 treated patients died, against 8 of the 45 controls. Chi-square gives 3.209 with 1 degree of freedom, P = 0.073. On the same data Fisher’s exact test gives a two-sided P of 0.140, while Yates’s correction brings the statistic down to 2.162 and pushes P up to 0.141.

In large samples chi-square, Fisher and Yates land close together; in small ones Fisher and Yates return higher, more cautious P values. There is no full agreement on which to prefer; Bewick and colleagues report that Bland recommends Fisher or Yates for a more conservative approach.

Paired data are a separate case, for instance the same patients measured twice. Chi-square ignores the pairing, so these tables go to McNemar’s test.