Bonferroni method for multiple comparisons with 1-way layout My statistics book describes the Bonferroni method of doing multiple comparisons with a one-way layout as essentially performing a t-test with each pair of interest with a significance level scaled down by the number of comparisons to be done. Then you might reject for certain pairs, indicating that the difference of means for the entire layout can be pinpointed as arising from certain pairs.
I would think that you would have different sized acceptance regions for each pair, coming from the pooled variance of that pair when running the standard t-test.  For instance, if two pairs had the same difference of means but one pair had much smaller individual sample variances than the other, it would seem to me easier to reject for the pair with small variances. 
All descriptions of the method that I have found though use the pooled variance for the entire layout (using the variance within groups statistic), making every pair have an acceptance region of the same width.  Morally, this makes sense to me -- you have the data for all of the groups, so why would you pretend like you only have the data for some of the groups when making an individual comparison of some pair.  
I think part of my problem is that my book doesn't give any justification for the validity of the method using the overall pooled variance, nor have a found a reasonable attempt at a justification in the online sources I have found.  Specifically, it is not intuitive to me that creating equal width confidence intervals using the total pooled variance leads to a combined coverage of the desired significance (or perhaps it is only less than or equal to the desired significance?)  If someone can point me to a reference to clarify my confusion, or provide some justifications in an answer, I would appreciate it.
(By the way, this issue seems to have come up in this post:
Two questions about Bonferroni adjustment
The accepted answer makes the same claim as my book and online sources, but again gives no reason for why it is wrong to just do a bunch of separate t-tests.  The comments by ttnphns on his own answer seem to be explaining the issue, but they are cryptic to me.)
 A: If you can believe the assumption that all variances are equal, then the tests based on the pooled SD are more powerful and hence more desirable. If you can't or don't believe that assumptio (but you can believe normality and independence), then it's is advisable to not use the pooled SD and instead truly take two samples at a time, using the Welch (or Satterthwaite) method for unequal SDs. It is still valid to adjust the significance levels of those tests using the Bonferroni inequality.
A: I'm going to expand on my April 2015 answer. 
Bonferroni method
The validity of the Bonferroni method has nothing to do with pooled variances or any other distributional assumptions. It is simply a probability inequality. Let $E_i$ denote the event that you made a type-I error in the $i$th test. Then the probability of making at least one type-I error in the first two tests is 
$$\Pr(E_1\cup E_2) = \Pr(E_1) + \Pr(E_2) - \Pr(E_1\cap E_2) \le \Pr(E_1)+\Pr(E_2)$$
Therefore, if you ensure that $\Pr(E_1)\le\alpha/2$ and $\Pr(E_2)\le\alpha/2$, you have that $\Pr(E_1\cup E_2)\le\alpha$. For $k$ tests, you can telescope the above derivation and show that
$$\Pr\{\bigcup_{i=1}^k E_i\} \le \sum_{i=1}^k \Pr(E_i)$$
and hence that, if each test is run at a significance level of $\alpha/k$, then the overall significance level is bounded above by $\alpha$. This is true no matter how you conduct those tests; they don't even all have to be done using the same method as long as each significance level is controlled at $\alpha/k$.
Pooling, acceptance-region widths, etc.
When you compare the $i$th and $j$th means under normality assumptions, you need an estimate of $\mathrm{Var}(\bar y_i-\bar y_j) = \sigma_i^2/n_i + \sigma_j^2/n_j$. If you know in advance that $\sigma_1^2=\sigma_2^2=\cdots=\sigma_k^2$, you may use the pooled variance $s^2$ as the estimate of their common value $\sigma^2$. If you don't think that the $\sigma_i^2$ are equal, and you don't know any relation between them, then the only usable information for estimating $\sigma_i^2/n_i + \sigma_j^2/n_j$ is in the sample variances $s_i^2$ and $s_j^2$. The other samples don't provide any useful information for estimating that quantity. Yes, it's true you're throwing away information, but you're not throwing away any relevant information for that particular comparison.
I hope this helps further answer the question.
