Inferenčna statistika
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1 Raziskovala metodologija v fizioterapiji Predavaje 3 Ifereča statistika Ištitut za biostatistiko i medicisko iformatiko Mediciska fakulteta, Uiverza v Ljubljai
2 Biomska porazdelitev! P(K = k, p) = # " k $ & p k (' p) 'k % parametra Bi(,p) število eot (eodvise) k slučaja spremeljivka (število uspeših dogodkov pri eotah) p verjetost uspeha Povpreča vredost (pričakovaa vredost, expected value) E(K)=ΣP(K=k)*k=*p Variaca Var(K)=*p*(-p)
3 Normala porazdelitev f (x µ,! ) = 2"! 2 e! (x!µ ) 2 2! 2 parametra N(µ, σ) 95% 2.5% 2.5% STANDARDIZACIJA X~N(µ, σ), Z=(X- µ)/σ Z~N(0, ) z 0,025 : P(Z<= z 0,025 ) =0, ,5 percetil z 0,025 =-z 0, ,5 percetil z =z -0.05/2 Tabele za N(0,) pr. glej Distributios.xls
4 Pomembost N(µ, σ) - CLI spremeljivka X merjea a eodvisih eotah Poavljaj velikokrat: izberemo vzorec velikosti, izraču Dobimo veliko povprečij Ta povprečja so porazdeljea po porazdelitvi ~ N(µ, σ/ ) Povprečje spremeljivke X je torej v populaciji eako µ. SE je odvisa od velikosti vzorca! X i X Stadarda apaka (stadard error of the mea, SE) PRIMER itervali zaupaja za pulz
5 Raziskovalo vprašaje??? µ s = 60??? Verjetost Ifereča statistika športiki x _ = i= x i Vzorčo povprečje: X s Name: Določiti, ali je povpreči pulz športikov eak 60 udarcev a miuto.
6 Statističo sklepaje _ X s ~ N(µ s, σ s / s ) SE Ničela hipoteza/domeva (H0): µ s = 60 Alterativa hipoteza/domeva (Ha): µ s 60 Če H0 drži _ (i je populacijska variaca zaa) X s ~ N(60, σ s / s ) Z = X! 60! ~ N(0,) Kolikša je verjetost, da bi lahko ta stadardizirai odmik od 0 (ali pa še večji odmik) astal po aključju (če velja ičela domeva)? p
7 Statističo sklepaje - primer Primer: 67, 42, 53, 40, 55, σ=5 _ X s ~ N(60, σ s / s ) Z = X! 60! p = P(X! "x! X # x H 0 ) = P(Z! "z! Z # z) ~ N(0,) Vzorčo povprečje=5,4, =5; SE= σ/ =5/ 5=2,24 z=(5,4-60)/2,24=-3,85: testa statistika Kakše je p (verjetost) pri tej testi statistiki?
8 Primer - adaljevaje p= P(X! "x X # x! H 0 ) = P(Z! "z Z # z! ) P(Z! "z! Z # z) = p = 2P(Z! "z) P(Z! "3,85) = 0, p = 2*0, = 0,0002 p=0,0002 Sklep: zavremo H0: µ s = 60 µ s povpreči pulz športikov v populaciji
9 Iterpretacija rezultatov Kaj pomei P=0,0002? če bi bil povpreče pulz v populaciji športikov eak 60 (i.e. ičela hipoteza drži) bi bilo zelo malo verjeto, da bi opazili a vzorcu tak ali večji odmik od 60, kot smo ga opazili. Kaj aredimo? Zavremo ičelo domevo i rečemo, da imajo športiki povpreče pulz, ki je različe od 60. Poavadi pravimo, da če je p<0,05 ali p<0,0, potem je rezultat statističo začile
10 Iterpretacija rezultatov Kaj pomei P=0,25? absece of evidece is ot evidece of absece Kaj bi aredili? Ne bi zavrili ičele domeve ( sprejeli!!!). Rekli bi, da a podlagi aših podatkov, e moremo trditi, da je povpreče pulz različe od 60. Rezultat i statističo začile.
11 Kaj lahko aredimo, če e pozamo populacijske variace? _ E vzorec: X ~ N(µ, σ/ ) Prej: Z = X µ ~ N(0,) σ Potem: X µ s ~ t t-porazdelitev z - stopijami prostosti Stopije prostosti (degrees of freedom, df) s = ( x i x) i= Variaco oceimo a podlagi vzorca N(0,) df=5 df=2 df=20 2
12 Z i t... Ali sta si podobi? 5 9 N(0,) t 4 N(0,) t N(0,) N(0,) t t Tudi za t-porazdelitev si bomo pomagali s tabelami za izraču verjetosti
13 Raziskovalo vprašaje??? µ s = µ s??? Verjetost Ifereča statistika športiki _ Vzorčo povprečje: X s x = i= x i ostali _ Vzorčo povprečje: X s Name: Primerjati povpreči pulz športikov i ostalih ljudi v populaciji
14 Statističo sklepaje?? µ v populaciji a vzorcu s -µ =0 s x s - X =0 s Ničela domeva: Kolikša je verjetost, da bi lahko ta odmik* (ali pa še večji odmik) astal po aključju? P-vredost (p) (stopja tvegaja) Če je verjetost majha, bomo zavrili (ičelo) domevo, da v populaciji i razlike Tipiča odločitev: p< 0,05 ---> µ s -µ s (zavremo domevo, da µ s -µ s odmik a vzorcu 0 ( statističo začila razlika ) = 0 i s 95 % gotovostjo trdimo, da gre za dejasko razliko v povprečem pulzu; razlika med pulzoma i 0) α=0,05 je stopja začilosti * Odmik bomo stadardizirali, glede a velikosti vzorca i a razpršeosti spremeljivke v populaciji
15 Ali imajo v povprečju dekleta i fatje v mirovaju isti pulz? Povprečje: 74, (m) 77,5 (f) Stadardi odklo: 3,8 (m) 2,6 (f) Populacija: študetje UQ µ f -µ m =0 Ničela domeva Vzorec: študetje Osov stat. x f - X m =3,4
16 Kako bomo primerjali vzorča _ povprečja dveh vzorcev? _ Testa statistika t _ E vzorec: X s ~ N(µ s, σ s / s ) Dva vzorca, če sta populacijski variaci zai X s X s ~ N(µ s µ s, (σ 2 s / s + σ2 s / s ) ) Dva vzorca, če populacijski variaci ista zai i sta eaki bolj realističa situacija Porazdelitev ~t +2-2 df SE Stadarda apaka (SE: stadard error) Skupa variaca (pooled variace)
17 Kako se porazdeli t, če µ = µ 2 i = 2 =50? X~N(µ, σ) -> Z=(X- µ)/ σ ~ N(0, ) stadarda ormala porazdelitev
18 Kako se porazdeli t, če µ = µ 2 i = 2 =0?
19 Kako se porazdeli t, če µ = µ 2 i = 2 =5? ~ ) ( + + = p t s x x t Če velja ičela domeva (µ = µ 2 ) zdaj vemo, da ) ( ) ( s x x t p + = µ µ
20 Skupa variaca () s p = (!)s 2 + ( 2!)s 2 2 (!)+ ( 2!) = povpr. SD (število študetov) študetke 77,5 2,6 50 = (50!)"2, 62 + (59!)"3,8 2 (50!)+ (59!) =3,3 študeti 74, 3,8 59 Stadarda apaka (2) SE = s p + 2 =3, Razlika povprečij (3) x! x2 = 77, 5! 74,= 3, 4 Stopije prostosti (4) df = + 2! 2 = ! 2 =07 = 2, 56 Testa statistika (5) ( x x2) ( µ µ 2) t ~ SE H 0 : µ = µ 2 ičela domeva = t + H a : µ µ 2 alterativa domeva t = 3, 4! 0 2, 56 =,32 ~H0 t
21 Kako pridemo iz teste statistike do p-vredosti? t 07 T=,32, P=0,92 x = x2 Pričakujemo 95% sredjih vredosti med -.99 i.99
22 Predpostavke za t-test za eodvisa vzorca. Porazdelitev spremeljivke v populaciji Majhe vzorec: spremeljivke so ormalo porazdeljee če iso obstajajo eparametriči testi (Ma-Whitey U / Wilcoxo rak-sum test) Velik vzorec: lahko uporabljamo test e glede a porazdelitev 2. Statističe eote so eodvise 3. Populacijski variaci sta eaki σ =σ 2 Lahko testiramo H0: σ =σ 2 (Leveov test) Če ista lahko uporabljamo Welchev t-test ~ t ν d.f. H0: σ f 2 / σ m 2 = Naš primer: s f 2 / s m 2 =,09, P=0,798 (Leveov test)
23 Postopek statističega sklepaja Izberite zastveo vprašaje Populacija Vzorec Spremeljivka/e Kaj primerjati Postavite ičelo domevo i alterativo domevo Izberite testo statistiko, (statističi test, ki ga boste boste uporabili) Določite porazdelitev teste statistike, če velja ičela domeva Izberite stopjo tvegaja (α) Izračuajte vredost teste statistike iz vašega vzorca Izračuajte P-vredost i odgovorite a zastveo vprašaje Ali imajo študetke višji pulz v mirovaju kot študeti? Populacija : študetje UQ Vzorec : študeti pri Osovah stat. Spremeljvke spol pulz v mirovaju Primerjava: povpreče pulz H 0 : µ f = µ m ičela domeva H a : µ f µ m alterativa domeva t = (x f! xm)! (µ f! µ m ) SE t = 3, 4! 0 =,32 ~H0 2, 56 t 07 ~t f + m!2 P=0,92, e zavremo ičele domeve
24 Primer 2: t-test za dva eodvisa vzorca
25 Test 2 Mea Stadard deviatio Group Group
26 Test 2 Mea Stadard deviatio H 0 :µ G = µ G2 Ha : µ G! µ G2 Group Group s p = 52 (30!)+6 2 (27!) ! 2 SE =5, = 4,9 pri " H 0 " velja t = 32! 23 4,9 =5, 52 = 9 4,9 = 2,5 ~ t 30+27!2 p = P(t 55 <! t! t 55 > t ) = 2P(t 55 > t ) = 2(t 55 > 2,5) = 0, 036 s p = s2 G( G ")+ s 2 G2( G2 ") G + G2 " 2 SE = s p G + G2 (X G " X G2)" (µ G " µ G2 ) ~ t G + SE G 2 "2 pri # H 0 # velja X G " X G2 SE ~ t G + G 2 "2 Kaj lahko sklepamo?
27 Ali bi lahko izračuali tudi 95% iterval zaupaja za razliko povprečij? 95% iterval zaupaja za µ µ 2 Iterval za katerega imamo 95% zaupaje, da vsebuje ezao populacijsko razliko povprečij 95% IZ za µ G! µ G2 : (xg! xg2)! t 55;!0,05/2 " SE,(xG! xg2)+ t 55;!0,05/2 " SE xg! xg2 = 9 SE = 4,9 9! 2 " 4,9 do " 4,9 0, 62 do 7,38 t 55 0,025 T 55;0,975 =2 0,025 S 95% zaupajem lahko trdimo, da je razlika populacijskih povprečij (µ G µ G2 ) med 0,62 i 7,38 t α ) 2 df, α / 2 : P( tdf = α
28 Predpostavka 2. Statističe eote so (e)odvise? Za isto skupio (iste eote) primerjamo 2 izhoda testa (takoj po kapi i 2 teda po jej. Uporabimo t-test za odvisa vzorca (pari t-test).
29 Pari t-test (t-test za dva odvisa vzorca) Ali se pulza pred i po obremeitvi statističo začilo razlikujeta. Razlika D (po pred) H0: µ po - µ pred = δ=0 s 68 s2 80 s3 63 s s46 26 Povprečje 5,4 SD 2, SE=SD/ 3, X µ s ~ t Testa statistika T=5,4/3,=6,52, df=45 p= 9,8*0-2 t 45; -0,05/2 =2,0 D δ s d ~ t 95% IZ za δ: 5,4-2,0*3, do 5,4+2,0*3, =45, do 57,7 udarcev a miuto
30 Predpostavke za t-test za odvisa vzorca. Porazdelitev spremeljivke v populaciji Majhe vzorec: spremeljivke so ormalo porazdeljee če iso obstajajo eparametriči testi (Wilcoxo siged-rak test) Velik vzorec: lahko uporabljamo test e glede a porazdelitev Statističe eote so odvise
31 Postopek statističega sklepaja Izberite zastveo vprašaje Populacija Vzorec Spremeljivka/e Kaj primerjati Postavite ičelo domevo i alterativo domevo Izberite testo statistiko, (statističi test, ki ga boste boste uporabili) Določite porazdelitev teste statistike, če velja ičela domeva Izberite stopjo tvegaja (α) Izračuajte vredost teste statistike iz vašega vzorca Izračuajte P-vredost i odgovorite a zastveo vprašaje Ali imajo študeti višji pulz po obremeitvi? Populacija : študetje UQ Vzorec : študeti pri Osovah stat. Spremeljvke pulz v mirovaju pulz po obremeitvi Primerjava: povpreča pulza istih študetov H 0 : µ pred = µ po ičela domeva H a : µ pred µ po alterativa domeva D δ s d ~ t t = 5, 4! 0 =6, 52 H 0 ~ 3, t 45 P<0,00, zavremo ičelo domevo
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