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Distribution of (p^_1p^_2)

E(p^_1p^_2)=p_1p_2
V(p^_1p^_2)=p_1(1p_1)/n1+ p_2(1p_2)/n2 Z=[(p^_1p^_2)(p_1p_2)]/sqrt(p_1(1p_1)/n1+ p_2(1p_2)/n2) 

Why approximate V(p^_1p^_2) using p^_1, p^_2?

since we don't know p_1, and p_2 beforehand


We approximate V(p^_1p^_2) using p^_1, p^_2 in 2 cases depending on H0

case1: If H0: p_1p_2 = 0, we use pooled p^=(x1+x2)/(n1+n2)
=> z=[(p^_1p^_2)(p_1p_2)]/sqrt(p^[1/n1+1/n2]) case2: If H0: p_1p_2 != 0, we use p^_1=x1/n1, p^_2=x2/n2 => Z=[(p^_1p^_2)(p_1p_2)]/sqrt(p^_1(1p^_1)/n1+ p^_2(1p^_2)/n2) 

How to tell if it is case 1?

=> Q asks us if p_1 [>,=,<] p2
> this implies H0: p1p2=0 

How to tell if it is case 2?

=> Q asks us if p_1p_2 [>,=,<] D
> this implies H0: p1p2=D 