Question

If the chi-square distribution is derived from a normal distribution, how can we use it to test the significance of expected and actual results for other types of distributions (Poisson, binomial). Please explain on an intuitive level (without formulas)?

Answer 1

I'll try my best to give an intuitive explanation, but it will be hard without mentioning any "formulas". I will use them though just to be sure that we are on the same page.

There is the Chi-Squared Goodness of Fit test (also known at the Person's Chi-Squared Test). If you have done contingency table analysis with binomial or multinomial data, then the Chi-Squared test statistic has the form:
\[\sum_{i=1}^I \sum_{j=1}^J \frac{\left(O_{i,j}-E_{i,j}\right)^2}{E_{i,j}}\], where \(i=1,...,I\) are the number of rows, and \(j=1,...,J\)are the number of columns,\(E\) are the expected values, and \(O\) are the observed values.

The binomial or multinomial distributions can be approximated by a normal distribution. And, of course this result becomes better with larger sample sizes. If turns out that by using the central limit theorem in this case, you can show that the chi-squared statistic above converges in distribution to a Chi-Squared distribution. This page shows a proof of this: http://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2003/lecture-notes/lec23.pdf

However, a more intuitive approach, and a simpler one, is to just consider the normal approximation of, say a binomial distribution \(X\sim \text{BIN}(n,p)\). If you have a large enough sample size (enough trials, \(n\) done), then we approximate the binomial with a normal distribution, that is \(X\sim N(np, np(1-p))\).
When standardizing, you get that (approximately):
\[\frac{X-np}{\sqrt{np(1-p)}}\sim N(0,1)\]
And the square of a normal(0,1) distribution is Chi-Squared distributed with 1 degree of freedom.
\[\frac{(X-np)^2}{np(1-p)}\sim \chi^2_{(1)}\]

\[\frac{(X-np)^2}{np(1-p)}=\frac{(X-np)^2}{np}+\frac{(X-np)^2}{n(1-p)}=\frac{(Y-np)^2}{np}+\frac{\left[(n-X) -n(1-p)\right]^2}{n(1-p)}\\
\, \\

=\sum_{i=1}^2\frac{(O_i-E_i)^2}{E_i}\] where \(X, n-X\) are the observed results or success, failiure \(O_i\), and \(np, n(1-p)\) are the expected values \(E_i\) of success, failure. The purpose of this test is testing a hypothesis that that you observe some given probability.

Now, the Chi-Squared Goodness of Fit test can be adapted to test whether your data can come from any specified distribution (Poisson, Gamma, etc.), which I'll denote as \(X\sim F(x)\), for\(F(x)\) is a CDF. . In this case,

This test can be used to see if your data (actual results) follow some distribution (expected results). You can divide your sample space into \(k\) cells such that your events are \(A_1,...,A_k\) and let the probability of occurrence be \(p_{i}=P(X \in A_i)\) where \(X\) is just any distribution. Now for a random sample of size \(n\), let \(O_i\) be the number of observed data values into the i-th cell, and \(E_i\) is the expected number in the i-th cell, so \(E_i=np_i\). This is back into the binomial/multinomial problem. So our hypothesis test here is

\(\text{H}_0\text{: your data has some distribution }X\sim F(x)\)
\(\text{H}_0\text{: your data does not have some distribution }X\sim F(x)\)

And so the chi-squared statistic is
\[ \sum_{i=1}^k \frac{(O_i-E_i)^2}{E_i}\sim \chi^2_{(k-1)}\]
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Now, there are other tests too that you can use that use pivotal quantities that are Chi-Square distributed, although they don't necessarily follow from the logic of your question. A very common one is likelihood ratio test statistic, which compares the fit of two models (which we call the null model, under the null hypothesis Ho, and the alternative model, under the alternative hypothesis H1). This one requires finding maximum likelihood estimators. The test statistic for this test is
:
\[ \Lambda= -2(\text{log likelihood null model}-\text{log likelihood alternative model})\sim \chi^2_{(q-p)}\]
where \(q\) is the number of free parameters under the alternative hypothesis, and \(p\) is the number of free parameters under the null hypothesis.

Answer 2

The hypotheses I stated there should read:
H0: The data has some probability distribution F
H1: The data does not have some probability distribution F (not H0!!)