How to Report Chi-Square Test Results in APA Format

You ran a chi-square test, SPSS gave you a table full of numbers, and now you need one sentence that satisfies your supervisor and APA 7. That sentence has five required parts, and most students leave out at least two of them. This guide walks through exactly what to report, with worked examples you can adapt directly.

What APA 7 Requires for Chi-Square Results

A complete chi-square report contains five pieces of information: the test statistic (χ²), the degrees of freedom, the sample size, the exact p value, and an effect size. APA 7 puts the degrees of freedom and sample size inside parentheses after the symbol, like this:

χ²(1, N = 200) = 6.85, p = .009, V = .19

A few notation rules trip students up. The symbol is the Greek letter chi, with a superscript 2. Statistical symbols such as p, N, and V are italicized. Report p values to two or three decimal places without a leading zero (.009, not 0.009), because p cannot exceed 1. Write p < .001 only when the value is genuinely smaller than .001; otherwise give the exact number. SPSS displays "Sig. = .000" when the value rounds to zero at three decimals, and copying ".000" into your paper is an error graders catch immediately.

Sample size deserves a word here. Unlike a t test, where df tells the reader roughly how many participants you had, chi-square degrees of freedom come from the number of categories, not the number of people. A χ² with 1 df could come from 40 participants or 4,000. That is why APA asks for N inside the parentheses.

Worked Example: Chi-Square Test of Independence

Suppose you surveyed 200 undergraduates about whether they prefer online or in-person classes, and you want to know whether preference is related to gender. (If you need a refresher on how the test itself works, see our chi-square test explained guide.) Here is the observed data:

Gender	Prefers online	Prefers in-person	Total
Female	62	48	110
Male	34	56	90
Total	96	104	200

Among women, 56.4% preferred online classes (62 of 110). Among men, the figure was 37.8% (34 of 90). The chi-square test asks whether a gap that size could plausibly arise by chance if gender and preference were unrelated in the population.

Running the test gives χ² = 6.85 with 1 degree of freedom (calculated as [rows − 1] × [columns − 1] = 1 × 1). The p value is .009. Since .009 falls below the conventional .05 threshold, the association is statistically significant.

Notice what belongs in the write-up beyond the statistics. APA expects you to describe the pattern in plain language, usually with the relevant percentages, so the reader knows the direction of the association. A bare "the test was significant" tells nobody anything useful. Which group preferred what?

Adding Cramér's V

Significance answers "is there an association?" Effect size answers "how strong is it?" For chi-square, the standard effect size is Cramér's V, calculated as:

V = √(χ² / (N × df*))

where df* is the smaller of (rows − 1) and (columns − 1). For a 2×2 table, df* = 1, and V reduces to the phi coefficient. Plugging in our numbers: V = √(6.85 / 200) = .19.

Interpretation benchmarks for a 2×2 table follow Cohen's guidelines for phi: .10 is small, .30 is medium, .50 is large. Our .19 sits between small and medium, so the honest interpretation is that gender and class-format preference are related, but gender explains only a modest portion of the variation in preference. Reporting V forces this honesty. A χ² of 6.85 on 20,000 participants would produce V = .02, an association too weak to matter for any practical decision, yet the p value would still look impressive. You can compute V in seconds with our Effect Size Calculator, and our effect size guide covers the interpretation benchmarks for other tests too.

Here is the complete APA sentence for this example:

A chi-square test of independence showed a significant association between gender and class format preference, χ²(1, N = 200) = 6.85, p = .009, V = .19. Women were more likely to prefer online classes (56.4%) than men (37.8%), though the effect was small to moderate.

Worked Example: Chi-Square Goodness-of-Fit Test

The goodness-of-fit version compares one categorical variable against a set of expected proportions, usually equal ones. Say you asked 120 students where they primarily study and got these responses:

Study location	Observed	Expected (equal)
Library	42	30
Home	38	30
Café	22	30
Campus lounge	18	30

If students had no location preference, each option would attract about 30 responses. The test gives χ² = 13.87 with df = 3 (categories minus one) and p = .003. Students clearly do not distribute themselves evenly; the library and home together account for two thirds of responses.

The reporting format is identical to the independence test, except the effect size changes. Cramér's V does not apply to a one-variable test. Use Cohen's w instead: w = √(χ²/N) = √(13.87/120) = .34, a medium effect by Cohen's benchmarks (.10 small, .30 medium, .50 large). Some papers skip the effect size for goodness-of-fit tests entirely, but including w costs you one calculation and strengthens the write-up.

A chi-square goodness-of-fit test indicated that study location preferences were not equally distributed, χ²(3, N = 120) = 13.87, p = .003, w = .34. The library was the most common choice (35.0%), followed by home (31.7%), café (18.3%), and campus lounge (15.0%).

When Expected Frequencies Fall Below 5

The chi-square test assumes that expected frequencies (not observed ones, a distinction students mix up constantly) are large enough for the approximation to hold. The usual rule: no more than 20% of cells should have an expected count below 5, and no cell should have an expected count below 1. SPSS prints this check as a footnote under the chi-square table, which is where you find the sentence "0 cells (0.0%) have expected count less than 5."

What if the assumption fails? You have three realistic options.

Fisher's Exact Test

For a 2×2 table, switch to Fisher's exact test, which computes the exact probability rather than relying on the chi-square approximation. SPSS produces it automatically for 2×2 tables. Fisher's test has no test statistic to report, so the format simplifies: Fisher's exact test indicated a significant association between treatment condition and dropout status (p = .021, two-tailed), with N = 38. You can still report Cramér's V from the crosstab as the effect size.

Combining Categories

For larger tables, merging sparse categories often solves the problem. If your "employment status" variable has 3 people in "self-employed" and 4 in "freelance," combining them into one category may lift the expected counts above the threshold. Only merge categories that make conceptual sense together, and say in your write-up that you did it and why. Silently collapsing categories after peeking at the results is a form of p-hacking.

Collecting More Data

Rarely available to students on a deadline, but worth stating: low expected counts are fundamentally a small-sample problem, and the cleanest fix is a bigger sample. Mention it as a limitation if the other two options were not workable.

Reporting Assumptions in Your Results Section

Chi-square has few assumptions compared to parametric tests, which makes them easy to report and easy to forget. State that observations were independent (each participant contributed exactly one response to one cell) and that the expected frequency condition was met. One sentence handles it:

All expected cell frequencies exceeded 5, satisfying the assumptions of the chi-square test.

If the assumption was violated and you responded to the violation, the reader needs to see both facts:

Because two cells (33.3%) had expected frequencies below 5, categories "self-employed" and "freelance" were combined into a single "independent worker" category prior to analysis.

Independence violations are more serious and cannot be patched with an alternative test from the same family. If the same participants appear in multiple cells, as happens when people select "all that apply" or when you measure the same group twice, chi-square is the wrong test altogether. Repeated measurements on the same people call for McNemar's test. Our guide to choosing a statistical test covers when each applies.

Putting It Together: The Full Results Paragraph

A complete results paragraph for the gender example moves through four beats: what test you ran and why, the assumption check, the statistical result with all five required elements, and the pattern in plain language.

A chi-square test of independence was conducted to examine the relationship between gender and preferred class format. All expected cell frequencies were greater than 5 (minimum expected count = 43.2). The association was statistically significant, χ²(1, N = 200) = 6.85, p = .009, with a small-to-moderate effect size, V = .19. Women preferred online classes at a higher rate (56.4%) than men (37.8%).

Four sentences. That is genuinely all a chi-square result needs, and padding it with restated hypotheses or software descriptions weakens it. Our broader guide on writing the results section covers how paragraphs like this fit into the full chapter. If you want the notation generated for you, paste your statistics into the APA Results Formatter and it returns the sentence with correct symbols, italics, and decimal formatting.

Should the contingency table itself go in the paper? For a 2×2, the percentages in the text usually suffice. Tables earn their space when you have three or more categories per variable, where prose descriptions of every cell become unreadable. If you include a table, do not repeat every number from it in the text; point to the table and describe the headline pattern.

Common Formatting Mistakes

These errors appear again and again in student drafts. Check your own write-up against each one.

Wrong symbol. "X² = 6.85" or "chi-square = 6.85" instead of χ². Insert the actual Greek character (in Word: Insert → Symbol, or type 03C7 then Alt+X).

Missing degrees of freedom or N. "χ² = 6.85, p = .009" is unverifiable. A reader cannot judge the result without knowing df and sample size.

No effect size. The single most common omission. APA 7 explicitly asks for effect sizes wherever possible, and many markers now deduct points for leaving them out.

Reporting p = .000. SPSS rounds; probabilities do not reach zero. Write p < .001.

Leading zeros on p and V. Write p = .009, not p = 0.009. Same for V and w, since neither can exceed 1.

Percentages without counts. "56.4% of women" means little if the reader does not know there were 110 women. Give N somewhere nearby.

Causal language. A significant chi-square shows association. "Gender influenced preference" claims more than the design supports; "gender was associated with preference" is what the test actually licenses.

Good vs. Poor Reporting, Side by Side

Poor	Good
The chi square test was significant (p<0.05), so gender affects class preference.	Gender was significantly associated with class format preference, χ²(1, N = 200) = 6.85, p = .009, V = .19.
X2 = 13.87, sig = .000. Students prefer the library.	Study locations were not chosen equally often, χ²(3, N = 120) = 13.87, p = .003, w = .34; the library was most popular (35.0%).
The result was not significant, so there is no relationship between the variables.	The association between residence type and meal plan enrollment was not significant, χ²(2, N = 150) = 3.41, p = .182, V = .15.

The third row deserves a comment. Non-significant results get reported with exactly the same five elements as significant ones, and "not significant" never means "proven unrelated." Your sample simply did not provide enough evidence to reject independence, which is a weaker claim.

Reporting conventions are the easy part to fix; interpretation errors cost more. When you can recite the template from memory, spend your remaining revision time asking whether your write-up says what the pattern means for your research question, because that is what separates a results section that lists numbers from one that earns marks.