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Slika - Shrani.si

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A. Juriˇ<strong>si</strong>ć in V. Batagelj: Verjetnostni račun in statistika 534<br />

✬<br />

✫<br />

H0 : π1 = π2 oziroma π1 − π2 = 0,<br />

H1 : π1 = π2 oziroma π1 − π2 = 0.<br />

Vemo, da se razlika vzorčnih deleˇzev porazdeljuje pribliˇzno normalno:<br />

<br />

p1 − p2 : N π1 − π2,<br />

π1(1 − π1)<br />

+<br />

n1<br />

π2(1<br />

<br />

− π2)<br />

n2<br />

.<br />

Seveda π1 in π2 nista znana. Ob predpostavki, da je ničelna domneva<br />

pravilna, je matematično upanje razlike vzorčnih deleˇzev hipotetična<br />

vrednost razlike deleˇzev, ki je v naˇsem primeru enaka 0. Problem pa je,<br />

kako oceniti standardni odklon. Ker velja domneva π1 = π2 = π, je<br />

disperzija razlike vzorčnih deleˇzev<br />

π1(1 − π1)<br />

n1<br />

+ π2(1 − π2)<br />

n2<br />

= π(1 − π)<br />

n1<br />

+ π(1 − π)<br />

Univerza v Ljubljani ▲<br />

❙ ▲<br />

▲<br />

n2<br />

<br />

1<br />

= π(1−π) +<br />

n1<br />

1<br />

<br />

.<br />

n2<br />

● ❙ ▲<br />

▲<br />

☛ ✖<br />

▲<br />

✩<br />

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