Between Groups Mean Square expectation Show that in the Normal one-way classification the Between Groups Mean Square has expectation $E(\frac {B}{k-1})=\sigma^2$.



$E(\frac {B}{k-1})=\frac {1}{k-1}E(\sum_1^k$$n_i$$(\bar{x_i}-\bar{x})^2)$=$\frac {1}{k-1}E(\sum_1^k$$n_i$$(\bar{x_i}^2-2\bar{x}\bar{x_i}+\bar{x}^2))$. And then I thought I could use the fact that $E\bar{x_i}^2=\sigma^2+\mu^2$ and $E\bar{x}^2=\mu^2+ \sigma^2/k$ and also $\sum_1^k\bar{x_i}=\bar{x}k$. But I am not sure how to deal with the $n_i$ in the sum and I can't get to $\sigma^2$ because of it. Do I need to use more definitions or it is possible to get the answer from this information?
Excuse me if I misunderstand your question, but, I think the expectation is not what you state.  If you take the simple case in which all $n_i$’s =n, then, using your terminology, $E(\frac {B}{k-1})=\sigma^2 + n\sigma_b^2$, where $\sigma_b^2$ is the between-means variance (subscript $_b$ for between-means).  This derives from a model in which an observation within a class is the sum of a between-means component and a within-class component and the variance of the latter is assumed to be $\sigma^2$.
Excuse me if I misunderstand your question, but, I think the expectation is not what you state.  If you take the simple case in which all $n_i$’s =n, then, using your terminology, $E(\frac {B}{k-1})=\sigma^2 + n\sigma_b^2$, where $\sigma_b^2$ is the between-means variance (subscript $_b$ for between-means).  This derives from a model in which an observation within a class is the sum of a between-means component and a within-class component and the variance of the latter is assumed to be $\sigma^2$.