I am trying to understand the unbalanced case of 1 way Anova. Suppose there are 3 groups of different sizes for the single factor say A. Then the overall mean is the WEIGHTED average of the means of the 3 groups.

Suppose we try to estimate $\mu$ + $a_i$ for i=1,2,3.

We impose a constraint, say sum of $a_i = 0.$

This makes $\mu$ the UNWEIGHTED mean of the $a_i's$. My query is how does this help ? Sure we can estimate the 4 parameters $\mu,a_i's$ but this does not match with $\mu=$ grand mean which is the weighted mean of the $a_i's$. Can someone say a few words on this topic?

I am trying to understand the unbalanced case of 1 way Anova. Suppose there are 3 groups of different sizes for the single factor say A. Then the overall mean is the WEIGHTED average of the means of the 3 groups. The one way anova then tests if the between group sum of squares where the mean of each group is compared with the WEIGHTED mean (= overall mean of all data points) significant compared to the within group sum of squares.

So the one way anova (if it is significant) tells us if at least one of the group means is higher or lower than the grand = weighted mean of the entire sample. Given the weighted mean and all the group means we can tell that at least one of them is above or below the weighted mean.

Now Suppose we try to estimate $\mu$ + $a_i$ for i=1,2,3.

We impose a constraint, say sum of $a_i = 0.$

This makes $\mu$ the UNWEIGHTED mean of the $a_i's$. My query is how does this help ? Sure we can estimate the 4 parameters $\mu,a_i's$ but this does not match with $\mu_{grand}$ the grand mean which is the weighted mean of the $a_i's$.

Given the weighted mean we can compare the a_i's to 0 to know if at least one of them is different to the weighted mean. But we cant compare with the unweighted mean because that is not what the one way anova is testing.

So then how does imposing the sum to zero constraint help ? What is the intuition that it conveys?