Bai Giang Thong Ke

TRNG I HC CNG NGHIP TP HCMKHOA KHOA HC C BN

HUNH HU DINH

BI GING TON THNG K

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TPHCM - Ngy 8 thng 3 nm 2015

Hunh Hu Dinh Trng i Hc Cng Nghip TPHCM

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Chng 1

C LNG THAM S

1.1 c lng im

Xt mt tp hp chnh v gi s ta quan tm ti bin lng X olng mt du hiu g ca c th trong tp hp chnh. V mt Tonhc, X c coi l mt BNN (gi tr ca n thay i t c th ny sangc th khc). Phn b xc sut ca X rt kh nm bt, v thng thngta gii hn vic xc nh mt s tham s c trng ca X nh gi trtrung bnh (k vng), phng sai, trung v (median), mode,. . . Cc thams ny khng th xc nh chnh xc c, m phi c lng t gi trcaX trn mt mu chn ngu nhin. Nh vy bi ton c lng thams c pht biu nh sau:

Gi s X l mt BNN c tham s c trng no (cha bit) mta ang quan tm. Vn t ra l: Cn c trn n gi tr x1, x2, . . . , xnca X o c trn mt mu kch thc n ly ra t tp hp chnh, cntm gi tr gn ng ca .

nh ngha 1.1. Mt hm = Sn(x1, x2, . . . , xn) ca n gi trx1, x2, . . . , xn c gi l mt c lng im cho .

cho gn ta s gi tt c lng im l c lng. kho stv mt Ton hc, ta s coi x1, x2, . . . , xn l gi tr quan st c (hay gitr thc nghim) ca mu tng qut X1, X2, . . . , Xn, trong cc BNNX1, X2, . . . , Xn c lp vi nhau v c cng phn b vi X.

Nh vy c lng = Sn l mt hm ca n BNN X1, X2, . . . , Xn vdo n cng l mt BNN. Gi tr ca cng thay i t mu quanst ny sang mu quan st khc.

Vic la chn mt c lng no l tt c cn c trn cc tiuchun di y.

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nh ngha 1.2. c lng Sn c gi l c lng khng chchcho nu ESn = .

c lng Sn c gi l c lng vng nu vi mi > 0 ta climn

P (|Sn | ) = 1.c lng Sn c gi l hiu qu nu Sn l c lng khng

chch v phng sai DSn l nh nht trong lp tt c cc c lngkhng chch.

Tnh cht khng chch c ngha l c lng Sn khng c sai s hthng.

Tnh cht vng m bo cho c lng Sn gn ty vi xc sutcao khi kch thc mu ln.

1.1.1 c lng gi tr trung bnh Gi s X l BNN vi EX = (cha bit). c gi l gi tr trungbnh ca tp hp chnh.

Nu ta c mt mu n gm gi tr x1, x2, . . . , xn ca X th trung bnhmu

x =x1 + x2 + + xn

n

s c dng lm c lng cho .

nh l 1.1. Trung bnh mu l c lng khng chch v vng chotrung bnh ca tp hp chnh.

Chng minh. Ta c x l gi tr quan st ca

X =X1 +X2 + +Xn

n

trong cc BNN X1, X2, . . . , Xn c lp vi nhau v c cng phn bvi X, suy ra

EX =EX1 + EX2 + + EXn

n=

n

n= .

Vy X l c lng khng chch ca . Hn na,

DX =DX1 +DX2 + +DXn

n2=

DX

n.

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p dng bt ng thc Chebyshev ta c

P(X > ) = P (X EX > ) DX

2=

DX

n2

suy ra

P(X ) = 1 P (X > ) 1 DX

n2.

Cho n + ta c ngay limn+

P(X ) = 1. Vy X l c

lng vng ca .

1.1.2 c lng phng sai 2

Gi s X l BNN vi DX = 2 (cha bit). 2 c gi l phng saica tp hp chnh. Nu ta c mt mu gm n gi tr x1, x2, . . . , xn ca Xth mt cch hp l phng sai mu cha hiu chnh

s2 =(x1 x)2 + (x2 x)2 + + (xn x)2

n

c xem xt dng c lng 2. Tuy nhin phng sai mu chahiu chnh s2 l mt c lng chch. Tht vy, s2 l gi tr quan stca BNN

S2 =

(X1 X

)2+(X2 X

)2+ +

(Xn X

)2n

trong cc BNN X1, X2, . . . , Xn c lp vi nhau v c cng phn bvi X.

t Yi = Xi , ta suy ra

EYi = EXi = 0.

EY = 0.

Xi X = Yi Y .

EY 2i = E(Xi )2 = E(Xi EXi)2 = DXi = 2.

DYi = DXi = 2.

EY 2 = E(Y EY )2 = DY = 2

n.

Do ,

1

n

ni=1

(Xi X

)2=

1

n

ni=1

(Yi Y

)2=

1

n

(n

i=1

Y 2i 2Yn

i=1

Yi + nY2)

=1

n

(n

i=1

Y 2i nY2)

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Vy

E(S2

)=

1

n

(n

i=1

EY 2i nEY2

)=

1

n

(n2 2

)=

n 1n

2 = 2.

Kt qu trn chng t S2 l mt c lng chch.Do , nu ta xt phng sai mu hiu chnh

s2 =(x1 x)2 + (x2 x)2 + + (xn x)2

n 1

th s2 l mt c lng khng chch ca 2.

1.1.3 c lng t lGi s ta quan tm n mt c tnh A no m mi c th ca tphp chnh c th mang hoc khng mang. Gi p l t l c th mangc tnh A trong . Chng mun c lng p da trn vic kho stmt mu gm n c th. Chng hn ta mun bit t l ph phm trongmt mt hng c nhp khu, t l sinh vin n t Min Ty trongtrng i hc Cng Nghip TP.HCM, v.v. . .

Xt bin lng X xc nh nh sau:

X =

{0 nu c th khng c c tnh A1 nu c th c c tnh A

T nh ngha ca X ta c

P (X = 0) = 1 p.

P (X = 1) = p.

Nu x1, x2, . . . , xn l mt mu gm n gi tr quan st ca X th x1 +x2 + . . .+ xn l s c th mang c tnh A ca mu v

f =x1 + x2 + + xn

n

chnh l tn sut xut hin c tnh A trong mu.Ta thy f l gi tr quan st ca BNN

F =X1 +X2 + +Xn

n

trong X1, X2, . . . , Xn l cc BNN c lp vi nhau v c cng phn bvi X. V EX = p nn ta d dng chng minh c f l mt c lngkhng chch v vng cho p.

Ch 1.1. T y tr v sau, tin cho vic trnh by phn l thuyt,cc mu c xem xt m mu tng qut.

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1.2 c lng khong

Bi ton tm c lng khong t ra nh sau: Cn c trn mu quanst X1, X2, . . . , Xn, hy xc nh mt khong (a; b) khong chatham s vi xc sut 1 cho trc (1 thng c chn l 0, 95hay 0, 99). Mt cch chnh xc hn, khong c lng c nh nghanh sau:

nh ngha 1.3. Khong c hai u mt a = a(X1, X2, . . . , Xn) vb = b(X1, X2, . . . , Xn) c gi l khong c lng vi tin cy 1nu P (a b) = 1 .

Ch 1.2. Hai u mt a, b ca khong c lng l hai BNN. Chngl hm caX1, X2, . . . , Xn nn thay i t mu ny c th ny sang muc th khc.

Khong c lng ch cho ta bit vi mt xc sut cao khong nycha ch ta khng chc chn c nm trong khong c lng haykhng (tr khi chng ta xc nh ton b tp chnh, m iu ny khngth thc hin trong thc t).

1.2.1 c lng khong cho k vng

Phng sai 2 bit

nh l 1.2. Gi s X N(, 2) trong 2 bit. Vi tin cy1 , gi z l gi tr tha mn (z) =

1 2

, y

(x) =12

x0

et2

2 dt.

Khi (X z n ;X + z

n

)l khong c lng cho vi

tin cy 1 , trong

X =X1 +X2 + +Xn

n

vi X1, X2, . . . , Xn l cc quan st c lp v X.

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Chng minh. V cc BNN Xi, i = 1, n c lp v c cng phn b viX N(, 2) nn X N(, 2

n), suy ra Z = (X)

n

N(0, 1). Khi

P

(X z

n< < X + z

n

)= P

(z 30 th ta c th xp x 2 bng phngsai hiu chnh s2 ca mu.

Khi , khong c lng trung bnh vi tin cy 1 s l(x z

sn;x+ z

sn

).

V d 1.2. Mt mu ngu nhin gm 100 sinh vin ca trng HCNTP.HCM c hi v qung ng h i t nh ti trng. Gi tr trungbnh v lch chun hiu chnh ca mu ny tng ng l 5km v0, 8km. Vi tin cy 95%, hy xc nh khong c lng qung ngtrung bnh i hc ca tt c sinh vin trng HCN TP.HCM.

Gii. T bi ta tnh c n = 100; x = 5km; s = 0, 8km; z = 1, 96.Do , khong c lng qung ng trung bnh i hc ca tt c sinhvin trng HCN l

(x z sn ; x+ z

sn

)= (4, 84; 5, 16).

V d 1.3. Trng HCN TP.HCM tin hnh mt cuc iu tra xemtrung bnh mt sinh vin ca trng tiu ht bao nhiu tin gi inthoi trong mt hc k. Mt mu ngu nhin gm 49 sinh vin cchn v s tin chi cho vic gi in thoi ca h nh sau (n v nghnng):

112 126 130 133 145 149 151155 157 161 167 169 171 173175 177 181 184 187 189 191195 197 200 201 205 208 210212 216 220 222 229 233 237240 243 247 249 255 260 263275 277 281 284 287 289 291

Vi tin cy 99%, hy xc nh khong c lng trung bnh s tingi in thoi ca sinh vin trng HCN trong mt hc k.

Gii. T bng s liu ta c n = 49; x = 206, 31; s = 48, 36; z = 2, 58.Khi , khong c lng trung bnh s tin gi in thoi ca

sinh vin trng HCN trong mt hc k l(x z sn ; x+ z

sn

)=

(188, 49; 224, 13).

V d 1.4. xc nh chiu cao trung bnh (n vm) ca cc cy bchn trong mt khu rng bch n rt ln, ngi ta chn ngu nhin 64cy o. Kt qu thu c nh sau:

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Khong chiu cao S cy5, 5 6, 5 66, 5 7, 5 157, 5 8, 5 208, 5 9, 5 139, 5 10, 5 10

Vi tin cy 98%, hy xc nh khong c lng chiu cao trungbnh ca cy bch n trong khu rng.

Gii. d tnh ton, mi khong chiu cao ta s ly trung im cakhong lm i din. T y, ta tnh c n = 64; x = 8, 09; s = 1, 2 vz = 2, 33.

Khi , khong c lng trung bnh chiu cao ca cy bch ngtrong khu rng l

(x z sn ;x+ z

sn

)= (7, 74; 8, 44).

Phng sai 2 cha bit v n < 30

C s cho vic xy dng khong c lng cho trng hp ny da vonh l sau:

nh l 1.3. Gi s X N(, 2) v X1, X2, . . . , Xn l cc BNN clp vi nhau v c cng phn b vi X. Khi , BNN T = (X)

n

Ss

c phn b Student vi bc t do n 1, tc T tn1, trong

S2 =

(X1 X

)2+(X2 X

)2+ +

(Xn X

)2n 1

.

Gi tn1 l gi tr sao cho S(tn1 ) = 12 , trong

S (x) =(n2

)(n 1)

(n12

) x0

(1 +

t2

n 1

)ndt.

Khi ,(X tn1 Sn ;X + t

n1

Sn

)l khong c lng cho vi

tin cy 1 .

V d 1.5. Mt phng php iu tr bnh mi ang c xem xtnghim thu. Mt ch tiu nh gi hiu qu ca phng php ls ngy trung bnh t lc iu tr cho n khi bnh nhn khi bnh.Mt mu ngu nhin gm 16 bnh nhn c theo di v s ngy iutr cho ti khi bnh nhn khi bnh c ghi li nh sau:

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4 4 5 8 6 10 3 92 6 4 7 9 11 6 8

Vi tin cy 95%, hy xc nh khong c lng trung bnh sngy cn thit bnh nhn c iu tr ht bnh.

Gii. T bi ta tnh c n = 16;x = 6, 375; s = 2, 630; t15 = 2, 131.Khi , khong c lng trung bnh s ngy cn thit cha khi

bnh l(x tn1 sn ;x+ t

n1

sn

)= (4, 974; 7, 776).

V d 1.6. Kho st mt mu gm 12 ngi mt a phng A chothy s ln h i xem phim trong 1 nm nh sau:

14 16 17 17 24 20 32 18 29 31 15 35

Vi tin cy 95%, xc nh khong c lng trung bnh s ln mtngi a phng A i xem phim trong thi gian 1 nm.

Gii. T bi ta tnh c n = 12;x = 22, 333; s = 7, 512; t11 = 2, 201.Khi , khong c lng trung bnh s ln mt ngi a phng

A i xem phim trong thi gian 1 nm l (17, 560; 27, 106).

1.2.2 c lng khong cho t lGi s trong tp hp chnh, mi c th ca n mang hay khng mangmt c tnh A no . Gi p l t l c th mang c tnh A trong tonb tp chnh (p cha bit). Ta mun c lng tham s p ny da trnmu iu tra.

Gi s trong mt mu kch thc n c k c th mang c tnh A.Chng ta bit tn xut mu f = k

nl mt c lng khng chch v

vng cho p. Bi ton t ra l xy dng khong c lng cho p vi tin cy 1 . thc hin iu ny ta tm hiu kt qu sau:

nh l 1.4. Cho Xi, i = 1, n l cc BNN Becnulli c lp vi nhauv Xi B(p). Khi , BNN

F =X1 +X2 + +Xn

n

c phn b xp x chun vi k vng v phng sai tng ng lEF = p;DF = p(1p)

nbit rng np > 5;n(1 p) > 5.

Chng minh. V X1, X2, . . . , Xn l cc BNN Becnulli c lp vi nhauv Xi B(p), i = 1, n nn nF = X1 +X2 + +Xn B(n, p).

Hn na, v np > 5 v n(1 p) > 5 nn B (n, p) N (np, np (1 p)).Do nF N(np, np(1 p)), suy ra F N(p, p(1p)

n).

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V F N(p, p(1p)n

) nn (Fp)n

p(1p) N(0, 1). Do , khong c lng

cho p vi tin cy 1 l(F z

p (1 p)

n;F + z

p (1 p)

n

)

Do ta khng bit p nn vi mu c th ta c th dng xp x

p(1p)n

f(1f)

nnu nf > 10;n(1 f) > 10. Khi , khong c lng cho p vi

tin cy 1 l(f z

f (1 f)

n; f + z

f (1 f)

n

).

V d 1.7. Trc ngy bu c tng thng, mt cuc thm d d lun c tin hnh. Ngi ta chn ngu nhin 400 ngi hi kin thc 240 ngi ni rng h s b phiu cho ng A. Tm khong c lngcho t l c tri b phiu cho ng A vi tin cy 95%.

Gii. T bi ta tnh c n = 400; f = 240400

= 35= 0, 6; z = 1, 96. V{

nf = 400 0, 6 = 240 > 10n (1 f) = 400 0, 4 = 160 > 10

nn khong c lng t l c tri b phiu cho ng A l(f z

f (1 f)

n; f + z

f (1 f)

n

)= (55, 2%; 64, 8%) .

V d 1.8. cy ngc trm th c tnh l phng hay l nhn l do mtgen c hai alen, A tri v a ln, quyt nh. Cc ng hp t AA v dhp t Aa c l phng, cn ng hp t aa c l nhn. Trong s 560 cyc c khi lai hai d hp t th c 110 cy l nhn. Tm khong clng cho xc sut p c cy l nhn khi lai hai d hp t vi tincy 95%. S liu trn c ph hp vi l thuyt ca Mendel hay khng (lthuyt ca Mendel cho rng p = 1

4)?

Gii. T bi ta tnh c n = 560; f = 110560

= 0, 1964; z = 1, 96. V{nf = 109, 984 > 10n (1 f) = 450, 016 > 10

nn khong c lng cho p l(f z

f (1 f)

n; f + z

f (1 f)

n

)= (16, 35%; 22, 93%) .

V p = 14khng thuc khong (16, 35%; 22, 93%) nn s liu trn

khng ph hp vi l thuyt ca Mendel.

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1.2.3 c lng khong cho phng saiGi s X N(, 2). Tp hp chnh y l tp hp tt c cc gi trca X. Mt mu c kch thc n bao gm cc gi tr X1, X2, . . . , Xn thuc t n quan st c lp t X. Vi mu c ly, ta mun tm khongc lng cho 2 vi tin cy 1 .

Trc ht, ta tm hiu kt qu quan trng sau:

nh l 1.5. Gi s X1, X2, . . . , Xn l n quan st c lp t BNNX N(, 2). Khi , T = (n1)S

2

2c phn b 2 vi bc t do n 1,

tc T 2(n 1), trong

S2 =

(X1 X

)2+(X2 X

)2+ +

(Xn X

)2n 1

l phng sai hiu chnh ca mu X1, X2, . . . , Xn.

Ta gi 2 l gi tr sao cho P (T > 2) = . Khi ,

P(21

2< T < 2

2

)= 1

2

2= 1 .

Hn na, ta c

212< T < 2

2 (n 1)S

2

22

< 2 30 tnh x v s. Sau , trong cng thc (1.1), ta s thay bng s. Khi , n =

(zs

)2 l c mu nh nht nu iu kin(zs

)2 30.

c tha mn.

V d 1.10. Ta mun xy dng mt khong c lng cho khi lngtrung bnh ca cc gi ng c ng bng my t ng vi tincy 99%. iu tra s b mt mu cho ta x = 11, 8kg, lch chun hiu chnh s = 0, 9kg. Hi cn phi ly kch thc mu ti thiu l baonhiu t c sai s khng vt qu 0, 1kg ?

Gii. T gi thit ta c x = 11, 8kg; s = 0, 9kg; = 0, 1kg; z = 2, 58. Cmu nh nht tha yu cu bi l n =

(zs

)2= 540.

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V d 1.11. Mt cuc nghin cu c tin hnh nhm xc nh lngtrung bnh cc lut s gii M da trn mt mu iu tra. Hi cn lymu vi kch thc ti thiu l bao nhiu sai s khng vt qu 100USD, vi tin cy c n nh l 95% ? Bit rng lch tiu chunca tp hp chnh l = 1000 USD.

Gii. T gi thit ta c = 1000 USD; = 100 USD; z = 1, 96. Cmu nh nht tha yu cu bi l n =

(z

)2= 385.

1.3.2 Trng hp c lng cho t lGi s ta mun c lng t l p vi sai s khng qu cho trc v tin cy 1 . Ta bit rng vi xc sut 1 th

|F p| z

p (1 p)

n

Ta cn c bt ng thc

z

p (1 p)

n n z

2p (1 p)

2

V p khng bit nn vi mu c th ta dng xp x

p(1p)n

f(1f)n

nu nf > 10;n(1 f) > 10. Do , bt ng thc trn c th vit li nhsau:

n z2f (1 f)

2. (1.2)

T (1.2) ta suy ra c mu nh nht c lng t l p vi sai skhng qu cho trc v tin cy 1 l n =

z2f(1f)

2

.

V d 1.12. Mt nh nng hc mun c lng t l ny mm ca mtloi ht ging.

1) Vi 1000 ht em gieo th c 640 ht ny mm. Tm khong clng t l ht ny mm vi tin cy 90%. Sai s y l bao nhiu ?

2) Nu mun c khong c lng t l ht ny mm vi tin cy90% v sai s khng vt qu 0, 02 th cn ly mu vi kch thc tithiu l bao nhiu ?

Gii. 1) T gi thit ta c n = 1000; f = 6401000

= 0, 64; z = 1, 64. V{nf = 640 > 10n (1 f) = 360 > 10

nn khong c lng t l ht ny mm l(f z

f (1 f)

n; f + z

f (1 f)

n

)= (59, 98%; 68, 02%)

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chnh xc z

f(1f)n

= 4, 02% = 0, 0402.2) T gi thit ta c f = 0, 64; = 0, 02; z = 1, 64. Do , c mu nh

nht tha yu cu bi l n =z2f(1f)

2

= 1550.

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Chng 2

KIM NH GI THITTHNG K

2.1 Nguyn l chungTrong chng ny chng ta s cp n mt vn rt quan trngtrong Thng k: l vn kim nh gi thit thng k. Ni dungca bi ton nh sau:

Cn c trn cc s liu thu c, hy cho kt lun v mt gi thitthng k no m ta quan tm.

Mt gi thit thng k l mt gi thit v s phn b ca tp hpchnh ang xt.

Nu phn b c c trng bi cc tham s (nh gi tr trungbnh, phng sai, . . . ) th gi thit thng k l gi thit v tham s caphn b n. Mt s th d v gi thit thng k:

Tp hp chnh c phn b chun vi k vng l 3.

Phng php iu tr A cha khi 90% bnh nhn.

Tui th trung bnh ca hai loi bng n A v B l nh nhau.

T nay tr i mt gi thit s c hiu l mt gi thit thng k.Mt qui tc hay mt th tc dn n vic chp nhn hay bc b githit nu gi l kim nh (test) thng k.

Gi thit c a ra kim nghim c k hiu l H0 v c gi lgi thit khng. l gi thit l ta nghi ng v mun bc b. Thngi km vi gi thit H0 l mt i thit, k hiu H1. H1 s c chpnhn khi H0 b bc b.

Cu hi t ra l: Chng ta chp nhn hay bc b mt gi thit bngcch no ? Cc nh thng k u nht tr vi nhau nguyn l sau y:

Nu mt bin c c xc sut rt nh th trong mt php thhay mt vi php th, bin c s khng xy ra.

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Nh vy, chng ta s quyt nh bc b gi thit H0 nu xc sutxut hin mt s kin quan st c, tnh trong iu kin gi thit H0ng, l nh.

Sau y ta s trnh by mt s v d minh ha ny.

V d 2.1. Gieo mt ng tin 1000 ln ta thy xut hin mt sp 700ln. Ta nghi ng xc sut xut hin mt sp cao hn mt nga v nhimv ca ta l kim tra iu . Gi p l xc sut xut hin mt sp. Nhvy, gi thit H0 l p = 0, 5 v i thit H1 l p > 0, 5. Nu gi thit H0ng, tc p = 0, 5, th xc sut gieo 1000 ln ng xu c 700 ln mtsp l C7001000

121000

= 5, 067 1038. Gi tr xc sut ny qu nh nn ta cth bc b H0 v chp nhn H1.

V d 2.2. Mi cuc nghin cu M cho bit tr em M tui ntrng tiu th trung bnh 19,4 OZ sa 1 ngy (OZ: ch vit tt caounce, n v o lng Anh: 1 OZ = 28,35g).

Trong mt mu ngu nhin gm 140 tr em, ngi ta tnh clng sa trung bnh chng ung l 18,5 OZ vi lch tiu chun l6,8 OZ. iu ny c cho php ta kt lun l trung bnh lng sa tiuth t hn 19,4 OZ hay khng ?

Gii. Gi l lng sa tiu th trung bnh ca mt a tr trong mtngy. Nh vy, gi thit H0 l = 19, 4 v i thit H1 l < 19, 4. Nugi thit H0 ng, ta s tnh xc sut trung bnh mu X b hn haybng 18, 5.

Nh bit, BBN X c phn b chun (hoc xp x chun) vi kvng l 19,4 v phng sai l s

n= 6,8

140= 0, 575. Khi ,

P(X 18, 5

)=

(18, 5 19, 4

0, 575

)+

1

2= 0, 0582.

Xc sut ny khng nh lm (thng thng xc sut b hn 0,05 mic xem l nh). Do , ta cha c c s bc b H0. Ni cch khc,chng ta cha th khng nh lng sa tiu th trung bnh ca tr thn 19,4 OZ.

Trong khi a ra quyt nh cho cc tnh hung tng t nh trn,phi la chn gia hai gi thit H0 v H1, ta c th phm hai loi sailm:

Bc b H0 trong khi H0 ng, m ta gi l sai lm loi I.

Chp nhn H0 trong khi H0 sai, m ta gi l sai lm loi II.

Sai lm loi I tng t nh sai lm ca quan ta khi kt n nhmngi v ti, cn sai lm loi II tng t nh tha bng ngi c ti.

Mt kim nh thng k c gi l l tng nu lm cc tiu c sailm loi I v sai lm loi II. Tic thay khng tn ti mt kim nh l

18


tng nh vy. Nu ta lm gim sai lm loi I th lm tng sai lm loiII v ngc li.

Trong mt x hi vn minh, ngi ta c xu hng tha nhn vic ktn nhm ngi v ti l mt sai lm nghim trng hn nhiu so vi sailm tha bng k c ti. Trong bi ton kim nh gi thit cng vy. Tacoi sai lm loi I l nghim trng hn sai lm loi II. Thnh th ngita c nh trc xc sut sai lm loi I. Xc sut ca vic mc sai lmloi I cn gi l mc ngha, k hiu . Xc sut mc sai lm loi IIc k hiu l . Con s 1 c gi l lc lng ca kim nh. Lclng ca kim nh l xc sut bc b H0 khi H0 sai. Thng thng c ly l 0, 05; 0, 02 v 0, 01. Trong tp hp cc kim nh thng k ccng mc ngha , thng k no c nh nht c xem l tt nht.Cc kim nh c s dng trong chng ny u c chng minhmt cch cht ch l cc kim nh tt nht.

Cn lu rng khi kim nh thng k dn ti vic chp nhn H0th bng bao nhiu th ta khng bit. Thnh th, vic chp nhn H0c hiu l cc chng c v s liu c cha c s bc b H0,cn phi c nghin cu tip.

Cc bc cn thit trong vic tin hnh mt kim nh gi thitthng k:

1. Pht biu gi thit H0 v i thit H1.

2. nh r mc ngha (xc sut mc sai lm loi I).

3. Chn test thng k.

4. Chn min bc b H0.

5. Tnh gi tr ca test thng k t mu quan st c.

6. Kt lun bc b hay chp nhn H0 ty theo gi tr ca test thngk c ri vo min bc b gi thit hay khng.

2.2 Kim nh gi thit v gi tr trung bnhGi s X N(, 2). Tp hp chnh y l tp hp tt c cc gi trca X. Mt mu c kch thc n bao gm cc gi tr X1, X2, . . . , Xn thuc t n quan st c lp t X. Ta mun kim nh gi thit v .

2.2.1 Phng sai 2 bitBi ton 2.1. Ta mun kim nh gi thit H0 vi i thit H1 nh sau:

H0 : = 0H1 : = 0

y 0 l gi tr cho trc.

19


Bi ton 2.1 c gi l bi ton kim nh hai pha.Test thng k c chn y l

T =(X 0)

n

,

vi mu c th ta dng k hiu t = (x0)n

.

Ta s bc b H0 nu T ln (hoc b) mt cch c ngha. Do , minbc b H0 c dng = {|T | > c} vi c ph thuc vo mc ngha .

Nu H0 ng, tc = 0, th T N(0, 1). Vy vi mc ngha cho, hng s c c tm t iu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = z

vi (z) = 12 .

V d 2.3. Nhng thng k trong nm 2008 cho thy mt ngi M idu lch Chu u trong vng 3 tun s chi ht trung bnh 1010 USD.Mt cuc nghin cu c tin hnh trong nm 2009 xc nh xemc s thay i g trong vic chi tiu mua sm khi du lch Chu u cangi M hay khng. Kho st 100 khch du lch cho thy s tin trungbnh h tiu l 1015 USD. Hy kim nh gi thit: S tin trung bnhmt ngi M chi tiu khi i du lch Chu u trong hai nm 2008v 2009 l nh nhau vi mc ngha 5%, bit lch chun qua tngnm l nh nhau, bng 300 USD.

Gii. Trc ht ta pht biu gi thit H0 v i thit H1:

H0 : = 1010H1 : = 1010

vi l s tin trung bnh mt ngi M chi tiu khi i du lch Chuu trong nm 2009.

Test thng k c chn l

t =(x 0)

n

vi x = 1015;0 = 1010;n = 100; = 300. Khi , ta tnh c t = 16 .Hn na, vi mc ngha = 5% th z = 1, 96. Ta thy |t| < z nn

gi thit H0 c chp nhn.

Bi ton 2.2. Ta mun kim nh gi thit H0 vi i thit H1 nh sau:

H0 : = 0H1 : > 0


20


Bi ton 2.2 c gi l bi ton kim nh mt pha.Test thng k c chn l

T =

(X 0

)n

,


.

Ta s bc b H0 nu T ln mt cch c ngha. Do , min bc bH0 c dng = {T > c} vi c ph thuc vo mc ngha .


P ({T > c}) = P ({T c}) = 1 c = z2.

V d 2.4. Theo thng bo ca Lu Nm Gc, qun i M b tr trungbnh 90 tn la mi cn c tn la. Mt t chc ha bnh quc t dnh kim tra 49 cn c xem thng bo ni trn c ng hay khng.on kim tra s dng mc ngha 0, 1.

1) Pht biu gi thit H0 v i thit H1.2) Gi s on kim tra tnh c s tn la b tr trung bnh trong

mi cn c l 92. H cn phi rt ra kt lun g ? Bit rng = 9.

Gii. 1) Mc d bi khng cp n, nhng ta thy mc ch lnnht ca on kim tra l xem Lu Nm Gc c b tr nhiu tn lahn so vi thng bo hay khng. Chnh v th, gi thit H0 v i thitH1 y l:

H0 : = 90H1 : > 90

vi l trung bnh s tn la m Lu Nm Gc b tr thc s cho tt ccn c qun s.

2) Test thng k c chn l

t =(x 0)

n

vi x = 92;0 = 90;n = 49; = 9. Khi , ta tnh c t = 1, 56.Hn na, vi mc ngha = 0, 1 th z2 = 1, 28. V t > z2 nn ta

bc b H0 v chp nhn H1, tc Lu Nm Gc b tr tn la dy c hnso vi thng bo.


H0 : = 0H1 : < 0


21


Bi ton 2.3 cng c gi l bi ton kim nh mt pha.Test thng k c chn l

T =

(X 0

)n

,


.

Ta s bc b H0 nu T nh mt cch c ngha. Do , min bc bH0 c dng = {T < c} vi c ph thuc vo mc ngha .


P ({T < c}) = c = z2.

V d 2.5. T mt tp hp chnh c phn b chun vi k vng (chabit) v lch chun = 10, ngi ta ly mt mu gm 121 quan stv tnh c x = 95. Vi mc ngha = 0, 01, hy kim nh gi thitH0 v i thit H1 nh sau:

H0 : = 100H1 : < 100

Gii. Test thng k c chn l

t =(x 0)

n

vi x = 95;0 = 100;n = 121; = 10. Khi , ta tnh c t = 5, 5.Hn na, vi mc ngha = 0, 01 ta c z2 = 2, 33. V t < z2

nn ta bc b H0 v chp nhn H1, tc trung bnh ca tp hp chnh bhn 100.

2.2.2 Phng sai 2 cha bit, kch thc mu n 30Trong trng hp ny ta vn dng cc test thng k nh trn nhng tathay lch chun ca tp chnh bng lch chun hiu chnh sca mu. Ch rng, theo nh l gii hn trung tm, test thng k Tc phn b xp x chun d tp chnh c phn b nh th no.

V d 2.6. Mt nghin cu cho rng trung bnh mt khch hng vosiu th A tiu ht 200 ngn ng. Ta mun kim nh khng nh trnbng cch chn ngu nhin 64 khch hng. Vi mu chn, ta tnhc s tin trung bnh h tiu l 220 ngn ng vi lch tiu chunl 50 ngn ng. Pht biu gi thit H0, i thit H1 v kim nh H0vi mc ngha = 5%.

22


Gii. Trc ht, ta pht biu gi thit H0 v i thit H1:

H0 : = 200H1 : = 200

vi l s tin trung bnh ca khch hng chi tiu trong siu th A.Test thng k c chn l

t =(x 0)

n

s

vi x = 220;0 = 200;n = 64; s = 50. Khi , ta tnh c t = 3, 2.Hn na, vi mc ngha = 5% ta c z = 1, 96. V |t| > z nn

ta bc b H0 v chp nhn H1.

V d 2.7. Trong nm hc trc, mc chi tiu trung bnh hng thngca sinh vin trng HCN l 1.400.000 ng. Trong nm hc ny, vimt mu ngu nhin 80 em, ta tm c mi thng h chi tiu trungbnh 1.460.000 ng vi lch chun hiu chnh l 100.000 ng.Vi mc ngha 5%, ta c th kt lun chi ph ca sinh vin nm naycao hn nm trc hay khng ?


H0 : = 1.400.000H1 : > 1.400.000

vi l chi ph chi tiu trung bnh ca ton b sinh vin trng HCN.Test thng k c s dng l

t =(x 0)

n

s

vi x = 1.460.000;0 = 1.400.000;n = 80; s = 100.000. Khi , ta tnh ct = 5, 37.

Hn na, vi mc ngha = 5% ta c z2 = 1, 65. V t > z2 nnta bc b H0 v chp nhn H1, tc sinh vin nm nay chi tiu nhiu hnnm trc.

V d 2.8. iu tr mt loi bnh R ngi ta thng dng loi thucA. Thi gian cha tr bng thuc A trung bnh ko di 30 ngy th bnhnhn khi bnh. V cha tr bng thuc A kh tn km nn cc nhkhoa hc c gng ch to ra mt loi thuc mi, m ta gi l B, viph cha tr thp hn v thi gian khi bnh nhanh hn. kim tratnh hiu qu tht s ca B, cc nh khoa hc chn ngu nhin 100bnh nhn mc bnh R v cho dng thuc B. Kt qu cho thy thi giankhi bnh trung bnh ca bnh nhn l 25 ngy vi lch chun hiu chnh l 5 ngy. Vi mc ngha 1%, ta c th kt lun thuc Bhiu qu hn thuc A hay khng ?

23


Gii. Trc ht, ta pht biu gi thit H0 v i thit H1 nh sau:

H0 : = 30H1 : < 30

vi l s ngy trung bnh ngi b bnh R c cha khi bngthuc B.

Test thng k c chn l

t =(x 0)

n

s

vi x = 25;0 = 30;n = 100; s = 5. Khi , ta tnh c t = 10.Hn na, vi mc ngha = 1% ta c z2 = 2, 33. V t < z2 nn

ta bc b H0 v chp nhn H1, tc thuc B hiu qu hn thuc A.

2.2.3 Phng sai 2 cha bit, kch thc mu n < 30Bi ton 2.4. Ta mun kim nh gi thit H0 vi i thit H1 nh sau:

H0 : = 0H1 : = 0


Test thng k c chn y l

T =(X 0)

n

S,


s.


Nu H0 ng, tc = 0, th T tn1. Vy vi mc ngha cho,hng s c c tm t iu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = tn1

vi S(tn1 ) = 12 .

V d 2.9. Mt cng ty sn xut pin tuyn b rng pin ca h c tuith trung bnh l 21, 5 gi. Mt c quan kim tra cht lng ca 6 pinv thu c s liu v tui th ca 6 pin ny (n v, gi):

19 18 22 20 16 25

Kt qu ny c xc nhn l qung co ca cng ty l ng hay khng ?Mc ngha c chn l = 5%.

24



H0 : = 21, 5H1 : = 21, 5

vi l tui th trung bnh ca pin trong thc t.Test thng k c chn l

t =(x 0)

n

s

vi x = 20;0 = 21, 5;n = 6; s = 3, 16. Khi , ta tnh c t = 1, 16.Hn na, vi mc ngha = 5% v bc t do n 1 = 5 ta c

t5 = 2, 571. V |t| < t5 nn ta cha c c s bc b H0.


H0 : = 0H1 : > 0


Test thng k c chn l

T =

(X 0

)n

S,


s.



P ({T > c}) = P ({T c}) = 1 c = tn12 .

V d 2.10. Trong nm hc trc, mc chi tiu trung bnh hng thngca sinh vin trng HCN l 1.400.000 ng. Trong nm hc ny, vimt mu ngu nhin 16 em, ta tm c mi thng h chi tiu trungbnh 1.460.000 ng vi lch chun hiu chnh l 100.000 ng.Vi mc ngha 5%, ta c th kt lun chi ph ca sinh vin nm naycao hn nm trc hay khng ?


H0 : = 1.400.000H1 : > 1.400.000

vi l chi ph chi tiu trung bnh ca sinh vin trng HCN.

25


Test thng k c s dng l

t =(x 0)

n

s

vi x = 1.460.000;0 = 1.400.000;n = 16; s = 100.000. Khi , ta tnh ct = 2.4.

Hn na, vi mc ngha = 5% v bc t do n 1 = 15 ta ct152 = 1, 753. V t > t152 nn ta bc b H0 v chp nhn H1, tc sinh vinnm nay chi tiu cao hn nm trc.

Bi ton 2.6. Ta mun kim nh gi thit H0 vi i thit H1 nhsau:

H0 : = 0H1 : < 0


Test thng k c chn l

T =

(X 0

)n

S,


s.

Ta s bc b H0 nu T nh mt cch c ngha. Do , min bc bH0 c dng = {T < c} vi c ph thuc vo mc ngha .


P ({T < c}) = c = tn12 .

V d 2.11. iu tr mt loi bnh R ngi ta thng dng loi thucA. Thi gian cha tr bng thuc A trung bnh ko di 30 ngy th bnhnhn khi bnh. V cha tr bng thuc A kh tn km nn cc nhkhoa hc c gng ch to ra mt loi thuc mi, m ta gi l B, viph cha tr thp hn v thi gian khi bnh nhanh hn. kim tratnh hiu qu tht s ca B, cc nh khoa hc chn ngu nhin 25bnh nhn mc bnh R v cho dng thuc B. Kt qu cho thy thi giankhi bnh trung bnh ca bnh nhn l 26 ngy vi lch chun hiu chnh l 5 ngy. Vi mc ngha 1%, ta c th kt lun thuc Bhiu qu hn thuc A hay khng ?


H0 : = 30H1 : < 30

vi l s ngy trung bnh ngi b bnh R c cha khi bngthuc B.

26


Test thng k c chn l

t =(x 0)

n

s

vi x = 26;0 = 30;n = 25; s = 5. Khi , ta tnh c t = 4.Hn na, vi mc ngha = 1% v bc t do n 1 = 24 ta c

t242 = 2, 492. V t < t242 nn ta bc b H0 v chp nhn H1, tc thuc Bhiu qu hn thuc A.

2.3 Kim nh gi thit v gi tr t lGi s trong tp hp chnh, mi c th ca n mang hay khng mangmt c tnh A no . Gi p l t l c th mang c tnh A trong tonb tp chnh, tt nhin ta khng bit c p. Mt mu c kch thc nbao gm cc gi tr X1, X2, . . . , Xn thu c t n quan st c lp (lu Xi B(p)). S dng mu thu c, ta s kim nh gi thit p = p0 vip0 l mt s cho.


H0 : p = p0H1 : p = p0

y p0 l gi tr cho trc.

Test thng k c chn l

T =(F p0)

n

p0(1 p0),

vi mu c th ta k hiu t = (fp0)n

p0(1p0).


NuH0 ng, tc p = p0, v np0 5;n(1p0) 5 th F N(p0, p0(1p0)n ),suy ra T N(0, 1). Vy vi mc ngha cho, hng s c c tm tiu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = z

vi (z) = 12 .

V d 2.12. a phng A ngi ta d on c 45% s h gia nh sdng my vi tnh. Kim tra ngu nhin 200 h ta thy c 80 h s dngmy vi tnh. Vi mc ngha 5%, ta c kt lun t l d on l chnhxc hay khng ?

27



H0 : p = 45%H1 : p = 45%

vi p l t l h gia nh s dng my vi tnh.V np0 = 90 > 5;n(1 p0) = 110 > 5 nn ta s dng test thng k

t =(f p0)

n

p0(1 p0)

vi f = 80200

= 0, 4; p0 = 0, 45;n = 200. Khi , ta tnh c t = 1, 42.Hn na, vi mc ngha = 5%, ta c z = 1, 96. V |t| < z nn

ta chp nhn gi thit H0.

Ch 2.1. Trng hp i thit H1 c dng p > p0 hoc p < p0 ta lmtng t nh cc bi ton 2.2 v 2.3.

V d 2.13. C mt bnh truyn nhim m khi dng thuc A th t lbnh nhn khi bnh hon ton l 60%. Mt loi thuc mi B ang cth nghim trn 121 ngi thy c 88 ngi c cha khi hon ton.Vi mc ngha = 1%, ta c th kt lun thuc B tt hn thuc A.


H0 : p = 60%H1 : p > 60%

vi p l t l bnh nhn c cha khi khi dng thuc B.V np0 = 72, 6 > 5;n(1 p0) = 48, 4 > 5 nn ta s dng test thng k

t =(f p0)

n

p0(1 p0)

vi f = 88121

; p0 = 0, 6;n = 121. Khi , ta tnh c t = 2, 86.Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t > z2 nn ta

chp nhn gi thit H1, tc thuc B tt hn thuc A.

V d 2.14. Mt cng ty sn xut bnh ko tuyn b rng 70% tr em tnh A thch bnh ko do cng ty sn xut. Mt mu gm 64 em chi th c 40 em thch bnh ko ca cng ty. Vi mc ngha 1%, ta cth ni cng ty tuyn b qu s tht khng ?


H0 : p = 70%H1 : p < 70%

vi p l t l tr em tnh A thch bnh ko do cng ty sn xut.

28


V np0 = 44, 8 > 5;n(1 p0) = 19, 2 > 5 nn ta s dng test thng k

t =(f p0)

n

p0(1 p0)

vi f = 4064

= 0, 625; p0 = 0, 7;n = 64. Khi , ta tnh c t = 1, 31.Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t > z nn

ta chp nhn H0, tc cng ty khng tuyn b qu s tht.

2.4 Kim nh gi thit v s bng nhauca hai trung bnh

Gi s X N(1, 21);Y N(2, 22), chng ta mun so snh 1 v 2da trn hai mu quan st c lp ca X v Y .

Mt mu ngu nhin c kch thc n bao gm cc gi trX1, X2, . . . , Xnc rt ra t tp hp chnh I, bao gm tt c cc gi tr ca BNN X;tng t, mt mu ngu nhin c kch thc m bao gm cc gi trY1, Y2, . . . , Ym c rt ra t tp hp chnh II, bao gm tt c cc gi trca BNN Y . Ta mun kim nh gi thit v 1 v 2.

2.4.1 Phng sai 21 v 22 bitBi ton 2.8. Ta mun kim nh gi thit H0 vi i thit H1 nh sau:

H0 : 1 = 2H1 : 1 = 2

Test thng k c chn l

T =X Y21n+

22m

,

vi mu c th ta dng k hiu t = xy21n+

22m

.


Nu H0 ng, tc 1 = 2, th T N(0, 1). Vy vi mc ngha cho, hng s c c tm t iu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = z

vi (z) = 12 .

29


V d 2.15. T hai tp hp chnh c phn b chun X v Y ta ly haimu c lp c kch thc tng ng l n = 40 v m = 50. Trung bnhmu tnh c l x = 160 v y = 155.

Bit tp hp chnhX c gi tr trung bnh l 1 (cha bit) v phngsai 21 = 90; tp hp chnh Y c gi tr trung bnh 2 (cha bit) vphng sai 22 = 100. Vi mc ngha 1%, hy kim nh gi thit H0 vi thit H1 c pht biu nh sau:

H0 : 1 = 2H1 : 1 = 2


t =x y21n+

22m

,

vi x = 160; y = 155;n = 40;m = 50; 21 = 90; 22 = 100. Khi , ta tnhc t = 2, 43.

Hn na, vi mc ngha = 1%, ta c z = 2, 58. V |t| < z nnta chp nhn H0. Bi ton 2.9. Ta mun kim nh gi thit H0 vi i thit H1 nh sau:

H0 : 1 = 2H1 : 1 > 2

Test thng k c chn l

T =X Y21n+

22m

,


22m

.



P ({T > c}) = P ({T c}) = 1 c = z2.



H0 : 1 = 2H1 : 1 > 2

30



t =x y21n+

22m

,

vi x = 100; y = 80;n = 100;m = 120; 21 = 200;22 = 240. Khi , ta tnhc t = 10.

Hn na, vi mc ngha = 5%, ta c z2 = 1, 65. V t > z2 nnta chp nhn gi thit H1. Bi ton 2.10. Ta mun kim nh gi thit H0 vi i thit H1 nhsau:

H0 : 1 = 2H1 : 1 < 2

Test thng k c chn l

T =X Y21n+

22m

,


22m

.

Ta s bc b H0 nu T b mt cch c ngha. Do , min bc b H0c dng = {T < c} vi c ph thuc vo mc ngha .


P ({T < c}) c = z2.



H0 : 1 = 2H1 : 1 < 2


t =x y21n+

22m

,

vi x = 80; y = 100;n = 64;m = 100; 21 = 128; 22 = 200. Khi , ta tnhc t = 10.

Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t < z2nn ta chp nhn gi thit H1.

31


2.4.2 Phng sai 21 v 22 cha bit, kch thc mun,m 30

Trong trng hp ny ta vn dng test thng k nh trn nhng tathay cc phng sai ca tp hp chnh l 21; 22 bng phng sai hiu chnh ca mu s21; s22.

V d 2.18. Mt cuc nghin cu c tin hnh so snh tui trungbnh ca ngi m khi sinh ra a con cui cng gia hai vng A v B.Mt mu ngu nhin gm 36 ph n vng A cho kt qu tui trungbnh khi sinh con ln cui l 33, lch chun hiu chnh l 5 nm;mt mu ngu nhin gm 49 ph n vng B cho kt qu tui trungbnh khi sinh con ln cui l 30, lch chun hiu chnh l 5,5nm. Vi mc ngha 5%, hy kim nh gi thit trung bnh tuisinh con ln cui ca ph n vng A v B l nh nhau.

Gii. Ta mun kim nh gi thit H0 v i thit H1 c pht biunh sau:

H0 : 1 = 2H1 : 1 = 2

vi 1, 2 ln lt l trung bnh s tui sinh con ln cui ca ph nvng A v B.

Test thng k c chn l

t =x ys21n+

s22m

,

vi x = 33; y = 30;n = 36;m = 49; s21 = 25; s22 = 30, 25. Ta tnh ct = 2, 62.

Hn na, vi mc ngha = 5%, ta c z = 1, 96. V |t| > z nnta chp nhn H1, tc trung bnh tui sinh con ln cui ca ph nvng A v B l khc nhau.

V d 2.19. Mt cuc nghin cu c tin hnh so snh tui trungbnh ca ngi m khi sinh ra a con cui cng gia hai vng A v B.Mt mu ngu nhin gm 36 ph n vng A cho kt qu tui trungbnh khi sinh con ln cui l 33, lch chun hiu chnh l 5 nm;mt mu ngu nhin gm 49 ph n vng B cho kt qu tui trungbnh khi sinh con ln cui l 30, lch chun hiu chnh l 5,5nm. Vi mc ngha 1%, hy kim nh gi thit trung bnh tuisinh con ln cui ca ph n vng A cao hn vng B.


H0 : 1 = 2H1 : 1 > 2

32


vi 1, 2 ln lt l trung bnh s tui sinh con ln cui ca ph nvng A v B.

Test thng k c chn l

t =x ys21n+

s22m

,

vi x = 33; y = 30;n = 36;m = 49; s21 = 25; s22 = 30, 25. Ta tnh ct = 2, 62.

Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t > z2 nnta chp nhn H1, tc trung bnh tui sinh con ln cui ca ph nvng A cao hn vng B.

V d 2.20. Mt cuc nghin cu c tin hnh so snh mc lngtrung bnh ca n gii v nam gii trong mt cng ty ln (n v, USD/gi). Mt mu ngu nhin gm 100 ph n c mc lng trung bnhl 8 USD/ gi, lch chun hiu chnh l 1,5 USD; mt mu ngunhin gm 64 n ng c mc lng trung bnh l 9,5 USD/ gi, lchchun hiu chnh l 2 USD. Vi mc ngha 1%, ta c th kt lunmc lng trung bnh ca nam gii cao hn n gii hay khng ?


H0 : 1 = 2H1 : 1 < 2

vi 1, 2 ln lt l lng trung bnh ca n gii v nam gii.Test thng k c chn l

t =x ys21n+

s22m

,

vi x = 8; y = 9, 5;n = 100;m = 64; s21 = 2, 25; s22 = 4. Khi , ta tnh ct = 5, 14.

Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t < z2 nnta chp nhn H1, tc mc lng trung bnh ca nam gii cao hn ngii.

2.4.3 Phng sai 21 = 22 (cha bit), kch thc mun,m < 30

Bi ton 2.11. Ta mun kim nh gi thit H0 vi i thit H1 cpht biu nh sau:

H0 : 1 = 2H1 : 1 = 2

33


Ta t

S2X =(X1 X

)2+(X2 X

)2+ +

(Xn X

)2n 1

.

S2Y =(Y1 Y

)2+(Y2 Y

)2+ +

(Ym Y

)2m 1

.

S2 = (n 1)S2X + (m 1)S2Y

n+m 2.

Test thng k c chn l

T =X Y

S

1n+ 1

m

,

vi mu c th ta dng k hiu t = xys

1n+ 1

m

.

Ta s bc b H0 nu |T | ln mt cch c ngha. Do , min bc bH0 c dng = {|T | > c} vi c ph thuc vo mc ngha .

Nu H0 ng, tc 1 = 2, th T tn+m2. Vy vi mc ngha cho, hng s c c tm t iu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = tn+m2

vi S(tn+m2 ) = 12 .

Ch 2.2. Vi trng hp i thit H1 c dng 1 > 2 hoc 1 < 2 talm tng t nh mc 2.4.1.

V d 2.21. so snh chiu cao trung bnh ca nam thanh nin tihai a phng A v B, ngi ta chn ngu nhin 9 thanh nin aphng A v 16 thanh nin a phng B. S o chiu cao ca hainhm ngi c cho bi bng sau (n v, cm):

a phng A:

154 156 157 159 161 161 165 166 170

a phng B:

153 160 163 165 165 167 167 169 171173 173 174 175 175 177 177

Vi mc ngha = 1%, ta c th kt lun chiu cao trung bnh canam thanh nin ti hai a phng A v B l nh nhau.

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H0 : 1 = 2H1 : 1 = 2

vi 1, 2 ln lt l trung bnh chiu cao nam thanh nin hai aphng A v B.

T s liu bi ta tnh c

n = 9; x = 161; sX = 5, 2.

m = 16; y = 169; sY = 6, 7.

s = 6, 22.

Test thng k c chn l

t =x y

s

1n+ 1

m

= 3, 09.

Vi mc ngha = 1% v bc t do n + m 2 = 23 ta tnh ct23 = 2, 807. V |t| > t23 nn ta chp nhn gi thit H1, tc chiu caotrung bnh ca nam thanh nin ti hai a phng A v B l khcnhau.

V d 2.22. so snh chiu cao trung bnh ca nam thanh nin tihai a phng A v B, ngi ta chn ngu nhin 9 thanh nin aphng A v 16 thanh nin a phng B. S o chiu cao ca hainhm ngi c cho bi bng sau (n v, cm):

a phng A:

154 156 157 159 161 161 165 166 170

a phng B:

153 160 163 165 165 167 167 169 171173 173 174 175 175 177 177

Vi mc ngha = 1%, ta c th kt lun chiu cao trung bnh canam thanh nin a phng A nh hn a phng B.


H0 : 1 = 2H1 : 1 < 2

vi 1, 2 ln lt l trung bnh chiu cao nam thanh nin hai aphng A v B.

T s liu bi ta tnh c

35


n = 9; x = 161; sX = 5, 2.

m = 16; y = 169; sY = 6, 7.

s = 6, 22.

Test thng k c chn l

t =x y

s

1n+ 1

m

= 3, 09.

Vi mc ngha = 1% v bc t do n + m 2 = 23 ta tnh ct232 = 2, 5. V t < t232 nn ta chp nhn gi thit H1, tc chiu cao trungbnh ca nam thanh nin ti a phng B ln hn a phng A.

V d 2.23. Sn lng ca hai ging la A v B c trng trong cngmt trm thc nghim trong 6 ma lin tip c kt qu sau (n v,tn/ha):

Ging la A: 7,2 7,3 7,3 7,5 7,8 8,0Ging la B: 6,5 6,9 6,9 7,0 7,2 7,5Vi mc ngha 5%, ta c th kt lun nng sut ging A cao hn

ging B.


H0 : 1 = 2H1 : 1 > 2

vi 1, 2 ln lt l trung bnh nng sut ca hai ging la A v B.T s liu bi ta tnh c

n = 6; x = 7, 52; sX = 0, 112.

m = 6; y = 7; sY = 0, 3347.

s = 0, 3269.

Test thng k c chn l

t =x y

s

1n+ 1

m

= 2, 76.

Vi mc ngha = 5% v bc t do n + m 2 = 10 ta tnh ct102 = 1, 812. V t > t102 nn ta chp nhn gi thit H1, tc nng sutging la A cao hn ging la B.

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2.5 Kim nh gi thit v s bng nhauca hai t l

Xt hai tp hp chnh I v II v mt c tnh A m mi c th ca haitp hp chnh c th c hoc khng. Ta mun so snh t l c th mangc tnh A ca tp hp chnh I, k hiu p1, vi t l c th mang c tnhA ca tp hp chnh II, k hiu p2.

Ta ly mt mu c kch thc n bao gm cc gi tr X1, X2, . . . , Xnt tp hp chnh I (lu Xi B(p1)); tng t, ta ly mt mu c kchthc m bao gm cc gi tr Y1, Y2, . . . , Ym t tp hp chnh II (lu Yj B(p2)). Da vo mu c chn ta s tin hnh kim nh gi thitv hai gi tr p1 v p2.

Bi ton 2.12. Ta mun kim nh gi thit H0 vi i thit H1 cpht biu nh sau:

H0 : p1 = p2H1 : p1 = p2

Trc ht, ta t

FX =X1 +X2 + +Xn

n, vi mu c th ta k hiu fX .

FY =Y1 + Y2 + + Ym

m, vi mu c th ta k hiu fY .

Nu np1, n(1 p1),mp2,m(1 p2) 5 th

FX N(p1,(1 p1)p1

n) v FY N(p2,

(1 p2)p2m

).

Test thng k c chn l

T =FX FY

p1(1p1)n

+ p2(1p2)m

.

V p1 v p2 khng bit, nn vi mu c th ta s thay p1(1p1)n +p2(1p2)

m

bng f(1f)( 1n+ 1

m), trong f = mfX+nfY

m+nv (n+m)f 10, (n+m)(1

f) 10. Khi , gi tr test thng k vi mu c th l t = fXfYf(1f)( 1

n+ 1

m)

Ta s bc b H0 nu |T | ln mt cch c ngha. Do , min bc bH0 c dng = {|T | > c} vi c ph thuc vo mc ngha .

Nu gi thit H0 ng, tc p1 = p2, th T N(0, 1). Vy vi mc ngha cho, hng s c c tm t iu kin

P ({|T | > c}) = P ({|T | c}) = 1 c = z.

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V d 2.24. iu tra 100 h gia nh a phng A th c 45 h sdng my vi tnh; 121 h a phng B th c 50 h s dng my vitnh. Vi mc ngha 5%, ta c th ni t l s dng my vi tnh haia phng A v B l nh nhau.


H0 : p1 = p2H1 : p1 = p2

vi p1, p2 ln lt l t l h gia nh s dng my vi tnh a phngA v B.

T s liu ca mu ta tnh c

n = 100; fX = 45100 = 0, 45.

m = 121; fY = 50121 = 0, 41.

m+ n = 221; f = nfX+mfYn+m

= 0, 43.

V (m+ n)f = 95 > 10; (m+ n)(1 f) = 126 > 10 nn ta s dng testthng k

t =fX fY

f(1 f)( 1n+ 1

m)= 0, 6.

Hn na, vi mc ngha = 5%, ta c z = 1, 96. V |t| < z nnta chp nhn gi thit H0.

Ch 2.3. Vi trng hp i thit H1 c dng p1 > p2 hoc p1 < p2 talm tng t nh mc 2.4.1.

V d 2.25. Kho st 120 sinh vin trng A c 35 em n t MinTy; 130 sinh vin trng B c 32 em n t Min Ty. Vi mc ngha1%, ta c th kt lun t l sinh vin n t Min Ty trng A caohn trng B.


H0 : p1 = p2H1 : p1 > p2

vi p1, p2 ln lt l t l sinh vin n t Min Ty trng A v B.T s liu ca mu ta tnh c

n = 120; fX = 35120 = 0, 29.

m = 130; fY = 31130 = 0, 24.


= 66250

= 0, 26.

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t =fX fY

f(1 f)( 1n+ 1

m)= 0, 9.

Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t < z2 nnta chp nhn gi thit H0.

V d 2.26. Kho st 120 sinh vin trng A c 35 em n t MinTy; 130 sinh vin trng B c 50 em n t Min Ty. Vi mc ngha1%, ta c th kt lun t l sinh vin n t Min Ty trng A thphn trng B.


H0 : p1 = p2H1 : p1 < p2

vi p1, p2 ln lt l t l sinh vin n t Min Ty trng A v B.T s liu ca mu ta tnh c

n = 120; fX = 35120 = 0, 29.

m = 130; fY = 50130 = 0, 38.


= 85250

= 0, 34.


t =fX fY

f(1 f)( 1n+ 1

m)= 1, 5.

Hn na, vi mc ngha = 1%, ta c z2 = 2, 33. V t > z2nn ta chp nhn gi thit H0.

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