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Let's assume we have X - a random variable, and its values are 1, 2, and 3. We also have frequencies for each of the values, which equal to 51, 40, and 65 respectively. So, this will look like this:

X 1 2 3
n 51 40 65

And the question is whether the variable X has uniform distribution. I am to use Pearson criterion to check it. I actually do know how to solve it, but I have stumbled across one problem: the number of degrees of freedom is equal to zero, hence, I cannot use tables with critical values for chi-squared, since there is no zero degrees of freedom. What do I do? Is it even possible at all to solve this task using this criterion?

Артем Макаров
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1 Answers1

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If your distribution was uniform you should have had $156/3=52$ observations each value. Now apply chi-squared test

$$\chi^2=\frac{(51-52)^2}{52}+\frac{(40-52)^2}{52}+\frac{(65-52)^2}{52}$$

and then check the result with your tables with 2 d.o.f.

tommik
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  • Alright, but why is the number of DOF equal to 2? I calculated it and got zero, because we are looking for two parameters of uniform distribution (a and b) for density function, so the number of DOF would be (3 - 2 - 1), where 3 is the number of observed values of X. Have I made an error in calculations? – Артем Макаров Nov 10 '21 at 18:05
  • @ Артем Макаров : your distribution is a discrete uniform and it is completely soecified thus you have no parameters to estimate – tommik Nov 10 '21 at 18:09