# Chi-squared test

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Chi-square distribution, showing X2 on the x-axis and P-value on the y-axis.

A chi-square test, also referred to as $χ²$ test (or chi-squared test), is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-square distribution when the null hypothesis is true. Also considered a chi-square test is a test in which this is asymptotically true, meaning that the sampling distribution (if the null hypothesis is true) can be made to approximate a chi-square distribution as closely as desired by making the sample size large enough. The chi-squared (I) test is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. Does the number of individuals or objects that fall in each category differ significantly from the number you would expect? Is this difference between the expected and observed due to sampling variation, or is it a real difference?

## Examples of chi-square tests with samples

The following are examples of chi-square tests where the chi-square distribution is approximately valid:

### Pearson's chi-square test

Pearson's chi-square test, also known as the chi-square goodness-of-fit test or chi-square test for independence. When the chi-square test is mentioned without any modifiers or without other precluding context, this test is often meant (for an exact test used in place of $χ²$, see Fisher's exact test).

### Yates's correction for continuity

Using the chi-square distribution to interpret Pearson's chi-square statistic requires one to assume that the discrete probability of observed binomial frequencies in the table can be approximated by the continuous chi-square distribution. This assumption is not quite correct, and introduces some error.

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 difference between each observed value and its expected value in a 2 × 2 contingency table.[1] This reduces the chi-square value obtained and thus increases its p-value.

## Exact chi-square distribution

One case where the distribution of the test statistic is an exact chi-square distribution is the test that the variance of a normally distributed population has a given value based on a sample variance. Such a test is uncommon in practice because values of variances to test against are seldom known exactly.

## Chi-square test for variance in a normal population

If a sample of size n is taken from a population having a normal distribution, then there is a result (see distribution of the sample variance) 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 n product items whose variation is to be tested. The test statistic T in this instance could be set to be the sum of squares about the sample mean, divided by the nominal value for the variance (i.e. the value to be tested as holding). Then T has a chi-square distribution with n − 1 degrees of freedom. For example if the sample size is 21, the acceptance region for T for a significance level of 5% is the interval 9.59 to 34.17.

## Chi-square test for independence and homogeneity in tables

Suppose a random sample of 650 of the 1 million residents of a city is taken, in which every resident of each of four neighborhoods, A, B, C, and D, is equally likely to be chosen. A null hypothesis says the randomly chosen person's neighborhood of residence is independent of the person's occupational classification, which is either "blue collar", "white collar", or "service". The data are tabulated:

$\begin{array}{l|c|c|c|c|c|c} & \text{A} & \text{B} & \text{C} & \text{D} & & \text{total} \\[6pt] \hline \text{Blue collar} & 90 & 60 & 104 & 95 & & 349 \\[6pt] \hline \text{White collar} & 30 & 50 & 51 & 20 & & 151 \\[6pt] \hline \text{Service} & 30 & 40 & 45 & 35 & & 150 \\[12pt] \hline \text{total} & 150 & 150 & 200 & 150 & & 650 \end{array}$

Let us take the sample proportion living in neighborhood A, 150/650, to estimate what proportion of the whole 1 million people live in neighborhood A. Similarly we take 349/650 to estimate what proportion of the 1 million people are blue-collar workers. Then the null hypothesis independence tells us that we should "expect" the number of blue-collar workers in neighborhood A to be

$\frac{150}{650}\times\frac{349}{650}\times650 \approx 80.54.$

Then in that "cell" of the table, we have

$\frac{(\text{observed}-\text{expected})^2}{\text{expected}} = \frac{(90-80.54)^2}{80.54}.$

The sum of these quantities over all of the cells is the test statistic. Under the null hypothesis, it has approximately a chi-square distribution whose number of degrees of freedom is

$(\text{number of rows}-1)(\text{number of columns}-1) = (3-1)(4-1) = 6. \,$

If the test statistic is improbably large according to that chi-square 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 service workers in the four neighborhoods are the same. However, the test is done in the same way.

## Applications

In cryptanalysis, chi-square test is used to compare the distribution of plaintext and (possibly) decrypted ciphertext. The lowest value of the test means that the decryption was successful with high probability.[2][3] This method can be generalized for solving modern cryptographic problems.[4]