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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