Pearson's chi-square test

Pearson's chi-square test (&chi;2) is one of a variety of chi-square tests – statistical procedures whose results are evaluated by reference to the chi-square distribution. It tests a null hypothesis that the relative frequencies of occurrence of observed events follow a specified frequency distribution. The events must be mutually exclusive. One of the simplest examples is the hypothesis that an ordinary six-sided die is "fair", i.e., all six outcomes occur equally often. Chi-square is calculated by finding the difference between each observed and theoretical frequency, squaring them, dividing each by the theoretical frequency, and taking the sum of the results:


 * $$ \chi^2 = \sum {(O - E)^2 \over E}$$

where:


 * O = an observed frequency
 * E = an expected (theoretical) frequency, asserted by the null hypothesis

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:


 * $$ \chi^2 = {(45 - 50)^2 \over 50} + {(55 - 50)^2 \over 50} = 1$$

If the null hypothesis is true (ie men and women are chosen with equal probability in the sample), the test statistic will be distributed with one degree of freedom. One might expect there to be two degrees of freedom, one for the male count and one for the female. However, in this case there is only one degree of freedom because the male and female count are constrained have a sum of 50 (the sample size), and this constraint reduces the number of degrees of freedom by one. 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, 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.

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.

Pearson's chi-square is the original and most widely-used chi-square test.

The null distribution of the Pearson statistic is only approximated as a chi-square distribution. 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. In the special case where there are only two cells in the table, the expected values follow a binomial distribution.


 * $$ E =^d \mbox{Bin}(n,p) $$

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.

When comparing the Pearson test statistic against a chi-squared distribution, the above binomial distribution is approximated as a Gaussian (normal) distribution:


 * $$ \mbox{Bin}(n,p) \approx^d \mbox{N}(np, np(1-p)) $$

Let O be the number of observations from the sample that are in the first cell. The Pearson test statistic can be expressed as


 * $$ \frac{(O-np)^2}{np} + \frac{(n-O-n(1-p))^2}{n(1-p)} $$

Which can in turn be expressed as:


 * $$ \left(\frac{(O-np)}{\sqrt{np(1-p)}}\right)^2 $$

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.

In the general case where there are $$k$$ cells in the contingency table, the Normal approximation results in a sum of $$k-1$$ standard normal variates, and is thus distributed as chi-square with $$k-1$$ degrees of freedom:

In cases whereby 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.

A more complicated, but more widely used form of Pearson's chi-square test arises in the case where the null hypothesis of interest includes unknown parameters. For instance we may wish to test whether some data follows a normal distribution but without specifying a mean or variance. In this situation the unknown parameters need to be estimated by the data, typically by maximum likelihood estimation, and these estimates are then used to calculate the expected values in the Pearson statistic. It is commonly stated that the degrees of freedom for the chi-square distribution of the statistic are then $$k-1-r$$, where $$r$$ is the number of unknown parameters. This result is valid when the original data was Multinomial and hence the estimated parameters are efficient for minimizing the chi-square statistic. More generally however, when maximum likelihood estimation does not coincide with minimum chi-square estimation, the distribution will lie somewhere between a chi-square distribution with $$k-r-1$$ and $$k-1$$ degrees of freedom (See for instance Chernoff and Lehmann 1954).