Chi-Square degrees of freedom confusion Wikipedia states that a $\chi^2$ distribution has $k$ degrees of freedom and is the sum of $k$ independent standard normal random variables (makes sense). But Newbold's book 'Statistics for Business and Economics' states that 
$\chi^2 \left( n - 1 \right) = \frac{ \left( n - 1 \right) \cdot s^2 } { \sigma^2 }$
and I quote '$\chi^2$ distribution with $n - 1$ degrees of freedom'.
So is it $k$ dof or $n - 1$ dof?
Am I possibly missing some notation common sense or?
Thank you.
 A: Many test statistics follow the $\chi^2$ distribution. There are quite a lot of $\chi^2$ tests, and all have different formula to calculate the value of $df$ (degrees of freedom). A few examples:


*

*Pearson's $\chi^2$ test (of goodness of fit): $df = k - 1$ (or $df = k - p - 1$), where $k$ is the number of classes (and $p$ is the number of estimated parameters of the investigated distribution, if any)

*$\chi^2$ test that the variance of a normally distributed population has a given value based on a sample variance: $df = n - 1$ (where $n$ is the sample size)

*$\chi^2$ test of independence: $df = \left( r - 1 \right) \cdot \left( c - 1 \right)$ (where $r$ is the number of rows, $c$ is the number of columns in the contingency table)


For more examples, please check the $\chi^2$-test wikipedia page.
So there isn't any contradiction here, everywhere there are $df$ independent normally distributed random variables in the background somehow, and $df$ sometimes equals $n - 1$, sometimes $k - 1$ (here $k$ is the $k$ from the Pearson's $\chi^2$ test above, not the $k$ in your question), sometimes some other formula, depends on the test.
