N TP XC SUT THNG K

THNG K M Tm ng (population) v mu (sample)Sp xp d liu theo nhm v biu tn sut (histogram)Phn b tn sut (frequency distribution)S bch phn (percentile) v s t phn (quartile)Sp xp d liu theo th t t nh n lnTm v tr (position) ca s bch phn/t phn:

i = (n+1)P/100Da vo v tr, tm gi tr ca s bch phn/t phnS t phn: l cc s bch phn th 25, 50, v 75

THNG K M TTrung v (median), yu v (mode), tr trung bnh (mean)Trung v (median) l s bch phn th 50.Yu v (mode) l phn t c tn sut xut hin cao nht trong mu/m ngTr trung bnh (mean): gi tr trung bnh ca tt c cc phn t trong mu/m ng.

THNG K M TPhng sai (variance) v lch chun (standard deviation):Phng sai (variance): trung bnh ca bnh phng lch gia cc phn t trong mu/m ng v tr trung bnh

lch chun (standard deviation): cn bc hai ca phng sai (variance)

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THNG K M TSkewness v Kurtosis:Skewness: o bt i xng ca mt phn bNghing v bn trii xngNghing v bn phiKurtosis: o phng/ nhn ca mt phn bPlatykurtic (tng i phng)Mesokurtic (bnh thng)Leptokurtic (tng i nhn)

THNG K M Tnh l Chebyshev:c p dng cho tt c cc phn b, dng xc nh mt chn di cho t l cc phn t nm trong khong k lch chun t tr trung bnh.C t nht l s phn t ca bt k phn b no s nm trong khong k lch chun t tr trung bnh

tnhtnm trong2

3

4 lch chun t tr trung bnh

THNG K M TQuy tc kinh nghim (Empirical rule):p dng cho cc phn b tng i c hnh chung (mound-shape) v i xng:

68%

1 lch chun t tr trung bnh

95%

nm trong khong

2 lch chun t tr trung bnh

Tt c

3 lch chun t tr trung bnh

THNG K M TBiu Thn v L (Stem-and-Leaf)Biu hnh hp (Box plot):Xc nh cc s t phn: s t phn di, trung v, s t phn trn.Xc nh chn trong (Inner fences), chn ngoi (Outer fences)Xc nh gi tr ln nht v nh nht nm trong chn trongV biu Xc nh cc im outliers v suspected outliers

XC SUTXc sut l php o v khng chc chn, v c hi xy ra, c biu din bi mt gi tr t 0 n 1 (hoc t 0% n 100%)Xc sut khch quan (objective probability - xc sut c in) v xc sut ch quan (subjective probability)

XC SUTTp hp: l tp hp cc phn t c quan tmTp hp rng: l tp hp khng cha phn t noUniversal set: l tp hp cha ton b cc phn tTp b (complement): tp b ca 1 tp hp A l tp hp cha tt c cc phn t khng nm trong tp hp A.Giao (intersection) : giao ca 2 tp hp A v B l tp hp cha tt c cc phn t nm trong c tp A v tp B.Hi (union) : hi ca 2 tp hp A v B l tp hp cha tt c cc phn t nm trong tp A hoc tp B hoc c trong tp A v B.

XC SUTTp loi tr ln nhau (mutually exclusive sets): l cc tp hp khng c chung phn t no, giao ca cc tp hp ny l mt tp hp rng.Partition: l tp hp cc tp loi tr ln nhau, chng cha tt c cc phn t, hi ca cc tp hp ny s to thnh universal set.

XC SUTSample space/event set: l tp hp cha tt c cc kt qu ca mt thc nghim.S kin (event): l tp hp cha cc kt qu c cng mt c im chung ca mt s kin Xc sut ca s kin: l tng ca tt c cc xc sut ca cc kt qu trong s kin

XC SUTNhng quy tc c bn ca xc sut:Khong gi tr: 0 P(A) 1Xc sut ca tp b: Xc sut ca tp giao:

i vi 2 tp loi tr ln nhau: Xc sut ca tp hi:

i vi 2 tp loi tr ln nhau:

XC SUTXc sut c iu kin:

Hai s kin c lp nhau:

XC SUTChnh hp (permutation): l s cch la chn c th t r phn t trong tng s n phn t

T hp (combination): l s cch la chn khng quan tm n th t r phn t trong tng s n phn t

XC SUTQuy lut tng xc sut:

nh l Bayes:

BIN NGU NHINBin ri rc v bin lin tc.Hm trng lng xc sut Bin ri rc

Hm tch ly (cumulative probability function):

BIN NGU NHINTr k vng (expected value) ca bin ngu nhin ri rc: l tr trung bnh c trng s ca cc gi tr ca bin ngu nhin v xc sut tng ng ca chng.

Tr k vng ca mt hm s ca mt bin ngu nhin ri rc:

BIN NGU NHINPhng sai v lch chun ca bin ngu nhin:Phng sai:

lch chun:

PHN B NH THCPhn b nh thc (Binomial distribution): bin ngu nhin nh thc X, l bin m s lt thnh cng trong n php th Bernoulli

Phn b nh thc:

Vi:p: xc sut thnh cngtrong 1 php th,q = 1- p,n: s lt thc hin php th, x: s lt thnh cng.

PHN B NH THCTr trung bnh, phng sai, lch chun ca bin ngu nhin nh thc:Tr trung bnh:

Phng sai:

lch chun:

PHN B SIU BIPhn b siu bi (hypergeometric distribution) dng xc nh xc sut xy ra x thnh cng trong n lt la chn trong mt m ng c N phn t gm c S thnh cng v khi ly mu khng thay th.

Tr trung bnh:

Phng sai:

PHN B POISSONPhn b Poisson xc nh xc sut c x lt xut hin trong mt khong thi gian hay trong mt phm vi nht nh.Phn b Possion cn c s dng xp x phn b nh thc khi xc sut ca thnh cng nh (p 0.05) v s lng php th ln (n 20).

PHN B HM MPhn b hm m (Exponential distribution) o xc sut ca thi gian gia 2 s kin tun theo phn b PoissonHm mt :

Tr trung bnh v phng sai u l 1/

Hm tch ly:

PHN B NORMALPhn b Normal c m t bi 2 tham s: tr trung bnh v phng sai 2

X N(, 2)Phn b chun (Standard Normal) l phn b Normal vi tr trung bnh = 0 v lch chun = 1.

Z N(0, 12)Bin i t phn b Normal ngu nhin thnh phn b chun:

Bin i ngc t phn b chun sang phn b Normal ngu nhin:

PHN B NORMALXp x phn b nh thc bng phn b Normal:i vi n 50 v gi tr ca p khng qu gn 0 v 1:

Khi 20 n 50:

Nu gi tr ca p gn bng 0 hoc gn bng 1 s dng phn b Poisson xp x phn b nh thc.

LY MU PHN B XC SUT MUPhn b xc sut ca tr trung bnh:Khi m ng (population) tun theo phn b Normal:Khi bit lch chun ca m ng

Khi khng bit lch chun ca m ng s dng lch chun ca mu s phn b t (phn b Student)

Khi ly mu ln (n 30) t m ng c tr trung bnh l v lch chun (nh l Gii hn trung tm):

LY MU PHN B XC SUT MUPhn b xc sut ca t l mu: Khi c mu ln, phn b xc sut ca t l ca mu tun theo phn b Normal:

Phn b xc sut ca phng sai mu: phn b Chi-square:Khi mu c ly t m ng tun theo phn b Normal, bin ngu nhin s tun theo phn b Chi-square vi bc t do l (n 1)

C LNG KHONGc lng khong:Khong gi tr c tin l c cha tham s ca m ngi cng vi khong gi tr l mt tin cy (1 )100%

C LNG KHONGc lng khong ca tr trung bnh:Khi m ng (population) tun theo phn b Normal:Khi bit lch chun ca m ng

Khi khng bit lch chun ca m ng s dng lch chun ca mu s phn b t (phn b Student)

Khi ly mu ln (n 30) t m ng c tr trung bnh l v lch chun (nh l Gii hn trung tm):

C LNG KHONGc lng khong ca t l:

c lng khong ca phng sai: phn b Chi-square:

KIM NH GI THUYT THNG KGi thuyt:Gi thuyt khng (H0): l gi thuyt ban u, c gi s l ng cho n khi c c s tin rng gi thuyt ny l sai.Gi thuyt i (H1): l gi thuyt ngc li vi gi thuyt khng.Ra quyt nh:ng: chp nhn mt gi thuyt ng hoc bc b mt gi thuyt sai.Sai: bc b mt gi thuyt ng hoc chp nhn 1 gi thuyt sai.

KIM NH GI THUYT THNG KCc loi kim nh:Kim nh mt pha: khi gi thuyt khng (H0) dng bt ng thcKim nh pha triKim nh pha phiKim nh hai pha: khi gi thuyt khng (H0) dng ng thcCc sai lm khi kim nh:Sai lm loi I: bc b mt gi thuyt thc s ngSai lm loi II: chp nhn mt gi thuyt thc s sai

KIM NH GI THUYT THNG KCc bc thc hin kim nh:

Thnh lp gi thuyt khng (H0) v gi thuyt i (H1)Thu thp thng tin lin quan n vn Tnh ton cc tr s thng k tng ngXc nh tr ti hn hoc p-valueSo snhTr thng k v tr ti hnp-value v

Ra quyt nh: bc b hoc khng bc b gi thuyt khngKt lun

KIM NH GI THUYT THNG KKim nh k vng ca m ng theo phn b Normal, phng sai bit

Kim nh hai phaKim nh mt phaGi thuytH0: = 0H1: 0H0: > 0 ( < 0 )H1: 0 ( 0 )Tr thng kTr ti hnRa quyt nhBc b H0 nu Bc b H0 nu

KIM NH GI THUYT THNG KKim nh k vng ca m ng theo phn b Normal, phng sai cha bit, mu nh

Kim nh hai phaKim nh mt phaGi thuytH0: = 0 H1: 0H0: > 0 ( < 0 )H1: 0 ( 0 )Tr thng kTr ti hnRa quyt nhBc b H0 nu Bc b H0 nu

KIM NH GI THUYT THNG KKim nh k vng ca m ng, phng sai cha bit bit, c mu ln

Kim nh hai phaKim nh mt phaGi thuytH0: = 0H1: 0H0: > 0 ( < 0 )H1: 0 ( 0 )Tr thng kTr ti hnRa quyt nhBc b H0 nu Bc b H0 nu

KIM NH GI THUYT THNG KKim nh v t l ca m ng, c mu ln

Kim nh hai phaKim nh mt phaGi thuytH0: p = p0H1: p p0H0: p > p0 ( p < p0 )H1: p p0 ( p p0 )Tr thng k vi q0 = 1- p0Tr ti hnRa quyt nhBc b H0 nu Bc b H0 nu

KIM NH GI THUYT THNG KKim nh v phng sai ca m ng theo phn b Normal

Kim nh hai pha

Kim nh mt pha

Gi thuyt

H0: 2 = 20

H1: 2 20

H0: 2 > 20 (2 < 20)

H1: 2 20 (2 ( 20)

Tr thng k

Tr ti hn

Ra quyt nh

Bc b H0 nu

Bc b H0 nu

KIM NH GI THUYT THNG KTnh ton p-value:Kim nh hai phaKim nh pha triKim nh pha phi

KIM NH GI THUYT THNG KKim nh v s bng nhau ca hai k vng ca hai m ng theo phn b Normal Ly mu theo cp

C mu nhC mu lnGi thuytH0: D = D (or D < D)H1: D D (or D D)[Usually D = 0]H0: D = D (or D < D)H1: D D (or D D)[Usually D = 0]Tr thng k

Ra quyt nhBc b H0 nu

Bc b H0 nu : trung bnh mu ca hiu s gia cc cp gi tr, sD : lch chun ca mu ca hiu s gia cc cp gi tr, c mu l n. D hiu s gia k vng ca hai m ng

KIM NH GI THUYT THNG KKim nh v s bng nhau ca hai k vng ca hai m ng theo phn b Normal Ly mu c lp

C mu nhC mu lnGi thuyt H0: 1 - 2 = 0 H0: 1 - 2 0 H0: 1 - 2 DH1: 1 - 2 0 H1: 1 - 2 0 H1: 1 - 2 DTr thng k

Ra quyt nhBc b H0 nuBc b H0 nu

KIM NH GI THUYT THNG KKim nh v s bng nhau ca hai k vng ca hai m ng theo phn b Normal Ly mu c lpKhi c gi thit l phng sai ca hai m ng l bng nhau (d khng bit gi tr chnh xc), c th tnh phng sai chung (pooled variance)

Tr thng k:

Bc t do: df = (n1+ n2 - 2)

KIM NH GI THUYT THNG KKim nh v s bng nhau ca t l ca hai m ng C mu ln

I. H0: p1 -p2 = 0H1: p1 -p20II.H0: p1 - p2 0H1: p1 - p2 > 0III. H0: p1 - p2 DH1: p1 - p2 > D

KIM NH GI THUYT THNG KKim nh v s bng nhau ca t l ca hai m ng C mu lnTr thng k trong trng hp I v II:

vi:

Tr thng k trong trng hp III:

KIM NH GI THUYT THNG KKim nh v s bng nhau ca phng sai ca hai m ng theo phn b Normal

Kim nh hai pha

Kim nh mt pha

Gi thuyt

H0: 12 = 22

H1: 12 22

H0: 12 < 22

H1: 12 ( 22

Tr thng k

t phng sai c gi tr ln

hn t s

Tr ti hn

Ra quyt nh

Bc b H0 nu

Bc b H0 nu

KIM NH PHI THAM S - PHI PHN BKim nh Chi-square:Kim nh tnh ph hp (Goodness-of-fit test):

H0: xc sut cho cc gi tr quan st ri vo lp th i = piH1: xc sut cho cc gi tr quan st ri vo lp th i piTr thng k:

vi:Ei: k vng ca phn t trong lp th i. Ei = npi Oi: s phn t thc t trong lp th in: tng s phn tk: s lpBc b H0 nu 2 > 2k-1,

KIM NH PHI THAM S - PHI PHN BKim nh Chi-square:Kim nh gi thuyt v tnh c lp gia hai thuc tnh ca m ng

H0: hai tiu chun c phn loi theo hng v ct c lp vi nhau H1: hai tiu chun c phn loi theo hng v ct ph thuc nhauTr thng k:

viEij: k vng ca phn t hng i, ct j.Oij : s phn t thc t hng i, ct jr: tng s hngc: tng s ctBc t do: df = (r 1)(c 1) Bc b H0 nu 2 > 2df,

PHN TCH PHNG SAI (ANOVA)Phn tch phng sai l mt phng php thng k dng xc nh s khc nhau gia cc tr trung bnh ca m ngTrong php phn tch phng sai, ta c:k mu c ly c lp t k m ngk tr trung bnh mu: x1, x2 , x3 , ... , xk k phng sai mu: s12, s22, s32, ...,sk2Tng s phn t: n = n1+ n2+ n3+ ...+nk Cc gi thit:Cc mu c ly ngu nhin, c lpCc m ng tun theo phn b Normal, vi tr trung bnh l i (c th khc nhau) nhng phng sai phi bng nhau

PHN TCH PHNG SAI (ANOVA)Gi thuyt:

H0: m1 = m2 = m3 = m4 = ... mk H1: Khng phi tt c mi (i = 1, ..., k) l bng nhauTr thng k: Ft = Fk 1, n - kTr ti hn: Fk 1, n k, Bc b H0 nu Ft > Fk 1, n k,

PHN TCH PHNG SAI (ANOVA)Bng ANOVA:

NgunTng bnh phng( SS )Bc t doTrung bnh bnh phng (MS)T s FNhn tk 1

Sai sn kTngSST = SSTR + SSEn 1

HI QUY BIM hnh c bn:

Y = 0 + 1X1 + 2X2 + . . . + kXk + Cc gi thit:~N(0,2), c lp vi cc sai s khcCc bin Xi c lp vi nhau

HI QUY BIKim nh gi thuyt v quan h hi quy:

H0: 1 = 2 = ...= k= 0H1: Khng phi tt c i (i=1,2,...,k) u bng 0

Bc b H0 nu Ft > Fk, n k 1,

NgunTng bnh phng( SS )Bc t doTrung bnh bnh phng (MS)T s FHi quySSRk

Sai sSSEn k 1 TngSST = SSR + SSEn 1

HI QUY BIH s tt nh R2:

H s tt nh iu chnh:

HI QUY BIKim nh cho tng h s ring r:

(1)H0: b1 = 0H1: b1 0(2)H0: b2 = 0Tr thng k:H1: b2 0 . . .(k)H0: bk = 0Bc b H0 nuH1: bk 0

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