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9 Chng 4

B MN TON GVGD: Nguyn nh Huy

Chng 2

P DNG MS-EXCEL TRONG THNG K SUY L

So snh gi tr trung bnh

Phng sai bit trc

D liu tng ng tng cp

Phng sai bng nhau

Phng sai khc nhau

So snh t s

So snh phng sai

10 Chng 4


A- SO SNH GI TR TRUNG BNH VI PHNG SAI BIT TRC

4.1 Khi nim thng k

Nu mu ln (N > 30) th phng sai ca mu , 2iS , c th c xem l phng sai ca

dn s , 2i

, khi y bn c th p dng trc nghim z so snh gi tr trung bnh ca hai mu

vi phng sai bit trc.

Gi thuyt:

Trc nghim bn phi

211

210

:H

:H

Trc nghim bn tri

211

210

:H

:H

Trc nghim hai bn

211

210

:H

:H

Chng 4 11


Gi tr thng k:

2

2

2

1

2

1

21

2

2

2

1

2

1

2121

NN

)XX(

NN

)()XX(z

Phn phi chun

Bin lun

Nu z < z (hai bn) hay z/2 (mt bn) Chp nhn gi thuyt H0

4.2 p dng Ms-EXCEL

Th d 6: Ngi ta chn hai mu, mi mu c 10 my, t hai l (I v II c sn xut vi

phng sai bit trc tng ng l 1 v 0,98) kho st thi gian hon thnh cng vic (pht)

ca chng:

I 6 8 9 10 6 15 9 7 13 11

II 5 5 4 3 9 9 6 13 17 12

Hi kh nng hon thnh cng vic ca hai my c khc nhau?

Nhp d liu vo bng tnh

Book1

A B C D E F G H I J K

1 I 6 8 9 10 6 15 9 7 13 11

2 II 5 5 4 3 9 9 6 13 17 12

Hnh 4.1: Hp thoi z-Test: Two Sample for Means

p dng z-test: Two Sample for Means

a. Nhp ln lt n lnh Tools v lnh Data Analysis.

b. Chn chng trnh z-Test: Two Sample for Means trong hp thoi DataAnalysis ri

nhn nt OK.

Chng 4 12


c. Trong hp thoi z-Test: Two Sample for Means, n ln lt cc chi tit cng ging nh

hai mu c d liu tng ng, song c thm cc chi tit:

- Phng sai ca d liu 1 (Variance 1 Variance),

- Phng sai ca d liu 2 (Variance 2 Variance),

z-Test: Two Sample for Means

I II

Mean 9.4 8.3

Known Variance 1 0.98

Observations 10 10

Hypothesized Mean Difference 0

z 2.472066162

P(Z

Chng 4 13


N/S

D

N/S

Dt

DD

D

Phn phi Student vi = N 1

Bin lun

Nu t < t hay t/2 ( = N 1) Chp nhn gi thuyt H0

4.4 p dng MS-EXCEL

Th d 7: Hm lng (mg) ca mt ch phm c xc nh trc v sau khi c lo ho

cp tc nh sau:

Trc 7,5 6,8 7,1 7,5 7,2 6,8 6,9 6,7 6,8 6,8

Sau 6,1 6,3 6,5 6,4 6,8 6,3 6,1 6,4 6,5 6,3

Hy cho bit hm lng hot cht c gim sau th nghim?



1 Trc 7.5 6.8 7.1 7.5 7.2 6.8 6.9 6.7 6.8 6.8

2 Sau 6.1 6.3 6.5 6.4 6.8 6.3 6.1 6.4 6.5 6.3

4.5 p dng t-Test: Paired Two Sample for Means


b. Chn chng trnh t-Test: Paired Two Sample for Means trong hp thoi Data Analysis

ri nhp nt OK.

c. Trong hp thoi t-Test: Paired Two Sample for Means, ln lt n nh cc chi tit:

- Phm vi ca d liu 1 (Variable 1 Range),

- Phm vi ca d liu 2 (Variable 2 Range),

- Nhn d liu (Labels),

- Ngng tin cy (Alpha),

- Sai bit gia hai gi tr trung bnh c tnh (Hypothesized Mean Difference),

- Phm vi u ra (Output Range).

Chng 4 14


Hnh 4.2: Hp thoi t-Test: Paired Two Sample for Means

t-Test: Paired Two Sample for Means

Trc Sau

Mean 7.01 6.37

Variance 0.089888889 0.042333333

Observations 10 10

Pearson Correlation 0.023415671


df 9

t Stat 5.627619665

P(T t0,05 = 1,833 Bc b gi thuyt H0.

Vy hm lng thuc gim sau th nghim.

Chng 4 15


C- SO SNH GI TR TRUNG BNH VI PHNG SAI BNG NHAU

4.6 Khi nim thng k

Trong trng hp hai mu nh (N < 30) c lp v c phng sai bng nhau*, bn c th

p dng trc nghim t ng phng sai (homoscedastic t-test) so snh gi tr trung bnh ca

hai mu y.

Gi thuyt

Nh trng hp hai mu c d liu tng ng tng cp

Gi tr thng k

21

2

2121

11

)()(

NNS

XX

p

21

2

p

21

N

1

N

1S

)XX(

Phn phi Student

2NN21

2NN

S)1N(S)1N(S

21

1

22

2

112

p

Bin lun

Nu t < t hay t/2 ( = N1 + N2 - 2) Chp nhn gi thuyt H0.

4.7 p dng MS-EXCEL

Th d 8: Ngi ta cho 10 bnh nhn ungthuc h cholesterol ng thi cho bnh nhn khc

ung gi c (placebo) ri xt nghim v nng cholesterol trong mu (g/L) ca c hai nhm:

Thuc 1,10 0,99 1,05 1,01 1,02 1,07 1,10 0,98 1,03 1,12

Gi c 1,25 1,31 1,28 1,20 1,18 1,22 1,22 1,17 1,19 1,21

Theo bng kt qu trn, thuc c tc dng h cholesterol trong mu?



1 Thuc 1,10 0,99 1,05 1,01 1,02 1,07 1,10 0,98 1,03 1,12

2 Gi c 1,25 1,31 1,28 1,20 1,18 1,22 1,22 1,17 1,19 1,21

4.8 p dng t-Test: Two-Sample Assuming Equal Variances


Chng 4 16


b. Chn chng trnh t-Test: Two-Sample Assuming Equa! Variances trong hp thoi Data

Analysis ri nhp nt OK.

c. Trong hp thoi t-Test: Two-Sample Assuming Equa! Variances, n nh ln lt cc

chi tit nh hai mu c d liu tng ng.

t-Test: Two-Sample Assuming Equal Variances

Thuc Gi c

Mean 1.047 1.223

Variance 0.002401111 0.002001111

Observations 10 10

Pooled Variance 0.002201111


df 18

t Stat -8.388352782

P(T

Chng 4 17


D- SO SANH GI TR TRUNG BNH VI PHNG SAI KHC NHAU

4..9 Khi nim thng k

Vi hai mu nh (N < 30) c lp v c phng sai khc nhau (hai mu phn bit), bn c

th p dng trc nghim t d phng sai (betero scedastic test) so snh gi tr trung bnh

ca hai mu y.

Gi thuyt

Tng t nh trng hp hai mu vi phng sai bng nhau.

Gi tr thng k

2

2

2

1

2

1

2121

N

S

N

S

)()XX(t

2

2

2

1

2

1

21

N

S

N

S

)XX(t

Phn phi Student

1N

N/S

1N

N/S

N

S

N

S

2

2

2

2

2

1

2

1

2

1

2

2

2

2

1

2

1

(Smith - Satterthwaite)

Bin lun

Nu t < t hay t/2 ( c tnh) Chp nhn gi thuyt H0.

4.10 p dng MS-EXCEL

Th d 9: Thi gian tan r (pht) ca mt loi vin bao t hai x nghip dc phm

(XNDP) khc nhau c kim nghim nh sau:

XNDP I 61 71 68 73 71 70 69 74

XNDP II 62 69 65 65 70 71 68 73

Thi gian tan r ca vin bao thuc hai XNDP c ging nhau?


A B C D E E G H I J

1 XNDP I 61 71 68 73 71 70 69 74

2 XNDP II 62 69 65 65 70 71 68 73

4.11 p dng t-Test: Two-Sample Assuming Unequal Variances


b. Chn chng trnh t-Test: Two-Sample Assuming Unequal Variances trong hp thoi

Data Analysis ri nhp nt OK.

Chng 4 18


c. Trong hp thoi t-Test: Two-Sample Assuming Unequal Variances, n nh ln lt cc

chi tit nh hai mu c d liu tng ng.

t-Test: Two-Sample Assuming Unequal Variances

XNDP I XNDP II

Mean 69.625 67.875

Variance 15.98214286 13.26785714

Observations 8 8


df 14

t Stat 0.915208631

P(T

Chng 4 19


E- SO SNH T S

4.12 Khi nim thng k

i vi mt th nghim c hai kt qu (binomial experiment) th d, i vi mt thuc

c k n: c hay khng - bn thng so snh hai t s vi nhau (thc nghim vi l thuyt hay

thc nghim vi thc nghim). Song i vi mt th nghim c nhiu kt qu (multinomial

experiment)-th d, bc s nh gi tnh trng ca cc bnh nhn c iu tr bi thuc trong

mt khong thi gian - bn cn so snh nhiu t s. Trc nghim khi bnh phng (X2) cho

php bn so snh khng nhng hai m cn nhiu t s (hay t l hoc xc sut) mt cch tin li.

X2 l phn phi v xc sut, khng c tnh i xng v ch c gi tr 0. Gi s bn c mt cng

trnh nghin cu vi N th nghim c lp, mi th nghim c k kt qu v mi kt qu mang

mt cc xc sut thc nghim l Pi(i = 1, 2, k). Nu gi Pi,0 l cc gi tr l thuyt tng ng

vi Pi th cc tn s l thuyt s l Ei = NPi,0. iu kin p dng trc nghim X2 mt cch

thnh cng l cc tn s l thuyt Ei phi 5.

Gi thuyt

H0 : P1 = P1,0, P2 = P2,0,..., Pk,0 Cc cp Pi v Pi,0 ging nhau.

H1: t nht c mt cp Pi v Pi,0 khc nhau.

Gi tr thng k

k

1i i

2

ii2

E

)EO(;

Oi: cc tn s thc nghim (observed frequency);Ei: cc tn s l thuyt (expected frequency)

Bin lun

Nu 2a

2 Bc b gi thuyt H0 (DF = k 1)

Trong chng trnh MS-EXCEL c hm s CHITEST c th tnh:

- Gi tr 2 theo biu thc:

r

1j

c

1j ij

2

ijij2

E

)EO(

Oij: tn s thc nghim ca thuc hng i v ct j;

Eij: tn s l thuyt ca thuc hng I vi ct j, r: s hng; v c: s ct.

- Xc sut P(X > 2 ) vi bc t do DF = (r 1)(c 1); trong , r l s hng v c l s ct

trong bng ngu nhin (contingency table).

Nu P(X > 2 ) > Chp nhn gi thuyt H0, v ngc li.

4.13 p dng MS-EXCEL

Th d 10: Kt qu iu tr trn hai nhm bnh nhn: rut nhm dng thuc v mt nhm

gi dc c tm tt nh sau:

Chng 4 20


Bin php iu tr S bnh nhn khi bnh S bnh nhn khng khi

Thuc 24 15

Gi dc 20 23

T l khi bnh do thuc (24/39 = 61%) v gi dc (20/43 = 46%) c khc nhau v mt

thng k?


B9 =CHIEST (B3:C4,B7:C8)

A B C D

1 THC NGHIM Khi bnh Khng khi Tng hng

2 iu tr 24 15 39

3 Thuc 20 23 43

4 Gi dc 44 38 82

5 Tng ct

6 L THUYT

7 Thuc 20.92682927 18.07317073

8 Gi dc 23.07317073 19.92682927

9 GI TR P 0.172954847

Sp xp d liu theo bng trc nghim hai mu c lp.

Tnh cc tng s

Tng hng (Row totals): chn D3 v nhp biu thc = SUM(B3:C3).

Dng con tr ko nt t in t D3 n D4.

Tng ct (Column totals): chn B5 v nhp biu thc = SUM(B3:B4).

Dng con tr ko nt t in t B5 n C5.

Tng cng (Grand total): chn D5 v nhp biu thc = SUM(D3:D4)

Tnh cc tn s l thuyt

Tn s l thuyt = (tng hng tng ct)/tng cng

Khi bnh do thuc: chn B7 v nhp biu thc D3*B5/D5

Khng khi bnh do thuc: chn C7 v nhp biu thc = D3*C5/D5

Khi bnh do gi dc: chn B8 v nhp biu thc = D4*B5/D5

Khng khi bnh do gi dc: chn C8 v nhp biu thc = D4*C5/D5

Hnh

4.13 p dng hm s CHITEST

Tnh xc sut P(X> 2 ) bng cch chn B9 v nhp biu thc nh trn hay s dng hp

thoi ca CHITEST.

Chng 4 21


Kt qu: P(X> 2 ) = 0,17 > = 0,05 nhn gi thuyt H0.

Vy t l khi bnh do thuc v do gi dc khng khc nhau.

F- SO SNH PHNG SAI

4.14 Khi nim thng k

Trc nghim so snh hai phng sai thng c p dng so snh chnh xc ca hai

phng php nh lng khc nhau.

Gi thuyt: 2

2

2

11

2

2

2

10

:H

:H

Gi tr thng k: 2

2

2

1

2

2

2

1

2

2

2

1

2

1

2

2

S

S

S

S

S

SF

Phn phi Fischer: 2N;1N2211

Bin lun

Nu F < F(1, 2) Chp nhn gi thuyt H0 vi xc sut (1 - )100%

4.15 p dng MS-EXCEL

Th d 11: Mt mu c phn tch bi hai phng php A v B vi kt qu c tm tt

trong bng sau:

A 6,4 5,2 4,8 5,2 4,3 4,4 5,1 5,8

B 2,6 3,5 3,4 3,2 3,4 2,8 2,9 2,8

Cho bit phng php no chnh xc hn?


A B C D E F G H I

1 6,4 5,2 4,8 5,2 4,3 4,4 5,1 5,8

2 2,6 3,5 3,4 3,2 3,4 2,8 2,9 2,8

p dng F-Test Two-Sample for Variances

Chng 4 22



b. Chn chng trnh F-Test Two-Sample for Variances trong hp thoi Data Analysis ri

nhp nt OK.

c. Trong hp thoi F-Test Two-Sample for Variances, ln lt n nh cc chi tit:

- Ta ca d liu 1 (Variable 1 Range),

- Ta ca d liu 2 (Variable 2 Range),

- Nhn d liu (Labels),

- Ngng tin cy (Alpha),

- Ta u ra (Output Range)

Hnh 4.3: Hp thoi F-Test Two-Sample for Variaces

F-Test Two-Sample for Variances

6.4 2.6

Mean 4.971428571 3.142857143

Variance 0.269047619 0.092857143

Observations 7 7

df 6 6

F 2.897435897

P(F F0,05 = 3,787 Bc b gi thuyt H0.

Vy chnh xc ca phng php B cao hn phng php A.

Chng 4 23


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