4.6 Show that in general, for Q defined as in (4.2.1), E(Q) E Xk i1 (Fi ei)2 ei # Xk i1

4.6 Show that in general, for Q defined as in (4.2.1), E(Q) ¼ E Xk i¼1

(Fi  ei)2 ei

" #

¼

Xk i¼1 nui(1  ui)

ei þ

(nui  ei From this we see that if the null hypothesis is true, nui ¼ ei and E(Q) ¼ k  1, the mean of the chi-square distribution.