Lam Sang Thong Ke

7/25/2019 Lam Sang Thong Ke

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Phng php c tnh cmucho mt nghin cu y hc

Nguyn Vn TunMt cng trnh nghin cu thng da vo mt mu (sample). Mt trong nhng

cu hi quan trng nht trc khi tin hnh nghin cu l cn bao nhiu mu hay baonhiu i tng cho nghin cu. i tng y l n vcn bn ca mt nghincu, l sbnh nhn hay stnh nguyn vin. c tnh slng i tng cn thit chomt cng trnh nghin cu ng vai tr cc k quan trng, v n c th l yu tquytnh sthnh cng hay tht bi ca nghin cu. Nu slng i tng khng th ktlun rt ra tcng trnh nghin cu khng c chnh xc cao, thm ch khng thktlun g c. Ngc li, nu s lng i tng qu nhiu hn s cn thit th ti

nguyn, tin bc v thi gian sbhao ph. Do , vn then cht trc khi nghin cul phi c tnh cho c mt si tng va cho mc tiu ca nghin cu. Slng i tng va ty thuc vo loi hnh nghin cu v hai thng schnh:

Phng php thit knghin cu v tiu ch lm sng (outcome measure).

Hsnh hng (effect size);

Sai lm m nh nghin cu chp nhn, cthl sai lm loi I v II (power);

Khng bit [hay cha quyt nh] c thit knghin cu v khng c sliu vhai thng s trn th khng thno c tnh cmu. Kinh nghim ca ngi vit cho

thy rt nhiu ngi khi tin hnh nghin cu thng khng c nim g vcc sliuny, cho nn khi n tham vn cc chuyn gia vthng k hc, hchnhn cu trli:khng thtnh c! Trong bi ny ti sbn qua hai thng strn v trnh by mtsv dnghin cu lm sng cthvc tnh cmu.

1. Thit knghin cu v tiu ch lm sng

1.1 Thit knghin cu

Thng tin thnht trong qui trnh c tnh cmu l thloi nghin cu, bi vyu tny c nh hng n phng php phn tch thng k v v thphng php ctnh cmu. C thphn bit cc thloi nghin cu ny da vo hai tiu ch: thi gianv c tnh. V thi gian, cc nghin cu thu thp d liu ti mt thi im hin ti(present) c gi l cross-sectional study (nghin cu tiu biu mt thi im); ccnghin cu c nh hng theo di tnh trng sc khe ca i tng trong mt thi gian,tc thu thp d liu tng i tng nhiu ln (hin ti v tng lai) c gi l


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prospective (hay longitudinal) study (nghin cu theo thi gian); v cc nghin cu ctin hnh hin ti nhng c nh hng tm hiu qu kh(past) c gi l retrospectivestudy.

Nghin cu ti mt thi im haycross-sectional study(c dch theo ngha en

l nghin cu ct ngang). y l mt thit km cc nh nghin cu chn mt qunthmt cch ngu nhin nhng tiu biu cho mt cng ng, ti mt thi im no .Ni cch khc, nh nghin cu thu thp dliu chmt ln duy nht ca cc i tngngay ti thi im (hin ti). Mc ch chnh ca cc nghin cu ny l tm hiu tlhin hnh (prevalence) ca mt bnh no , hay tm hiu mi tng quan gia mt yutnguy cv mt bnh.

Nghin cu i chng hay case-control study. Trong cc nghin cu ny, mcch chnh l tm hiu mi lin hgia mt (hay nhiu) yu tnguy c(risk factors) v

mt bnh rt c th. tin hnh nghin cu ny, nh nghin cu bt u bng mtnhm bnh nhn v mt nhm i tng khng bnh (i chng), v i ngc thigian tm hiu nhng yu tnguy cm chai nhm phi nhim trong qu kh.

Nghin cu xui thi gian (longitudinal studies hay prospective study). Ngcli vi nghin cu i chng (trng hp nh nghin cu bit ai mc bnh v ai khngmc bnh), vi cc nghin cu theo thi gian nh nghin cu bt u bng mt nhmkhng mc bnh, v theo di mt thi gian sau quan st ai mc bnh hay khng mcbnh trong thi gian . Ngc li vi nghin cu i chng (trng hp nh nghincu i ngc vqu khtm hiu ai bphi nhim yu tnguy c), vi cc nghin

cu theo thi gian, nh nghin cu bit ngay tlc ban u ai bphi nhim hay khngphi nhim yu tnguy c. Mc ch ca cc nghin cu xui thi gian thng l ctnh tlpht sinh (incidence) bnh trong mt thi gian (iu ny khc vi mc ch canghin cu ti mt thi im l c tnh t lhin hnh tc prevalence ca bnh).Ngoi ra, cc nghin cu theo thi gian cn cho php nh nghin cu tm hiu mi linhgia mt hay nhiu yu tnguy cv nguy cpht sinh bnh tt. Khc vi nghincu cross-section ch ghi nhn s kin ti mt thi im, cc nghin cu longitudinalphi theo di i tng trong mt thi gian c thl nhiu nm thng.

1.2 Tiu ch lm sng

Sau khi xc nh thloi nghin cu, nh nghin cu cn phi quyt nh chnmt tiu ch lm sng chnh (primary outcome measure) cn cvo m c tnh cmu. Quyt nh chn tiu ch lm sng l mt quyt nh va mang tnh lm sng, vamang tnh khoa hc. Bi v mc tiu ti hu ca nghin cu y khoa l em li li ch chobnh nhn hay cng ng, cho nn tiu ch c chn phi c ngha thc ti vi


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bnh nhn. Chng hn nhtrong vic thm nh hiu quca cc phng php truy tmung th, th tlpht hin ung thv iu trkhng phi l tiu ch c ngha thc t,nhng t l tvong v thi gian sng st sau khi truy tm ung thmi l tiu ch c ngha lm sng v thc t. Mc khc, tiu ch phi p ng cc tiu chun khoa hc vtin cy v chnh xc. Nu mt nghin cu c mc tiu tm hiu hiu quca mt

loi thuc phng chng bnh xva ng mch, th cholesterol trong mu khng thc xem l tiu ch c ngha lm sng, d n p ng yu cu khoa hc tnh. Do ,vic chn mt tiu ch lm sng cho nghin cu cn phi cn nhc rt cn thn.

Quyt nh chn tiu ch lm sng l mt quyt nh quan trng, bi v n c nhhng n cmu rt ln. Chng hn nhtrong cc nghin cu long xng, cc nhnghin cu c thso snh mt xng hay tlgy xng gia hai nhm can thip bit hiu quca thuc. Nu chn mt xng lm tiu ch lm sng th slng cmu c thsl con svi trm bnh nhn, nhng nu chn tlgy xng con sc

mu c thln n vi chc ngn i tng.

2. Khi nim vhsnh hng (effect size)

Hsnh hng, ni mt cch n gin, l mt chsvnh hng ca mtthut can thip. V phn nh mc khc bit, hsnh hng cho php chng ta trnhkhi cch din dch gii hn bi ngn ng nh phn (nh c hay khng c nhhng?), v tp trung vo mt cch din dch mang tnh khoa hc hn (nhmc nh hng cao hay thp cno?) Ba trng hp n gin sau y sminh ha cho khi

nim vhsnh hng:

Trng hp 1: Trong mt nghin cu gm 50 bnh nhn cao huyt p c iutrbng mt thuc trong nhm beta-blocker. Trc khi iu tr, huyt p tm thu (SBP)trung bnh cho cnhm l 140 mmHg v lch chun l 22 mmHg. Sau khi iu tr,huyt p tm thu gim xung cn 125 mmHg.

Trng hp 2: Mt nghin cu khc thm nh hiu quca mt thuc chnglong xng trong nhm bisphosphonate. Nghin cu c tin hnh trn 50 bnh nhn.

Trc khi iu tr, mt xng cxng i (femoral neck bone mineral density, vittt l BMD) trung bnh l 0.68 g/cm2vi lch chun 0.12 g/cm2. Sau 6 thng iu tr,BMD trung bnh cho cnhm tng ln 0.72 g/cm2 vilch chun 0.13 g/cm2.

Trng hp3: Mt nghin cu bnh chng (case-control study) nhm thmnh nh hng ca thi quen ht thuc l n glucose trong mu. Nhm ht thuc lgm 30 ngi c glucose trung bnh l 130 mg/dL vi lch chun 35 mg/dL.


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Nhm khng ht thuc l gm 70 ngi c glucose trung bnh l 110 mg/dL vi lch chun 50 mg/dL.

Trong trng hp 1, chng ta c thc tnh mc nh hng bng cch lyhuyt p sau khi iu trtrcho huyt p trc khi iu tr: d1= 125 140 = -15 mmHg.

Tng t, nh hng ca thuc bisphosphonate cho trng hp 2 l d2= 0.72 0.68 =0.04 g/cm2. V trng hp 3, nh hng ca ht thuc l c thc tnh bng d3 =130 110 = 20 mg/dL.

Kh khn trong cch c tnh nh hng trn y l khng th so snh trctip c nh hng, bi v n vo lng khc nhau. V, quan trng hn na, dao ng (phn nh bng lch chun) gia 3 trng hp cng rt khc nhau. Phngphp so snh trc tip nh hng l tng l hon chuyn sao cho cba trng hp ccng mt n vo lng. t c mc ch ny, cch n gin nht l ly nh

hng chia cho lch chun. Tsny c tn ting Anh l effect size(c khi cn gi lstandardized difference) m ti tm dch l hsnh hng. Cng thc chung cho ctnh hsnh hng (svit tt bng ES) l:

1 0

0

x xES

s

= [1]

Trong :

1x l strung bnh ca nhm can thip;

0x l strung bnh ca nhm i chng; v

0s l lch chun ca nhm i chng.

Hsnh hng ca 3 trng hp trn l:

Trng hp 1:ES1= -15 / 22 = 0.68

Trng hp 2:ES2= 0.04 / 0.12= 0.33

Trng hp 3:ES3= 20 / 50 = 0.40

Nn nhrng lch chun c cng n vo lng vi nh hng trung bnh, cho

nn hsnh hng khng c n v. Ni cch khc, n vo lng nh hng bygil lch chun. Chng hn nhtrong trng hp 1, thuc beta-blocker c tc dnggim huyt p tm thu khong 0.68 lch chun, cn trong trng hp 2, thucbisphosphonate tng mt xng ch0.33 lch chun. V c cng n vso snh, cth ni [n gin] rng h s nh hng ca thuc beta-blocker cao hn thucbisphosphonate.


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Theo mt qui c [khng c cskhoa hc my], mt hsnh hng bng 0.2c xem l thp, 0.5 l trung bnh, v >0.8 l cao [1]. Mt hsnh hng 0.2tng ng vi khc bit vchiu cao ca mt em b 15 tui v mt em b 16 tui.Mt hsnh hng 0.5 tng ng vi khc bit vchiu cao ca mt em b 14tui v mt em b 18 tui. Mt hsnh hng 0.8 tng ng vi khc bit vch

sthng minh (IQ) ca mt sinh vin nm thnht v mt tin s.

3. Sai lm loi I, II v khi nim vpower

3.1 Sai lm loi I v II

Thng k hc l mt phng php khoa hc c mc ch pht hin, hay i tmnhng ci c thgp chung li bng cm tcha c bit (unknown). Ci cha c

bit y l nhng hin tng chng ta khng quan st c, hay quan st c nhngkhng y . Ci cha bit c thl mt n s(nhchiu cao trung bnh ngi VitNam, hay trng lng mt phn t), hiu qu ca mt thut iu tr, t l lu hnh(prevalence), t lpht sinh (incidence) ca bnh, v.v Chng ta c tho chiu cao,hay tin hnh xt nghim bit hiu quca thuc, nhng cc nghin cu nhthchc tin hnh trn mt nhm i tng, chkhng phi ton bqun thca dn s.Vn l sdng kt quca mt nhm i tng suy lun cho mt qun th lnhn. Mc ch ca c tnh cmu l tm slng i tng sao cho suy lun t chnh xc cao nht v y nht.

mc n gin nht, nhng ci cha bit ny c thxut hin di hai hnhthc: hoc l c, hoc l khng. Chng hn nhmt thut iu trc hay khng c hiuquchng gy xng. Bi v khng ai bit hin tng mt cch y , chng ta phit ra githit. Githit n gin nht l githit o(hin tng khng tn ti, k hiuHo) v githit chnh(hin tng tn ti, k hiu Ha).

Chng ta sdng cc phng php kim nh thng k (statistical test) nhkim

nh t, F, z, 2, v.v nh gi khnng ca gi thit. Kt quca mt kim nhthng k c thn gin chia thnh hai gi tr: hoc l c ngha thng k (statistical

significance), hoc l khng c ngha thng k(non-significance). C ngha thng ky thng da vo trsP: thng thng, nu P < 0.05, chng ta pht biu kt quc ngha thng k; nu P > 0.05 chng ta ni kt qukhng c ngha thng k. Cng cthxem c ngha thng k hay khng c ngha thng k nhl c tn hiu hay khngc tn hiu. Hy tm t k hiu T+ l kt quc ngha thng k, v T- l kt qukimnh khng c ngha thng k.


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Hy xem xt mt v dc th: bit thuc risedronate c hiu quhay khngtrong vic iu tr long xng, chng ta tin hnh mt nghin cu gm 2 nhm bnhnhn (mt nhm c iu tr bng risedronate v mt nhm ch s dng gi dcplacebo). Chng ta theo di v thu thp sliu gy xng, c tnh tlgy xng chotng nhm, v so snh hai tlbng mt kim nh thng k. Kt qukim nh thng

k hoc l c ngha thng k (P0.05). Xinnhc li rng chng ta khng bit risedronate tht sc hiu nghim chng gy xnghay khng; chng ta chc tht githit. Do , khi xem xt mt githit v kt qukim nh thng k, chng ta c bn tnh hung:

(a)Githuyt Ha ng (thuc risedronate c hiu nghim) v kt qukim nh thngk P


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Bng 1. Cc tnh hung trong vic thnghim mt githit khoa hc

Githuyt HaKt qukim nh thngk

ng(thuc c hiu nghim)

Sai

(thuc khng c hiu nghim)

C ngha thng k(p0,05)

Sai lm loi II (type II error)= P(NS | Ha)

m tnh tht (true negative)1-= P(NS | Ho)

Ch thch: k hiu S trong bng ny c ngha l significant (tc p0.05). Do , c thm t4 tnh hung trn bng ngn ngxc sut c iu

kin nhsau: Power = 1 = P(S| Ha); = P(NS| Ha); v = P(S | Ho). Xin nhc li rng khiu ton hc P(A | B) c ngha l mt xc sut c iu kin, cthhn k hiu P(S| Ha) cngha l xc sut S xy ra nu (hay vi iu kin) Ha l ng.

3.2 Kim nh githit thng k v chn on y khoa

C lnhng l gii trn y, i vi mt sbn c, vn cn kh tru tng. Mt

cch minh ha cc khi nimpower v trsP l qua chn on y khoa. Tht vy, cthv nghin cu khoa hc v suy lun khoa hc nhl mt qui trnh chn on bnh.Trong chn on, thot u chng ta khng bit bnh nhn mc bnh hay khng, v phithu thp thng tin (nh tm hiu tin sbnh, cch sng, thi quen, v.v) v lm xtnghim (nhquang tuyn X, nhsiu m, phn tch mu, nc tiu, v.v) i n ktlun.

C hai githit: bnh nhn khng c bnh (k hiu Ho) v bnh nhn mc bnh(Ha). mc n gin nht, kt quxt nghim c th l dng tnh (+ve) hay mtnh (-ve). Trong chn on cng c 4 tnh hung v ti s bn trong phn di y,

nhng vn r rng hn, chng ta hy xem qua mt v dcthnhsau:

Trong chn on ung th, bit chc chn c ung thhay khng, phng phpchun l dng sinh thit (tc gii phu xem xt m di ng knh hin vi xc nhxem c ung thhay khng c ung th. Nhng sinh thit l mt phu thut c tnh cchxm phm vo cthbnh nhn, nn khng thp dng phu thut ny mt cch i trcho mi ngi. Thay vo , y khoa pht trin nhng phng php xt nghim khng


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mang tnh xm phm th nghim ung th. Cc phng php ny bao gm quangtuyn X hay thmu. Kt quca mt xt nghim bng quang tuyn X hay thmu cthtm tt bng hai gi tr: hoc l dng tnh (+ve), hoc l m tnh (-ve).

Nhng khng c mt phng php thnghim gin tip no, d tinh vi n u

i na, l hon ho v chnh xc tuyt i. Mt sngi c kt qudng tnh, nhngthc skhng c ung th. V mt sngi c kt qum tnh, nhng trong thc tlic ung th. n y th chng ta c bn khnng:

Bnh nhn c ung th, v kt quthnghim l dng tnh. y l trng hpdng tnh tht(danh tchuyn mn l nhy, ting Anh gi l sensitivity);

bnh nhn khng c ung th, nhng kt qu th nghim l dng tnh. y ltrng hp dng tnh gi(false positive);

bnh nhn khng c ung th, nhng kt quthnghim l m tnh. y l trnghp ca m tnh tht (specificity); v,

bnh nhn c ung th, v kt quthnghim l m tnh. y l trng hp mtnh gihay c hiu(false negative).

C thtm lc 4 tnh hung trong Bng 2 sau y:

Bng 2. Cc tnh hung trong vic chn on y khoa: kt quxt nghim v bnhtrng

Bnh trngKt quxt nghim C bnh Khng c bnh

+ve (dng tnh) nhy hay dng tnh tht(sensitivity),

Dng tnh gi(false positive)

-ve (m tnh) m tnh gi(false negative), c hiu hay m tnh tht(Specificity),

n y, chng ta c th nhn ra mi tng quan song song gia chn on ykhoa v kim nh mt githit khoa hc. Trong chn on y khoa c chsdng tnh


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tht, tng ng vi khi nim power trong nghin cu khoa hc. Trong chn on ykhoa c xc sut dng tnh gi, v xc sut ny chnh l trsp trong suy lun khoa hc.Bng sau y scho thy mi tng quan :

Bng 3. Tng quan gia chn on y khoa v suy lun trong khoa hc

Chn on y khoa Kim nh githit khoa hcChn on bnh Thnghim mt githit khoa hcBnh trng (c hay khng) Githit khoa hc (Ha hay Ho)Phng php xt nghim Kim nh thng kKt quxt nghim +ve Trsp < 0.05 hay c ngha thng kKt quxt nghim ve Trsp > 0.05 hay khng c ngha thng

kDng tnh tht (sensitivity) Power; 1-; P(s | Ha)

Dng tnh gi(false positive) Sai lm loi I; trsp; ; P(S | Ho)m tnh gi(false negative) Sai lm loi II; ; = P(NS | Ha)m tnh tht (c hiu, hay specificity) m tnh tht; 1-= P(NS | Ho)

Cng nh cc phng php xt nghim y khoa khng bao gi hon ho, ccphng php kim nh thng k cng c sai st. V do , kt qunghin cu lc nocng c bt nh (nhsbt nh trong mt chn on y khoa vy). Vn l chngta phi thit knghin cu sao cho sai stloi I v II thp nht.

4. Phng php c tnh cmu

Nh cp trong phn u ca bi vit, c tnh si tng cn thit chomt cng trnh nghin cu, ngoi thloi nghin cu, chng ta cn phi c 3 sliu: xcsut sai st loi I v power, v hsnh hng. Slng cmu l hm sca ba thng

s ny. Gi n l slng cmu cn thit, l sai st loi I, l sai st loi II (tc 1-l power), hsnh hng lES, th cng thc chung c tnh cmu l:

( )( )2

2

2/ES

zzn b+=

Trong , 2/z v z l nhng hng s(tht ra l slch chun) tphn phi chun

(standardized normal distribution) cho xc sut sai st v . Bi v, trong cng thctrnES l mu s, cho nn nu ESthp th slng cmu stng; ngc li, nuEScao th slng cmu sgim.


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V nh hng nh th, hsnh hng phi c ginh trc khi tnh ton.y l thng skhng phi lc no cng c sn, cho nn nh nghin cu cn phi xemxt cc nghin cu trc hay nh hng c ngha lm sng tnh ton cmu.

Vxc sut sai st, thng thng mt nghin cu chp nhn sai st loi I khong

1% hay 5% (tc= 0.01 hay 0.05), v xc sut sai st loi II khong = 0.1 n = 0.2

(tc power phi t0.8 n 0.9). Mi trng hp gn lin vi mt hng s 2/z v z

nhva cp. Hai hng sny c th tm gn bng cng thc ( )22/ bzzC += . C

c xc nh bi lut phn phi chun nhtrnh by trong Bng 3 di y. Chng hn

nhnu mun = 0.05 v power = 0.80, th hng sC l 7.85.

Bng 3: Hng sC lin quan n sai st loi I v II

= = 0.20(Power = 0.80) = 0.10(Power = 0.90) = 0.05(Power = 0.95)0.10 6.15 8.53 10.790.05 7.85 10.51 13.000.01 13.33 16.74 19.84

4.1 Cc nghin cu vi tiu ch l bin lin tc (continuous variable)

4.1.1 Trong trng hp nghin cu chc mt nhm i tng, v mc tiu l

c tnh mt chstrung bnh (k hiu) vi mt sai snh trc l . Vi nghin cunhth, hsnh hng c thc tnh bng ES =/. V si tng (n) cn thitcho nghin cu c thtnh ton theo cng thc sau y:

( )2

Cn

ES= [2]

Trong , C l hng stBng 3.

4.1.2 Trong trng hp nghin cu trc-sau (before-after studies).

Nhiunghin cu can thip trn mt nhm bnh nhn, m theo tiu ch lm sng mi bnhnhn c o lng hai ln: trc khi can thip v sau khi can thip. Trong thut ngdch thc, ngi ta gi l nghin cu trc-sau (before-after study). Chng hn nhnh gi hiu qu ca mt loi thuc iu tr cao huyt p, cc nh nghin cu c thchn mt nhm bnh nhn thch hp, sau o lng huyt p trc khi iu trv sau


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khi iu tr. Hsnh hng c thtnh tkhc bit gia hai thi im, nhng ycn mt thng slin quan khc: l hstng quan gia hai ln o lng.

Gi o lng trc khi iu tr ca bnh nhn i l iX v sau khi iu tr l iY.

nh hng ca thut iu trc thc tnh cho mi bnh nhn ibng i i iY X = . T, chng ta c thtnh nh hng trung bnh v lch chun ca

i . Trong thc t,

chng ta khng bit i , cho nn phi da vo mt mu. Nu gi c smu ca i

lid , chng ta c thc tnh nh hng trung bnh v lch chun ca id . Gi ch

strung bnh l d v lch chun l s. Hsnh hng c thc tnh bng cngthc:

dES

s=

Ngoi ra, gi r l hstng quan gia hai o lng. Vi cc thng sny, slng cmu cn thit cho nghin cu l:

( )

( )2

2 1C rn

ES

= [3]

4.1.3 Trong trng hp nghin cu vi hai nhm i tng, mc tiu thng

l so snh hai chstrung bnh. Gi chstrung bnh ca nhm 1 v 2 l 1 v 2 . Gi

lch chun ca hai nhm l 1 v 2 . Nu hai lch chun khng khc nhau, hs

nh hng c thc tnh tcng thc [1] nhsau:

1 2

1

ES

=

Slng i tng cho mi nhm(n) cn thit cho nghin cu c thtnh ton nhsau(gi tr ca hng sC c xc nh txc sut sai st loi I v II (hay power) trongBng 3):

( )

2

2Cn

ES

= [4]

4.1.4 Trong trng hp nghin cu vi hai nhm i tng nhng mc tiukim nh nh hng tng ng (equivalence studies). Trong nhiu nghincu, chng ta mun nh gi xem hai thut can thip hay iu trc hiu qunhnhau.

Gi chstrung bnh ca nhm 1 v 2 l 1 v 2 . Nu | 1 2 | < d (trong d l

khc bit khng c ngha lm sng), th chng ta tuyn brng hai thut iu trc nh


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hng tng ng. Trong trng ny, hsnh hng sl (ti sdng k hiuHthayv ES khng nhm ln vi cng thc [1]):

1 2 dH

=

V slng cmu cn thit cho mi nhml:

2

2Cn

H= [5]

4.2 Cc nghin cu vi tiu ch l bin nhphn (binomial variable)

Trong phn trc chng ta lm quen vi phng php c tnh cmu so

snh hai strung bnh bng kim nh t. Nhng c nghin cu bin skhng lin tc mmang tnh nhphn (nhc / khng, sng / cht, dt bnh / khng dt bnh, v.v), chstm lc [dnhin] khng thl strung bnh, m l tl (proportion). Nhiu nghincu m t c mc ch kh n gin l c tnh mt t l. Chng hn nh gii y tthng hay tm hiu t l lu hnh bnh trong mt cng ng. Trong trng hp ny,chng ta khng c nhng o lng mang tnh lin tc, nhng kt quchl nhng gi trnhphn nhc / khng. Phng php c tnh cmu cng khc vi cc phng phpcho cc nghin cu vi bin slin tc.

Nm 1991, mt cuc thm d kin Mcho thy 45% ngi c hi sn sngkhuyn khch con hnn hin mt quthn cho nhng bnh nhn cn thit. Khong tincy 95% ca t l ny l 42% n 48%, tc mt khong cch n 6%! Kt qu ny[tng i] thiu chnh xc, d s lng i tng tham gia ln n 1000 ngi. Tisao? trli cu hi ny, chng ta thxem qua mt vi l thuyt vc tnh cmucho mt tl.

4.2.1 Trong trng hp nghin cu chc mt nhm i tng, v mc tiu l

c tnh mt tl(k hiu) vmt bin clm sng. Qua l thuyt xc sut, chng ta

bit rng nu trong n i tng, c k bin c th c sca l p = x / n, vi sai s

chun ( ) ( ) 1 /SE p p p n= . Khong tin cy 95% ca mt tl [trong qun th] l:

( ) 1.96p SE p .

By gi, th lt ngc vn : chng ta mun c tnh sao khong tin cy

( )2 1.96 SE p khng qu mt hng sm. Ni cch khc, chng ta mun:


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14

( ) ( )

( )

1 1 2 2

2

1 2

2 1 1C p p p pn

p p d

+ =

[8]

4.3 Cc nghin cu c tnh h s tng quan (coefficient ofcorrelation)

4.3.1 Trng hp ch c hai bin lin tc. Trong cc nghin cu quan st

(observational studies), kccc nghin cu mt thi im (cross-sectional studies), i

khi mc tiu chnh l c tnh mt h s tng quan gia hai bin lin tc (chng hn

nhhstng quan gia tui v nng cholesterol). Gi hstng quan giahai bin l, githit t ra l: Ho:= 0 hoc H1 0: . (Nu 0= , hai bin hon ton

c lp vi nhau, tc khng c mi lin h).

Trong thc t, chng ta khng bit, nhng c thc tnh qua hstng quan

quan st c l r, c khi cn gi l hsPearson. Githit c thkim nh bng ch

sthng k tnhsau:

31

1log

2

1

+= n

r

rt e

Trong nl scmu. Chst phn phi theo lut phn phi chun vi trung bnh 0

v phng sai 1. Do , vn l tm n sao cho t c ngha thng k, v p sca nl:

23

1 1log

4 1e

Cn

= + +

[9]

4.3.2 Trng hp nghin cu c nhiu bin lin tc. Vi nhng nghin cu

c mt bin ph thuc (dependent variable) v nhiu bin c lp (independent

variables), mc tiu thng l xc nh cc bin c lp c th gii thch bao nhiuphn trm phng sai ca bin phthuc. Phng php phn tch chnh l m hnh hi

qui tuyn tnh a bin (multiple linear regression). Trong m hnh ny, chsphn nh

mi lin ha chiu ny l hsxc nh bi(coefficient of determination), k hiu R2.


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Phng php c tnh cmu cho hsR2tng i phc tp, v thng phi s

dng n thut m phng (simulation). Tuy nhin, mt squi c kh tin cy c thp

dng nhsau:

Vi power = 0.80 v = 0.05, nghin cu cn ti thiu 50 i tng c tnh R2 0.23; hay ti thiu 100 c tnh R2 0.12 [2].

Vi m bin c lp v 1 bin ph thuc, s lng cmu cn thit ti

thiu l n > 104 + m [3].

Vi m 5, slng cmu cn thit ti thiu l n > 50 + m [4].

4.4 Cc nghin cu c tnh tsnguy c(odds ratio)

Trong cc nghin cu i chng (case-control study), nh nghin cu thngmun tm hiu mi lin hgia mt yu tnguy c(risk factor) v mt bnh cth. Milin hny thng c o lng bng odds ratio (OR) m ti tm dch l tsnguyc (chkhng phi tschnh m ng nghip trong nc hay sdng). Chng hnnhnu tsnguy cgia ht thuc l v gy xng l 2, th iu ny c ngha l nhngngi ht thuc l c nguy cbgy xng tng khong 2 ln so vi nhng ngi khnght thuc l.

Do , c tnh cmu thnghim mt githit vmi lin hgia mt yu

tnguy cv bnh thng da vo tsnguy c. c tnh cmu cho cc nghincu nhth, nh nghin cu cn phi c trong tay 3 sliu:

Tllu hnh (prevalence) ca yu tnguy ctrong mt qun th(gi tt lp); Tsnguy cm nh nghin cu mun bit; v

Cc sai sthng k thhin qua xc sut av power.

Vi cc sliu trn, cng thc sau y scung cp cho nh nghin cu mt c tnhslng i tng cn thit cho nghin cu (N):

( )

( ) ( )

2

2

1

ln 1

r CN

r OR p p

+=

[10]

Trong , r l t s cmu gia hai nhm (v trong cc nghin cu i chng,khng nht thit hai nhm phi c cng cmu). Nu r = 1 (tc hai nhm c cng slng cmu), th cng thc trn sn gin thnh:


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( ) ( )2

4

ln 1

CN

OR p p=

[11]

4.5 Cc nghin cu vi bin phthuc l thi gian dn n skin(survival studies)

Trong nhiu nghin cu khoa hc, kcnghin cu lm sng, cc nh nghin cuthng theo di i tng trong mt thi gian, c khi ln n vi mi nm. Bin cxyra trong thi gian nhc bnh hay khng c bnh, sng hay cht, v.v l nhng bincc ngha lm sng nht nh, nhng thi gian dn n bnh nhn mc bnh hay chtcn quan trng hn cho vic nh gi nh hng ca mt thut iu trhay mt yu tnguy c. Nhng thi gian ny khc nhau gia cc bnh nhn. Chng hn nhthi imtlc iu trung thn thi im bnh nhn cht rt khc nhau gia cc bnh nhn, v

do tiu ch lm sng thng l thi gian sng st ca bnh nhn tnh tkhi c iutr(hay tkhi c chn on bnh).

Nghin cu tiu biu thng c 2 nhm bnh nhn: mt nhm i chng v mtnhm can thip. Phng php tnh cmu cho cc nghin cu thloi ny kh phc tp,nhng mt cch tnh n gin cng c thng dng. Nu thi gian theo di i tng

c nh trc, v tlpht sinh ca hai nhm trong thi gian l 1p v 2p , th ts

nguy c(hazards ratio) c thc tnh nhsau [5,6]:

( )( )2

1

loglog

pph

e

e=

V scmu cn thit cho tng nhm l:

( )

( )( )221

2

12

1

+=

hpp

hCn [12]

4.5 Cc nghin cu vchn on (diagnostic studies)

Nghin cu vchn on thng xoay quanh hai chs: nhy (sensivity) vc hiu (specificity) nh trnh by trong Bng 2. Mt phng php chn on cxem l ng tin cy v c thsdng trong thc hnh lm sng cn phi t nhy vc hiu ti thiu 0.75 (hay tt hn na l 0.80). Vic pht hin bnh qua chn on cnty thuc vo tllu hnh (prevalence) ca bnh trong mt qun th. Do , phngphp c tnh cmu phi da vo cc chsny. Mt cch cth, nh nghin cu cnphi xc nh cc sliu sau y:


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Xc sut dng tnh tht (hay nhy k hiu sep ) ti thiu l bao nhiu?

Xc sut m tnh tht (hay c hiu k hiu spp ) ti thiu l bao nhiu?

Sai sca hai xc sut dng tnh tht v m tnh tht l bao nhiu (k hiu w)?

Tllu hnh ca bnh trong qun thl bao nhiu (k hiu disp )

Vi cc thng sny, slng cmu c tnh nhy c thc tnh bngcng thc sau y [7]:

Trc ht, c tnh TP+FN (tc l sdng tnh tht true positive v m tnhgi- false negative)

( )2

2 1

w

ppZFNTP sese

=+

Trong , 2Z l hng sca phn phi chun. Nu = 0.05, hng s2Z bng

1.96. Sau , c tnh slng cmu (ti sdng k hiu sen chr y l

scmu cho nhy):

dis

sep

FNTPn

+= [13]

Tng t, slng cmu c tnh c hiu c thc tnh qua hai bc

nhsau:

Trc ht, c tnh FP+TN (tc l sdng tnh gi- false positive v m tnhtht true negative)

( )2

2 1

w

ppZTNFP

spsp =+

Sau , c tnh slng cmu (ti sdng k hiu spn chr y l sc

mu cho nhy):

dis

spp

TNFPn

+=

1 [14]

5. V d


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Trong phn ny, ti snu nhiu v dvc tnh cmu minh ha cho phnl thuyt va trnh by trong phn trn. Ti stp trung cc v dlin quan n nghincu lm sng bn c tp ch dtheo di.

5.1 c tnh cmu cho mt chstrung bnh

V d1 c tnh mt chstrung bnh: Chng ta mun c tnh chiu cao n ng ngi Vit, v chp nhn sai strong vng 1 cm (d = 1) vi khong tin cy 0.95

(tc =0.05) v power = 0.8 (hay = 0.2). Cc nghin cu trc cho bit lch chunchiu cao ngi Vit khong 4.6 cm. Nhvy, hsnh hng l:ES = 1/4.6 = 0.217,v hng sC = 7.85. Chng ta c thp dng cng thc [2] c tnh cmu cn thitcho nghin cu:

( ) ( ) 166217.0

85.722 === ES

Cn

Ni cch khc, chng ta cn phi o chiu cao 166 i tng c tnh chiucao n ng Vit vi sai strong vng 1 cm.

Nu sai schp nhn l 0.5 cm (thay v 1 cm), s lng i tng cn thit l:

( )2

7.85664

0.5/4.6n = = . Nu sai sm chng ta chp nhn l 0.1 cm th slng i

tng nghin cu ln n 16610 ngi! Qua cc c tnh ny, chng ta ddng thy cmu ty thuc rt ln vo sai s m chng ta chp nhn. Mun c c tnh cngchnh xc, chng ta cn cng nhiu i tng nghin cu.

V d2 c tnh cmu cho nghin cu trc sau: Mt loi thuc iutrc khnng tng alkaline phosphatase bnh nhn long xng. lch chunca alkaline phosphatase l 15 U/l. Mt nghin cu mi stin hnh trong mt qun thbnh nhn Vit Nam, v cc nh nghin cu mun bit bao nhiu bnh nhn cn tuynchng minh rng thuc c thalkaline phosphatase t60 n 65 U/l sau 3 thng iu

tr, vi sai s = 0.05 v power = 0.8.

y l mt loi nghin cu trc sau (before-after study); c ngha l trcv sau khi iu tr. y, chng ta chc mt nhm bnh nhn, nhng c o hai ln(trc khi dng thuc v sau khi dng thuc). Chtiu lm sng nh gi hiu nghimca thuc l thay i valkaline phosphatase. Trong trng hp ny, chng ta c thc tnh hsnh hng nhsau:


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19

3333.015

5==ES

V l nghin cu trc sau, chng ta cn mt thng tin khc na: l h s tngquan gia hai ln o lng alkaline phosphatase. Chng ta khng bit hsny, nhngc thginh n dao ng khong 0.6 n 0.8. Vi hstng quan 0.6, v sdngcng thc [3], chng ta c thc tnh scmu nhsau:

( )

( )

( )

( )56

3333.0

6.0185.721222

=

=

=ES

rCn

Nhng nu hstng quan l 0.8, th scmu trthnh:

( )

( )

28

3333.0

8.0185.722

=

=n

Ni cch khc, khi hstng quan cng cao (tc tin cy ca o lng cao), slngcmu cng thp.

5.2 c tnh cmu cho so snh hai strung bnh (hai nhm)

V d3 Nghin cu so snh hai chstrung bnh: Mt nghin cu c thitkthnghim thuc alendronate trong vic iu trlong xng phnsau thi kmn kinh. C hai nhm bnh nhn c tuyn: nhm 1 l nhm can thip (c iu trbng alendronate), v nhm 2 l nhm i chng (tc khng c iu tr). Tiu ch nh gi hiu quca thuc l mt xng (bone mineral density BMD). Sliu tnghin cu dch thc cho thy gi trtrung bnh ca BMD trong phnsau thi k mnkinh l 0.80 g/cm2, vi lch chun l 0.12 g/cm2. Vn t ra l chng ta cn phinghin cu bao nhiu i tng chng minh rng sau 12 thng iu trBMD canhm 1 tng khong 5% so vi nhm 2?

Trong v d trn, tm gi tr s trung bnh ca nhm 2 l 2 v nhm 1 l 1,

chng ta c:2

= 0.8*1.05 = 0.84 g/cm2(tc tng 5% so vi nhm 1), v do , = 0.84

0.80 = 0.04 g/cm2. lch chun l = 0.12 g/cm2. Nhvy, hsnh hng l:

3333.012.0/04.0 ==ES . Vi power = 0.90 v = 0.05, hng sC = 10.51, v scmu cn thit l:

( ) ( )189

333.0

51.102222

=

==ES

Cn


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Kt qutrn cho bit chng ta cn 190 bnh nhn cho mi nhm(hay 380 bnh

nhn cho cng trnh nghin cu). Trong trng hp ny, power = 0.90 v = 0.05 cngha l g ? Trli: hai thng s c ngha l nu chng ta tin hnh tht nhiu nghincu (v d1000) v mi nghin cu vi 380 bnh nhn, sc 90% (hay 900) nghin cuscho ra kt qutrn vi trsp < 0.05.

V y l thloi nghin cu thng dng, cho nn c ngi vmt biu (xembiu 1 di y) c tnh cmu cho nhng ai khng thch tnh ton. Biu nyi hi ngi sdng phi bit c hsnh hng (m biu vit l standardiseddifference) v power.

Biu 1. Biu (nomogram) cho c tnh cmu v power cho ccnghin cu hai nhm. (Ngun: British Medical Journal, 1980, 281,13361338).Cch sdng: Ly v d3, chng ta c standardised difference l 0.33(tc ES), power = 0.9. nh du 0.33 ct tri, 0.9 ct phi; kni haiim nh du bng mt thc thng. im giao cho gia ng kthng v ct gia chnh l scmu cn thit cho = 0.05 hay 0.01.


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V d4 c tnh cmu chng minh hai thut iu trc hiu qutng ng nhau: Mt nghin cu c thit kchng minh rng hiu qucathuc A v B tng ng nhau. Cc nh nghin cu chn mt xng (BMD) lmtiu ch lm sng. Nghin cu gm 2 nhm bnh nhn long xng c phn chia ngunhin: nhm 1 c iu trbng A, v nhm 2 c iu trbng B. Cc nghin cutrc cho thy sau 6 thng iu tr, A c th tng BMD khong 7%, v B c tc dngtng BMD khong 4%. Theo cc nghin cu , lch chun ca tng BMD l 10%.Cc nh nghin cu quyt nh rng nu khc bit vBMD gia hai nhm trong vng2 g/cm2th sxem l hai loi thuc c hiu qunh nhau. Vn t ra l cn bao nhiu

i tng cho nghin cu vi = 0.05 v power = 0.8?

Vi cc sliu trn, chng ta c thc tnh hsnh hng (xem phn 4.1.4)

nhsau: 1.010

24721=

=

=

dH . Vi = 0.05 v power = 0.8, hng sC =

7.85 (Bng 3). Slng cmu cn thit cho mi nhm (theo cng thc [5]) l:

( )1570

1.0

85.72222

=

==H

Cn

Ni cch khc, cng trnh cn tuyn chn 3140 i tng t c yu cu vmc tiu ca nghin cu.

5.3 c tnh cmu cho cc nghin cu c tnh hstng quan

V d 5 Nghin cu tm hiu tng quan. Mi tng quan gia lngng trong mu (fasting plasma glucose - FPG) v ttrng cth(body mass index BMI) thng khng nht qun gia cc sc dn, v ngay cmt sc, cng khng nhtqun gia cc qun th. Mt nghin cu c thit kc tnh hstng quan giaBMI v FPG. S liu t cc nghin cu trc cho thy h s tng quan thng daong t0.08 n 0.30. Vn t ra l trong dao ng , nghin cu cn o lng

bao nhiu i tng c kt quvi tin cy 99% (tc = 0.01) v power = 0.80?

Cng thc [9] c thng dng c tnh cmu cho nghin cu. Gidnh

hstng quan tht l 0.15, v vi= 0.01 v power = 0.80, hng sC = 13.33 (Bng3). Scmu do l:

499

15.01

15.01log25.0

33.113

1

1log25.0

322

=

+

+=

++=

r

r

Cn


22/229

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Ni cch khc, cng trnh ny cn phi tuyn khong 500 i tng t c yu cuv mc tiu ca nghin cu. Bng sau y cho thy s lng cmu c thdao ngkh cao ty thuc vo hstng quan.

Bng 4. Scmu cn thit c tnh hstng quan vi = 0.01 hay 0.05 v

power = 0.80

Scmu cn thit cho power = 0.80 vHstng quan

= 0.01 = 0.05

0.05 4527 31380.10 1128 7830.15 499 347

0.20 279 1940.25 177 1230.30 121 850.35 88 620.40 66 470.45 51 360.50 41 29

5.3 c tnh cmu c tnh mt tl

V d6 Nghin cu c tnh tllu hnh: Chng ta mun c tnh tlnng ht thuc l Vit Nam sao cho c skhng cao hn hay thp hn 2% so vi tltht trong ton dn s. Mt nghin cu trc y cho thy tlht thuc trong n ngngi Vit c thln n 70%. Cu hi t ra l chng ta cn nghin cu trn bao nhiun ng t yu cu trn.

Trong v dny, chng ta c sai sm = 0.02, p = 0.70, v slng cmu cn

thit cho nghin cu (theo cng thc [6]) l:

21.96

0.7 0.30.02

n

Ni cch khc, chng ta cn nghin cu t nht l 2017. Nu chng ta mun gim sai st2% xung 1% (tc m = 0.01) th s lng i tng s l 8067! Chcn thm chnh xc 1%, s lng mu c th thm hn 6000 ngi. Do , vn c tnh c


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mu phi rt thn trng, xem xt cn bng gia chnh xc thng tin cn thu thp v chiph.

5.4 c tnh cmu cho so snh hai tl

V d 7 Nghin cu so snh hai t l pht sinh (incidence rate): Mt thnghim lm sng i chng ngu nhin c thit knh gi hiu quca mt loithuc chng gy xng sng. Hai nhm bnh nhn sc tuyn. Nhm 1 c iu trbng thuc, v nhm 2 l nhm i chng (khng c iu tr). Cc nh nghin cu githit rng t lgy xng trong nhm 2 l khong 10%, v thuc c th lm gim t lny xung khong 6%. Nu cc nh nghin cu mun thnghim githit ny vi sai

st I l = 0.01 v power = 0.90, bao nhiu bnh nhn cn phi c tuyn m chonghin cu?

y, chng ta c = 0.10 0.06 = 0.04, v p = (0.10 + 0.06)/2 = 0.08. Vi

= 0.01, / 2z = 2.57 v vi power = 0.90, z= 1.28. Do , slng bnh nhn cn thit

cho mi nhm (theo cng thc [7]) l:

( )( )

2

2

2.57 2 0.08 0.92 1.28 0.1 0.90 0.06 0.941361

0.04n

+ + = =

Nhvy, cng trnh nghin cu ny cn phi tuyn t nht l 2722 (1361 x 2) bnh nhnkim nh githit trn.

V d8 Nghin cu chng minh hai tltng ng: Quay li vi v d4vnghin cu nhm chng minh hai loi thuc A v B c hiu qunh nhau. Nhngln ny, tiu ch lm sng nh gi hiu quca thuc l tlpht sinh gy xng ctsng (incidence of vertebral fracture), chkhng phi l sthay i mt xng. Sliu tcc nghin cu trc y cho thy tlgy xng mi cc bnh nhn c iutrbng A l khong 2% v B l 3%. ng trn quan im lm sng, cc nh nghin cucho rng nu hai tlkhc nhau trong vng 0.5% th c thxem l tng ng. Vn

t ra l cn bao nhiu i tng cho nghin cu t c = 0.05 (tc tin cy0.95) v power = 0.80?

Vi cc sliu trn (p1= 0.02,p2= 0.03, v d = 0.005 tc 0.5%) v p dng cngthc [8], chng ta c thc tnh scmu cn thit cho mi nhm nhsau:


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24

( ) ( )[ ]

( )( ) ( )[ ]

( )=

+=

+=

22

21

2211

005.003.002.0

97.003.098.002.085.72112

dpp

ppppCn 15291

Do , cng trnh ny cn tuyn 30582 i tng t yu cu ca nghin cu. Nhc

ththy c trong cng thc trn, yu tquan trng trong cch tnh cmu cho nghincu loi ny l khc bit d xem l hiu quhai loi thuc tng ng. Scmutrn da vo tiu chun khc bit 0.5% kt lun tng ng. Nhng nu tiuchun ddi hn mt cht (nh1%) th slng cmu gim xung cho mi nhmgim xung cn 7646 i tng vn l mt con sln. So snh vi kt quc tnh tv d4, chng ta thy cng mt mc tiu nghin cu, nhng vn chn tiu ch lmsng y rt quan trng v c nh hng ln n scmu.

5.5 c tnh cmu cho mt tsnguy c(odds ratio)

V d9 Nghin cu bnh-chng (case-control study): Nh nghin cu muntm hiu mi lin hgia ht thuc l v nguy cgy xng ct sng (vertebral fracture).Hai nhm i tng c chn: Nhm 1 l nhng bnh nhn mi gy xng ct sng, vnhm 2 l nhng i tng khng gy xng, nhng c cng tui, gii vi nhmbnh nhn. Sau khi c hai nhm i tng, cc nh nghin cu sphng vn xem tronghai nhm, c bao nhiu ngi ht thuc l. Cc nh nghin cu githit rng tsnguyc gy xng nhng ngi ht thuc l l 2. Nu cc nh nghin cu mun th

nghim githit ny vi sai st I l= 0.05 v power = 0.80, bao nhiu i tng cnphi c tuyn mcho nghin cu? p dng cng thc [11], chng ta c:

( ) ( ) ( )349

75.025.02ln

85.74

1ln

422

=

=

=

ppOR

Cn

Cng trnh nghin cu cn phi tuyn t nht l 350 i tng (175 bnh nhn v175 i chng) kim nh githit trn.

5.6 c tnh cmu cho nghin cu vsng st

V d10 Nghin cu so snh thi gian sng st: Nhcp trong mc 4.5,nhiu nghin cu y khoa c mc ch so snh thi gian sng st (survival time) gia hainhm. Cm tsng st y phi c hiu rng hn, khng chphn nh tvong vcn sng, m bao gm thi gian dn n mt skin lm sng (c thl ung th, t qu,gy xng, i tho ng, v.v) Chng hn nhnghin cu tm hiu xem mt thucmi c thko di thi gian sng ca bnh nhn hay khng, cc nh nghin cu theo di2 nhm bnh nhn (nhm 1 c iu trv nhm 2 l nhm i chng) trong vng 2


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Mt khc, cc sliu trn, chng ta c thc tnh FP+TN:

( ) ( )( )

=

=

=+2

2

2

2

05.0

05.095.096.11

w

ppZTNFP

spsp 73

Vi tlhin hnh ca bnh l 20%, slng cmu cn thit c tnh c hiu(theo cng thc [14]) l: 73 / (1 0.20) = 91.

Trong trng hp ny, chng ta c hai c tnh kh khc nhau. Nu da vo tiuch nhy, nghin cu cn n 1229 i tng, nhng nu ly tiu ch c hiu nghincu chcn 91 i tng. Vn cn li l phi xc nh gia hai ch snhy v chiu, chsno quan trng hn ly lm tiu ch cho nghin cu. y l mt quytnh lm sng cn phi cn nhc trc khi c tnh cmu!

6. Mt svn vc tnh cmu

iu chnh cho hin tng bcuc. Cc phng php c tnh cmu trnhby trn y da vo mt ginh quan trng l nghin cu stin hnh sung s, tc lkhng c i tng bcuc v sliu thu thp y cho tt cmi i tng. Nhngtrong thc t, chng ta bit rng khng c mt nghin cu no hon ho nhthc. Hintng bcuc (withdrawal) rt phbin trong nghin cu lm sng. V nhiu l do, itng khng ththam gia tu n cui cng trnh nghin cu, v phi bcuc. V

bcuc, cho nn sliu ca i tng khng y . Ty theo thloi v tnh can thiphay xm phm ca cng trnh nghin cu, tlbcuc v khng y sliu c thdao ng t10% n 30%.

V th, c tnh c mu cng phi xem xt n kh nng trn bng cch iuchnh cho tlbcuc. Nu theo l thuyt c tnh, nghin cu cn ni tng, v nutlbcuc l q th slng i tng thc tcn phi l n/(1-q). Chng hn nhtrongv d8, slng bnh nhn cn thit cho nghin cu theo l thuyt l 30582 ngi, nu tlbcuc l 20%, th trong thc tnh nghin cu phi cn tuyn n 30582/(1-0.20) =34280 bnh nhn.

iu chnh cho trng hp khng cn i gia hai nhm. Cc cng thc tnhton cmu ti trnh by trong phn trc cn ginh rng trong cc nghin cu gmhai nhm th hai nhm phi c cng s lng i tng. Nhng trong thc t, nhiunghin cu (nht l nghin cu bnh chng case-control study) slng bnh nhn kh


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m bng s lng i chng, nu bnh thuc vo dng him trong cng ng. V th,nhm i chng cn phi c nhiu i tng hn nhm bnh.

Trong trng hp mt cn i gia hai nhm, s lng c mu cng cn phic iu chnh. Gi tng scmu l thuyt (ca hai nhm cng li) lN; gi tng scmu iu chnh lN*, nu chng ta k vng rng tscmu ca nhm 1 v nhm 2l k, th N*c thxc nh bng cng thc sau y:

( )k

kNN

4

1*

2+

=

Dthy rng nu hai nhm cn i (c cng cmu, hay k = 1) thN =N*. Quay li vd 10, s lng c mu cn thit cho mi nhm l 195 (do N= 390). Nhng nuchng ta k vng rng scmu ca nhm 1 phi hn nhm 2 khong 1.5 ln, th tng slng i tng cn phi tuyn chn l: N = [390 (1+1.5)2] / (41.5) = 406, tc tngkhong 16 i tng so vi c tnh l thuyt.

c tnh c mu l bc u trong nghin cu. Trc khi kt thc bi vitny, ti mun nhn chi nhn mnh mt ln na, c tnh cmu cho nghin cu lmt bc cc k quan trng trong vic thit kmt nghin cu cho c ngha khoa hc,v n c thquyt nh thnh bi ca nghin cu. Trc khi c tnh cmu nh nghincu cn phi bit trc (hay t ra l c vi githit cth) vvn mnh quan tm.

c tnh cmu cn mt sthng snhcp n trong phn u ca chng,

v nu cc thng sny khng c th khng thc tnh c. Trong trng hp mtnghin cu hon ton mi, tc cha ai tng lm trc , c thcc thng svnhhng v dao ng o lng skhng c, v nh nghin cu cn phi tin hnh mtsm phng (simulation) hay mt nghin cu skhi c nhng thng scn thit.Cch c tnh cmu bng m phng l mt lnh vc nghin cu kh chuyn su, khngnm trong ti ca sch ny, nhng bn c c th tm hiu thm phng php nytrong cc sch gio khoa vthng k hc cp cao hn.


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Ti liu tham kho:

1.

Cohen J. Statistical power analysis for the behavioral science. NY: Academic Press,1969.

2.

Hair JF, Anderson RE, et al. Multivariate data analysis, 5th

Ed. New Jersey:Prentice-Hall, 1998.

3.

Green SB. How many subjects does it take to do a regression analysis. MultivariatBehav Res 1991; 26:499-510.

4. Harris RJ. A primer of multivariate analysis, 2nd Ed. New York: Academic Press,1985.

5. Freeman LS. Tables of the number of patients required in clinical trials using thelogrank test. Stat Med 1982; 1:121-129.

6. Lee ET. Statistical methods for survival analysis. Page 320. New York: Wiley, 1992.

7.

Jones SJ, Carley S, Harrison M. An introduction to power and sample size estimation.Emerg Med J 2003; 20:453-458.

Ti liu c thm: Cc cng thc trnh by trong bi vit ny c thtm thy trong ccsch gio khoa vdch thc v thng k hc. Ba cun sch sau y c thxem nhloisch dn nhp:

1. Machin JM. Biostatistical Methods The Assessment of Relative Risks. New York:John Wiley & Sons, 2000.

2. Kahn HA, Sempos CT. Statistical Methods in Epidemiology. New York: OxfordUniversity Press, 1989.

3. Phn tch s liu v to biu bng R - hng dn thc hnhca ti (tc gibiny) do Nh xut bn Khoa hc v Kthut pht hnh, Thnh phHCh Minh, 2006.Trong sch c hng dn cch tnh cmu (v phn tch sliu) bng my tnh qua ngnngthng k R.

Ngoi ra, bn c mun tm hiu thm vcc phng php tnh cmu c thtm ccc bi bo quan trng sau y:

1. Florey CD. Sample size for beginners.BMJ1993;306(6886):1181-4.


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2. Day SJ, Graham DF. Sample size and power for comparing two or more treatmentgroups in clinical trials.BMJ1989;299(6700):663-5.

3. Kieser M, Hauschke D. Approximate sample sizes for testing hypotheses about theratio and difference of two means. J Biopharm Stat1999;9(4):641-50.

4. Miller DK, Homan SM. Graphical aid for determining power of clinical trialsinvolving two groups.BMJ1988;297(6649):672-6.

5. Campbell MJ, Julious SA, Altman DG. Estimating sample sizes for binary, orderedcategorical, and continuous outcomes in two group comparisons. BMJ1995;311(7013):1145-8.

6. Sahai H, Khurshid A. Formulae and tables for the determination of sample sizes and

power in clinical trials for testing differences in proportions for the two-sample design: areview. Stat Med1996;15(1):1-21.


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1

Lm sng thng k

lch chun hay sai schun?

Nguyn Vn Tun

Trong vi nm qua, ti nhn kh nhiu email hi vnhng vn cn bn trong

thng k sinh hc v phng php dch thc. Ti c nh mmcLm sng thng

k (Statistical Clinic) trao i vi bn c vcc vn m ti thy quan trng ny.Ti hn hoan cho n cc cu hi ca bn c c cm hng trli.

Trong hng trm thhi v tham vn trong thi gian 3 nm qua, ti m c n 5

thhi vvn m ti ly lm ta cho bi vit ny. Chng hn nhmt bn c

H Ni vit email n ti hi: Tha thy! Em c thy trong cc tp san y hc ngi ta

th

ng hay trnh by strung bnh km theo SEM, nh

ng c

ng c bi bo trnh by s

trung bnh km theo SD. Xin hi Thy cch trnh by no ng?

y l mt cu hi n gin nhng ti thy c ngha ng dng kh rng, nn

mun nhn ct boLm sng thng ktrli bn c.

***

Trong cc tp san y hc, chng ta thng thy nhng ct sdi hnh thcx y,trong x l strung bnh, cn y th c khi l lch chun (standard deviation SD)

hay sai s chun (standard error SE). Cng c tc gi vit SEM (vit tt t cm tstandard error of the mean). Cch trnh by nhththng dng n ni mt schuyngia v cc ban bin tp tp san y hc phi ln ting khuyn co. Theo khuyn co chungv cng l qui c nghin cu y hc: m tmt bin s lm sng tun theo lutphn phi chun, cc nh nghin cu nn cch trnh by strung bnh v km lch chun (khng phi sai s chun; m tmt bin s lm sng khng tuntheo lut phn phi chun, nn trnh by strung vv svtr 25% v 75% (tcl interquartile range).

hiu qui c ny, chng ta cn phi tm hiu ngha ca lch chun v sai

schun. Ti thy iu ny cn thit, bi v hu ht sch gio khoa thng k (ngay csch gio khoa do ngi Ty phng vit) u khng gii r nhng khc bit v nghaca hai chsthng k ny.

M tmt bin stheo lut phn phi chun


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2

Xin nhc li thut ng: cm t phn phi chun y chnh l Normaldistribution (hay c sch cn gi l Gaussian distribution, ly ttn ca nh ton hcvi ngi c Frederick Gauss). Mt bin stun theo lut phn phi chun, khi vbng biu , ging nhhnh mt ci chung cn i (Biu 1). Phn phi ny cxc nh bng hai thng s: strung bnh v lch chun. tit kim chngha, ti

sly k hiu m thhin strung bnh, vs thhin lch chun.

Ti sao chng ta cn lch chun? trli cu hi ny, chng ta thxem quav dsau y:

V d 1. Mt bin s phn nh tnh trng ca mt bnh trong hai nhm bnhnhn (nhm A gm 6 bnh nhn, v nhm B gm 4 bnh nhn) nhsau:

Nhm A: 6, 7, 8, 4, 5, 6

Nhm B: 10, 2, 3, 9

C thddng thy rng strung bnh ca nhm A l 6, bng vi strung bnhca nhm B. Tuy c cng s trung bnh, chng ta kh c th kt lun hai nhm nytng ng nhau, bi v khc bit trong nhm B cao hn trong nhm A. Tht vy,khc bit gia sln nht v snhnht trong nhm B l 8 (tc 10 trcho 2) gp hailn so vi nhm A vi khc bit l 4 (ly 8 trcho 4).

Chng ta cn mt ch sphn nh skhc bit gia cc bnh nhn (hay nitheo thut ngl bin thin). Cch lm hin nhin nht l ly kt quca tng bnh nhn

trcho strung bnh v cng chung li. Gi chsny lD, v phn bit hai nhm Av B, chng ta dng k hiu di dng (subscript):

Nhm A: AD = (6-6) + (7-6) + (8-6) + (4-6) + (5-6) + (6-6) = 0

Nhm B: BD = (10-6) + (2-6) + (3-6) + (9-6) = 0

Nhthy trn, vn y l tng skhc bit caDl 0. NhvyD vn chaphn nh c bin thin m chng ta mun. Mt cch lm cho D c hn hn lchng ta ly bnh phng ca tng c nhn v cng sbnh phng li vi nhau. Gi ch

smi ny l 2D , chng ta c:

Nhm A: 2AD = (6-6)2+ (7-6)2+ (8-6)2+ (4-6)2+ (5-6)2+ (6-6)2= 10

Nhm B: 2BD = (10-6)2+ (2-6)2+ (3-6)2+ (9-6)2= 50


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3

By gi th 2D r rng cho thy nhm B c bin thin cao hn nhm A.

Nhng cn mt vn , v 2D l tng s, tc l chu nh hng scmu trong tng

nhm. Mt cch iu chnh hp l nht l chia 2D cho scmu. Gi chsmi ny lS

2, chng ta c:

Nhm A: 2S = 10 / 6 = 1.67

Nhm B: 2BS = 50 / 4 = 12.5

Nhng khch quan hn na, chng ta cn phi iu chnh cho sthng ss

dng trong tnh ton. Ch rng khi tnhD hay 2D , chng ta trkt qumi bnh nhn

cho s trung bnh (tc l tn mt thng s). V th, thay v chia 2D cho s c mu,

chng ta phi chia cho scmu tr1. Gi chsmi nht l 2s , chng ta c:

Nhm A: 2 10 25 1A

s = =

Nhm B: 250

16.74 1B

s = =

Chs 2s y chnh lphng sai.

Nhng cn mt vn nhna: bi v n vphng sai l bnh phng, khcvi n v ca s trung bnh. V th, cch hon chuyn tt nht l chuyn gi tr ca

phng sai sao cho c cng n vvi strung bnh bng cch ly cn sbc hai, v ychnh l lch chun(k hius).

Nhm A: 2 1.41As = =

Nhm B: 16.7 4.08Bs = =

n y, chng ta c th thy nhm B c bin thin cao hn nhm A. Mtcch nh lng ha lch chun tng quan vi strung bnh l ly lch chunchia cho strung bnh (v nu cn, nhn cho 100). Kt quca tnh ton ny c tn l h

sbin thin(coefficient of variation CV):

Nhm A: CVA= 1.41 / 6 100 = 23.5%Nhm B: CVB= 4.08 / 6 100 = 68.3%


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4

Li th ca h s bin thin l n cho chng ta mt php so snh cc bin skhng c cng n v. Chng hn nhchng ta c thso snh bin thin ca p sutmu v cholesterol trong mt qun th, v hsbin thin c cng n vphn trm.

n y, chng ta c th tm lc sphn phi ca hai nhm bnh nhn bng

bng sau y:

Nhm Si tng(N)

Trung bnh lch chun Hsbinthin

A 6 6.0 1.41 23.5%B 4 6.0 4.08 68.3%

M tsbin thin ca strung bnh: sai schun

Cc sch gio khoa thng k thng m tcch tnh sai schun trong phn mu, nhng khng gii thch n c ngha l g v ti sao phi cn n chsthng k ny.Cng thc tnh sai schun (k hiu bng SE vit tt tstandard error) rt n gin:ly lch chun chia cho cn sbc hai ca scmu (n):

sSE

n=

p dng cng thc trn cho v d, SEca nhm A v B ln lc l:

Nhm A: 1.41/ 6 0.58ASE = =

Nhm B: 4.08 / 4 2.04ASE = =

Ti sao chng ta cn tnh SE? Xin nhc li nguyn l v mc ch ng sau cathng k hc l c tnh nhng thng sca mt qun th(population). Trong thc tchng ta khng bit cc thng sny, m chda vo nhng c tnh tmt hay nhiumu suy lun cho gi trca qun thm cc mu c chn. Chng hn nhchngta khng bit chiu cao ca ngi Vit l bao nhiu (bi v u c ai o lng chiu cao

ca 82 triu dn); chng ta phi chn mt mu gm n i tng tnh trstrung bnhca mu ny, v dng trstrung bnh ca mu suy lun cho ton dn s.

Nhng chn mu phi ngu nhin th mi mang tnh i din cao. Cmi lnchn mu, chng ta c mt nhm i tng khc. V, cmi mu, chng ta c mt strung bnh mi. Cu hi t ra l: nu chn mu nhiu ln (nhiu y c ngha lhng triu hay tln) th cc strung bnh ny dao ng cno.


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5

V d2. Hy ly mt v dcth(nhng n gin) minh ha cho tng va

trnh by. Gischng ta c mt qun thch10 ngi, v chiu cao tnh bng cm ca10 ngi ny l:

Qun th: 130, 189, 200, 156, 154, 160, 162, 170, 145, 140

Nhvy chiu cao trung bnh ca qun th(chng ta bit) l 160.6 cm. Gi ch

sny l = 160.6 cm.

By gi, gischng ta khng c iu kin v ti lc o chiu cao ca ton bqun th, m chc khnng ly mu 5 ngi tqun thny c tnh chiu cao.Chng ta c thly nhiu mu ngu nhin, mi ln 5 ngi:

Ln th1: 140, 160, 200, 140, 145 x1= 157.0Ln th2: 154, 170, 162, 160, 162 x2= 161.6Ln th3: 145, 140, 156, 140, 156 x3= 147.4Ln th4: 140, 170, 162, 170, 145 x4= 157.4Ln th5: 156, 156, 170, 189, 170 x5= 168.2Ln th6: 130, 170, 170, 170, 170 x6= 162.0Ln th7: 156, 154, 145, 154, 189 x7= 159.6Ln th8: 200, 154, 140, 170, 170 x8= 166.8Ln th9: 140, 170, 145, 162, 160 x9= 155.4Ln th10: 200, 200, 162, 170, 162 x10= 178.8

.

Ch trong dy trn, cc sx1, x2, x3, l s trung bnh cho mi mu cchn. Chng ta thy cmi ln chn mu, strung bnh chiu cao c tnh khc nhau,v bin thin t147.4 cm n 178.8 cm. Cc strung bnh ny dao ng chung quanh strung bnh ca qun th(tc l 160.6 cm).

Nu chng ta chn muNln (mi ln vi n i tng), th chng ta scN strung bnh. lch chun caN strung bnh ny chnh l sai schun. (Nen nhNy l hng triu hay tln). Do , sai schun phn nh dao ng hay bin thinca cc strung bnh mu (sample averages).

Mt s sch gio khoa thng k dng danh t Standard error of the mean(SEM), nhng y l mt cch dng tsai. Nhti va trnh by trn, khng c ci gil standard error of the mean, m ch l standard deviation of the means (ch ch


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Strung bnh ca mu: = 157.0 cmlch chun:s= 25.4 cm

Sai schun: SE = 25.4/ 5 = 11.36 cm

Nhvy, theo l thuyt xc sut, chng ta c thni rng xc sut 95% l strungbnh ca ton qun thdao ng t1571.9611.36= 139 cm n 157+1.9611.36=179 cm. (Trong thc t, chng ta bit rng strung bnh ca ton qun thl 160.6 cm).

Tm tt

Cn phi ni ngay rng khng mt bin s lm sng no c thc m tchbng mt c s. c mt bc tranh chung vmt bin slm sng, chng ta nn sdng ba c schnh: scmu, strung bnh, v lch chun. Sai schun khng

cung cp thng tin vbin thin ca mt qun th, cho nn c sny khng nn sdng cho vic m tmt chslm sng.

Nhng trong thc t, v hiu sai hay nhp nhng vlch chun v sai schunnn cc bi bo y hc c trnh by thiu thng nht. Lc th cc tc gitrnh by lch chun, li c khi cung cp sai schun. y khng phi l vn gian ln khoahc, m chn gin l thiu hiu bit. Chnh v thm ban bin tp cc tp san y hcquc tra chdn khuyn co tc gichnn trnh by lch chun km theo strungbnh v cmu.

Bi v mu sca sai schun l scmu, cho nn sai schun thng thphn lch chun. Chnh v thm c khi tc gic lngi trnh by lch chun qucao (ngi ngi bnh duyt cht vn v c thbi bo btchi) nn hctnh trnh bybng lch chun m khng ghi ch thch! Tnh trng nhp nhng ny mi l gian lnkhoa hc nhng l mt gian ln trnh thp.

Hi vng rng nhng gii thch trn y ca ti cung cp cho bn c mt cchhiu su hn v r rng hn vkhc bit gia lch chun v sai schun.

Ch thch: Bi vit ny thc cht l da vo mt bi ging vphng php dch thcm ngi vit thc hin Bmn ni tit (i hc Y dc, Thnh phHCh Minh)

vo thng 7 nm 2006, v bui tp hun vnghin cu khoa hc ti Bnh vin a khoaKin Giang vo thng 2 nm 2007. Thnh tht cm n cc bc s, hc vin v bn c

ykhoa.net t nhiu cu hi lm cm hng cho bi vit.


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8

Thut ngsdng trong bi vit

Ting Vit Ting AnhStrung bnh Meanlch chun Standard deviation (SD)Sai schun Standard error (SE)Khong tin cy 95% 95% confidence intervalStrung v MedianPhn phi chun Normal distribution (Gaussian distribution)Bin thin VariationPhng sai VarianceHsbin thin Coefficient of variation (CV)Qun th Population

Sample MuThng s ParameterEstimate c s


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Chng trnh hun luyn y khoa YKHOA.NET Training Nguyn Vn Tun 1

Lm sng thng k

c tnh khong tin cy 95%cho mt bin s hon chuyn sang n vlogart

Hi: Nhiu bin s lm sng khng tun theo lut phn phi Gaussian, do cch tnh khong tin cy 95% theo phng php thng thng khng thp dng. Nu

phi bin i bin ssang logart th cch tnh khong tin cy 95% snhthno?

Rt nhiu bin slm sng (v trong sinh hc ni chung) nhlng ng trongmu, cholesterol trong mu, v nhiu chssinh hc khc khng tun theo lut phnphi chun. Trong trng hp ny, phng php m t bin s thng l s trung v(median), v cc im t phn v 25% v 75% (tc l 25th quartile v 75th quartile).Nhng cng c trng hp phn tch, chng ta cn phi hon chuyn cc bin s nysang mt n vkhc sao cho tun theo lut phn phi chun. Mt trong nhng hm shon chuyn l logart. Khi mt bin s hon chuyn sang mt n vkhc th tt ccc strung bnh v lch chun cng thay i, cch din dch cng thay i. Bi vitny strnh by mt cch tnh rt n gin duy tr ngha sinh hc ban u ca bins.

Hy ly mt v dcth. Chng ta o lng SHBG 50 bnh nhn nam tui60 trln, v kt qunhsau:

53.6, 87.1, 35.2, 40.7, 74.5, 35.6, 82.9, 50.2, 33.8, 40.6,

110.5, 147.6, 35.8, 52.5, 72.5, 90.5, 37.8, 76.0, 48.5, 44.7,

53.2, 32.6, 39.3, 49.4, 34.6, 99.3, 46.4, 73.2, 57.7, 24.9,45.5, 46.7, 45.9, 50.8, 69.2, 57.2, 30.0, 31.5, 50.8, 46.6,

70.8, 64.4, 34.2, 51.9, 49.8, 78.3, 52.1, 33.4, 35.5, 67.4

Mt vi chsthng k cho bin sSHBG c thc tnh nhsau:

Strung bnh 55.46lch chun 23.42Trung v 50

Nu tnh theo lut phn phi chun, khong tin cy 95% ca SHBG l: 55.461.9623.42= 9.55 v 101.37 mmol/L. Nhng trc khi chp nhn shp l ca khong tin cy ny,chng ta phi xem qua phn phi ca bin SHBG (Biu 1) di y.


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Histogram of shbg

shbg

Frequency

20 40 60 80 100 120 140

0

2

4

6

8

10

12

Biu 1: Phn phi ca bin SHBG

Nhc th thy, phn ln bnh nhn c SHBG thp hn 80 mmol/L, v rt t bnhnhn vi SHBG cao hn 80 mmol/L. Ni cch khc, phn phi ca SHBG xin lch vnhng gi trthp, khng cn i, tc l khng tun theo lut phn phi chun (Normaldistribution). Do , khong tin cy 95% v strung bnh va c tnh trn khng c ngha v vi phm mt qui lut thng k hc.

Cch khc phc cho tnh trng ny l hon chuyn SHBG sang mt n vsaocho tun theo lut phn phi chun. V lch vmt pha (pha tri) chng ta c thp

dng hm slogart hon chuyn. Chng hn nhthay v 53.6, chng ta chuyn thnhlog(53.6) = 3.98. Tip tc hon chuyn nhth, chng ta sc mt dy smi nhsau:

3.982 4.467 3.561 3.706 4.311 3.572 4.418 3.916 3.520 3.704 4.705 4.995 3.578 3.961

4.284 4.505 3.632 4.331 3.882 3.800 3.974 3.484 3.671 3.900 3.544 4.598 3.837 4.293

4.055 3.215 3.818 3.844 3.826 3.928 4.237 4.047 3.401 3.450 3.928 3.842 4.260 4.165

3.532 3.949 3.908 4.361 3.953 3.509 3.570 4.211

By gichng ta thxem phn phi ca log(SHBG) (Biu 2):


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Histogram of log(shbg)

log(shbg)

Frequency

3.5 4.0 4.5 5.0

0

5

10

15

Biu 2: Phn phi ca bin log(SHBG)

Phn phi ny vn cha thong, v vn cn xin lch. Chng ta thy gi trcaonht ca log(SHBG) l khong 5, cho nn chng ta c th p dng mt hm s honchuyn mi: log(SHBG + 5). Chng hn nh nu SHBG = 53.6, th log(SHBG+5) =log(53.6 + 5) = 4.07. Gi trmi ny cho 50 bnh nhn v biu phn phi nhsau:

4.071 4.523 3.694 3.822 4.376 3.704 4.476 4.011 3.658 3.820 4.749 5.028 3.709 4.052

4.350 4.559 3.757 4.394 3.980 3.906 4.064 3.627 3.791 3.996 3.679 4.647 3.940 4.359

4.138 3.398 3.922 3.945 3.930 4.022 4.307 4.130 3.555 3.597 4.022 3.944 4.328 4.240

3.669 4.041 4.004 4.422 4.045 3.648 3.701 4.282

Histogram o f log(shbg + 5)

log(shbg + 5)

Frequency

3.5 4.0 4.5 5.0

0

2

4

6

8

10

Biu 3: Phn phi ca bin log(SHBG+5)


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By gith chng ta thnh cng hon chuyn SHBG sang phn phi chun. Do , chngta c thc tnh strung bnh v lch chun cho bin smi:

Trung bnh log(SHBG+5): 4.041

lch chun (SD) ca log(SHBG+5): 0.3427

Nh vy, khong tin cy 95% ca bin s mi l: 4.041-1.960.3427 = 3.369 n4.041+1.960.3427 = 4.712.

Vn t ra l chng ta cn phi hon chuyn ngc li n v mmol/L, v mt n vlogart rt kh hiu v kh din dch. hon chuyn ngc li, chng ta tm gilog(SHBG+5) = y, v mc tiu l chng ta tm SHBG:

Log(SHBG + 5) =y

Do , SHBG +5 = ey

Hay, cthhn:SHBG = e

y 5

Do , strung bnh v khong tin cy 95% SHBG c thc tnh nhsau:

Trung bnh SHBG: e4.041 5 = 51.86 mmol/L

V khong tin cy 95%: e3.369 5 = 24.05 n e4.712 5 = 106.3 mmol/L.

Chng ta c thso snh kt quc tnh sai v kt quc tnh ng nhsau:

c tnh khng hon chuyn c tnh da vo honchuyn logart

Strung bnh 55.46 51.86Khong tin cy 95% CI 23.42 101.37 24.05 106.3

Nhn vo Biu 1, chng ta sthy ngay rng cc c sda vo hon chuyn logart hp

l h

n l nh

ng

c skhng hon chuy

n, v chng ph

nnh

y

h

n s

phn ph

i c

aSHBG.

V dtrn y cho thy trc khi phn tch bng bt cm hnh no, chng ta cn phi xemxt cn thn phn phi ca bin s. Bi v phn ln cc phng php phn tch thng k davo ginh lut phn phi chun, vi phm ginh ny cng c ngha l kt qukhng c ngha khoa hc cao.


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Ghi ch:

Cc tnh ton trn y c ththc hin bng my tnh cm tay hay Excel. Nhng i vi bnc quen sdng ngn ngthng k R, th cc tnh ton v biu trn c thc hin bngcc m sau y. (Bn c c thct tt cm v dn vo R tmnh kim nghim).

# nhp sliu 50 bnh nhn vo bin c tn l shbg

shbg


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Mun bit thm cch sdng R cho phn tch thng k, cc bn c ththam kho cun schPhn tch sliu v to biu bng R ca ti, do Nh xut bn Khoa hc Kthut phthnh u nm 2007.

Nguyn Vn Tun


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Lm sng thng k

Khong tin cy 95% ca trung v

Nguyn Vn Tun

Hi: Em o mt bin s lm sng, nhng v bin sny khng tun theo lutphn phi chun, nn em phi dng s trung vm tbin s. Em mun bit cchtnh khong tin cy 95% ca n. Tm trong sch gio khoa khng thy sch no chcch

tnh ny. Mong thy chcch tnh khong tin cy 95% ca strung v.

y l mt vn th v! i vi cc bin khng tun theo lut phn phi chun,chng ta khng thsdng strung bnh v lch chun m tbin. Thay vo ,chng ta phi p dng cc phng php thng k phi tham s (non-parametric statistics)

tnh. Mt trong nhng chsm ttrung bnh ca bin l strung v(median).

ng nhbn c vit, cc sch gio khoa khng m tcch tnh khong tin cy95% ca strung v. n gin v khng c cng thc no tnh. Tuy nhin, trongba thp nin trli y, vi spht trin ca my tnh, mt cuc cch mng thng k xy ra. Phng php cch mng c tn l bootstrap method do nh thng k hcBradley Efron pht trin vo nm 1979. Phng php bootstrap c ng dng rngri trong nhiu lnh vc khoa hc, v n nay c th xem l mt phng php chun.Trong bi ny, ti sli dng cu hi gii thiu phng php ny. V phi sdngmy tnh, cho nn bn c cn phi bit qua mt ngn ngthng k, chng hn nhR

tin vic theo di. Chng ta sbt u bng mt v dcth.

Phng php c tnh strung v

V d1. Sliu vchsau (pain index) 11 bnh nhn thp khp nhsau:

0.05, 0.15, 0.35, 0.25, 0.20, 0.05, 0.10, 0.05, 0.30, 0.05, v 0.25

(Ch chscng cao, au cng nghim trng). Strung bnh ca 11 bnh nhn l0.163 v lch chun 0.112. V strung bnh thp hn 2 ln lch chun, chng ta cthkt lun rng bin sny khng tun theo lut phn phi chun. Cch tnh median cthtin hnh qua hai bc n gin sau y:

Bc 1: Sp xp dliu theo thttthp nht n cao nht:

0.05, 0.05, 0.05, 0.05, 0.10, 0.15, 0.20, 0.25, 0.25, 0.30, 0.35

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11)


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Ch : hng th2 (strong ngoc) l sthttthp n cao.

Bc 2: Xc nh sgia. V c 11 bnh nhn, sgia phi l shng th6. Shng th6 l 0.15 v y chnh l strung v:

0.05, 0.05, 0.05, 0.05, 0.10, 0.15, 0.20, 0.25, 0.25, 0.30, 0.35

Phng php bootstrap

Vn by gil xc nh khong tin cy 95% ca strung v. Ni cch khc,nu nghin cu c lp li 1000 ln, v mi ln chn 11 i tng, th khong tin cyca strung vra sao. Phng php bootstrap rt c ch gii quyt vn . Phngphp ny c tin hnh nhsau:

Bc 1: Bt u bng mu gcx1,x2,x3, ,xn. Trong v dtrn:

0.05, 0.05, 0.05, 0.05, 0.10, 0.15, 0.20, 0.25, 0.25, 0.30, 0.35

Bc 2: Chn ngu nhin n c nhn tmu gc vi qui trnh ly mu c hon li(replacement sample). Mi ln chn mu, tnh strung vv tm gi sny l mi.

Cn gii thch thm y vphng php ly mu c hon li c ngha l mt cnhn c thc hn mt ln trong mt ln chn mu. Chng hn nhtqun

th2, 3, 4, 5, ly mu c hon li c ngha l ln chn mu thnht c thl 2, 4,5, 2 (tc i tng thhai c chn hai ln); ln thhai c thl 4, 4, 2, 2, 5(tc i tng thhai v thtc chn hai ln); ln thba c thl 2, 5, 2, 3;v.v...

Bc 3: Lp li bc haiNln (Nthng l 1000 hay 10000 hay thm ch 1 triu ty theo nhu cu). Trong trng hp trn, 10 mu u tin c thl:

Mu 1: 0.05 0.05 0.10 0.05 0.20 0.20 0.05 0.25 0.10 0.10 0.30 0.10

Mu 2: 0.05 0.25 0.30 0.05 0.30 0.30 0.05 0.05 0.25 0.05 0.35 0.25

Mu 3: 0.35 0.10 0.05 0.25 0.05 0.05 0.20 0.25 0.15 0.25 0.10 0.15

Mu 4: 0.05 0.05 0.10 0.25 0.15 0.05 0.20 0.05 0.10 0.25 0.05 0.10

Mu 5: 0.30 0.25 0.05 0.25 0.25 0.05 0.20 0.05 0.25 0.05 0.05 0.20

Mu 6: 0.05 0.25 0.10 0.05 0.05 0.15 0.25 0.05 0.05 0.05 0.05 0.05

Mu 7: 0.05 0.15 0.25 0.05 0.05 0.30 0.20 0.25 0.30 0.05 0.35 0.20

Mu 8: 0.05 0.05 0.20 0.05 0.10 0.05 0.05 0.10 0.20 0.10 0.05 0.05

Mu 9: 0.05 0.05 0.10 0.25 0.20 0.25 0.25 0.20 0.35 0.25 0.35 0.25

Mu 10: 0.05 0.05 0.05 0.25 0.35 0.25 0.25 0.15 0.20 0.20 0.15 0.20

v.v


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(Ch : spha sauc ngha l strung vcho tng mu)

Bc 4: n y chng ta cN strung v. Sp xp N stthp n cao v nhs: 1, 2, 3, ,N. Chn strung vhng 2.5% v 97.5% caN strung v, v chnh l khong tin cy 95%. Chng hn nhnuN = 1000 ln, th khong tincy 95% chnh l strung vhng th25 v 975.

Cc bc tnh ton trn c ththc hin bng ngn ngR (hay mt ngn nghayphn mm no m bn c quen thuc) rt ddng. i vi R, cc m sdng (v giithch km theo) nhsau:

# nhp cc sliu gc vo mt vector c tn l x

x


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V phng php bootstrap i hi c my tnh, v do , ngi sdng phi amhiu mt ngn nghay phn mm thng k. Trong bi ny, ti sdng ngn ngR thc hin phng php bootstrap, v R l mt ngn ngtng i dsdng nhng rtlinh hot tnh ton cc vn kh trong thc tnghin cu lm sng. Bn c munbit thm vngn ngR c thtm c cun sch Phn tch sliu v to biu bngR ca ti, do Nh xut bn Khoa hc Kthut pht hnh u nm 2007. Trong cphn hng dn cch chn mu nhsdng trong bi vit ny.

Thut ngsdng trong bi vit

Ting Vit Ting AnhThng k phi tham s Non-parametric statisticsTrung v Median

Khong tin cy 95% 95% confidence interval


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Lm sng thng k

Lm cch no chn ngu nhin

Hi: Em mun lm mt nghin cu trong nhng bnh nhn em khm hng tun,

Thy ni phi chn ngu nhin th kt qumi c ngha khoa hc. Vy xin Thy chcch chn ngu nhin. Nu em chn mi bnh nhn th3 hay thnm c thxem l

ngu nhin khng?

y l mt cu hi lin quan n vn thit knghin cu. Lin quan n phnhai ca cu hi, trli ngn gn l: khng. Cch chn theo thtbnh nhn th3, 6,9, (hay 5, 10, 15, 20, ) th khng thxem l ngu nhin c, bi v cch chn ni ln rng y l cch chn c hthng!

Thth nh ngha chn ngu nhin l g? Chn ngu nhin c ngha l chni tng sao cho tt c cc i tng trong mt qun th c xc sut c chn nhnhau. Nu chng ta c 10 i tng, th mi i tng c xc sut c chn l 1/10.Nu chng ta c hai nhm A v B, v chn ngu nhin c ngha l i tng c chnvo nhm A c xc sut bng vi i tng c chn vo nhm B (tc l 50%).

ngha ca vic chn ngu nhin rt quan trng trong nghin cu y hc v trit lca nghin cu khoa hc. Tt ccc m hnh phn tch thng k u ginh rng muc chn phi l mu ngu nhin. Chkhi no mu ngu nhin th kt quphn tchmi c gi trkhoa hc cao. Ngoi ra, trong cc nghin cu bnh chng (case-control

study) khi so snh hai nhm, chng ta cn phi m bo hai nhm tng ng nhau vcc yu t lm sng c thc nh hng n kt qunghin cu. Chng hn nhnuchng ta mun tm hiu nh hng ca thi quen ht thuc l n nguy cung thphi,chng ta c thso snh tlung thgia nhm ht thuc l v nhm khng ht thuc l.Nhng nhthvn cha , v cc yu tkhc nhtui, hormone, mi trng sng,v.v (gi chung bng thut ngcovariates) cng c thgy ung th. Do , vn lphi chn hai nhm tng ng nhau vnhng covariates ny. Chkhi no hai nhmc cng (hay tng ng) vcc yu tcovariates th kt lun vmi lin hgia htthuc l v ung thmi ng tin cy.

Nhng cch phn chia i tng sao cho hai nhm tng ng nhau rt kh lmbng phng php th cng, bi v chng ta hon ton c th chn hai nhm tngng nhau v tui, nhng c th li khc nhau v hormone. Hay chng ta c thphn chia i tng sao cho hai nhm tng ng nhau vtui v hormone, nhngc th hai nhm khng tng ng v mi trng sng. S lng covariates cngnhiu, cch phn chia cng phc tp. Ch c cch duy nht l ngu nhin ha(randomization) th mi m bo tng ng gia hai nhm.


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Mi chng ta (trong thgii 4 tngi) u l nhng c thduy nht, hiu theongha khng c ai ging ai, v s c nht v nh c nh ngha bng nhngnhng c im v nhng c tnh lin quan n mi c nhn. C thhai ngi c cngchiu cao, cng cn nng, cng tui, nhng hai ngi c thkhc nhau vcc c

im lm sng khc, v nht l khc nhau vmi trng sng. V th, nu chng ta chni tng da vo mt hay hai c tnh th vn cha , m phi chn sao cho hon tonngu nhin. y l trit l ng sau ca cc nghin cu lm sng i chng ngu nhin(randomized clinical trial). Qua nhiu nm kinh nghim, y hc hon thin v chngminh rng cch ngu nhin ha thc stng ng ha cc nhm.

My tnh c thgip chng ta chn hay phn chia ngu nhin. iu cn thit lchng ta phi c mt phn mm thng k. y, ti ssdng phn mm R ngunhin ha. Bn c mun bit thm vRc ththam kho cun sch Phn tch sliu

v to biu bng R ca ti do Nh xut bn Khoa hc K thut va mi pht hnhnm 2007.

Phng php chn ngu nhin

Quay li cu hi trn, gisbn c bit rng mi thng sbnh nhn n khml 500 ngi, v cng trnh nghin cu cn 100 ngi. Cch chn ngu nhin 100 bnhnhn c thtin hnh tng bc nhsau:

Bc 1: ln danh sch t1 n 500 (tc qun thnghin cu). i vi Rvic

ny cc k n gin vi lnh:

population


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selected

v Rscho chng ta bit:

[1] 42 172 31 22 234 432 75 190 386 183 64 291 139 323 356 68 462 485

[19] 61 253 456 484 337 363 488 136 498 113 117 197 378 406 256 476 466 351

[37] 95 1 218 300 219 69 28 43 250 239 326 303 84 210 3 162 493 36

[55] 425 368 182 233 57 311 51 282 93 100 130 70 18 74 446 376 321 103

[73] 125 344 500 391 34 161 78 349 252 265 147 289 9 342 231 395 73 13

[91] 180 400 6 414 367 137 81 155 360 187

(Bn c c thkhng cn lu n nhng snh[1], [19], [37], v.v v y lnhng scho chng ta bit vtr khi u ca tng dng sliu).

Theo kt qutrn, chng ta nn chn cc bnh nhn s42, 172, 31, v.v Nhng

danh sch ny kh sdng, v chng ta bit rng bnh nhn n khm theo thtvi m 1, 2, 3, , 500. V th, cn phi sp xp bin selectedtheo tht, v

hm sortgip chng ta lm vic ny rt hu hiu:

sort(selected)

v Rscho chng ta bit:

[1] 1 3 6 9 13 18 22 28 31 34 36 42 43 51 57 61 64 68

[19] 69 70 73 74 75 78 81 84 93 95 100 103 113 117 125 130 136 137

[37] 139 147 155 161 162 172 180 182 183 187 190 197 210 218 219 231 233 234

[55] 239 250 252 253 256 265 282 289 291 300 303 311 321 323 326 337 342 344

[73] 349 351 356 360 363 367 368 376 378 386 391 395 400 406 414 425 432 446

[91] 456 462 466 476 484 485 488 493 498 500

By gith chng ta c mt danh sch ngu nhin. Theo danh sch ny, bnhnhn u tin (s 1), tip theo l bnh nhn s3, 6, v 500 nn c chn.

Cn ch rng v y l cch chn hon ton ngu nhin, cho nn c mi lnchng ta ra 3 lnh trn th R cung cp mt danh sch hon ton mi. Bn c c thkim tra pht biu ny bng cch ra 3 lnh trn nhsau:

population


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Phng php phn nhm ngu nhin

Trong cc nghin cu lm sng i chng ngu nhin, chng ta thng c hai

nhm i tng. Vi slng cmu nh trc l n, mc tiu l chia n/2 i tng vonhm 1 v n/2 vo nhm 2. C vi phng php chia ngu nhin. Cch n gin nhtl ly s chn hay lquyt nh phn nhm. Chng hn nu i tng c chn[ngu nhin] l schn th scho vo nhm 1 v slvo nhm 2 (hay ngc li). ViRchng ta c thtin hnh phn nhm cc k n gin.

V d1: Phn nhm tng th. Gischng ta c 100 bnh nhn v mun phn50 vo nhm can thip (A) v 50 vo nhm i chng (P). Chng ta tin hnh theo trnhtsau y:

Bc 1: Cho bit chng ta c 100 i tng v to 100 m sv cho vo bin id.

n


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Bc 3: Xc nh int l s chn hay l bng hm %% v cho vo bin odd.

Dng hm replacechia nhm: nu oddl sl, cho vo nhm A; nu odd

l schn, cho vo nhm P, v gi nhm bng tn mi l group:

odd


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table(group)

V kt qul c 45 bnh nhn c phn chia vo nhm A, 55 vo nhm P.

group

A P

45 55

V phn chia hon ton ngu nhin, cho nn slng bnh nhn khng hon toncn i, nht l nhng nghin cu c s lng i tng khng nhiu. Nhng i vinhng nghin cu vi hng ngn i tng th phn chia theo ngu nhin ha c thcni rt hu hiu.

Tt nhin, chng ta c thchy (lp li cc lnh trn) qui trnh trn cho n khino slng i tng ca hai nhm cn bng th ngng (bn c chn gin ct (cut)

ton blnh v dn (paste) vo Rcc lnh di y):

n


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nn mt cuc cch mng trong vic thm nh cc thut iu tr trong vng na thkqua. ng trn phng din kthut, vic chn ngu nhin v phn nhm ngu nhinrt n gin nu bn c c sn phn mm R(c thti vmy tnh c nhn hon tonmin ph). Bn c nn tmnh kim tra cc lnh trn y bng cch thay i cc thngshiu thm vcchca chn v phn nhm ngu nhin.

Ghi ch kthut:

Cc lnh Rtrong v d1 c thn gin ha thnh mt hm (function). Gi hm bngtn grp, chng ta c thvit nhsau:

grp


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Lm sng thng k

Kim nh t v hon chuyn sliu

Hi: Ti nghe ni rng khi nh gi skhc bit gia hai nhm bng t-test cnphi chuyn i sliu. Ti sao?

nh gi khc bit gia hai nhm, chng ta thng sdng phng phpkim nh t (hay t-test). Kim nh t c l l mt trong nhng phng php n ginnht trong thng k hc, v c thtnh ton mt cch thcng, m khng cn n mytnh hay phn mm phn tch sliu (nhng nu c th tt hn!)

Tuy n gin, nhng phng php kim nh t cng rt dsai lm. Sai lm thngthng nht l khng n nhng ginh ng sau phng php ny. Phng php

kim nh t chthch hp nu sliu p ng nhng iu kin hay ginh sau y:

Hai nhm so snh phi hon ton c lp nhau;

Bin so snh phi tun theo lut phn phi chun (Gaussian distribution);

Phng sai ca hai nhm bng nhau, hay gn bng nhau; v

Cc i tng phi c chn mt cch ngu nhin (random sample).

Thno l c lp? Khi ni n c lp y l ni n hai nhm khng ctng quan nhau. Chng hn nhmt nhm 1 gm bnh nhn A, B, C v D; nhm 2

gm bnh nhn E, F, G v H, th hai nhm ny c lp nhau. Nhng nu c mt nhmbnh nhn m o hai ln, th hai bin sca hai ln o khng c lp vi nhau. clp cng c ngha l khng lin hnhau. Chng hn nhnu 2 bnh nhn trong nhm 1(A v C) c lin hhuyt thng, v nu bin m chng ta phn tch c yu tdi truyn tho lng ca hai bnh nhn khng c xem l c lp.

1. L thuyt ca kim nh t

Cho hai qun thc lp 1 v 2, vi chstrung bnh 1 v 2 , v phng sai2

. Chng ta mun nh gi khc bit gia hai qun th. Nhng chng ta khng bitcc gi trny.

tm hiu xem 1 v 2 c khc nhau hay khng, chng ta ly mu t hai

qun th. Gischng ta ly ngu nhin 1n i tng tqun th1, v 2n i tng

tqun th2. Sau khi o lng bin s, chng ta c kt qunhsau:


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Nhm 1 Nhm 2Si tng

1n 2n

Trung bnh1x 2x

Phng sai 21s

22s

lch chun 1s 2s

Xin nhc li, chng ta mun tm hiu khc bit gia hai qun th(chkhngphi gia hai nhm mu). Mc ch ny c thpht biu bng hai githuyt nhsau:

Githuyt v hiu Ho: 1 2 =

Githuyt chnh H1: 1 2

Gi = 1 2 , hai githuyt trn cng c thpht biu nhsau:

Ho: = 0

H1: 0

Trong iu kin khng bit cc gi tr ca qun th 1 v 2 , c s thch hp nht

qun thchnh l hai s trung bnh 1x v 2x tnh tmu 1 v mu 2. V, c tnh

khc bit chnh l khc bit gia hai strung bnh:

d = 1x 2x [1]

Nhng v ly mu, cho nn d c thbin thin tmu ny sang mu khc, v vn ltm phng sai ca d. L thuyt xc sut cho chng ta bit rng phng sai ca khc bitgia hai bin bng tng phng sai ca hai bin trcho 2 ln hip bin, tc l:

var(a b) = var(a) + var(b) 2cov(a,b)

Trong , var l vit tt ca variance (phng sai), v covar l vit tt ca covariance(hip bin). Hip bin phn nh tng quan gia hai bin. Nhng nu hai bin honton c lp, th hip bin sl 0, v cng thc trn n gin thnh:

var(a b) = var(a) + var(b)


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p dng cng thc ny, chng ta c thc tnh phng sai cho d trong [1] nh sau(Ti sk hiu phng sai bng sbnh phng):

22

21

2 sssd += [2]

T, lch chun ca d l:22

21 sssd += [3]

Nhng v nhng c su da vo scmu, cho nn chng ta phi iu chnh bngcch chia phng sai cho scmu:

2

22

1

21

n

s

n

sSEd += [4]

Nu phng sai ca hai nhm bng nhau (tc 2 2 21 2s s s= = ), phng trnh [4] n gin

thnh:

1 2

1 1dSE s

n n= + [5]

Kim nh t n gin l tsca d trn SEd, hay cthhn:

2

2

2

1

2

1

ns

ns

dt

+

= [6]

C thxem cng thc [5] nh l t sca tn hiu (signal) v nhiu (SEd).Tht vy, d phn nh khc bit gia hai nhm, v SEd phn nh nhiu ca d.Thnh ra, nu tstcao, chng ta c bng chng ni tn hiu nhiu hn nhiu (tc c ngha thng k); nu tstthp di 1 chng hn, chng ta c bng chng pht biutn hiu thp hn nhiu v do khc bit khng c ngha thng k.

Nhng cao l cao bao nhiu c thni l c ngha thng k? trli cuhi ny, chng ta quay trvvi githuyt. Nu gi thuyt v hiu Ho l s tht (tckhng c khc bit gia 2 qun th), th sphn phi ngu nhin ca t nhth no. Maymn thay, c nh thng k hc trli cu hi ny: l ng William Gossett, ngipht kin kim nh t. Theo chng minh ca Gossett, nu hai qun thkhng khc nhau,th gi trca t ty thuc vo scmu (hay ni theo ngn ngthng k hc l bc tdo degrees of freedom). Sbc tdo (k hiu) c tnh bng cng thc sau y:


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df = 1n + 2n 2

Bng 1sau y trnh by tst cho tng bc tdo v khong xc sut m tst c th

dao ng ngu nhin:

Bng 1. Tstcho tng bc tdo nu githuyt v hiu Ho ng

Bc tdo (df) Xc sut 95% tst sdao ng trong khong

Xc sut 99% tst sdao ng trong khong

5 -2.57 n 2.57 -4.03 n 4.0310 -2.23 n 2.23 -3.17 n 3.1714 -2.14 n 2.14 -2.98 n 2.9816 -2.12 n 2.12 -2.92 n 2.92

18 -2.10 n 2.10 -2.88 n 2.8820 -2.08 n 2.08 -2.84 n 2.8424 -2.06 n 2.06 -2.80 n 2.8030 -2.04 n 2.04 -2.75 n 2.7534 -2.03 n 2.03 -2.73 n 2.7340 -2.02 n 2.02 -2.70 n 2.7050 -2.01 n 2.01 -2.68 n 2.6860 -2.00 n 2.00 -2.66 n 2.6670 -2.00 n 2.00 -2.65 n 2.65

80 -2.00 n 2.00 -2.64 n 2.6490 -1.99 n 1.99 -2.64 n 2.64100 -1.98 n 1.98 -2.62 n 2.62500 -1.96 n 1.96 -2.58 n 2.581000 -1.96 n 1.96 -2.58 n 2.58

Do , nu tsttnh ton tcng thc [6] nm ngoi khong tin cy trn y, chng tac thni rng khc bit gia hai qun thc ngha thng k (thut ngting Anh lstatistically significant).

2. Kim nh t vi bin c hon chuyn logart

V d1. Mt nghin cu nhm so snh nng lysozyme gia hai nhm bnhnhn (tm gi l nhm 1 v nhm 2). Nhm 1 gm 29 bnh nhn, v nhm 2 gm 30bnh nhn, tui t20 n 60. Nng lysozyme (mg/L) nh sau v c th tm lctrong Bng 2:


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Nhm 1: 0.2, 0.3, 0.4, 1.1, 2.0, 2.1, 3.3, 3.8, 4.5, 4.8, 4.9, 5.0, 5.3,7.5, 9.8, 10.4, 10.9, 11.3, 12.4, 16.2, 17.6, 18.9, 20.7, 24.0, 25.4,

40.0, 42.2, 50.0, 60.0

Nhm 2: 0.2, 0.3, 0.4, 0.7, 1.2, 1.5, 1.5, 1.9, 2.0, 2.4, 2.5, 2.8, 3.6,4.8, 4.8, 5.4, 5.7, 5.8, 7.5, 8.7, 8.8, 9.1, 10.3, 15.6, 16.1, 16.5,

16.7, 20.0, 20.7, 33.0

Bng 2. Nng lysozyme bnh nhn nhm 1 v nhm 2

Nhm 1 Nhm 2Si tng

1n = 29 2n = 30

Trung bnh1x = 14.31 2x = 7.68

Phng sai21s = 247.8

22s = 61.6

lch chun1s = 15.7 2s = 7.8

p dng cng thc [6], chng ta c tst nhsau:

2

22

1

21

n

s

n

s

dt

+

= =14.31 7.68

14.31 7.6829 30

+

= 2.03

Vi bc tdo df = 29+30-2 = 57, v nu hai nhm khng khc nhau, chng ta k vngrng tst dao ng t-2.00 n 2.00 (theo Bng 1). Nhng tst quan st c nmngoi khong tin cy ny, nn chng ta c thpht biu rng lysozyme ca hai nhmkhc nhau.

Nhng kt quv kt lun trn c thsai! Nhn qua tm tt trnh by trong Bng2, chng ta ch phng sai ca nhm 1 cao gp 4 ln so vi nhm 1. Ngoi ra, phngsai c xu hng bin thin theo strung bnh: nhm c strung bnh cao cng l nhmc phng sai cao. lch chun ca nhm 1 cao hn nhm 2 gp hai ln.

Chng ta cng ch rng lch chun ca hai nhm cao hn s trung bnh.iu ny hm cho bit sliu lysozyme khng tun theo lut phn phi chun, v phntch trn vi phm ginh thng k. Chng ta thxem qua phn phi ca lysozymetrong nhm 1 v nhm 2 nhsau:


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Histogram of group1

group1

Frequency

0 10 20 30 40 50 60

0

5

10

15

Histogram of grou p2

group2

Frequency

0 5 10 15 20 25 30 35

0

5

10

15

Biu 1. Phn phi lysozyme ca nhm 1 (biu bn phi) v nhm 2 (biu bnphi)

R rng lysozyme c xu hng lch vcc gi trnh. Vi xu hng ny, chng ta cth s dng hm logart hon chuyn s liu. Sau khi hon chuyn bng logart,chng ta c sliu mi cho nhm 1 v 2 nhsau (v bng tm lc 3)

Nhm 1:-1.60943791 -1.20397280 -0.91629073 0.09531018 0.69314718 0.74193734

1.19392247 1.33500107 1.50407740 1.56861592 1.58923521 1.60943791

1.66770682 2.01490302 2.28238239 2.34180581 2.38876279 2.42480273

2.51769647 2.78501124 2.86789890 2.93916192 3.03013370 3.17805383

3.23474917 3.68887945 3.74242022 3.91202301 4.09434456

Nhm 2:-1.6094379 -1.2039728 -0.9162907 -0.3566749 0.1823216 0.4054651

0.4054651 0.6418539 0.6931472 0.8754687 0.9162907 1.0296194

1.2809338 1.5686159 1.5686159 1.6863990 1.7404662 1.7578579

2.0149030 2.1633230 2.1747517 2.2082744 2.3321439 2.7472709

2.7788193 2.8033604 2.8154087 2.9957323 3.0301337 3.4965076

Bng 3. Nng lysozyme bnh nhn nhm 1 v nhm 2

Nhm 1 Nhm 2Si tng

1n = 29 2n = 30

Trung bnh1x = 1.92 2x = 1.41

Phng sai 21s = 2.19

22s = 1.73

lch chun1s = 1.48 2s = 1.32


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By gith hai phng sai tng ng nhau, v chng ta c thp dng kim nh t quacng thc [6] nhsau:

2

22

1

21

n

s

n

s

d

t+

= =1.92 1.41

2.19 1.7329 30

+= 1.406

Nhvy, tstnm trong khong -2.00 n 2.00, tc l khong dao ng hon ton dongu nhin. Do , chng ta kt lun rng lysozyme ca hai nhm tng ng nhau.

3. Kim nh t vi bin c hon chuyn cn sbc 2

Nhiu nghin cu lm sng, tiu ch nh gi kt qu(outcome measure) chn gin l sm, v trc khi tin hnh kim nh t, sliu cn phi hon chuyn bngcn sbc 2 lm cho sliu tun theo lut phn phi chun.

V d2. Trong nghin cu trnh by di y, cc nh khoa hc m slng vikhun lactobacilli trong nc bt ca hai nhm bnh nhn. Nhm 1 gm c 7 bnh nhnc tim vc-xin, v nhm 2 gm 6 i tng khng c tim vc-xin. Kt qunghin cu nhsau:

Nhm 1 Nhm 2

Svi khunlactobacilli (k)

Hon chuynk

Svi khunlactobacilli (k)

Hon chuynk

7925 89.02 3158 56.2015643 125.07 3669 60.5717462 132.14 5930 77.0110805 103.95 5697 75.489300 96.44 8331 91.277538 86.82 11822 108.736297 79.35

Sliu ny c thtm lc trong Bng 4sau y:

Bng 4. Tm lc sliu lactobacilli

Nhm 1 Nhm 2Si tng

1n = 7 2n = 6


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Trung bnh (x)1x = 10710 2x = 6434

lch chun (sd)1s = 4266 2s = 3219

Ts sd/ x 41.2 40.1

Chng ta ch rng tslch chun trn cn sbc 2 ca strung bnh ca hai nhml khong 40 n 41 (tc tng ng nhau). iu ny cho thy, chng ta cn phi honchuyn sliu bng hm cn sbc 2, v kt quc trnh by trong ct 2 (mu ) catng nhm trong bng sliu gc trn. Sau khi hon chuyn chng ta c mt bng tmlc mi nhsau:

Bng 5. Tm lc sliu hon chuyn lactobacilli bng cn sbc 2

Nhm 1 Nhm 2

Si tng 1n = 7 2n = 6Trung bnh (x)

1x = 101.8 2x = 78.2

lch chun (sd)1s = 20.0 2s = 19.5

Nu phn tch da vo sliu hon chuyn, chng ta c tstnhsau:

2

22

1

21

n

s

n

s

dt

+

= =( ) ( )

2 2

101.8 78.2

20 19.5

7 6

+

= 2.05

Vi bc tdo = 7+6-2 = 11, v nu hai nhm khng khc nhau, chng ta k vng tstsdao ng trong khong -2.23 n 2.23 (Bng 1) vi xc sut 95%. y, chng ta c tst quan st l 2.05, nm trong khong xc sut ngu nhin ny, chng ta phi kt lunrng cha c bng chng kt lun rng hai nhm bnh nhn khc nhau vslng vikhun lactobacilli. (Bn c c thtlm phn tch trn sliu cha c hon chuynv sthy kt qukhc vi kt lun va trnh by!)

4. Kim nh t vi bin l tl

V d3. Bng sliu sau y l kt quca mt nghin cu lm sng i chngngu nhin, vi mc tiu so snh hai phng php tp luyn bnh nhn vi chng mt trv tui gi. Nhm mt gm 11 bnh nhn c tp luyn, v nhm hai gm 8 bnh nhni chng (khng tp luyn). Sau hai tun tp luyn, mi bnh nhn c cho 20 cu hi


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vnhng vic trong i sng hng ngy (nhkha ca, buc giy, qut dn, mc quno, v.v). Scu trli ng c ghi nhn v chia cho 20 (tc tnh tltrli ng).

Tlthnh cng trong 20 cu hi cho 2 nhm bnh nhn mt tr

Nhm 1: 0.05, 0.15, 0.35, 0.25, 0.20, 0.05, 0.10, 0.05, 0.30, 0.05, 0.25

Nhm 2: 0.0, 0.15, 0.0, 0.05, 0.0, 0.0, 0.05, 0.10

Bng 6. Tm lc sliu ca bnh nhn mt tr

Nhm 1 Nhm 2Si tng 11 8Trung bnh (x) 0.164 0.044

lch chun (sd) 0.112 0.056

Trong trng hp ny, chng ta thy lch chun bng hay cao hn s trungbnh, v l tn hiu cho thy bin skhng tun theo lut phn phi chun.

Mt trong nhng hm hon chuyn kh hu hiu cho cc sliu mang tnh t l

(proportion) l hm lng gic arsin ca cn sbc 2 (tc arcsin x , trong xl tl).

Chng hn nh nu x = 0.05, th arcsin arcsin 0.05x = = 0.2255. Sau khi hon

chuyn bng hm arcsin x , chng ta c sliu mi nhsau.

Sliu hon chuyn bng hm arcsin x

Nhm 1:0.2255134 0.3976994 0.6330518 0.5235988 0.4636476 0.2255134 0.3217506

0.2255134 0.5796397 0.2255134 0.5235988

Nhm 2:0.0000000 0.3976994 0.0000000 0.2255134 0.0000000 0.0000000 0.2255134

0.3217506

Bng 7. Tm lc sliu ca bnh nhn mt tr sau khi hon chuyn

Nhm 1 Nhm 2Si tng 11 8


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Trung bnh (x) 0.395 0.146lch chun (sd) 0.158 0.166

p dng cng th[6] cho sliu hon chuyn, chng ta c:

2

22

1

21

n

s

n

s

dt

+

= =( ) ( )

2 2

0.395 0.146

0.158 0.146

11 8

+

= 3.30

Vi bc tdo 17 (df = 11 + 8 2), v nu khng c khc bit gia hai nhm bnhnhn, chng ta k vng tst dao ng trong khong -2.10 n 2.10 vi xc sut 95%.Tuy nhin, y tst = 3.30, nm ngoi khong dao ng ngu nhin trn, chng ta cbng chng pht biu rng khc bit hay nh hng ca tp luyn c ngha thngk. Tht ra, trsP ca tst trn l 0.005.

5. Tm lc

Nhva m ttrong 3 v dtrn, chng ta thy rng vic phn tch sliu bngphng php kim nh t cc k n gin, khng cn n my tnh. Logic ng sau caphng php kim nh t (cng nhca nhiu phng php khc) l kim nh mt githuyt v hiu (Ho) nhsau:

Githuyt Ho : Khng c khc nhau gia hai nhm;

Tnh ton tst(khc bit gia 2 nhm chia cho dao ng) Nu Ho ng, xc nh bin thin ca t0 trong vng 95% hay 99%

Nu t nm ngoi khong bin thin ca t0, chng ta loi githuyt Ho.

D phng tnh v logic n gin nh th, nhng phng php kim nh tthng b p dng sai, do khng ch n cc gi nh ng sau ca phng php.Trong nhiu trng hp, sai phng php dn n kt lun sai. Do , nh hng cavic bt cn trong phn tch c khi rt nghim trng. Hi vng qua cc v dny, bn c bit qua vi phng php hon chuyn sliu, v c mt ci nhn mi hn vphngphp kim nh t.

Nguyn Vn Tun

Ch thch:

Tt ccc phn tch trn c thtin hnh rt n gin bng ngn ngthng k R. Diy l cc m R m ti dng cho cc phn tch v biu trn. Bn c c th t


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mnh kim tra bng cch ct tng phn v dn vo R hiu thm. (Cch hc hay nht lbt chc). Nu mun tm hiu thm vR, bn c c thtm mua quyn sch Phntch s liu v to biu bng R ca ti do Nh xut bn Khoa hc K thut phthnh nm 2007.

# M R tm tst cho Bng 1

# bc tdo degrees of freedom

df


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hist(group1)

hist(group2)

# Hon chuyn sliu bng hm logart

log.group1


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Lm sng thng k

Cn thn vi confounder

Mt trong nhng e da n vic din gii cc nghin cu dch thc v lm sng

l vn confounder (c l con-phao-). Thut ng confounder rt kh dch sangting Vit, nhng nh ngha ca n c thtm tt nhsau: confounder l mt bin trunggian c nh hng n bin c lp v bin phthuc. Chng hn nhchng ta bit rnghormone testosterone c nh hng n nguy cgy xng n ng; ngoi ra, nguy cgy xng cng tng dn theo tui; nhng testosterone cng suy gim theo tui.Nu mt nghin cu bo co rng mi lin hgia testosterone v gy xng m khng cp n tui, chng ta c th nghi ng rng mi lin h quan st c gia

testosterone v gy xng c phi bnh hng bi tui. Do , khi phn tch ccnghin cu y hc, chng ta phi ht sc cn thn vi nh hng ca confounder.

minh ha cho nh hng ca confounder, chng ta hy xem v d sau y:nng hemoglobin (g/dL) v chiu cao (cm) c o lng 31 sinh vin y khoa tuit19 n 30, v kt qunhsau:

Sinh vin Chiu cao (cm) Hb (g/dL)

1 168 11.0

2 165 14.5

3 166 14.0

4 167 13.5

5 167 13.8

6 167 14.1

7 170 13.8

8 171 14.7

9 172 15.0

10 172 15.1

11 173 13.8

12 177 13.1

13 178 13.8

14 166 16.0

15 172 16.5

16 175 18.9

17 173 16.1

18 174 16.3

19 176 16.2

20 176 17.0

21 178 16.5

22 179 16.4

23 178 16.2

24 180 16.3


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25 180 17.0

26 185 17.1

27 186 17.3

28 186 15.8

29 189 16.3

30 193 15.9

Gischng ta mun tm hiu mi tng quan gia chiu cao v Hb. Biu sau y scho thy mu tng quan :

165 170 175 180 185 190

12

14

16

18

Height and hemo globin

height

hb

Biu 1: Chiu cao v hem

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Lam Sang Thong Ke