I am getting a chi-square value of 1600 for a data, is it possible to get this big value for a chi-square? for 5% risk acceptable risk and 5 degree of freedom the critical value should be 11.070 and I am getting 1600. Does that mean I reject null hypothesis which means data is uniformly distributed?
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$\begingroup$ The data and a discussion of it might be helpful in a way that simply quoting an unusual chi-squared statistic is not. $\endgroup$– BruceETCommented Feb 4, 2021 at 10:22
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This is by design: when the null hypothesis is false, you want the test statistic to be unusual. Chi-squared ($\chi^2$) statistics can become extraordinarily large when the data do not conform to your hypotheses. There is no upper limit to how large they can become.
I will explain.
Five degrees of freedom implies you estimated the expected counts in at least six categories (and maybe many more). The $\chi^2$ statistic is the sum of contributions from all categories (which I will also refer to as "cells" or "bins"). Each contribution is a ratio: on the top, the numerator is the squared difference between the observed count and the expected count; on the bottom, the denominator is the expected count. We can noodle around with the arithmetic a little to see what might be going on.
Usually $\chi^2$ tests are considered suspect whenever any expected count is less than $5$ (although this can be relaxed a little), so let's initially suppose you checked this condition and it applies.
In order to produce a $\chi^2$ statistic of $1600,$ the $6+$ cells would have to contribute an average amount of $1600/6 \approx 270$ each. Since each denominator is at least $5,$ this means the numerators must average approximately $5\times (1600/6) \approx 1300.$ Because the numerators are squared differences, we deduce the differences are on the order of $\sqrt {1300} \approx 36.$ Finally (whew!), this means at least some observed values must be around $36+5\approx 40$ or greater.
It is also possible for one cell to contribute most of the value to the total of $1600.$ For instance, reasoning as before, if its expectation were $5$ and its count were $k,$ we would need
$$1600\approx \frac{(k-5)^2}{5},$$
implying
$$k\approx 95.$$
Finally I will note that when a cell has a tiny expectation, it doesn't take much to create a huge value of $\chi^2.$ For instance, suppose the expected count in a particular cell is just $0.1.$ Then a count of $13$ in this cell would contribute
$$\frac{(13-0.1)^2}{0.1} = 1664,$$
already ensuring that $\chi^2$ exceeds $1600.$ (This sheds light on why cells with tiny expected values might create problems for a $\chi^2$ test.)
From these considerations we can construct examples.
For instance, suppose your theory told you a set of $225$ numbers should have a standard Normal distribution. To test this, you created six bins separated by the breakpoints $-2,-1,0,1,2.$ The standard Normal distribution says the expected counts in these bins are (approximately) $5.1, 30.6, 76.8, 76.8, 30.6, 5.1.$ Suppose, though, that reality departed substantially from the theory: specifically, the process generating the data is Normal, but actually has a mean of $1.741$ and as a result the counts you observed happened to be $0, 1, 9, 42, 84, 89.$ That last count of $89,$ where only $5.1$ values are expected, alone contributes almost $1400$ to the $\chi^2$ statistic. (The $\chi^2$ statistic for these data is $1606.8$ with five degrees of freedom.)
It is in this fashion that the $\chi^2$ test detects shifts of probability among the cells and will easily grow large when the data are inconsistent with the expected values.
