F-test formula under robust standard error I am attempting to write a program that will (among other things) use the F-test in multivariate regression under standard robust errors. I am having trouble finding a specific  formula for the F-statistic using robust errors. 
I understand the purpose of robust errors. I simply need the modified formula for the F-statistic.  
Can anyone help?
 A: In the linear regression model
$$\mathbf y = \mathbf X\beta + \mathbf u$$
with $K$ regressors and a sample of size $N$, if we have $q$ linear restrictions on the parameters that we want to test, 
$$\mathbf R\beta = \mathbf r$$
where $\mathbf R$ is $q \times K$, then we have the Wald statistic
$$W= (\mathbf R\hat \beta - \mathbf r)'(\mathbf R \mathbf {\hat V} \mathbf R')^{-1}(\mathbf R\hat \beta - \mathbf r) \sim_{asymp.} \mathcal \chi^2_q$$
and where $\mathbf {\hat V}$ is the consistently estimated heteroskedasticity-robust asymptotic variance-covariance matrix of the estimator,
$$\mathbf {\hat V}=(\mathbf X'\mathbf X)^{-1}\left(\sum_{i=1}^n \hat u_i^2\mathbf x_i'\mathbf x_i\right)(\mathbf X'\mathbf X)^{-1}$$
or
$$\mathbf {\hat V}=\frac {N}{N-K}(\mathbf X'\mathbf X)^{-1}\left(\sum_{i=1}^n \hat u_i^2\mathbf x_i'\mathbf x_i\right)(\mathbf X'\mathbf X)^{-1}$$
(there is some evidence that this degrees-of-freedom correction improves finite-sample performance).
If we divide the statistic by $q$ we obtain an approximate $F$-statistic
$$W/q \sim_{approx}  F_{q, N-K}$$
but why add one more layer of approximation?
ADDENDUM 2-8-2014 
The reason why we obtain an approximate $F$-statistic if we divide a chi-square by its degrees of freedom is because
$$\lim_{N-K \rightarrow \infty} qF_{q, N-K} = \chi^2_q$$
see this post.
