nguyễn văn tuấn - wordpress.com
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NguyễnVănTuấnGarvanIns)tuteofMedicalResearch
UNSWMedicine,UNSWSydney,AustraliaUniversityofTechnologySydney,Australia
SchoolofMedicineSydney,UniversityofNotreDameAustraliaBoneandMuscleResearchLab,TonDucThangUniversity,Vietnam
Workshopvề"ThiếtkếnghiêncứuvàphânNchdữliệu"�ĐHTônĐứcThắng25/12/2017–7/1/2018 ©TuanV.Nguyen
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Chuyệnthờisự
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Consố30huyềnthoại!
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Vàinghiêncứutrướcđây
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LídoướcGnhcỡmẫu
• Ngânsáchnghiêncứu
• Đạođứckhoahọc
• Kiểmđịnhgiảthuyếtkhoahọc
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Chẩnđoánvànghiêncứukhoahọc
Bệnh trạng Có Không
Xét nghiệm +ve True +ve False +ve
(sensitivity) -ve False -ve True -ve
(Specificity)
Có ảnh Không có hưởng ảnh hưởng
Test thống kê Có ý nghĩa OK Sai sót I
1 – β α Không có Sai sót II OK
β 1 – α
“Cóýnghĩa”:P<α“Khôngcóýnghĩa”:P>α
Chẩn đoán (diagnosis) Nghiên cứu (research)
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Haisailầmtrongkiểmđịnhgiảthuyết
Cóảnhhưởng(liênquan)
Khôngcóảnhhưởng(liênquan)
Kếtquảcóýnghĩathốngkê
OK SailầmloạiI(alpha)
Khôngcóýnghĩathốngkê
SailầmloạiII(beta)
OK
Trong nghiên cứu khoa học, chúng ta chấp nhận: alpha = 0.05, 0.01 beta = 0.10, 0.20 Power = 1 – beta
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"Chấtliệu"đểướcGnhcỡmẫu
• Chỉsốthốngkê:trungbìnhhaytỉlệ?
• Độdaođộng:độlệchchuẩn,saisốmẫu(marginalerror)
• SailầmIvàII(alphavàbeta)
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Bamôhìnhnghiêncứu...
• ƯớcGnhmộttỉlệ– điềutraxãhội– nghiêncứuvềquimôbệnhtrongcộngđồng
• Sosánh2tỉlệ(biếnoutcomelànhịphân)– nghiêncứucanthiệp– sosánhhainhóm
• Sosánh2sốtrungbình(biếnoutcomelàliêntục)– nghiêncứucanthiệp– sosánhhainhóm
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CỡmẫunghiêncứuướcGnhmộttỉlệ
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Vídụ
• Mục)êu:xácđịnhtỉlệbệnh)ểuđường
• Theoyvăn,tỉlệbệnh)ểuđườngtrongcộngđồngngườitrungniênlà10%(p=0.10)
• Chấpnhậnxácsuất95%làtỉlệcóthểdaođộngtrongkhoảng8%đến12%
• Nóicáchkhác,e=(0.12–0.08)/4=0.01
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Líthuyết
Líthuyết:plàướcsốkháchquannhấtchoπ.Khoảng)ncậy95%củaπ:
p–2e<π<p+2eTrongđó,elàsaisốchuẩn:
e =p 1− p( )
nNóicáchkhác,
p− 2p 1− p( )
n
"
#$$
%
&''< π < p+ 2
p 1− p( )n
"
#$$
%
&''
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Đặtngượcvấnđề
TìmnđểướcNnhπvớimộtsaisốe
Haynóichung:
n = 2e!
"#$
%&2
p 1− p( )
n = zα /2e
!
"#
$
%&2
p 1− p( )
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DùngR
• CáchNnhcỡmẫubằngRrấtđơngiản!
• CóthểviếtvàimãR,hoặcdùnghàmn.for.surveytrongpackageepiDisplay
alpha=0.05; z = qnorm(0.05/2)
p=0.10; e = 0.01
n = (z/e)^2*p*(1-p)
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PackageepiDisplay:n.for.surveylibrary(epiDisplay)
n = n.for.survey(p=0.10, delta=0.01, alpha=0.05)
> n Sample size for survey. Assumptions: Proportion = 0.1 Confidence limit = 95 % Delta = 0.01 from the estimate. Sample size = 3457
e
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Cỡmẫunghiêncứusosánhhaitỉlệ
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Vídụ
• Mục)êu:đánhgiáhiệuquảthuốcđiềutrịungthưvú
• Môhìnhnghiêncứu:RCT,chiathành2nhóm:
– Nhómchứng
– Nhómđiềutrị
• Outcome:tỉlệsốngsótsau3năm
– Nhómchứng:75%(p1=0.75)
– Nhómđiềutrịtăng20%=75*1.2=90%(p2=0.90)
• alpha=0.05,power=0.9(beta=0.10)
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Líthuyết
p = p1 + p22
• Gọitỉlệcủahainhómlàp1vàp2• Tỉlệtrungbìnhlàp
• Cỡmẫucầnthiếtchomỗinhómlà(xấpxỉ):
n =2p 1− p( ) zα /2 + zβ( )
2
p1 − p2( )2
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MỗigiátrịalphavàbetagắnliềnvớizZlàchỉsốcủaphânbốchuẩn
α Za/2
0.05 1.96
0.01 2.81
β Power Z1-β
0.20 0.80 0.84
0.10 0.90 1.28
α β (Zα/2 +Z1-β)2
0.05 0.80 7.84
0.05 0.90 10.50
0.01 0.80 13.32
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DùngpackageepiDisplay:n.for.2p
library(epiDisplay)
n.for.2p(p1=0.75, p2=0.90, power=0.9,
alpha=0.05, ratio=1)
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DùngpackageepiDisplay> n.for.2p(p1=0.75, p2=0.90, power=0.9, alpha=0.05, ratio=1)
Estimation of sample size for testing Ho: p1==p2 Assumptions: alpha = 0.05 power = 0.9 p1 = 0.75 p2 = 0.9 n2/n1 = 1 Estimated required sample size: n1 = 146 n2 = 146 n1 + n2 = 292
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Cỡmẫunghiêncứusosánhhaisốtrungbình
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Vídụ
• Mục)êu:đánhgiáhiệuquảcanthiệptăngmậtđộxương
• Môhìnhnghiêncứu:RCT,chiathành2nhóm:
– Nhómchứng
– Nhómđiềutrị
• Outcome:mậtđộxươngtrungbình
– Nhómchứng:m1=0.80
– Nhómđiềutrịtăng5%:m2=0.80x1.05=0.84
• Độlệchchuẩn:s=0.12g/cm2
• alpha=0.05,power=0.9(beta=0.10)
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Líthuyết• Gọigiátrịtrungbìnhcủahainhómlàm1vàm2
• Cỡmẫucầnthiếtchonghiêncứulà:
n =2s2 zα /2 + zβ( )
2
m1 −m2( )2
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DùngpackageepiDisplay:n.for.2means
library(epiDisplay)
n.for.2means(mu1=0.8, mu2=0.84, sd1=0.12,
sd2=0.12, power=0.9, alpha=0.05, ratio=1)
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> n.for.2means(mu1=0.8, mu2=0.84, sd1=0.12, sd2=0.12, power=0.9, alpha=0.05, ratio=1)
Estimation of sample size for testing Ho: mu1==mu2 Assumptions: alpha = 0.05 power = 0.9 n2/n1 = 1 mu1 = 0.8 mu2 = 0.84 sd1 = 0.12 sd2 = 0.12 Estimated required sample size: n1 = 190 n2 = 190 n1 + n2 = 380
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Tómtắt
• ƯớcNnhcỡmẫu:mộtbướcquantrọngtrongnghiêncứukhoahọc
• Khôngcóconsố30huyềnthoại!
• Cỡmẫuphụthuộcvào
– saisótloạiIvàII(αvàβ)– loạibiếnoutcomevàmứcđộảnhhưởng
– độlệchchuẩnhaysaisốmẫu