# Pearson's chi-square test

Pearson's chi-square test

Pearson's chi-square (&chi;2) test is the best-known of several chi-square tests – statistical 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.

The first step in the chi-square test is to calculate the chi-square 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.

:$X^2 = sum_\left\{i=1\right\}^\left\{n\right\} \left\{\left(O_i - E_i\right)^2 over E_i\right\}$

where

:$X^2$ = the test statistic that asymptotically approaches a &chi;2 distribution.:$O_i$ = an observed frequency;:$E_i$ = an expected (theoretical) frequency, asserted by the null hypothesis;:$n$ = the number of possible outcomes of each event.

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 possible outcomes, minus 1.

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.

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 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

:$X^2 = \left\{\left(45 - 50\right)^2 over 50\right\} + \left\{\left(55 - 50\right)^2 over 50\right\} = 1.$

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, 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 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 had 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. Statistics for Applications. "MIT OpenCourseWare". [http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-443Fall2003/4226DF27-A1D0-4BB8-939A-B2A4167B5480/0/lec23.pdf Lecture 23] . Retrieved 21 March 2007.]

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,

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

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 Pearson test statistic approximates a chi-squared distribution,

:$mbox\left\{Bin\right\}\left(n,p\right) approx^d mbox\left\{N\right\}\left(np, np\left(1-p\right)\right).$

Let "O1" be the number of observations from the sample that are in the first cell. The Pearson test statistic can be expressed as

:$frac\left\{\left(O_1-np\right)^2\right\}\left\{np\right\} + frac\left\{\left(n-O_1-n\left(1-p\right)\right)^2\right\}\left\{n\left(1-p\right)\right\},$

which can in turn be expressed as

:$left\left(frac\left\{\left(O_1-np\right)\right\}\left\{sqrt\left\{np\left(1-p\right) ight\right)^2.$

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

:$frac\left\{\left(O_1-mu\right)^2\right\}\left\{sigma^2\right\}.$

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.

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" − 1 − "r" and "k" − 1 degrees of freedom (See for instance Chernoff and Lehmann 1954).

ee also

*Median test
*Chi-squared nomogram

Notes

References

*Chernoff H, Lehmann E.L. "The use of maximum likelihood estimates in $chi^2$ tests for goodness-of-fit." The Annals of Mathematical Statistics 1954; 25:579-586.
*cite journal
author = Plackett, R.L.
year = 1983
title = Karl Pearson and the Chi-Squared Test
journal = International Statistical Review
volume = 51
issue = 1
pages = 59–72

* [http://www.psychstat.missouristate.edu/introbook/SBK28.htm CHI-SQUARE AND TESTS OF CONTINGENCY TABLES]
* [http://www.socr.ucla.edu/htmls/ana/ChiSquareCT_Analysis.html Chi-Square Applet Calculator]
* [http://stat-www.berkeley.edu/~stark/Java/Html/chiHiLite.htm Sampling Distribution of the Sample Chi-Square Statistic] &mdash; a Java applet showing the sampling distribution of the Pearson test statistic.
* [http://jumk.de/statistic-calculator/ Online Chi-Square Test for uniform distribution]
* [http://www.statsoft.com/textbook/sttable.html Statistic distribution tables including chi]
* [http://www.celiagreen.com/charlesmccreery/statistics/chisquare.pdf A tutorial on the chi-square test devised for Oxford University psychology students]

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