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### 5 Cards in this Set

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 Distribution of (p^_1-p^_2) -E(p^_1-p^_2)=p_1-p_2 -V(p^_1-p^_2)=p_1(1-p_1)/n1+ p_2(1-p_2)/n2 -Z=[(p^_1-p^_2)-(p_1-p_2)]/sqrt(p_1(1-p_1)/n1+ p_2(1-p_2)/n2) Why approximate V(p^_1-p^_2) using p^_1, p^_2? since we don't know p_1, and p_2 beforehand We approximate V(p^_1-p^_2) using p^_1, p^_2 in 2 cases depending on H0 case1: If H0: p_1-p_2 = 0, we use pooled p^=(x1+x2)/(n1+n2) => z=[(p^_1-p^_2)-(p_1-p_2)]/sqrt(p^[1/n1+1/n2]) case2: If H0: p_1-p_2 != 0, we use p^_1=x1/n1, p^_2=x2/n2 => Z=[(p^_1-p^_2)-(p_1-p_2)]/sqrt(p^_1(1-p^_1)/n1+ p^_2(1-p^_2)/n2) How to tell if it is case 1? => Q asks us if p_1 [>,=,<] p2 --> this implies H0: p1-p2=0 How to tell if it is case 2? => Q asks us if p_1-p_2 [>,=,<] D --> this implies H0: p1-p2=D