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5 Cards in this Set
- Front
- Back
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Distribution of (p^_1-p^_2)
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-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) |
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Why approximate V(p^_1-p^_2) using p^_1, p^_2?
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since we don't know p_1, and p_2 beforehand
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We approximate V(p^_1-p^_2) using p^_1, p^_2 in 2 cases depending on H0
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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) |
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How to tell if it is case 1?
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=> Q asks us if p_1 [>,=,<] p2
--> this implies H0: p1-p2=0 |
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How to tell if it is case 2?
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=> Q asks us if p_1-p_2 [>,=,<] D
--> this implies H0: p1-p2=D |
