Statistics [14]: Chi-Squared Test

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Published: January 14, 2021

Basics of $\chi^2$ test.

Pearson Chi-Squared Test

Pearson’s $\chi^2$ test is a statistical test applied to sets of categorical data to evaluate how likely it is that any observed difference between the sets arose by chance.

It tests a null hypothesis stating 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.

To be specific, assume the population follows the following discrete distribution

$X$	$x_1$	$x_2$	$\cdots$	$x_k$
$P$	$p_1$	$p_2$	$\cdots$	$p_k$

Now make $n$ independent observations, and the frequencies of the $k$ values are respectively $N_i (i=1,2,...,k$ , then

$X = {\displaystyle \sum_{i=1}^k\dfrac{(N_i-np_i)^2}{np_i}}$

follows the $\chi^2$ distribution with $k-1$ degree of freedom approximately. (This can be proved using the central limit theorem, which can be found here.)

If there are $s$ parameters that need to be estimated, then

$X = {\displaystyle \sum_{i=1}^k\dfrac{(N_i-np_i)^2}{np_i}}$

follows the $\chi^2$ distribution with $k-s-1$ degree of freedom approximately.

Example 1

In 1910, Rutherford and Geiger observed the number of particles emitted by the radioactive materials. They observed 2608 times in 7.5 s intervals and recorded the number of particles reaching a certain area. Below is the recording, they recorded 10094 particles in total, and $n_k$ is the times that exactly $k$ particles are observed.

$k$	0	1	2	3	4	5	6	7	8	9	$\geq 10$
$n_k$	57	203	383	525	532	408	273	139	45	27	16
$n\cdot \hat{p}_k$	54	211	407	525	508	394	254	140	68	29	17

Test whether the data follows the Poisson distribution.

Solution. Let the number of particles in one observation be the random variable $X$ , as there are 2608 observations, so the probability of one particle falls within one observation is $\dfrac{1}{2608}$ . And as there are 10094 particles recorded, so the distribution of $X$ is

$X \sim B\left(10094, \dfrac{1}{2608}\right)$

On the other hand, according to Poisson theorem, $X\sim P(\hat{\lambda})$ , where $\hat{\lambda} = \dfrac{10094}{2608} = 3.87$ . Hence,

$\hat{p}_i = \dfrac{\hat{\lambda}^i}{i!}e^{-\hat{\lambda}} (i=0,1,...,9),\hat{p}_{10} = 1 - \sum_{i=0}^{9}\hat{p}_i$

Therefore, the $\chi^2$ statistics under the observation would be

$Y = {\displaystyle \sum_{i=0}^10\dfrac{(N_i-N\hat{p}_i)}{N\hat{p}_i}}=12.88$

Using the lookup table,

$Y\sim \chi^2(9) \Rightarrow p = P(Y > 12.88) > 0.1$

Therefore, the hypothesis that the data follows the Poisson distribution can be accepted.

Independence Test

$\chi^2$ can also be used to test the correlations between different properties. Suppose $A$ and $B$ have the following relations.

$A$ \ $B$	$1$	$2$	$\cdots$	$j$	$\cdots$	$t$	SUM
$1$	$n_{11}$	$n_{12}$	$\cdots$	$n_{1j}$	$\cdots$	$n_{1t}$	$c_1$
$2$	$n_{21}$	$n_{22}$	$\cdots$	$n_{2j}$	$\cdots$	$n_{2t}$	$c_2$
$\vdots$	$\vdots$	$\vdots$	$\ddots$	$\vdots$	$\ddots$	$\vdots$	$\vdots$
$i$	$n_{i1}$	$n_{i2}$	$\cdots$	$n_{ij}$	$\cdots$	$n_{it}$	$c_i$
$\vdots$	$\vdots$	$\vdots$	$\ddots$	$\vdots$	$\ddots$	$\vdots$	$\vdots$
$s$	$n_{s1}$	$n_{s2}$	$\cdots$	$n_{sj}$	$\cdots$	$n_{st}$	$c_s$
SUM	$d_1$	$d_2$	$\cdots$	$d_j$	$\cdots$	$d_t$	$n$

Then,

$Y = {\displaystyle \sum_{i=1}^s\sum_{j=1}^t \dfrac{\left(n_{ij} - n\cdot\dfrac{c_i}{n}\cdot\dfrac{d_j}{n}\right)}{n\cdot\dfrac{c_i}{n}\cdot\dfrac{d_j}{n}} = \sum_{i=1}^s\sum_{j=1}^t\dfrac{\left(n n_{ij} - c_id_j\right)}{nc_id_j} }$

follows $\chi^2$ distribution with $(s-1)(t-1)$ degrees of freedom.

Example 2

Test the independence of income with respect to the number of kids in a family.

kids \ income	0-1	1-2	2-3	3	SUM
0	2161	3577	2184	1636	9558
1	2755	5081	2222	1052	11110
2	936	1753	640	306	3635
3	225	419	96	38	778
4	39	98	31	14	182
SUM	6116	10928	5173	3046	25263

Solution.

$Y = {\displaystyle \sum_{i=1}^s\sum_{j=1}^t\dfrac{\left(n n_{ij} - c_id_j\right)}{nc_id_j} } = 568.6$

As $P(\chi^2(12) > 32.9) = 0.001$ , hence, $p$ value is far smaller than 0.001, thus, the income of a family is strongly related with the number of kids.

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Chao Huang

Statistics [14]: Chi-Squared Test

Pearson Chi-Squared Test

Example 1

Independence Test

Example 2

Table of Contents

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