Pearson's chi-squared test or Pearson's
\chi2
It is a p-value test. The setup is as follows:[2] [3]
N
(O1,O2,...,On)
\sumiOi=N
Multinomial(N;p1,...,pn)
Categorical(p1,...,pn)
\chi2:=\sumi
| |||||||||||||
Npi |
A simple example is testing the hypothesis that an ordinary six-sided is "fair" (i. e., all six outcomes are equally likely to occur). In this case, the observed data is
(O1,O2,...,O6)
Multinomial(N;1/6,...,1/6)
\chi2:=
6 | |
\sum | |
i=1 |
(Oi-N/6)2 | |
N/6 |
\chi2>11.07
p<0.05
Pearson's chi-squared test is used to assess three types of comparison: goodness of fit, homogeneity, and independence.
For all three tests, the computational procedure includes the following steps:
\chi2
\chi2
\chi2
\chi2
\chi2
H0
H1
\chi2
In this case
N
n
E | ||||
|
,
p=1
Oi
N
One specific example of its application would be its application for log-rank test.
When testing whether observations are random variables whose distribution belongs to a given family of distributions, the "theoretical frequencies" are calculated using a distribution from that family fitted in some standard way. The reduction in the degrees of freedom is calculated as
p=s+1
s
p=4
p=3
p=2
n-p
n
The degrees of freedom are not based on the number of observations as with a Student's t or F-distribution. For example, if testing for a fair, six-sided, there would be five degrees of freedom because there are six categories or parameters (each number); the number of times the die is rolled does not influence the number of degrees of freedom.
Upper-tail critical values of chi-square distribution[5] | ||||||
---|---|---|---|---|---|---|
Degrees of freedom | Probability less than the critical value | |||||
0.90 | 0.95 | 0.975 | 0.99 | 0.999 | ||
1 | 2.706 | 3.841 | 5.024 | 6.635 | 10.828 | |
2 | 4.605 | 5.991 | 7.378 | 9.210 | 13.816 | |
3 | 6.251 | 7.815 | 9.348 | 11.345 | 16.266 | |
4 | 7.779 | 9.488 | 11.143 | 13.277 | 18.467 | |
5 | 9.236 | 11.070 | 12.833 | 15.086 | 20.515 | |
6 | 10.645 | 12.592 | 14.449 | 16.812 | 22.458 | |
7 | 12.017 | 14.067 | 16.013 | 18.475 | 24.322 | |
8 | 13.362 | 15.507 | 17.535 | 20.090 | 26.125 | |
9 | 14.684 | 16.919 | 19.023 | 21.666 | 27.877 | |
10 | 15.987 | 18.307 | 20.483 | 23.209 | 29.588 | |
11 | 17.275 | 19.675 | 21.920 | 24.725 | 31.264 | |
12 | 18.549 | 21.026 | 23.337 | 26.217 | 32.910 | |
13 | 19.812 | 22.362 | 24.736 | 27.688 | 34.528 | |
14 | 21.064 | 23.685 | 26.119 | 29.141 | 36.123 | |
15 | 22.307 | 24.996 | 27.488 | 30.578 | 37.697 | |
16 | 23.542 | 26.296 | 28.845 | 32.000 | 39.252 | |
17 | 24.769 | 27.587 | 30.191 | 33.409 | 40.790 | |
18 | 25.989 | 28.869 | 31.526 | 34.805 | 42.312 | |
19 | 27.204 | 30.144 | 32.852 | 36.191 | 43.820 | |
20 | 28.412 | 31.410 | 34.170 | 37.566 | 45.315 | |
21 | 29.615 | 32.671 | 35.479 | 38.932 | 46.797 | |
22 | 30.813 | 33.924 | 36.781 | 40.289 | 48.268 | |
23 | 32.007 | 35.172 | 38.076 | 41.638 | 49.728 | |
24 | 33.196 | 36.415 | 39.364 | 42.980 | 51.179 | |
25 | 34.382 | 37.652 | 40.646 | 44.314 | 52.620 | |
26 | 35.563 | 38.885 | 41.923 | 45.642 | 54.052 | |
27 | 36.741 | 40.113 | 43.195 | 46.963 | 55.476 | |
28 | 37.916 | 41.337 | 44.461 | 48.278 | 56.892 | |
29 | 39.087 | 42.557 | 45.722 | 49.588 | 58.301 | |
30 | 40.256 | 43.773 | 46.979 | 50.892 | 59.703 | |
31 | 41.422 | 44.985 | 48.232 | 52.191 | 61.098 | |
32 | 42.585 | 46.194 | 49.480 | 53.486 | 62.487 | |
33 | 43.745 | 47.400 | 50.725 | 54.776 | 63.870 | |
34 | 44.903 | 48.602 | 51.966 | 56.061 | 65.247 | |
35 | 46.059 | 49.802 | 53.203 | 57.342 | 66.619 | |
36 | 47.212 | 50.998 | 54.437 | 58.619 | 67.985 | |
37 | 48.363 | 52.192 | 55.668 | 59.893 | 69.347 | |
38 | 49.513 | 53.384 | 56.896 | 61.162 | 70.703 | |
39 | 50.660 | 54.572 | 58.120 | 62.428 | 72.055 | |
40 | 51.805 | 55.758 | 59.342 | 63.691 | 73.402 | |
41 | 52.949 | 56.942 | 60.561 | 64.950 | 74.745 | |
42 | 54.090 | 58.124 | 61.777 | 66.206 | 76.084 | |
43 | 55.230 | 59.304 | 62.990 | 67.459 | 77.419 | |
44 | 56.369 | 60.481 | 64.201 | 68.710 | 78.750 | |
45 | 57.505 | 61.656 | 65.410 | 69.957 | 80.077 | |
46 | 58.641 | 62.830 | 66.617 | 71.201 | 81.400 | |
47 | 59.774 | 64.001 | 67.821 | 72.443 | 82.720 | |
48 | 60.907 | 65.171 | 69.023 | 73.683 | 84.037 | |
49 | 62.038 | 66.339 | 70.222 | 74.919 | 85.351 | |
50 | 63.167 | 67.505 | 71.420 | 76.154 | 86.661 | |
51 | 64.295 | 68.669 | 72.616 | 77.386 | 87.968 | |
52 | 65.422 | 69.832 | 73.810 | 78.616 | 89.272 | |
53 | 66.548 | 70.993 | 75.002 | 79.843 | 90.573 | |
54 | 67.673 | 72.153 | 76.192 | 81.069 | 91.872 | |
55 | 68.796 | 73.311 | 77.380 | 82.292 | 93.168 | |
56 | 69.919 | 74.468 | 78.567 | 83.513 | 94.461 | |
57 | 71.040 | 75.624 | 79.752 | 84.733 | 95.751 | |
58 | 72.160 | 76.778 | 80.936 | 85.950 | 97.039 | |
59 | 73.279 | 77.931 | 82.117 | 87.166 | 98.324 | |
60 | 74.397 | 79.082 | 83.298 | 88.379 | 99.607 | |
61 | 75.514 | 80.232 | 84.476 | 89.591 | 100.888 | |
62 | 76.630 | 81.381 | 85.654 | 90.802 | 102.166 | |
63 | 77.745 | 82.529 | 86.830 | 92.010 | 103.442 | |
64 | 78.860 | 83.675 | 88.004 | 93.217 | 104.716 | |
65 | 79.973 | 84.821 | 89.177 | 94.422 | 105.988 | |
66 | 81.085 | 85.965 | 90.349 | 95.626 | 107.258 | |
67 | 82.197 | 87.108 | 91.519 | 96.828 | 108.526 | |
68 | 83.308 | 88.250 | 92.689 | 98.028 | 109.791 | |
69 | 84.418 | 89.391 | 93.856 | 99.228 | 111.055 | |
70 | 85.527 | 90.531 | 95.023 | 100.425 | 112.317 | |
71 | 86.635 | 91.670 | 96.189 | 101.621 | 113.577 | |
72 | 87.743 | 92.808 | 97.353 | 102.816 | 114.835 | |
73 | 88.850 | 93.945 | 98.516 | 104.010 | 116.092 | |
74 | 89.956 | 95.081 | 99.678 | 105.202 | 117.346 | |
75 | 91.061 | 96.217 | 100.839 | 106.393 | 118.599 | |
76 | 92.166 | 97.351 | 101.999 | 107.583 | 119.850 | |
77 | 93.270 | 98.484 | 103.158 | 108.771 | 121.100 | |
78 | 94.374 | 99.617 | 104.316 | 109.958 | 122.348 | |
79 | 95.476 | 100.749 | 105.473 | 111.144 | 123.594 | |
80 | 96.578 | 101.879 | 106.629 | 112.329 | 124.839 | |
81 | 97.680 | 103.010 | 107.783 | 113.512 | 126.083 | |
82 | 98.780 | 104.139 | 108.937 | 114.695 | 127.324 | |
83 | 99.880 | 105.267 | 110.090 | 115.876 | 128.565 | |
84 | 100.980 | 106.395 | 111.242 | 117.057 | 129.804 | |
85 | 102.079 | 107.522 | 112.393 | 118.236 | 131.041 | |
86 | 103.177 | 108.648 | 113.544 | 119.414 | 132.277 | |
87 | 104.275 | 109.773 | 114.693 | 120.591 | 133.512 | |
88 | 105.372 | 110.898 | 115.841 | 121.767 | 134.746 | |
89 | 106.469 | 112.022 | 116.989 | 122.942 | 135.978 | |
90 | 107.565 | 113.145 | 118.136 | 124.116 | 137.208 | |
91 | 108.661 | 114.268 | 119.282 | 125.289 | 138.438 | |
92 | 109.756 | 115.390 | 120.427 | 126.462 | 139.666 | |
93 | 110.850 | 116.511 | 121.571 | 127.633 | 140.893 | |
94 | 111.944 | 117.632 | 122.715 | 128.803 | 142.119 | |
95 | 113.038 | 118.752 | 123.858 | 129.973 | 143.344 | |
96 | 114.131 | 119.871 | 125.000 | 131.141 | 144.567 | |
97 | 115.223 | 120.990 | 126.141 | 132.309 | 145.789 | |
98 | 116.315 | 122.108 | 127.282 | 133.476 | 147.010 | |
99 | 117.407 | 123.225 | 128.422 | 134.642 | 148.230 | |
100 | 118.498 | 124.342 | 129.561 | 135.807 | 149.449 |
\chi2=
n | |
\sum | |
i=1 |
| |||||||||||||
Ei |
=N
n | |
\sum | |
i=1 |
| |||||||||||||
pi |
where
\chi2
\chi2
Oi
N
Ei=Npi
pi
n
The chi-squared statistic can then be used to calculate a p-value by comparing the value of the statistic to a chi-squared distribution. The number of degrees of freedom is equal to the number of cells
n
p
The chi-squared statistic can be also calculated as
\chi2=
n | |
\sum | |
i=1 |
| |||||||
Ei |
-N.
This result is the consequence of the Binomial theorem.
The result about the numbers of degrees of freedom is valid when the original data are multinomial and hence the estimated parameters are efficient for minimizing the chi-squared statistic. More generally however, when maximum likelihood estimation does not coincide with minimum chi-squared estimation, the distribution will lie somewhere between a chi-squared distribution with
n-1-p
n-1
The chi-squared test indicates a statistically significant association between the level of education completed and routine check-up attendance (chi2(3) = 14.6090, p = 0.002). The proportions suggest that as the level of education increases, so does the proportion of individuals attending routine check-ups. Specifically, individuals who have graduated from college or university attend routine check-ups at a higher proportion (31.52%) compared to those who have not graduated high school (8.44%). This finding may suggest that higher educational attainment is associated with a greater likelihood of engaging in health-promoting behaviors such as routine check-ups.
In Bayesian statistics, one would instead use a Dirichlet distribution as conjugate prior. If one took a uniform prior, then the maximum likelihood estimate for the population probability is the observed probability, and one may compute a credible region around this or another estimate.
In this case, an "observation" consists of the values of two outcomes and the null hypothesis is that the occurrence of these outcomes is statistically independent. Each observation is allocated to one cell of a two-dimensional array of cells (called a contingency table) according to the values of the two outcomes. If there are r rows and c columns in the table, the "theoretical frequency" for a cell, given the hypothesis of independence, is
Ei,j=Npi ⋅ p ⋅ ,
where
N
pi ⋅ =
Oi ⋅ | |
N |
=
c | |
\sum | |
j=1 |
Oi,j | |
N |
,
is the fraction of observations of type i ignoring the column attribute (fraction of row totals), and
p ⋅ =
O ⋅ | |
N |
=
r | |
\sum | |
i=1 |
Oi,j | |
N |
is the fraction of observations of type j ignoring the row attribute (fraction of column totals). The term "frequencies" refers to absolute numbers rather than already normalized values.
The value of the test-statistic is
\chi2=
r | |
\sum | |
i=1 |
c | |
\sum | |
j=1 |
{(Oi,j-Ei,j)2\overEi,j
=N\sumi,jpi ⋅ p ⋅ \left(
(Oi,j/N)-pi ⋅ p ⋅ | |
pi ⋅ p ⋅ |
\right)2
Note that
\chi2
Oi,j=Ei,j\foralli,j
Fitting the model of "independence" reduces the number of degrees of freedom by p = r + c − 1. The number of degrees of freedom is equal to the number of cells rc, minus the reduction in degrees of freedom, p, which reduces to (r − 1)(c − 1).
For the test of independence, also known as the test of homogeneity, a chi-squared probability of less than or equal to 0.05 (or the chi-squared statistic being at or larger than the 0.05 critical point) is commonly interpreted by applied workers as justification for rejecting the null hypothesis that the row variable is independent of the column variable.[6] The alternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified.
The chi-squared test, when used with the standard approximation that a chi-squared distribution is applicable, has the following assumptions:[7]
A test that relies on different assumptions is Fisher's exact test; if its assumption of fixed marginal distributions is met it is substantially more accurate in obtaining a significance level, especially with few observations. In the vast majority of applications this assumption will not be met, and Fisher's exact test will be over conservative and not have correct coverage.[11]
The null distribution of the Pearson statistic with j rows and k columns is approximated by the chi-squared distribution with(k − 1)(j − 1) degrees of freedom.[12]
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,
O \sim 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 by a Gaussian (normal) distribution and thus the Pearson test statistic approximates a chi-squared distribution,
Bin(n,p) ≈ N(np,np(1-p)).
Let O1 be the number of observations from the sample that are in the first cell. The Pearson test statistic can be expressed as
| |||||||
np |
+
| |||||||
n(1-p) |
,
which can in turn be expressed as
\left( | O1-np |
\sqrt{np(1-p) |
By the normal approximation to a binomial this is the squared of one standard normal variate, and hence is distributed as chi-squared with 1 degree of freedom. Note that the denominator is one standard deviation of the Gaussian approximation, so can be written
| |||||||
\sigma2 |
.
So as consistent with the meaning of the chi-squared 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-squared distribution is then integrated on the right of the statistic value to obtain the P-value, which is equal to the probability of getting a statistic equal or bigger than the observed one, assuming the null hypothesis.
When the test is applied to a contingency table containing two rows and two columns, the test is equivalent to a Z-test of proportions.
Broadly similar arguments as above lead to the desired result, though the details are more involved. One may apply an orthogonal change of variables to turn the limiting summands in the test statistic into one fewer squares of i.i.d. standard normal random variables.[13]
Let us now prove that the distribution indeed approaches asymptotically the
\chi2
Let
n
m
pi
1\lei\lem
\{ki\}
ki
m | |
\sum | |
i=1 |
ki=n and
m | |
\sum | |
i=1 |
pi=1.
Let
2 | |
\chi | |
P(\{k |
i\},\{pi\})
2 | |
\chi | |
P(\{p |
i\})
\chi2
m-1
n\toinfty.
For any arbitrary value T:
2 | |
P(\chi | |
P(\{p |
i\})>T)=
\sum | |
\{ki\ |
2 | |
|\chi | |
P(\{k |
i\},\{pi\})>T}
n! | |
k1! … km! |
m | |
\prod | |
i=1 |
ki | |
{p | |
i} |
We will use a procedure similar to the approximation in de Moivre–Laplace theorem. Contributions from small
ki
n
n
n!
ki!
2 | |
P(\chi | |
P(\{p |
i\})>T)\sim
\sum | |
\{ki\ |
2 | |
|\chi | |
P(\{k |
i\},\{pi\})>T}
m | |
\prod | |
i=1 |
\left(
npi | |
ki |
ki | ||
\right) | \sqrt{ |
2\pin | |||||||||
|
By substituting for
xi=
ki-npi | |
\sqrt{n |
we may approximate for large
n
ki
xi
km=npm-\sqrt{n}
m-1 | |
\sum | |
i=1 |
xi,
we arrive at
2 | |
\begin{align} P(\chi | |
P |
(\{pi\})>T)&\sim\sqrt{
2\pin | |||||||||
|
By expanding the logarithm and taking the leading terms in
n
2 | |
P(\chi | |
P(\{p |
i\})>T)\sim
1 | ||||||||||||
|
Pearson's chi,
2 | |
\chi | |
P(\{k |
i\},\{pi\})=
2 | |
\chi | |
P(\{\sqrt{n} |
xi+npi\},\{pi\})
(km-n
2/(n | |
p | |
m) |
pm)
This argument can be written as:
- | 1 |
2 |
m-1 | |
\sum | |
i,j=1 |
xiAijxj, i,j=1, … ,m-1, Aij=\tfrac{\deltaij
A
(m-1) x (m-1)
\{xi\}
m-1
\{yi\}
m-1 | |
\sum | |
i,j=1 |
xiAijxj=
m-1 | |
\sum | |
i=1 |
2. | |
y | |
i |
This linear change of variables merely multiplies the integral by a constant Jacobian, so we get:
2 | |
P(\chi | |
P(\{p |
i\})>T)\simC
\int | ||||||||||||||||
|
m-1 | |
\left\{\prod | |
i=1 |
dyi\right\}
m-1 | ||
\prod | \exp\left[- | |
i=1 |
1 | |
2 |
m-1 | |
\left(\sum | |
i=1 |
2 | |
y | |
i |
\right)\right]
Where C is a constant.
This is the probability that squared sum of
m-1
\chi2
m-1
We have thus shown that at the limit where
n\toinfty,
m-1
A 6-sided die is thrown 60 times. The number of times it lands with 1, 2, 3, 4, 5 and 6 face up is 5, 8, 9, 8, 10 and 20, respectively. Is the die biased, according to the Pearson's chi-squared test at a significance level of 95% and/or 99%?
The null hypothesis is that the die is unbiased, hence each number is expected to occur the same number of times, in this case, = 10. The outcomes can be tabulated as follows:
i | Oi | Ei | Oi-Ei | (Oi-
| ||||||
---|---|---|---|---|---|---|---|---|---|---|
1 | 5 | 10 | -5 | 25 | ||||||
2 | 8 | 10 | -2 | 4 | ||||||
3 | 9 | 10 | -1 | 1 | ||||||
4 | 8 | 10 | -2 | 4 | ||||||
5 | 10 | 10 | 0 | 0 | ||||||
6 | 20 | 10 | 10 | 100 | ||||||
Sum | 134 |
We then consult an Upper-tail critical values of chi-square distribution table, the tabular value refers to the sum of the squared variables each divided by the expected outcomes. For the present example, this means
{\chi2}=25/10+4/10+1/10+4/10+0/10+100/10=13.4
This is the experimental result whose unlikeliness (with a fair die) we wish to estimate.
Degrees of freedom | Probability less than the critical value | |||||
---|---|---|---|---|---|---|
0.90 | 0.95 | 0.975 | 0.99 | 0.999 | ||
5 | 9.236 | 11.070 | 12.833 | 15.086 | 20.515 |
The experimental sum of 13.4 is between the critical values of 97.5% and 99% significance or confidence (p-value). Specifically, getting 20 rolls of 6, when the expectation is only 10 such values, is unlikely with a fair die.
See main article: Goodness of fit.
In this context, the frequencies of both theoretical and empirical distributions are unnormalised counts, and for a chi-squared test the total sample sizes
N
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 44 men in the sample and 56 women, then
\chi2={(44-50)2\over50}+{(56-50)2\over50}=1.44.
If the null hypothesis is true (i.e., men and women are chosen with equal probability), the test statistic will be drawn from a chi-squared distribution with one degree of freedom (because if the male frequency is known, then the female frequency is determined).
Consultation of the chi-squared 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 0.23. This probability is higher than conventional criteria for statistical significance (0.01 or 0.05), 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 (i.e., we would consider our sample within the range of what we would expect for a 50/50 male/female ratio.)
The approximation to the chi-squared distribution breaks down if expected frequencies are too low. It will normally be acceptable so long as no more than 20% 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 obtained by reducing the absolute value of each difference between observed and expected frequencies by 0.5 before squaring; this is called Yates's correction for continuity.
In cases where the expected value, E, is found to be small (indicating a small underlying population probability, and/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. When 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. This test uses the conditional distribution of the test statistic given the marginal totals, and thus assumes that the margins were determined before the study; alternatives such as Boschloo's test which do not make this assumption are uniformly more powerful.
It can be shown that the
\chi2
\Psi
\chi2