2
$\begingroup$

Disclaimer: I'm new to stats so please bear with me

Everywhere I look, the Chi-Square Distribution is explained with a (Z-Score)^2 i.e $$ ((X-\mu )/\sigma )^2 $$ and based on a random variable from a STANDARD NORMAL Distribution in which case it becomes $$X^2$$

However, the Chi Square Goodness of Fit Test is explained with the following formula: $$(Oi-Ei)^2/Ei$$ which can be reinterpreted as $$ (X-\mu )^2/\mu $$

What I'm trying to understand is how can the $\chi^2$ Critical Value for a given P-Value that is derived from $$\sum ((Oi-Ei)^2/Ei) $$

be equal to the $\chi^2$ Critical Value that is derived from a $\chi^2$ distribution based on a random variable that follows a Standard Normal Distribution

$$ \sum ((X-\mu )/\sigma )^2 => \sum ((X-0 )/ 1 )^2 => \sum X^2 $$

The reason I'm asking is because I was watching this video https://youtu.be/ZNXso_riZag?t=620 where the guy just plugs in the P-Value and the degrees of freedom without specifying any sort of $\mu$ which leads to me to believe the Critical Value he got was from a $\chi^2$ distribution whose random variable was based on a STANDARD Normal Distribution and yet the actual normal distribution of his problem is different from a Standard Normal Distribution. So shouldn't he be getting a different $\chi^2$ critical value ?

Basically,

$$\sum ((X-\mu )/\sigma )^2 \neq \sum ((Oi-Ei)^2/Ei) $$

$\endgroup$

1 Answer 1

5
$\begingroup$

It can be shown that $$\chi^2 :=\sum_{j=1}^k\left[\frac{(\text{obs}_j-\text{exp}_j)^2}{\text{exp}_j}\right]\overset{\mathscr L}{\to} \chi^2_{k-1}.\tag 1\label 1$$

That is, the asymptotic distribution of the goodness-of-fit or more formally Pearson $\chi^2$ statistic is chi-square distribution with degrees of freedom equal to the number of cells minus one.

What the video showed is the usage of $\eqref 1$ in the calculation. Implicit is that the total frequency must be reasonably large.

$\endgroup$
2
  • $\begingroup$ What is the symbol on top of the arrow? Also, is this trying to say that it becomes a $\chi^2$ distribution with k-1 degrees of freedom whose random variable is based on a STANDARD Normal Distribution? $\endgroup$
    – mandem
    Oct 30 at 20:51
  • 3
    $\begingroup$ It's a curly "L", $\mathscr{L}$. Presumably (when taken along with the arrow) it's standing for something along the lines of "in the limit as $n\to \infty$, is distributed as" $\endgroup$
    – Glen_b
    Oct 30 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.