Báo cáo bài pttksl (nhóm trang + Mạnh)

8/3/2019 Bo co bi pttksl (nhm trang + Mnh)

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B CNG THNGTRNG I HC CNG NGHIP H NI

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BI TP LN MNPhn tch v thng k s liu

Nhm Sinh vin:

1. Nguyn s Mnh2. Bi Th Trang

Lp:C Tin3Kho:10

Gio vin hng dn: Th.S. Nguyn Phng Nga H ni, thng 12 nm 2010


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Li Ni uTrc kia vic nghin cu, tnh ton..cho mt tp d liu l mt cng vitng i kh khn v mt thi gian. Nhng ngy nay, vi s tr gip cCNTT th cc cng vic n gin hn rt nhiu.c bitSPSS for Windowscung cp mt h thng qun l d liu, v phntch thng k trong mt mi trng ha, s dng cc trnh n m t(menu) v cc hp thoi (dialogue box) n gin thc hin hu ht ccng vic cho bn. Phn ln cc nhim v c th c hon thin bng cr v nhp chut.Trong phn trnh by ny, nhm chng ti xin a ra mt s bi tp, phntch cc tham s m t c trng bng phng php thch hp, v biu histogram, tm ng hi qui, tm khong tin cy i vi dc, tm mkhong tin cy i vi gi tr trung bnh ca bin ph thuc,thc hin tng hi qui v nh gi nng lc d bo ca m hnh,Bng vic sdng phn mm SPSS.Do thi gian nghin cu v s hiu bit cn nhiu hn ch. Chnh v vy bTrnh by khng trnh c nhng sai st, rt mong nhn c s ng gpca c gio v cc bn.V chng em xin cm n c gio Nguyn Phng Nga tn tnh hngdn nhm chng em lm bo bo co ny

Nhm chng em xin chn thnh cm n!


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I. Mt s bi tp s dng phn mn SPSSBi1:Cho mt tp s liu kt qu thc nghim:

33 32 30 31 22 29 32 24 3433 33 25 34 26 29 35 33 34

A. Hy tnh cc i lng c trng ca tp s liu trnB. Phn tch, nh gi tp s liu

Bi Lm:

Sp xp tp s liu theo gi tr tng dn

X1 X2 X3 X4 X5 X6 X7 X8 X922 24 25 26 29 29 30 31 32X10 X11 X12 X13 X14 X15 X16 X17 X1832 33 33 33 33 34 34 34 35

A. Tnh cc i lng c trng ca tp s liu


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Dng kim nh AlalyzeDescriptivesfrequencies1. Bng phn b tn sut

Xi

Frequency Percent Valid PercentCumulative

Percent

Valid 22 1 5.6 5.6 5.6

24 1 5.6 5.6 11.1

25 1 5.6 5.6 16.7

26 1 5.6 5.6 22.2

29 2 11.1 11.1 33.3

30 1 5.6 5.6 38.931 1 5.6 5.6 44.4

32 2 11.1 11.1 55.6

33 4 22.2 22.2 77.8

34 3 16.7 16.7 94.4

35 1 5.6 5.6 100.0

Total 18 100.0 100.0

2. Cc i lng c trng ca tp s liu


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Statistics

Xi

N Valid 18

Missing 0

Mean ( Trung bnh) 30.5000

Std. Error of Mean .91555

Median (Trung v) 32.0000

Mode (S tri) 33.00

Std. Deviation ( lch chun) 3.88436

Variance (Phng sai) 15.088

Range (khong cch) 13.00

Minimum 22.00

Maximum 35.00

Sum 549.00

Percentiles 25 28.2500

50 32.0000

75 33.2500

- M0 = 33

- Med = * [Xn/2 + X(n/2+1)] = 32- X = 30,5- S2 = 15,088. c trng cho s sai bit ca tp kt qu v c trng

cho phn tn ca tp s liu vi gi tr trung bnhPhng sai cng ln th sai bit v phn tn xung quanh cng ln- Sf = S2 = 3,9. c ngha nh phng sai- Cv = 12,7%

B. Phn tch nh gi tp s liu mu- Khong chnh xc tin cy ca tp s liu l: x (p,f) = X- = t(p,f)* Sx

= 1,94- Khong gii hn tin cy ca tp s liu l:

X - t(p,f) * Sf < Xi < X + t(p,f) * Sf


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28,57 < Xi < 32,43 (vi tin cy thng k l 95%)

Bi2:S dng 4 phng php nghin cu A, B, C v D. Kt qu lm lp li theomi phng php 6 ln thu c bng

N Ph2A Ph2B Ph2C Ph2D1 18,00 18,55 17,65 19,102 18,05 17,60 17,70 18,40

3 17,95 18,00 17,90 18,104 18,15 18,30 17,65 18,705 17,95 18,25 17,85 18,806 18,20 17,90 17,75 18,50

A. Tnh gi tr trung bnh, phng sai ca mi phng php v nhn xtB. Bit gi tr tht l 18,1. Phn tch nh gi sai s ca mi phng php

Bi lm:A.

Dng phn mm SPSS


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Statistics

ph2A ph2B ph2C ph2D

N Valid 6 6 6 6

Missing 0 0 0 0

Mean 18.0500 18.1000 17.7500 18.6000

Std. Error of Mean .04282 .13723 .04282 .14142

Std. Deviation .10488 .33615 .10488 .34641

Variance .011 .113 .011 .120

Sum 108.30 108.60 106.50 111.60

Percentiles 25 17.9500 17.8250 17.6500 18.3250

50 18.0250 18.1250 17.7250 18.6000

75 18.1625 18.3625 17.8625 18.8750

Ta cPh2A Ph2B Ph2C Ph2D

X 18,05 18,10 177,75 18,60S2 0.012 0.112 0.018 0.120Theo kt qu phn tch ta thy Nu nghin cu theo c 4 phng php A, B , C v D th s sai bit ca ccs liu trong kt qu thc nghim v phn tn xung quanh gi tr trung bnh l tng i nh. Tuy nhin vi phng php A th nh nht

B. Phn tch nh gi sai s ca mi phng php- Phng php Ata= 0,354 > = 2,57x sai s ngu nhin. Xi phn b u hai phi ca gi tr thc trntrc s. Sai s ny bao gi cng gp phi v ta ch c th gim sai s ngunhin

- Phng php BtB= 0 < t b = 2,57 mc sai s ngu nhin- Phng php Ctc= 6,48 >> t bng(95,5)= 2,57x sai s h thng. Xi Tp trung v mt pha ca gi tr thc trn trcs. Ta c th loiaj b c sai s ny nu tm ra c nguyn nhn


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tD = 3,54 mc sai s h thng

Bi 3:Hai nghin cu A v B thu c kt qu nh sau

A 4,40 4,56 4,42 4,59 4,55 4,45 4,55 4,39B 4,42 4,47 4,70 4,72 4,53 4,55 4,60 4,64A 4,75 4,72 4,53 4,66 4,90 4,50 4,45 4,66B 4,29 4,52 4,57 4,56 4,66 - - -A 4,60 4,36 4,75 4,22B - - - -

A. Phn tch cc i lng c trng ca tp kt qu thc nghim trnB. So snh gi tr trung bnh v gi tr phng sai ca 2 thc nghim A

v B

Bi lm:A.Cc i lng c trng ca tp s liuTheo phn tch SPSS ta c

B.Ta thy F = 2,12 < F b(95,12,19)= 2,54Vy ta chp nhn Ho bc b Ha. Tc l chp nhn S2A = S2B

Theo kt qu phn tch ta cKt qu A:- Khong ca tp s liu:R= 0,68- Gi tr trung bnh: X = 4,56- lch chun Sf = 0,17- Phng sai S2 = 0,03


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Kt qu B:- Khong ca tp s liu:R= 0,43- Gi tr trung bnh: X = 4,54- lch chun Sf = 0,12- Phng sai S2 = 0,14

B. So snh phng saiTa thy F= S2A/S2B= 2,12


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* Khng lin quan vi nhau tng i mtH0 = 2 kt qu nghin cu khng khc nhauHa = 2 kt qu khc nhaud = 33,01-31,89 = 1,12tt = 1,9 < t b(0,95,18)= 2,101Vy chp nhn H0 tc l hai kt qu nghin cu khng khc nhau.* Lin quan vi nhau i mtd = 0,83Sd = 0,667t = 3,2 > tt(0,95,9)= 2,26. Vy hai kt qu nghin cu l khc nhau.

Bi 5:Xc nh nh hng ca nhn t A ln th nghim ngi ta thu ckt qu sau. Hy phn tch phng sai xem A c nh hng ln kt qu thnghim khng?

Mcnghincu

Ln lp

1 2 3a1 1 4 9a2 4 9 16

a3 9 16 23

Ta c: SS1 = 12 + 42 + 92 + 42 + 92 + 162 + 92 + 162 + 232 =1812SS2 = 1/3 (142+ 292 + 482) = 1113,7SS3 = 1/6 (14 + 29 + 48)2 = 1380,2C S2A= 133,25S2TN = 71,97 Ftnh = 1,85F bng(95, 2,6)= 5,14Ta thy Ftnh < F bng. Vy nhn t A khng nh hng ti kt qu thcnghim


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Bi 6:Xc nh nh hng ca nhn t A ln th nghim ngi ta thu ckt qu sau. Hy phn tch phng sai xem A c nh hng ln kt qu thnghim khng?

Mcnghincu

Ln lp

1 2 3 4A 3 6 1 2B 5 7 4 6C 2 3 2 2

Ta c: SS1 = 32 + 62 + 12 + 22 + 52 + 72 + 42 + 62 + 22 + 32 + 22 + 22 =197SS2 = 1/4 (122+ 222 + 92) = 177,25SS3 = 1/9 (12 + 22 + 9)2 = 200,67C S2A= 11,7S2TN = 0,41 Ftnh = 28,54F bng(95, 2,6)= 5,14Ta thy Ftnh >> F bng. Vy nhn t A nh hng mnh ti kt qu thcnghim

Bi 7:Trung tm nghin cu la go quc t ti Philippines mun xem xts lin quan gia nng xut la m Y, vi mt gieo trng X. H thc hicc th nghim v a ra kt qu bng. Tm ng hi quy tuyn tnh giX v Y (quan h hm gia chng)


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Nng sut la m Mt trng4,862 1605,244 175

5,128 1925,052 1955,298 2385,410 2405,234 2525,608 282

Bi lm:Phn tch SPSS s dng cng c AnalyzeRegressionCurve EstimationTa c h s tng quan r = b =728.0 = 0.853Ta chn X l bin ph thucY l bin c lp

Statistics

X Y

N Valid 8 8

Missing 0 0

Mean 216.75 5.22950

a = X - bY = 216,75 0,853 *5,23 = 212,29Vy phng trnh ng hi quy s l X = 212,29 + 0,853*Y


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Model Summary and Parameter Estimates

Dependent Variable:X

Equation

Model Summary Parameter Estimates

R Square F df1 df2 Sig. Constant b1

Linear .728 16.036 1 6 .007 -618.657 159.749

The independent variable is Y.

Nhn th ta thy cc im khng nm trn ng hi quy. iu ny cngha l kt qu nghin cu gia nng xut la m y vi mt gieo trng l khng lin quan ti nhau.


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Bai 8: Xt quan h thn nhit ( nhit x) v nhp p ca tim(mchp y) H thc hin cc th nghim v a ra kt qu bng. Tm ng hquy tuyn tnh gia X v Y (quan h hm gia chng)

Tr em Nhit (x) Mach p(y)1 68 22 65 53 70 14 62 105 60 96 55 137 58 108 65 39 69 410 63 6

Bi lm:Phn tch SPSS s dng cng c AnalyzeRegressionCurve EstimationTa c h s tng quan r = b = 886,0 = 0,94Ta chn x l bin c lpY l bin ph thuc


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Statistics

nhietdo machdap

N Valid 10 10

Missing 0 0

Mean 63.50 6.30

Ta ca =Y - b X = 63,5 0,886*6.3 = 57,9Vy phng trnh ng hi quy s lY = 57,9 X + 0,886

Nhn th ta thy c mt im nm trn ng hi quy. iu ny c nghal kt qu nghin cu gia nhit v mch p gn nh l khng tngquan ti nhau.


Dependent Variable:nhietdo

Equatio

n


R Square F df1 df2 Sig. Constant b1Linear .886 62.265 1 8 .000 70.736 -1.149

The independent variable is machdap.


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II. Chn 2 tp d liu (Thuc 1 ct no ca file cho), 1tp c n30. Vi mi tp hy phn tch cctham s m t c trng bng phng php thch hp v vbiu histogram ca n.T kt qu, c nhn xt g v mi tp d liu?

Bi lm:Ta tin hnh chn tp A v B thuc ct B ca file cho


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1 2 3 4 5 6 7 8A 5 9 10 7 7 7 6 6B 7 9 8 7 5 9 19 11

9 10 11 12 13 14 15 16

A 9 7 5 10 11 7 5 7B 11 4 7 5 5 9 11 1017 18 19 20 21 22 23 24

A 4 17 5 9 6 8 5B 11 9 6 7 5 4 4 5

25 26 27 28 29 30 31 32AB 7 6 8 5 12 7 9 7

Ta c bng phn b tn sut ca tp d liu A

A


Percent

Valid 4 1 3.2 4.3 4.3

5 4 12.9 17.4 21.7

6 3 9.7 13.0 34.8

7 6 19.4 26.1 60.98 1 3.2 4.3 65.2

9 3 9.7 13.0 78.3

10 2 6.5 8.7 87.0

11 1 3.2 4.3 91.3

15 1 3.2 4.3 95.7

17 1 3.2 4.3 100.0

Total 23 74.2 100.0

Missing System 8 25.8

Total 31 100.0

Bng phn b tn sut ca tp d liu B


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B


Percent

Valid 4 3 9.7 9.7 9.7

5 6 19.4 19.4 29.0

6 2 6.5 6.5 35.5

7 6 19.4 19.4 54.8

8 2 6.5 6.5 61.3

9 5 16.1 16.1 77.4

10 1 3.2 3.2 80.6

11 4 12.9 12.9 93.5

12 1 3.2 3.2 96.8

19 1 3.2 3.2 100.0

Total 31 100.0 100.0

Ta c cc tham s c trng ca 2 tp s liu


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A B

N Valid 23 31

Missing 8 0

Mean 7.91 7.81

Std. Error of Mean .659 .565

Median 7.00 7.00

Mode 7 5a

Std. Deviation 3.161 3.146

Variance 9.992 9.895

Range 13 15

Sum 182 242

Percentiles 25 6.00 5.00

50 7.00 7.00

75 9.00 9.00

a. Multiple modes exist. The smallest value is shown

T kt qu phn tch ta thyFtnh= S2A/ S2B = 1 < F bng. Vy chp nhn H0 bc b HaS2A = S2B ng tin cy


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Biu Histogram ca tp A


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Biu histogram ca tp B

III. Chn 1 ct no trong s hn 300 thuc tnh m t v 1ct thuc tnh ph thuc, hy tm ng hi qui v phn tch


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cht lng ng hi qui qua vic tnh cc h s xc nh,h s tng quan v phn tch qua vic kim nh gi thuyti vi dc B ca m hnh. T kt qu nhn c, hy chobit m hnh c dng d bo c khng? ti sao?

Hy tm khong tin cy 95% i vi dc B. ngha cakhong ny l g?Hy tm mt khong tin cy 95% i vi gi tr trung bnh cabin ph thuc khi cho bin c lp x=1

Bi lm:Ta chn ct thuc tnh m t S (X)v ct thuc tnh ph thuc T(Y)

Statistics

X Y

N Valid 816 816

Missing 1 1

Mean 1.53 5.90

Std. Deviation 1.282 3.821

Variance 1.643 14.598

Sum 1248 4817


Dependent Variable:X

Equation



Linear .134 126.473 1 814 .000 .803 .123The independent variable is Y.


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- H s xc nh: r 2 = 0,134- Ta c h s tng quan r = b =134,0 = 0,37a =Y b X = 5,90 0,37 * 1,53 = 5,33Vy ng hi quy tuyn tnh s l Y = 5,33 + 0,37X

- Ta tin hnh kim tra nng lc ca m hnh bng kim tra gi thuytH0: B = 0Ha: B 0n = 816 v = 0,05 (chn tin cy l 95%)

df = 816-2 = 814t/2 = 160, s = S 2= 1,28

SSxx = )( X xi 2= Xi 2 ( Xi )2/N = - (1248)2/816


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T ta tnh c ttnh = SSxxS b

/=

Ta so snh t vi t/2= 160 Nu ttnh< t/2 ta bc b H0 v kt lun dc B 0Chng t rng mc = 0,05 cc d liu mu cung cp bng chng kt lun rng Y ng gp thng tin hu ch cho d on ca mc X bncch s dng m hnh tuyn tnh. Trong trng hp ny m hnh c thdng d bo cCn ngc li nu ttnh> t/2

- Khong tin cy 95% i vi dc B l b t/2 [

SSxx

S ] = 0,37 160*

Nu khong trn ln hn 0 v nh hn 1 th X (y),E(y) tng khi X tng(thun) Nu khong trn ln hn -1 v nh hn 0 th X (y),E(y) tng khi Xgim(nghch)- Khi X= 1 khong tin cy 95% i vi gi tr trung bnh ca bin ph thukhi cho bin c lp x l Y ^ t/2 SSxx

X

n

2)1(1 +

Trong Y^ = 5,33 + 0,37*1 = 5,71 - X = 0,53

IV. Ly 5 ct bin c lp v 1 ct bin ph thuc, thc hintm ng hi qui v nh gi nng lc d bo ca m hnh.Hy tm khong tin cy 95% i vi dc Bi. ngha cakhong ny l g?

Bi lm:


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Ta chn tp d liu sauc lp1 c lp2 c lp3 c lp4 c lp5 Phthuc4 2 2 1 3 05 1 4 1 4 4

5 0 3 2 1 03 4 6 6 0 11 2 5 3 0 23 2 4 5 2 6

A. Ta thy h s tng quanr = b = 019,0 = 0,138- Ta c m hnh tuyn tnh gi thuyt hnh thc: Y = B0 + B1X1 + B2X2 +B3X3 + B4X4 + B5X5 +e* S dng cc d liu mu tm phng trnh tnh ton da trn bn phng cc tiu.-S dng cng thc bnh phng cc tiu tm B0, B1, B2,B3,B4,B5Ta c

B = (X`X)XY =

108

98

12

23

67

54

Vy B0 = 54, B1 =67, B2 = 23, B3 = 12, B4 = 98, B5 = 108Ta c phng trnh tnh ton da trn bnh phng cc tiuY^ = 54 + 67X1 + 23X2 +12X3 + 98X4 + 108X5

* Ta kim nh nng lc ca m hnh xem c ph hp vi d liu khngTc l ta tnh h s xc nh R 2 = 0,019


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Dependent Variable:doclap1

Equation



Linear .019 .077 1 4 .795 3.688 -.087

The independent variable is phuthuoc.

* Xt tnh hu dng ca m hnhTa cH0: B1=B2= B3= B4= B5=0Ha: c t nht mt Bi 0

F = ))1(/()21(/2

+ k n R

k R

= )]1(/[/(mod)

+ k nSSE

k SS

= 0,077F bng = 0 (v nm di ct P)C ngha l chp nhn Ha bc b H0. M hnh ny rt hu dng

B.Ta c sai tnh cho Bi l : bi t/2, t/2= 3,182(df = 3)H0 :Bi = 0

Ha: Bi 0Kim nh thng k: ta thy t/2= 3,182

Model Summary

R R SquareAdjusted R

SquareStd. Error of the

Estimate

.066 .004 -.327 1.928

The independent variable is phuthuoc.

Ta c t = 928,1bi

= 928,1138,0 = 0,072

Kim nh thng k: ta thy t/2= 3,182 > t = 0,072


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Vy bc b H0.iu c ngha l tp thuc tnh m t c ng gp thngtin cho d bo ca tp thuc tnh ph thuc


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