Pearson's chi-squared test orPearson's${\displaystyle {\chi }^2}$ test is astatistical test applied to sets ofcategorical data to evaluate how likely it is that any observed difference between the sets arose by chance. It is the most widely used of manychi-squared tests (e.g.,Yates,likelihood ratio,portmanteau test in time series, etc.) –statistical procedures whose results are evaluated by reference to thechi-squared distribution. Its properties were first investigated byKarl Pearson in 1900.^[1] In contexts where it is important to improve a distinction between thetest statistic and its distribution, names similar toPearson χ-squared test or statistic are used.It is ap-value test.

A simple example is testing the hypothesis that an ordinary six-sided die is "fair" (i. e., all six outcomes are equally likely to occur). In this case, the observed data is${\displaystyle (O_1,O_2,...,O_6)}$, the number of times that the dice has fallen on each number. The null hypothesis is${\displaystyle \mathrm{M}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{a}\mathrm{l}(N;1/6,...,1/6)}$, and${\chi }^2:=\sum _{i=1}^6\frac{{\left(O_i-N/6\right)}^2}{N/6}$. As detailed below, if${\displaystyle {\chi }^2>11.07}$, then the fairness of dice can be rejected at the level of.

The setup is as follows:^[2][3]

• Before the experiment, the experimenter fixes a certain number${\displaystyle N}$ of samples to take.
• Theobserved data is${\displaystyle (O_1,O_2,...,O_n)}$, the count number of samples from a finite set of given categories. They satisfy$\sum _i{O}_i=N$.
• Thenull hypothesis is that the count numbers are sampled from amultinomial distribution${\displaystyle \mathrm{M}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{a}\mathrm{l}(N;p_1,...,p_n)}$. That is, the underlying data is sampledIID from acategorical distribution${\displaystyle \mathrm{C}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}(p_1,...,p_n)}$ over the given categories.
• The Pearson's chi-squaredtest statistic is defined as${\chi }^2:=\sum _i\frac{{\left(O_i-N{p}_i\right)}^2}{N{p}_i}$. The p-value of the test statistic is computed either numerically or by looking it up in a table.
• If the p-value is small enough (usually p < 0.05 by convention), then the null hypothesis is rejected, and we conclude that the observed data does not follow the multinomial distribution.

Pearson's chi-squared test is used to assess three types of comparison:goodness of fit,homogeneity, andindependence.

• A test of goodness of fit establishes whether an observedfrequency distribution differs from a theoretical distribution.
• A test of homogeneity compares the distribution of counts for two or more groups using the same categorical variable (e.g. choice of activity—college, military, employment, travel—of graduates of a high school reported a year after graduation, sorted by graduation year, to see if number of graduates choosing a given activity has changed from class to class, or from decade to decade).^[4]
• A test of independence assesses whether observations consisting of measures on two variables, expressed in acontingency table, are independent of each other (e.g. polling responses from people of different nationalities to see if one's nationality is related to the response).

For all three tests, the computational procedure includes the following steps:

1. Calculate the chi-squared teststatistic,${\displaystyle {\chi }^2}$, which resembles anormalized sum of squared deviations between observed and theoreticalfrequencies (see below).
2. Determine thedegrees of freedom,df, of that statistic.
   1. For a test of goodness-of-fit,df = Cats − Params, whereCats is the number of observation categories recognized by the model, andParams is the number of parameters in the model adjusted to make the model best fit the observations: The number of categories reduced by the number of fitted parameters in the distribution.
   2. For a test of homogeneity,df = (Rows − 1)×(Cols − 1), whereRows corresponds to the number of categories (i.e. rows in the associated contingency table), andCols corresponds to the number of independent groups (i.e. columns in the associated contingency table).^[4]
   3. For a test of independence,df = (Rows − 1)×(Cols − 1), where in this case,Rows corresponds to the number of categories in one variable, andCols corresponds to the number of categories in the second variable.^[4]
3. Select a desired level of confidence (significance level,p-value, or the correspondingalpha level) for the result of the test.
4. Compare${\displaystyle {\chi }^2}$ to the critical value from thechi-squared distribution withdf degrees of freedom and the selected confidence level (one-sided, since the test is only in one direction, i.e. is the test value greater than the critical value?), which in many cases gives a good approximation of the distribution of${\displaystyle {\chi }^2}$.
5. Sustain or reject the null hypothesis that the observed frequency distribution is the same as the theoretical distribution based on whether the test statistic exceeds the critical value of${\displaystyle {\chi }^2}$. If the test statistic exceeds the critical value of${\displaystyle {\chi }^2}$, the null hypothesis (${\displaystyle H_0}$ = there isno difference between the distributions) can be rejected, and the alternative hypothesis (${\displaystyle H_1}$ = thereis a difference between the distributions) can be accepted, both with the selected level of confidence. If the test statistic falls below the threshold${\displaystyle {\chi }^2}$ value, then no clear conclusion can be reached, and the null hypothesis is sustained (we fail to reject the null hypothesis), though not necessarily accepted.

In this case${\displaystyle N}$ observations are divided among${\displaystyle n}$ cells. A simple application is to test the hypothesis that, in the general population, values would occur in each cell with equal frequency. The "theoretical frequency" for any cell (under the null hypothesis of adiscrete uniform distribution) is thus calculated as$${\displaystyle E_i=\frac{N}{n},}$$and the reduction in the degrees of freedom is${\displaystyle p=1}$, notionally because the observed frequencies${\displaystyle O_i}$ are constrained to sum to${\displaystyle N}$.

One specific example of its application would be its application for log-rank test.

When testing whether observations are random variables whose distribution belongs to a given family of distributions, the "theoretical frequencies" are calculated using a distribution from that family fitted in some standard way. The reduction in the degrees of freedom is calculated as${\displaystyle p=s+1}$, where${\displaystyle s}$ is the number of parameters used in fitting the distribution. For instance, when checking a three-parameterGeneralized gamma distribution,${\displaystyle p=4}$, and when checking a normal distribution (where the parameters are mean and standard deviation),${\displaystyle p=3}$, and when checking a Poisson distribution (where the parameter is the expected value),${\displaystyle p=2}$. Thus, there will be${\displaystyle n-p}$ degrees of freedom, where${\displaystyle n}$ is the number of categories.

The degrees of freedom are not based on the number of observations as with aStudent's t orF-distribution. For example, if testing for a fair, six-sided die, there would be five degrees of freedom because there are six categories or parameters (each number); the number of times the die is rolled does not influence the number of degrees of freedom.

The value of the test-statistic is

$${\displaystyle {\chi }^2=\sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}=N\sum _{i=1}^n\frac{{\left(O_i/N-p_i\right)}^2}{p_i}}$$

where

• ${\displaystyle {\chi }^2}$ = Pearson's cumulative test statistic, which asymptotically approaches a${\displaystyle {\chi }^2}$ distribution.
• ${\displaystyle O_i}$ = the number of observations of typei.
• ${\displaystyle N}$ = total number of observations
• ${\displaystyle E_i=N{p}_i}$ = the expected (theoretical) count of typei, asserted by the null hypothesis that the fraction of typei in the population is${\displaystyle p_i}$
• ${\displaystyle n}$ = the number of cells in the table.

The chi-squared statistic can then be used to calculate ap-value bycomparing the value of the statistic to achi-squared distribution. The number ofdegrees of freedom is equal to the number of cells${\displaystyle n}$, minus the reduction in degrees of freedom,${\displaystyle p}$.

The chi-squared statistic can be also calculated as

$${\displaystyle {\chi }^2=\sum _{i=1}^n\frac{O_i^2}{E_i}-N.}$$

This result is the consequence of the Binomial theorem.

The result about the numbers of degrees of freedom is valid when the original data are multinomial and hence the estimated parameters are efficient for minimizing the chi-squared statistic. More generally however, when maximum likelihood estimation does not coincide with minimum chi-squared estimation, the distribution will lie somewhere between a chi-squared distribution with${\displaystyle n-1-p}$ and${\displaystyle n-1}$ degrees of freedom (See for instance Chernoff and Lehmann, 1954).

The chi-squared test indicates a statistically significant association between the level of education completed and routine check-up attendance (chi2(3) = 14.6090, p = 0.002). The proportions suggest that as the level of education increases, so does the proportion of individuals attending routine check-ups. Specifically, individuals who have graduated from college or university attend routine check-ups at a higher proportion (31.52%) compared to those who have not graduated high school (8.44%). This finding may suggest that higher educational attainment is associated with a greater likelihood of engaging in health-promoting behaviors such as routine check-ups.

InBayesian statistics, one would instead use aDirichlet distribution asconjugate prior. If one took a uniform prior, then themaximum likelihood estimate for the population probability is the observed probability, and one may compute acredible region around this or another estimate.

In this case, an "observation" consists of the values of two outcomes and the null hypothesis is that the occurrence of these outcomes isstatistically independent. Each observation is allocated to one cell of a two-dimensional array of cells (called acontingency table) according to the values of the two outcomes. If there arer rows andc columns in the table, the "theoretical frequency" for a cell, given the hypothesis of independence, is

$${\displaystyle E_{i,j}=N{p}_{i\cdot }{p}_{\cdot j},}$$

where${\displaystyle N}$ is the total sample size (the sum of all cells in the table), and

$${\displaystyle p_{i\cdot }=\frac{O_{i\cdot }}{N}=\sum _{j=1}^c\frac{O_{i,j}}{N},}$$

is the fraction of observations of typei ignoring the column attribute (fraction of row totals), and$${\displaystyle p_{\cdot j}=\frac{O_{\cdot j}}{N}=\sum _{i=1}^r\frac{O_{i,j}}{N}}$$

is the fraction of observations of typej ignoring the row attribute (fraction of column totals). The term "frequencies" refers to absolute numbers rather than already normalized values.

The value of the test-statistic is

Note that${\displaystyle {\chi }^2}$ is 0 if and only if${\displaystyle O_{i,j}=E_{i,j}\mathrm{\forall }i,j}$, i.e. only if the expected and true number of observations are equal in all cells.

Fitting the model of "independence" reduces the number of degrees of freedom byp =r +c − 1. The number ofdegrees of freedom is equal to the number of cellsrc, minus the reduction in degrees of freedom,p, which reduces to (r − 1)(c − 1).

For the test of independence, also known as the test of homogeneity, a chi-squared probability of less than or equal to 0.05 (or the chi-squared statistic being at or larger than the 0.05 critical point) is commonly interpreted by applied workers as justification for rejecting the null hypothesis that the row variable is independent of the column variable.^[6]Thealternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified.

The chi-squared test, when used with the standard approximation that a chi-squared distribution is applicable, has the following assumptions:^[7]

Simple random sample

The sample data is a random sampling from a fixed distribution or population where every collection of members of the population of the given sample size has an equal probability of selection. Variants of the test have been developed for complex samples, such as where the data is weighted. Other forms can be used such aspurposive sampling.^[8]

Sample size (whole table)

A sample with a sufficiently large size is assumed. If a chi squared test is conducted on a sample with a smaller size, then the chi squared test will yield an inaccurate inference. The researcher, by using chi squared test on small samples, might end up committing aType II error. For small sample sizes theCash test is preferred.^[9][10]

Expected cell count

Adequate expected cell counts. Some require 5 or more, and others require 10 or more. A common rule is 5 or more in all cells of a 2-by-2 table, and 5 or more in 80% of cells in larger tables, but no cells with zero expected count. When this assumption is not met,Yates's correction is applied.

Independence

The observations are always assumed to be independent of each other. This means chi-squared cannot be used to test correlated data (like matched pairs or panel data). In those cases,McNemar's test may be more appropriate.

A test that relies on different assumptions isFisher's exact test; if its assumption of fixed marginal distributions is met it is substantially more accurate in obtaining a significance level, especially with few observations. In the vast majority of applications this assumption will not be met, and Fisher's exact test will be over conservative and not have correct coverage.^[11]

An alternative derivation is on themultinomial distribution page.

A 6-sided die is thrown 60 times. The number of times it lands with 1, 2, 3, 4, 5 and 6 face up is 5, 8, 9, 8, 10 and 20, respectively. Is the die biased, according to the Pearson's chi-squared test at a significance level of 95% and/or 99%?

The null hypothesis is that the die is unbiased, hence each number is expected to occur the same number of times, in this case,⁠60/n⁠ = 10. The outcomes can be tabulated as follows:

${\displaystyle i}$	${\displaystyle O_i}$	${\displaystyle E_i}$	${\displaystyle O_i-E_i}$	${\displaystyle (O_i-E_i{)}^2}$
1	5	10	−5	25
2	8	10	−2	4
3	9	10	−1	1
4	8	10	−2	4
5	10	10	0	0
6	20	10	10	100
Sum				134

We then consult anUpper-tail critical values of chi-square distribution table, the tabular value refers to the sum of the squared variables each divided by the expected outcomes. For the present example, this means$${\displaystyle {\chi }^2=\frac{25}{10}+\frac{4}{10}+\frac{1}{10}+\frac{4}{10}+\frac{0}{10}+\frac{100}{10}=\mathrm{13.4.}}$$

This is the experimental result whose unlikeliness (with a fair die) we wish to estimate.

The experimental sum of 13.4 is between the critical values of 97.5% and 99% significance or confidence (p-value). Specifically, getting 20 rolls of 6, when the expectation is only 10 such values, is unlikely with a fair die.

In this context, thefrequencies of both theoretical and empirical distributions are unnormalised counts, and for a chi-squared test the total sample sizes${\displaystyle N}$ of both these distributions (sums of all cells of the correspondingcontingency tables) have to be the same.

For example, to test the hypothesis that a random sample of 100 people has been drawn from a population in which men and women are equal in frequency, the observed number of men and women would be compared to the theoretical frequencies of 50 men and 50 women. If there were 44 men in the sample and 56 women, then

$${\displaystyle {\chi }^2=\frac{{\left(44-50\right)}^2}{50}+\frac{{\left(56-50\right)}^2}{50}=\mathrm{1.44.}}$$

If the null hypothesis is true (i.e., men and women are chosen with equal probability), the test statistic will be drawn from a chi-squared distribution with onedegree of freedom (because if the male frequency is known, then the female frequency is determined).

Consultation of thechi-squared distribution for 1 degree of freedom shows that theprobability of observing this difference (or a more extreme difference than this) if men and women are equally numerous in the population is approximately 0.23. This probability is higher than conventional criteria forstatistical significance (0.01 or 0.05), so normally we would not reject the null hypothesis that the number of men in the population is the same as the number of women (i.e., we would consider our sample within the range of what we would expect for a 50/50 male–female ratio.)

The chi-squared test for homogeneity of proportions (that is, comparing proportions across groups in a contingency table) is frequently used to verify if the rows of a given nonnegative${\displaystyle m\times n}$ contingency matrix${\displaystyle A}$ are proportional.

Consider${\displaystyle T}$ outcomes of multinomial trials classified according to two criteria: membership in one of${\displaystyle n}$ groups and assignment to one of${\displaystyle m}$ categories. The outcomes may be arranged in an${\displaystyle m\times n}$ contingency table, where${\displaystyle O_{ij}}$ denotes the observed frequency in row${\displaystyle i}$ and column${\displaystyle j}$, with row totals${\displaystyle O_{i\cdot }}$, column totals${\displaystyle O_{\cdot j}}$ and grand total${\displaystyle T}$.

Let${\displaystyle O_{ij}}$ denote the observed frequency in the cell corresponding to row${\displaystyle i}$ and column${\displaystyle j}$, for${\displaystyle i=1,2,\dots ,m}$ and${\displaystyle j=1,2,\dots ,n}$. Define the row sums

${\displaystyle R_i=\sum _{j=1}^n{O}_{ij},i=1,\dots ,m,}$

the column sums

${\displaystyle C_j=\sum _{i=1}^m{O}_{ij},j=1,\dots ,n,}$

and the grand total

${\displaystyle T=\sum _{i=1}^m\sum _{j=1}^n{O}_{ij}.}$

The null hypothesis is

${\displaystyle H_0:{\pi }_{1j}={\pi }_{2j}=\cdots ={\pi }_{mj},\mathrm{\forall }j\in 1,\dots ,n,}$

where${\displaystyle {\pi }_{ij}\in [0,1]}$ denotes the proportion of individuals in group${\displaystyle i}$ falling into category${\displaystyle j}$, with${\displaystyle \sum _{j=1}^n{\pi }_{ij}=1}$ for each${\displaystyle i}$.

Under${\displaystyle H_0}$, the expected frequency in cell${\displaystyle (i,j)}$ is

${\displaystyle E_{ij}=\frac{R_i{C}_j}{T},i=1,\dots ,m;;j=1,\dots ,n.}$

The Pearson chi-squared test statistic is then

${\displaystyle {\chi }_{\text{stat}}^2=\sum _{i=1}^m\sum _{j=1}^n\frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}.}$

For example, suppose there are two groups of students (Group 1 and Group 2) and three study preferences (Alone, With peers, Tutoring). Let${\displaystyle O_{ij}}$ denote the observed frequency in row${\displaystyle i}$ (preference) and column${\displaystyle j}$ (group).

The observed frequencies are as follows:

Under the null hypothesis${\displaystyle H_0}$, the rows are proportional across groups.

The expected frequencies are computed using:

${\displaystyle E_{ij}=\frac{R_i\cdot C_j}{T},}$

where${\displaystyle R_i}$ is the row total,${\displaystyle C_j}$ is the column total, and${\displaystyle T}$ is the grand total.

For example, for the first row and first column:

${\displaystyle E_{11}=\frac{20\cdot 40}{100}=8.}$

The expected frequencies are:

The Pearson chi-squared statistic is then:

${\displaystyle {\chi }_{\text{stat}}^2=\sum _{i=1}^3\sum _{j=1}^2\frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}=\mathrm{7.5.}}$

With${\displaystyle (3-1)(2-1)=2}$ degrees of freedom, this value can be compared to the chi-squared distribution to test the null hypothesis.

The approximation to the chi-squared distribution breaks down if expected frequencies are too low.^[14][15]

It will normally be acceptable so long as no more than 20% of the events have expected frequencies below 5.^[7] Where there is only 1 degree of freedom, the approximation is not reliable if expected frequencies are below 10. In this case, a better approximation can be obtained by reducing the absolute value of each difference between observed and expected frequencies by 0.5 before squaring; this is calledYates's correction for continuity.

In cases where the expected value, E, is found to be small (indicating a small underlying population probability, and/or a small number of observations), the normal approximation of the multinomial distribution can fail, and in such cases it is found to be more appropriate to use theG-test, alikelihood ratio-based test statistic. When the total sample size is small, it is necessary to use an appropriate exact test, typically either thebinomial test or, forcontingency tables,Fisher's exact test. This test uses the conditional distribution of the test statistic given the marginal totals, and thus assumes that the margins were determined before the study; alternatives such asBoschloo's test which do not make this assumption areuniformly more powerful.

In Pearson's test of homogeneity, if all entries of a matrix${\displaystyle A}$ are multiplied by a positive constant${\displaystyle c}$, the Pearson chi-squared statistic is also multiplied by${\displaystyle c}$:

${\displaystyle {\chi }_{\text{stat}}^2(cA)=c{\chi }_{\text{stat}}^2(A).}$

Therefore, if all rows of${\displaystyle A}$ are exactly proportional,

${\displaystyle {\chi }_{\text{stat}}^2(cA)=c{\chi }_{\text{stat}}^2(A)=0}$

for any${\displaystyle c}$ and any significance level${\displaystyle \alpha }$. Otherwise,${\displaystyle {\chi }_{\text{stat}}^2(cA)}$ can become arbitrarily large or small as${\displaystyle c}$ increases or decreases. Hence, at a fixed significance level${\displaystyle \alpha }$, the null hypothesis${\displaystyle H_0}$ will be rejected with confidence${\displaystyle 1-\alpha }$ when${\displaystyle c}$ is sufficiently large, and not rejected when${\displaystyle c}$ is sufficiently small.^[15] That is, the chi-squared statistic increases linearly when the entire contingency table is multiplied by a constant factor, reflecting the proportional scaling of observed and expected frequencies.

It can be shown that the${\displaystyle {\chi }^2}$ test is a low order approximation of the${\displaystyle \mathrm{\Psi }}$ test.^[16] The above reasons for the above issues become apparent when the higher order terms are investigated.

• Chi-squared nomogram
• Cramér's V – a measure of correlation for the chi-squared test
• Degrees of freedom (statistics)
• Deviance (statistics), another measure of the quality of fit
• Fisher's exact test
• G-test, test to which chi-squared test is an approximation
• Lexis ratio, earlier statistic, replaced by chi-squared
• Mann–Whitney U test
• Median test
• Minimum chi-square estimation
• Reduced chi-squared statistic
• Two-proportion Z-test

• Statistical tests for contingency tables
• Normality tests
• Statistical approximations

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