 Artikel ieu keur dikeureuyeuh, ditarjamahkeun tina basa Inggris. Bantosanna diantos kanggo narjamahkeun.

Tes Pearson's chi-kuadrat2) salah sahiji variasi tina tes chi-kuadrat – procedure statistik nu hasilna di-evaluasi dumasar kana sebaran chi-kuadrat. Tes ieu mimiti dipaluruh ku Karl Pearson.

It tests a null hypothesis that the relative frequencies of occurrence of observed events follow a specified frequency distribution. The events are assumed to be independent and have the same distribution, and the outcomes of éach event must be mutually exclusive. A simple example is the hypothesis that an ordinary six-sided die is "fair", i.e., all six outcomes occur equally often. Péarson's chi-square is the original and most widely-used chi-square test.

Chi-square is calculated by finding the difference between éach observed and théoretical frequency for éach possible outcome, squaring them, dividing éach by the théoretical frequency, and taking the sum of the results. The number of degrees of freedom is equal to the number of possible outcomes, minus 1:

$\chi _{n-1}^{2}=\sum _{i=1}^{n}{(O_{i}-E_{i})^{2} \over E_{i}}$ where

$O_{i}$ = an observed frequency;
$E_{i}$ = an expected (theoretical) frequency, asserted by the null hypothesis;
$n$ = the number of possible outcomes of each event.

Péarson'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 théoretical distribution. A test of independence assesses whether paired observations on two variables, expressed in a contingency table, are independent of éach other – for example, whether péople from different regions differ in the frequency with which they report that they support a political candidate.

A chi-square probability of 0.05 or less 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 alternate hypothesis is not rejected when the variables have an associated relationship.

## Example

For example, to test the hypothesis that a random sample of 100 péople 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 théoretical frequencies of 50 men and 50 women. If there were 45 men in the sample and 55 women, then

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

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 éach 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, 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.

## Problems

The approximation to the chi-square distribution bréaks 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 had by reducing the absolute value of éach 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 Péarson 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,

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

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.

If n is sufficiently large, the above binomial distribution may be approximated as by Gaussian (normal) distribution and thus the Péarson 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 Péarson 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 méaning of the chi-square distribution, we are méasuring how probable the observed number of standard deviations away from the méan 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 modél.

### Many cells

Similar arguments as above léad to the desired result. (TODO: details) éach cell (except the final one, whose value is completely determined by the others) is tréated as an independent binomial variable, and their contributions are summed and éach contributes one degree of freedom.