# Chi-square value

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?

• The data and a discussion of it might be helpful in a way that simply quoting an unusual chi-squared statistic is not. Commented Feb 4, 2021 at 10:22

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.