Determine distribution of test statistic I have a set of normally distributed parameters, $\hat{b}_{i}$, with different but known variances $\sigma_{i}^2$, that is 
$\hat{b}_{i}\sim N(b_{i},\sigma_i^2)$
where the $\hat{}$ indicates a measured value and the paramter without the $\hat{}$ the true value. I want to test if the $b_i$'s, $i=1,\dots,k$ are equal. To do this I think I should use a test statistic like
$z=\sum_{i=1}^{k}\frac{b_i-\bar{b}}{\sigma_i^2}$,
where $\bar{b}$ is a weighed avarage, $\bar{b}=\sum_{i=1}^k\frac{b_i\cdot 1/\sigma_i^2}{1/\sigma_i^2}$. I know that if the variances are equal the test statistic $z$ is $\chi^2$ distributed with $k-1$ degrees of freedom.
My problem is that the variances are not equal, so how do I determine the distribution $z$ is taken from? My guess is that there should be some $k$-dependent factor times the $\chi^2$ distribution, but I don't know how to find it. I hope someone can help. 
 A: After the clarification of the OP in the comments, we have a set of unbiased and normally distributed estimators $\hat b_i$ each with is own and known variance (so the Behrens-Fisher problem does not apply here). The OP wants to test whether the unknown parameters under estimation are in reality all equal to some $b$. Then, under this null hypothesis, the variables
$$z_i= \frac {\hat b_i - b}{\sigma_i} \sim N(0,1)\;\; \forall i=1,...,k$$
Note that $\forall k$ we write the same $b$ since the existence of this common value is what we want to test. Then, to remain close to the OP's attempts, and if we assume that the estimators are independent (because, say they have been obtained each from a different and independent sample), the statistic
$$q= \sum_{i=1}^kz_i^2 = \sum_{i=1}^k\left(\frac {\hat b_i - b}{\sigma_i}\right)^2 \sim \mathcal \chi^2_k$$
under the null hypothesis. Implementation of this test requires of course that we provide a specific value for the postulated common $b$. One could wonder here what happens to the behavior of the test statistic if say, the majority of the $b_i$'s are indeed equal, but a few of them are not. Will it "show", if pooled together?  
If we have no value for $b$, then one would perhaps want to run pairwise tests for equality of means, which do not require knowledge of the common mean. In this case one examines separately the statistics
$$w_{ij} = \frac{\hat b_i - \hat b_j}{\sqrt {\sigma^2_i + \sigma^2_j}} \sim N(0,1) \;\; \forall i,j$$
under the null, and there is no need to square and go for a chi-square test. Is this sequential testing, I wonder.
