constant Example 13.2 at p 292 of Greene's, where N=6, k=3 and T=15(see Ex. 7.2 on Exercise Reproduce first, third and fourth rows of the results at Table 13. 1 on p. 292 of 1.2 The Within and Between Groups Estimators We could formulate a pooled regression model in three ways. First, the original formulation is xtB+t,i=1,2,…,N;t=1,2,…,T. In term of deviations from the group means vt-折=(xt-x)+ε-,i=1,2,…,N;t=1,2,…,!T and in terms of the group means =a+又B+E,i=1,2,…,N To estimate B by OLS, in(1) we would use the total sum of squares and cross pI ∑∑(xt-卖(xt-天) and ∑∑(xt-m-列 where=M∑∑1 Xit and=M∑1∑1mt In(2), the moments matrices we use are within-group (i.e, deviations from the group means )sums of squares and cross products within ∑∑(xt一)(x一 1t=1constant. Example: Example 13.2 at p.292 of Greene’s, where N=6, k=3 and T=15 (see Ex. 7.2 on p.118). Exercise: Reproduce first, third and fourth rows of the results at Table 13.1 on p.292 of Greene. 1.2 The Within and Between Groups Estimators We could formulate a pooled regression model in three ways. First, the original formulation is yit = α + x 0 itβ + εit, i = 1, 2, ..., N; t = 1, 2, ..., T. (1) In term of deviations from the group means, yit − y¯i = (xit − x¯i) 0β + εit − ε¯i , i = 1, 2, ..., N; t = 1, 2, ..., T, (2) and in terms of the group means, y¯i = α + x¯ 0 iβ + ε¯i , i = 1, 2, ..., N. (3) To estimate β by OLS, in (1) we would use the total sum of squares and cross products, S total xx = X N i=1 X T t=1 (xit − x¯¯)(xit − x¯¯) 0 and S total xy = X N i=1 X T t=1 (xit − x¯¯)(yit − y¯¯), where x¯¯ = 1 NT PN i=1 PT t=1 xit and y¯¯ = 1 NT PN i=1 PT t=1 yit. In (2), the moments matrices we use are within-group (i.e., deviations from the group means) sums of squares and cross products, S Within xx = X N i=1 X T t=1 (xit − x¯i)(xit − x¯i) 0 8