Chi-squared test
A chi-square test, also written as test, is a statistical hypothesis test that is valid to perform when the test statistic is chi-square distributed under the null hypothesis, specifically Pearson's chi-square test and variants thereof. Pearson's chi-square test is used to determine whether there is a statistically significant difference between the expected frequencies and the observed frequencies in one or more categories of a contingency table.
In the standard applications of this test, the observations are classified into mutually exclusive classes. If the null hypothesis is true, the test statistic computed from the observations follows a frequency distribution. The purpose of the test is to evaluate how likely the observed frequencies would be assuming the null hypothesis is true.
Test statistics that follow a distribution occur when the observations are independent and normally distributed, which assumptions are often justified under the central limit theorem. There are also tests for testing the null hypothesis of independence of a pair of random variables based on observations of the pairs.
Chi-square tests often refers to tests for which the distribution of the test statistic approaches the distribution asymptotically, meaning that the sampling distribution of the test statistic approximates a distribution more and more closely as sample sizes increase.
History
In the 19th century, statistical analytical methods were mainly applied in biological data analysis and it was customary for researchers to assume that observations followed a normal distribution, such as Sir George Airy and Professor Merriman, whose works were criticized by Karl Pearson in his 1900 paper.At the end of 19th century, Pearson noticed the existence of significant skewness within some biological observations. In order to model the observations regardless of being normal or skewed, Pearson, in a series of articles published from 1893 to 1916, devised the Pearson distribution, a family of continuous probability distributions, which includes the normal distribution and many skewed distributions, and proposed a method of statistical analysis consisting of using the Pearson distribution to model the observation and performing a test of goodness of fit to determine how well the model really fits to the observations.
Pearson's test
In 1900, Pearson published a paper on the test which is considered to be one of the foundations of modern statistics. In this paper, Pearson investigated a test of goodness of fit.Suppose that observations in a random sample from a population are classified into mutually exclusive classes with respective observed numbers , and a null hypothesis gives the probability that an observation falls into the th class. So we have the expected numbers for all, where
Pearson proposed that, under the circumstance of the null hypothesis being correct, as the limiting distribution of the quantity given below is the distribution.
Pearson dealt first with the case in which the expected numbers are large enough known numbers in all cells assuming every may be taken as normally distributed, and reached the result that, in the limit as becomes large, follows the distribution with degrees of freedom.
However, Pearson next considered the case in which the expected numbers depended on the parameters that had to be estimated from the sample, and suggested that, with the notation of being the true expected numbers and being the estimated expected numbers, the difference
will usually be positive and small enough to be omitted. In a conclusion, Pearson argued that if we regarded as also distributed as distribution with degrees of freedom, the error in this approximation would not affect practical decisions. This conclusion caused some controversy in practical applications and was not settled for 20 years until Fisher's 1922 and 1924 papers.
Other examples of tests
One test statistic that follows a chi-square distribution exactly is the test that the variance of a normally distributed population has a given value based on a sample variance. Such tests are uncommon in practice because the true variance of the population is usually unknown. However, there are several statistical tests where the chi-square distribution is approximately valid:Fisher's exact test
For an exact test used in place of the 2 x 2 test for independence, see Fisher's exact test.Binomial test
For an exact test used in place of the 2 x 1 test for goodness of fit, see Binomial test.Other tests
- Cochran–Mantel–Haenszel chi-square test.
- McNemar's test, used in certain tables with pairing
- Tukey's test of additivity
- The portmanteau test in time-series analysis, testing for the presence of autocorrelation
- Likelihood-ratio tests in general statistical modelling, for testing whether there is evidence of the need to move from a simple model to a more complicated one.
Yates's correction for continuity
To reduce the error in approximation, Frank Yates suggested a correction for continuity that adjusts the formula for Pearson's chi-square test by subtracting 0.5 from the absolute difference between each observed value and its expected value in a contingency table. This reduces the value obtained and thus increases its p-value.
Chi-square test for variance in a normal population
If a sample of size is taken from a population having a normal distribution, then there is a result which allows a test to be made of whether the variance of the population has a pre-determined value. For example, a manufacturing process might have been in stable condition for a long period, allowing a value for the variance to be determined essentially without error. Suppose that a variant of the process is being tested, giving rise to a small sample of product items whose variation is to be tested. The test statistic in this instance could be set to be the sum of squares about the sample mean, divided by the nominal value for the variance. Then has a distribution with degrees of freedom. For example, if the sample size is 21, the acceptance region for with a significance level of 5% is between 9.59 and 34.17.Example test for categorical data
Suppose there is a city of 1,000,000 residents with four neighborhoods:,,, and. A random sample of 650 residents of the city is taken and their occupation is recorded as "white collar", "blue collar", or "no collar". The null hypothesis is that each person's neighborhood of residence is independent of the person's occupational classification. The data are tabulated as:Let us take the sample living in neighborhood, 150, to estimate what proportion of the whole 1,000,000 live in neighborhood. Similarly, we take to estimate what proportion of the 1,000,000 are white-collar workers. By the assumption of independence under the hypothesis, we should "expect" the number of white-collar workers in neighborhood to be
Then in that "cell" of the table, we have
The sum of these quantities over all of the cells is the test statistic; in this case,. Under the null hypothesis, this sum has approximately a distribution whose number of degrees of freedom are
If the test statistic is improbably large according to that distribution, then one rejects the null hypothesis of independence.
A related issue is a test of homogeneity. Suppose that instead of giving every resident of each of the four neighborhoods an equal chance of inclusion in the sample, we decide in advance how many residents of each neighborhood to include. Then each resident has the same chance of being chosen as do all residents of the same neighborhood, but residents of different neighborhoods would have different probabilities of being chosen if the four sample sizes are not proportional to the populations of the four neighborhoods. In such a case, we would be testing "homogeneity" rather than "independence". The question is whether the proportions of blue-collar, white-collar, and no-collar workers in the four neighborhoods are the same. However, the test is done in the same way.
Applications
In cryptanalysis, the test is used to compare the distribution of plaintext and decrypted ciphertext. The lowest value of the test means that the decryption was successful with high probability. This method can be generalized for solving modern cryptographic problems.In bioinformatics, the test is used to compare the distribution of certain properties of genes belonging to different categories.