Chi-Squared Formula I have several nonlinear curves of the transmission spectrum of a semiconductor. I got a fit to each of the curve. To get the chi squared for each my teacher gave me the formula $$\chi^2=\sum_{i=1}^{N}\frac{(T_{exp}-T_{fit})^2}{N\sigma^2_{exp}} $$ where $$\sigma_{exp}=\sum_{i=1}^{N}\frac{\sigma_{i}}{N}.$$ However, reading some statisics book I haven't seen this formula anywhere and have only seen $$\chi^2=\sum_{i=1}^{N}\frac{(T_{exp}-T_{fit})^2}{\sigma^2_{i}}.$$ Can someone explain the top, or is the formula just wrong?
 A: Let's unpack the formula.  Ignoring the controversial factor of $1/N$, it is a sum of expressions of the form
$$\frac{(T_{exp}-T_{fit})^2}{\sigma^2_{exp}}.$$
These are squares of ratios.  The numerators, $T_{exp}-T_{fit},$ are the residuals: they (additively) compare the observed values (presumably $T_{exp}$) to the fitted values $T_{fit}$.  This makes sense when the hypothesized explanation for any differences between the fit and the observations is some random, uncorrelated additive variation.
The denominators, $\sigma_{exp},$ are estimates of the expected amount of variation (based on the fit).  In some circumstances they vary from point to point; in other situations they are taken as constant.  No matter: the point is that the ratio
$$\frac{T_{exp}-T_{fit}}{\sigma_{exp}}$$
re-expresses each residual as a multiple of the expected amount of variation.  (Notice that it is unitless.)  Intuitively, then, these ratios ought to be around $\pm 1$ in size, although we would hope many of them would be closer to $0$.  If we make a number of statistical assumptions, have sufficient amounts of data, perform least-squares or maximum likelihood fitting, and invoke the Central Limit Theorem, we may justify considering these ratios as being approximately, sort of, perhaps, Normally distributed.
The sum, then--without the offending factor of $1/N$ that would turn it into a mere average--looks like the sum of squares of uncorrelated standard Normal variables.  This is the definition of $\chi^2$.  That is, when all these assumptions hold, we may interpret the sum 
$$\sum_i\frac{(T_{exp}(i)-T_{fit}(i))^2}{\sigma^2_{exp}(i)} = \sum_i\left(\frac{T_{exp}(i)-T_{fit}(i)}{\sigma_{exp}(i)}\right)^2$$
(where the index $i$ is used to enumerate the observations) as if it were a single random draw from a $\chi^2$ distribution.  Actually that's not quite right, because these residuals are correlated.  They are connected by virtue of the common underlying fit, which typically relies on estimates of a small number of parameters, say $p$ of them.  Because of this, the sum--qua random variable--actually behaves more like a sum of $N-p$ squared standard Normal variates.  This value, $N-p$, is the number of "degrees of freedom."
Were we to divide the sum by $N$, we would obtain a scaled $\chi^2$ distribution.  There's nothing wrong about this, but it isn't too helpful, either: tables and software compute values of $\chi^2$ distributions, not values divided by $N$.  After all, $N$ isn't even directly relevant: only $N-p$ matters in calculating $\chi^2$.  Leaving out the $1/N$ factor therefore makes the calculation depend only on one number, the degrees of freedom, rather than a pair of numbers.  That--the second version in the question--is the one to use.
