Chi-square table of critical values

Q: Why does the chi-square table have no two-tail column?

Because χ² squares every difference, it measures how far the observed counts sit from the expected counts without recording which direction. Both kinds of disagreement push the statistic up, so all the evidence against the null sits in the upper tail. Every column of a chi-square table is already an upper-tail area, and you never halve α before looking one up.

Q: What if my df is not listed in the table?

Printed tables usually run through df = 30 and then jump to 40, 50, and higher. If your exact df is missing, use the next higher listed df. Chi-square critical values grow as df grows, so the higher row is the larger cutoff and the harder test to pass, which keeps you from claiming significance you have not earned. Note this runs opposite to the habit from a t table, where the smaller df is the conservative one. Technology avoids the choice by computing the exact value.

By Jude Wallis · Published July 30, 2026

A chi-square table lists critical values by degrees of freedom and significance level. Find your df row and your alpha column, then reject the null hypothesis when your chi-square statistic is larger than the value you read. For a two-way table with df = 2 at alpha = 0.05, that value is 5.991.

AP Statistics: Unit 3 (topics 3.14 Setting Up a Chi-Square Test for Homogeneity or Independence, 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence). Chi-square tests for homogeneity and independence are Unit 3 topics 3.14 and 3.15 in the Fall 2026 AP Statistics course. The exam supplies a formula sheet and tables for both sections, so you read the critical value rather than memorize it. The goodness-of-fit test was removed from the AP exam with the Fall 2026 course but remains standard college content.

What a chi-square critical value is

A critical value is the cutoff your test statistic has to beat. For a chi-square test the statistic is

\chi^2 = \sum \frac{(O - E)^2}{E}

where $O$ is an observed count and $E$ is the expected count under the null hypothesis. Every term is squared, so $\chi^2$ can never be negative, and it only gets larger as the observed counts drift further from what the null predicts.

The critical value, written $\chi^{2*}$ , is the point that leaves exactly $\alpha$ (the significance level) in the upper tail of the chi-square curve. If your statistic lands past it, the data are further from the null than $\alpha$ of samples would be by chance alone, and you reject. A chi-square table, sometimes called a chi-square distribution table, is a grid of those cutoffs with degrees of freedom down the side and $\alpha$ across the top.

Finding the degrees of freedom

Degrees of freedom ( $df$ ) pick the row. The chi-square distribution is a family of curves, one per $df$ , and each has its own set of cutoffs, so the wrong row gives the wrong answer even when the arithmetic is right.

For the two chi-square tests on the AP exam, both of which run on a two-way table:

df = (\text{rows} - 1)(\text{columns} - 1)

Count only the category rows and columns. The total row and total column are not part of the table for this purpose. A 2 by 3 table gives $df = (2 - 1)(3 - 1) = 2$ , and a 3 by 4 table gives $df = (3 - 1)(4 - 1) = 6$ .

A goodness-of-fit test instead uses $df = (\text{number of categories}) - 1$ . That test was removed from the AP exam starting with the Fall 2026 course, though it is still standard in a college introductory statistics course, so the row rule is worth knowing. The full comparison of the three chi-square tests covers when each one applies.

How to read the chi-square table

Three steps, in this order:

**Work out $df$ ** from the shape of your table, using the rule above.
**Pick the $\alpha$ column.** Most AP questions use $\alpha = 0.05$ . Unlike a t table, there is no separate one-tail and two-tail column here, for the reason in the next section.
Read across and down. Where the row and column meet is $\chi^{2*}$ . Reject the null hypothesis when your statistic is greater than that number.

The five values students meet most often, all at $\alpha = 0.05$ : $df = 1$ gives 3.841, $df = 2$ gives 5.991, $df = 3$ gives 7.815, $df = 4$ gives 9.488, and $df = 6$ gives 12.592.

Remember that the comparison runs the opposite way from a p-value comparison. A large statistic and a small p-value are the same evidence, so $\chi^2 > \chi^{2*}$ and $p < \alpha$ always agree.

Why there is only one tail

A t table gives you one-tail and two-tail columns. A chi-square table gives you neither, and that catches people out.

The reason is in the formula. Because every term is $(O - E)^2 / E$ , the statistic measures distance from the null, not direction. Observed counts far above what was expected and observed counts far below both push $\chi^2$ up. A small $\chi^2$ means the counts sit close to the null prediction, which is agreement, not evidence against it.

So the evidence lives entirely in the upper tail, and every column in the table is an upper-tail area. There is no such thing as a two-tailed chi-square test at this level, and you never halve $\alpha$ before looking it up.

Conditions to check before you use the table

The table describes a chi-square curve, and your statistic only follows that curve when the test's conditions hold. Check all three first:

Random: the counts come from a random sample or a randomized experiment.
Independent: when sampling without replacement, the sample is under 10% of the population.
Large counts: every expected count is at least 5.

That last one is the one to watch. The threshold applies to expected counts, not observed counts, so a cell with 2 observed is fine as long as its expected count clears 5. If some expected counts fall short, combine categories or collect more data rather than reading the table anyway. The chi-square calculator builds the expected-count table for you so the check is quick.

Look up the value without the table

A printed table only carries a handful of $\alpha$ columns, so it cannot answer "what is the exact p-value." It tells you which side of the cutoff you are on and nothing more.

When you want the actual probability, the chi-square calculator takes your two-way table and returns the statistic, the degrees of freedom, and the exact p-value. The p-value calculator does the same from a statistic and $df$ you already have. On the exam you get a chi-square table in the formula packet, so knowing how to read it stays worth the few minutes.

The table on this page is generated from the same tested distribution code the calculators use, so the two always agree.

Read the critical value for a test of homogeneity

You sample 50 students at each of three schools and record whether each one has taken a statistics course. That gives a 2 by 3 table of counts. Which critical value do you compare your $\chi^2$ statistic against at $\alpha = 0.05$ ?

Count the category rows and columns. The responses (yes, no) give 2 rows, and the three schools give 3 columns. Ignore the total row and total column.
Apply the two-way rule: $df = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(3 - 1) = 2$ .
Go to the row for $df = 2$ .
Go to the column for $\alpha = 0.05$ . There is no one-tail or two-tail choice to make, because the whole test lives in the upper tail.
Read the value where the row and column meet.

$\chi^{2*} = 5.991$ . Reject the null hypothesis of a common distribution if your statistic is greater than 5.991.

Carry the test through with real counts

The three schools give these counts. School A: 30 yes, 20 no. School B: 25 yes, 25 no. School C: 15 yes, 35 no. Test at $\alpha = 0.05$ whether the three schools have the same distribution of responses.

Find the totals. Row totals are $30 + 25 + 15 = 70$ yes and $20 + 25 + 35 = 80$ no. Each column total is 50, and the grand total is $n = 150$ .
Build the expected counts with $E = (\text{row total})(\text{column total}) / n$ . Every yes cell has $E = 70 \times 50 / 150 = 23.333$ , and every no cell has $E = 80 \times 50 / 150 = 26.667$ . All six clear 5, so the large-counts condition holds.
Compute the six terms of $(O - E)^2 / E$ . Yes row: $(30 - 23.333)^2 / 23.333 = 1.9048$ , $(25 - 23.333)^2 / 23.333 = 0.1190$ , $(15 - 23.333)^2 / 23.333 = 2.9762$ . No row: $(20 - 26.667)^2 / 26.667 = 1.6667$ , $(25 - 26.667)^2 / 26.667 = 0.1042$ , $(35 - 26.667)^2 / 26.667 = 2.6042$ .
Add them: $\chi^2 = 1.9048 + 0.1190 + 2.9762 + 1.6667 + 0.1042 + 2.6042 = 9.375$ .
Degrees of freedom are $df = (2 - 1)(3 - 1) = 2$ , so the critical value at $\alpha = 0.05$ is 5.991.
Compare: $9.375 > 5.991$ , so the statistic falls in the rejection region.

$\chi^2 = 9.375$ with $df = 2$ beats the critical value 5.991, so reject the null hypothesis. There is convincing evidence the three schools do not share the same distribution of responses. The exact p-value is 0.0092, which is below 0.05 and agrees with the critical-value decision.

Frequently asked questions

Why does the chi-square table have no two-tail column?

Because $\chi^2$ squares every difference, it measures how far the observed counts sit from the expected counts without recording which direction. Both kinds of disagreement push the statistic up, so all the evidence against the null sits in the upper tail. Every column of a chi-square table is already an upper-tail area, and you never halve $\alpha$ before looking one up.

Do I use rows times columns or categories minus 1 for df?

It depends on the test. A two-way table, meaning a test for independence or homogeneity, uses $df = (\text{rows} - 1)(\text{columns} - 1)$ . A goodness-of-fit test on a single row of categories uses $df = (\text{number of categories}) - 1$ . Only the two-way tests are on the AP exam under the Fall 2026 course.

What if my df is not listed in the table?

Printed tables usually run through $df = 30$ and then jump to 40, 50, and higher. If your exact $df$ is missing, use the next higher listed $df$ . Chi-square critical values grow as $df$ grows, so the higher row is the larger cutoff and the harder test to pass, which keeps you from claiming significance you have not earned. Note this runs opposite to the habit from a t table, where the smaller $df$ is the conservative one. Technology avoids the choice by computing the exact value.

Does the AP exam give me a chi-square table?

Yes. AP Statistics supplies a formula sheet and tables for both sections of the exam, including a chi-square table, so you do not memorize critical values. The exam is fully digital in the Bluebook app starting with the May 2027 administration, and the tables come with it.

My statistic is smaller than the critical value. What do I write?

You fail to reject the null hypothesis. Say it that way rather than "accept the null," because a small $\chi^2$ only shows the observed counts are consistent with the null, not that the null is true. Write the conclusion in context: there is not convincing evidence of an association (or of a difference between the groups).

Chi-square critical values

Each entry is the value that leaves the column’s area in the upper tail. Reject the null hypothesis when your statistic is larger than the entry in your row.

Chi-square critical values by degrees of freedom and upper-tail probability
df	0.1	0.05	0.025	0.01	0.005
1	2.706	3.841	5.024	6.635	7.879
2	4.605	5.991	7.378	9.210	10.597
3	6.251	7.815	9.348	11.345	12.838
4	7.779	9.488	11.143	13.277	14.860
5	9.236	11.070	12.833	15.086	16.750
6	10.645	12.592	14.449	16.812	18.548
7	12.017	14.067	16.013	18.475	20.278
8	13.362	15.507	17.535	20.090	21.955
9	14.684	16.919	19.023	21.666	23.589
10	15.987	18.307	20.483	23.209	25.188
11	17.275	19.675	21.920	24.725	26.757
12	18.549	21.026	23.337	26.217	28.300
13	19.812	22.362	24.736	27.688	29.819
14	21.064	23.685	26.119	29.141	31.319
15	22.307	24.996	27.488	30.578	32.801
16	23.542	26.296	28.845	32.000	34.267
17	24.769	27.587	30.191	33.409	35.718
18	25.989	28.869	31.526	34.805	37.156
19	27.204	30.144	32.852	36.191	38.582
20	28.412	31.410	34.170	37.566	39.997
21	29.615	32.671	35.479	38.932	41.401
22	30.813	33.924	36.781	40.289	42.796
23	32.007	35.172	38.076	41.638	44.181
24	33.196	36.415	39.364	42.980	45.559
25	34.382	37.652	40.646	44.314	46.928
26	35.563	38.885	41.923	45.642	48.290
27	36.741	40.113	43.195	46.963	49.645
28	37.916	41.337	44.461	48.278	50.993
29	39.087	42.557	45.722	49.588	52.336
30	40.256	43.773	46.979	50.892	53.672
40	51.805	55.758	59.342	63.691	66.766
50	63.167	67.505	71.420	76.154	79.490
60	74.397	79.082	83.298	88.379	91.952
80	96.578	101.879	106.629	112.329	116.321
100	118.498	124.342	129.561	135.807	140.169

Column headings are upper-tail areas, so the 0.05 column is the one to use for a test at the 5% significance level. There is no separate two-tail column, because a chi-square test only ever puts evidence in the upper tail.