Currently trying to prove that the sum of squares error can be partitioned into SSPE and SSLF.
$ SSE = \sum_{j=1}^c \sum_{i=1}^{nj} (Y_{ij}- \hat Y_{ij})^2 = \sum_{j=1}^c \sum_{i=1}^{nj} (Y_{ij}- \bar Y_{j})^2 + \sum_{j=1}^c \sum_{i=1}^{nj} (\bar Y_{j}- \hat Y_{ij})^2 $
Ive gotten to the point where i have to show $ \sum_{i=1}^{nj} (Y_{ij}- \bar Y_{j}) =0 $
Is this correct? $ \sum_{i=1}^{nj} Y_{ij} - \sum_{i=1}^{nj} \bar Y_j = n_j \bar Y_j - n_j \bar Y_j = 0 $
This is probably very simple but I cannot seem to figure it out.