# Pearson's chi-square test

Pearson's chi-square2) test is the best-known of several chi-square testsstatistical procedures whose results are evaluated by reference to the chi-square distribution. Its properties were first investigated by Karl Pearson. In contexts where it is important to make a distinction between the test statistic and its distribution, names similar to Pearson X-squared test or statistic are used.

It tests a null hypothesis that the frequency distribution of certain events observed in a sample is consistent with a particular theoretical distribution. The events considered must be mutually exclusive and have total probability 1. A common case for this is where the events each cover an outcome of a categorical variable. A simple example is the hypothesis that an ordinary six-sided die is "fair", i.e., all six outcomes are equally likely to occur. Pearson's chi-square is the original and most widely-used chi-square test.

## Definition

Pearson's chi-square is used to assess two types of comparison: tests of goodness of fit and tests of independence. A test of goodness of fit establishes whether or not an observed frequency distribution differs from a theoretical distribution. A test of independence assesses whether paired observations on two variables, expressed in a contingency table, are independent of each other – for example, whether people from different regions differ in the frequency with which they report that they support a political candidate.

The first step in the chi-square test is to calculate the chi-square statistic. In order to avoid ambiguity, the value of the test-statistic is denoted by X2 rather than χ2: this also serves as a reminder that the distribution of the test statistic is not exactly that of a chi-square random variable. However some authors do use the χ2 for the test statistic. The chi-square statistic is calculated by finding the difference between each observed and theoretical frequency for each possible outcome, squaring them, dividing each by the theoretical frequency, and taking the sum of the results. A second important part of determining the test statistic is to define the degrees of freedom of the test: this is essentially the number of squares errors involving the observed frequencies adjusted for the effect of using some of those observations to define the "theoretical frequencies".

### Test for fit of a distribution

In this case N observations are divided among n cells. A simple application is where it is required to test the hypothesis that, in the general population, values would occur in each cell with equal frequency. Then the "theoretical frequency" for any cell (under the null hypothesis) is calculated as $E_i=N/n \, ,$

and the reduction in the degrees of freedom is p=1: notionally because the observed frequencies Oi are constrained to sum to N. 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 p=s+1, where s is the number of parameters used in fitting the distribution. In other words, there will be (categories - 1) degrees of freedom. It should be noted that the degrees of freedom are not based on the number of observations as with a Student's t or F-distribution. For example, if testing for a fair, six-sided die, there would be five degrees of freedom because there are six categories/parameters (each number). The number of times the die is rolled will have absolutely no effect on the number of degrees of freedom.

The value of the test-statistic is $X^2 = \sum_{i=1}^{n} {(O_i - E_i)^2 \over E_i} ,$

where

X2 = the test statistic that asymptotically approaches a χ2 distribution.
Oi = an observed frequency;
Ei = an expected (theoretical) frequency, asserted by the null hypothesis;
n = the number of possible outcomes of each event.

The chi-square statistic can then be used to calculate a p-value by comparing the value of the statistic to a chi-square distribution. The number of degrees of freedom is equal to the number of cells n, minus the reduction in degrees of freedom, p.

### Test of independence

In this case, an "observation" consists of the values of two outcomes and the null hypothesis is that the occurrence of these outcomes is statistically independent. Each outcome is allocated to one cell of a two-dimensional array of cells (called a table) according to the values of the two outcomes. If there are r rows and c columns in the table, the "theoretical frequency" for a cell, given the hypothesis of independence, is $E_{i,j}=\frac{\sum_{k=1}^{c} O_{i,k} \sum_{k=1}^{r} O_{k,j}}{N} \, ,$

and fitting the model of "independence" reduces the number of degrees of freedom by p = r + c − 1. The value of the test-statistic is $X^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} {(O_{i,j} - E_{i,j})^2 \over E_{i,j}} .$

The number of degrees of freedom is equal to the number of cells rc, minus the reduction in degrees of freedom, p, which reduces to (r − 1)(c − 1).

For the test of independence, a chi-square probability of less than or equal to 0.05 (or the chi-square 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 unrelated (that is, only randomly related) to the column variable. The alternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified.

## Example

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 45 men in the sample and 55 women, then $X^2 = {(45 - 50)^2 \over 50} + {(55 - 50)^2 \over 50} = 1.$

If the null hypothesis is true (i.e., men and women are chosen with equal probability in the sample), the test statistic will be drawn from a chi-square distribution with one degree of freedom. Though one might expect two degrees of freedom (one each for the men and women), we must take into account that the total number of men and women is constrained (100), and thus there is only one degree of freedom (2 − 1). Alternatively, if the male count is known the female count is determined, and vice-versa.

Consultation of the chi-square distribution for 1 degree of freedom shows that the probability of observing this difference (or a more extreme difference than this) if men and women are equally numerous in the population is approximately 0.3. This probability is higher than conventional criteria for statistical significance (.001-.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'd expect for a 50/50 male/female ratio.)

## Problems

The approximation to the chi-square distribution breaks down if expected frequencies are too low. It will normally be acceptable so long as no more than 10% of the events have expected frequencies below 5. 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 called Yates' correction for continuity.

In cases where the expected value, E, is found to be small (indicating either a small underlying population probability, 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 the G-test, a likelihood ratio-based test statistic. Where the total sample size is small, it is necessary to use an appropriate exact test, typically either the binomial test or (for contingency tables) Fisher's exact test; but note that this test assumes fixed and known marginal totals.

## Distribution

The null distribution of the Pearson statistic with j rows and k columns is approximated by the chi-square distribution with (k − 1)(j − 1) degrees of freedom. 

This approximation arises as the true distribution, under the null hypothesis, if the expected value is given by a multinomial distribution. For large sample sizes, the central limit theorem says this distribution tends toward a certain multivariate normal distribution.

### Two cells

In the special case where there are only two cells in the table, the expected values follow a binomial distribution,

E = dBin(n,p),

where

p = probability, under the null hypothesis,
n = number of observations in the sample.

In the above example the hypothesised probability of a male observation is 0.5, with 100 samples. Thus we expect to observe 50 males.

If n is sufficiently large, the above binomial distribution may be approximated by a Gaussian (normal) distribution and thus the Pearson test statistic approximates a chi-squared distribution, $\mbox{Bin}(n,p) \approx^d \mbox{N}(np, np(1-p)).$

Let O1 be the number of observations from the sample that are in the first cell. The Pearson test statistic can be expressed as $\frac{(O_1-np)^2}{np} + \frac{(n-O_1-n(1-p))^2}{n(1-p)},$

which can in turn be expressed as $\left(\frac{(O_1-np)}{\sqrt{np(1-p)}}\right)^2.$

By the normal approximation to a binomial this is the square of one standard normal variate, and hence is distributed as chi-square with 1 degree of freedom. Note that the denominator is one standard deviation of the Gaussian approximation, so can be written $\frac{(O_1-\mu)^2}{\sigma^2}.$

So as consistent with the meaning of the chi-square distribution, we are measuring how probable the observed number of standard deviations away from the mean is under the Gaussian approximation (which is a good approximation for large n).

The chi-square distribution is then integrated on the right of the statistic value to obtain the probability that this result or worse were observed given the model.

### Many cells

Similar arguments as above lead to the desired result. Each cell (except the final one, whose value is completely determined by the others) is treated as an independent binomial variable, and their contributions are summed and each contributes one degree of freedom.