34 Bài tập xử lý số liệu trên phần mềm R

BI TP TIN HC NG DNG

BI TP TIN HC NG DNG

B MN: THC HNH X L S LIU THC NGHIM TRONG PHNG TH NGHIM

BI TP TIN HC NG DNG

GVHD: PHM MINH TUN

NHM: 4

LP: HTP8B

DANH SCH NHM

STT

H V TN

MSSV

1

Nguyn Th Vn

12064671

2

ng Th Nhung

12089271

3

L Th Hng Lin

12054971

4

Hong Hu Trung

12128641

5

Trng Cng V

12001685

BI 1:

p-value = 0.8701>0,05 cho thy khng c s khc bit v hiu sut trch ly ca 2 loi dung mi mc ngha =5%.

C th chn c 2 loi dung mi ny (diethyl eter, cn) trch ly polyphenol. Tuy nhin theo kin c nhn, ta nn chn cn v gi thnh r v khng gy nh hng nhiu ti sc khe.

PH LC:

>hieusuatdungmoi data1plot(data1)

> shapiro.test(hieusuat)

# p-value = 0.7311 >0.05 nn hiu sut trch ly ca 2 dung mi tun theo lut phn phi chun.

>var.test(hieusuat~dungmoi)

# p-value =0.06226 >0.05 nn khng c s khc bit v phng sai ca hai loi dung mi.

> t.test(hieusuat~dungmoi,var.equal=T)

# p-value = 0.8701

BI 2:

p-value = 1.319e-07 < 0.05 nn thi gian bn bt hki s dng CMC c ngha thng k mc 0.05. Khi s dng ph gia, thi gian bn bt ca t 1.22h cao hn thi gian bn bt ca mu i chng 0.05h. Nn ph gia CMC c tc dng ko di thi gian bn bt. V vy c th s dng loi ph gia ny trong ch bin

PH LC:

>dc=c(1.10,0.99,1.05,1.01,1.02,1.07,1.10,0.98,1.03,

1.12)

>CMC=c(1.25,1.31,1.28,1.20,1.18,1.22,1.22,1.17,1.19,

1.21)

> t.test(dc,CMC,paired=TRUE)

Paired t-test

data: dc and CMC

t = -8.7467, df = 9, p-value = 1.078e-05

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.2215188 -0.1304812

sample estimates:

mean of the differences

-0.176

BI: 3

Loi Enzyme

Lng acid amin tng s

A

18.17c +- 1.17

B

14.80 a +- 0.84

C

19.00 c +- 0.82

D

16.25 ab +- 1.17

Ch thch: cc k t a,b,c trn cng mt ct ch s khc bit lng acid amin tng s trong nc mm khi s dng cc loi enzyme thy phn khc nhau.

BIU : kh nng thy phn ca tng loi enzyme(Trong 1=A;2=B; 3=C;4=D)

Qua kt qu p-value=7.463e-05acidamin=c(17,18,17,20,19,18,14,15,16,15,14,19,20,18, 19,16,15,16,18)

>group=c(rep("A",6),rep("B",5),rep("C",4),rep("D",4))

> group=as.factor(group)

> analysis=lm(acidamin~group)

> anova(analysis)

Analysis of Variance Table

Response: acidamin

Df Sum Sq Mean Sq F value Pr(>F)

group 3 50.564 16.8547 15.431 7.463e-05 ***

Residuals 15 16.383 1.0922

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

> av=aov(acidamin~group)

> TukeyHSD(av)

diff lwr upr p adj

B-A -3.3666667 -5.1905943 -1.5427391 0.0004459

C-A 0.8333333 -1.1109800 2.7776467 0.6151972

D-A -1.9166667 -3.8609800 0.0276467 0.0539882

C-B 4.2000000 2.1794103 6.2205897 0.0001307

D-B 1.4500000 -0.5705897 3.4705897 0.2079384

D-C -2.7500000 -4.8798886 -0.6201114 0.0098498

BI 4:

Nng ph gia(%)

ng knh ca bnh

0.5

71.14 a +_ 6.89

0.3

63.71 ab +_ 6.63

0.1

61.14 b +_ 7.15

Ch thch: cc k t a,b,c trn cng mt ct ch s khc bit kh nng trng n ca bnh ti cc nng ph gia s dng khc nhau

Biu th hin s ph thuc gia kh nng trng n v nng ph gia

Theo kt qu ta thy gi tr p-value=0.037phugia=c(68,80,69,76,68,77,60,71,62,58,74,65,59, 57,58,60,70,51,57,71,61)

>group=gl(3,7)

>groupdataanalysisanova(analysis)

# p- value =0.037resTukeyHSD(res)

2-1 -7.428571 -16.83138 1.9742350 0.1369187

3-1 -10.000000 -19.40281 -0.5971936 0.0361071

3-2 -2.571429 -11.97423 6.8313778 0.7677005

BI 5:

Da vo php kim nh t.test gia mu th nghim v mu i chng chng ta c p-value = 4.327e-05 dc=c(3.45,3.58,3.59,3.62,3.59,3.57,3.35,3.74,3.29,3.48,3.45,3.58,3.59,3.62,3.59,3.57,3.35,3.74,3.29,3.48,3.45,3.58,3.59,3.62,3.59,3.57,3.35,3.74,3.29,3.48,3.45,3.58,3.59,3.62,3.59)

>tn=c(3.57,3.57,3.59,3.58,3.67,3.69,3.74,3.58,3.74,3.75,3.61,3.78,3.67,3.69,3.35,3.58,3.68,3.59,3.58,3.58,3.68,3.59,3.58,3.74,3.75,3.61,3.78,3.67,3.69,3.74,3.58,3.68)

>t.test(dc,tn)

# p-value = 4.327e-05

BI 6:

> khongthich=c(15,33)

> thich=c(132,145)

> t.test(khongthich,thich,paired=T)

> # p-value = 0.0139 (data6=matrix(c(15,132,33,145),2,2,dimnames=list(c("Khongthich","Thich"),c("Huongchanhday","Huongvani")))

> barplot(data6,col="green")

KT LUN:

Mc a thch i vi hng chanh dy: 132/(15+132)*100 = 87,79%

Mc a thch i vi hng vani:

145/(33+145)*100 = 81.46%

Suy ra chanh dy c mc yu thch cao hn nn s dng hng chanh dy.

Biu th hin mc yu thch gia hng vani v hng chanh dy.

Biu th hin mc yu thch gia hng chanh dy v hng vani qua s lng ngi

BI 7:

VNG

Hm lng saponin (%)

I

7.19c +_ 0.43

II

4.22 a +_ 2.88

III

5.48 b +_ 2.42

Ch thch: cc k t a,b,c trn cng mt ct ch s khc bit hm lng saponin khi ly nhn sm cc vng khc nhau

Qua gi tr p-value=0.584> 0.05 cho thy c s khc bit v hm lng saponin thu c gia 3 vng mc ngha 0.05. thu c hm lng saponin cao nht v hiu qu nht chng ti ngh ly nhn sm vng I

Biu th hin hm lng trung bnh ca nhn sm ty vo tng vng trng

PH LC:

>saponin=c(7.53,6.87,7.12,7.53,6.84,6.67,7.81,5.87,

5.64,6.14,6.07,5.79,6.13,6.35,6.5,6.49,6.55,6.63)

>group data3shapiro.test(saponin)

# p-value = 0.584> 0.05 nn bng s liu tun theo phn phi chun.

>analysisanova(analysis)

# p- value resTukeyHSD(res)

diff lwr upr p adj

2-1 -1.2937143 -1.76514063 -0.8222879 0.0000098

3-1 -0.7540476 -1.20197156 -0.3061237 0.0014894

3-2 0.5396667 0.05214608 1.0271873 0.0293357

BI 8:

Nu ti l nhn vin R & D ti s chn cc tnh cht sau

trong sn phm 1

mu sn phm 1

Hng sn phm 1 hoc

Hng sn phm 2

V mn sn phm 2

PH LC:

> dotrong1 = c(7,6,7,8,9,7,8,9,7,8)

> dotrong2= c(6,5,5,6,4,5,6,7,5,6)

>t.test(dotrong1,dotrong2)

# p-value = 6.895e-05 trung bnh ca dotrong2 =5.5. Nn ti chn dotrong1( trong ca sn phm 1)

> domau1 domau2 t.test(domau1,domau2)

# p-value = 2.074e-05 trung bnh ca domau2= 5.4 nn ti chn domau1( mu ca sn phm 1).

> huong1 huong2 t.test(huong1,huong2)

#p-value = 0.445 >0.05 khng c s khc bit c ngha thng k mc ngha anpha 5% gia huong1 v huong2 .Vy chn huong1 hoc huong2 u c.

> viman1 viman2 t.test(viman1,viman2)

#p-value = 0.004765 0.71 l t l ngi yu thch SP B,nn ti s chn SP A s dng.

BI 10:

Qua gi tr p value=0.1066>0.05 ta thy khng c s khc bit gia sn phm ang bn v sn phm ci tin. Cng ty khng nn tung sn phm ci tin ra th trng.PH LC:

dbF)=6.491e-06 < 0.05 ta thy c s khc bit hiu sut trch ly khi thi gian trch ly khc nhau. Hiu sut trch ly thu c khng c s khc bit khi thi gian trch ly l 100, 115 pht nhng li c s khc bit khi thi gian l 55, 70, 85 pht. hiu sut trch ly hiu qu chng ti ngh s dng enzyme vi thi gian trch ly l 100 pht

PH LC :

>group =c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5)

>y=c(16.77,18.56,17.83,21.52,20.42,21.27,22.16,24.73,23.01,24.92,24.27,23.96,24.73,24.41,25.82)

>groupanalysisanova(analysis)

# p-value= 6.491e-06< 0.05.C s khc bit gia cc nhm thi gian mc ngha anpha=5%.

>resTukeyHSD(res)

diff lwr upr p adj

2-1 3.3500000 1.05617893 5.643821 0.0050152

3-1 5.5800000 3.28617893 7.873821 0.0000889

4-1 6.6633333 4.36951226 8.957154 0.0000185

5-1 7.2666667 4.97284560 9.560488 0.0000084

3-2 2.2300000 -0.06382107 4.523821 0.0576303

4-2 3.3133333 1.01951226 5.607154 0.0054137

5-2 3.9166667 1.62284560 6.210488 0.0016040

4-3 1.0833333 -1.21048774 3.377154 0.5541443

5-3 1.6866667 -0.60715440 3.980488 0.1868802

5-4 0.6033333 -1.69048774 2.897154 0.9030810

Biu th hin hiu sut trch ly ph thuc vo thigian trch ly.

BI 12:

GING LA

Nng sut

1

7,00 c +- 1,00

2

8,60 c +- 1,14

3

4,60 a +- 1,14

4

4,80 ab +- 0,84

Ch thch: cc k t a,b,c trn cng 1 ct ch s khc bit nng sut la khi trng cc ging la khc nhau

KT LUN: Nng sut ca cc ging khc nhau khng phi do ngu nhin m do phm cht ging( do Pr(>F)= 3.229e-05).nng sut la khng c s khc bit ging 1, 2 nhng li c s khc bit ging 3,4. t c hiu qu tt nht, chng ti ngh chn ging 2 ph bin rng ri trong sn xut

BIU : biu hin nng sut trung bnh ca cc ging la.

PH LC:

> thi.nghiem=gl(4,5)

>giong.lua=c(8,7,6,6,8,9,10,7,9,8,5,5,4,3,6,5,4,5,4,6)

> bai12=data.frame(thi.nghiem,giong.lua)

> group=as.factor(thi.nghiem)

> analysis=lm(giong.lua~group)

> anova(analysis)


group 3 54.55 18.183 16.915 3.229e-05 ***

> av=aov(giong.lua~group)

> TukeyHSD(av)

diff lwr upr p adj

2-1 1.6 -0.2760962 3.4760962 0.1091981

3-1 -2.4 -4.2760962 -0.5239038 0.0102311

4-1 -2.2 -4.0760962 -0.3239038 0.0189329

3-2 -4.0 -5.8760962 -2.1239038 0.0000822

4-2 -3.8 -5.6760962 -1.9239038 0.0001458

4-3 0.2 -1.6760962 2.0760962 0.9897700

BI 13:

Nng enzyme

Hm lng vitamin C

0

49.83 a +_ 2.03

0.05

78.12a +_ 1.51

0.1

89.03b +_ 1.33

0.15

91.74b,c +_ 0.90

o.2

94.32c,d +_ 0.72

0.25

96.75d +_ 1.38

0.3

97.57d +_ 1.16

Ch thch:cc k t a,b,c trn cng 1 ct ch s khc bit hm lng vitamin C((mg/g) tnh theo hm lng cht kh) thu c khi s dng cc nng enzyme(%)khc nhau.

Qua kt qu p-value=2.389*10-15vtm=c(49.02,48.33,52.14,76.43,78.64,79.31,88.65,90.52,87.94,90.75,92.51,91.96,93.56,94.42,94.99,95.23,97.12,97.91,96.41,97.56,98.73)

>group= c(gl(7,3))

>groupdataanalysisanova(analysis)

#P-value= 2.389e-15 C s khc nhau mc ngha 5% gia cc nhm

>resTukeyHSD(res)

$group

diff lwr upr p adj

2-1 28.2966667 24.5355155 32.057818 0.0000000

3-1 39.2066667 35.4455155 42.967818 0.0000000

4-1 41.9100000 38.1488488 45.671151 0.0000000

5-1 44.4933333 40.7321822 48.254484 0.0000000

6-1 46.9233333 43.1621822 50.684484 0.0000000

7-1 47.7366667 43.9755155 51.497818 0.0000000

3-2 10.9100000 7.1488488 14.671151 0.0000018

4-2 13.6133333 9.8521822 17.374484 0.0000001

5-2 16.1966667 12.4355155 19.957818 0.0000000

6-2 18.6266667 14.8655155 22.387818 0.0000000

7-2 19.4400000 15.6788488 23.201151 0.0000000

4-3 2.7033333 -1.0578178 6.464484 0.2468338

5-3 5.2866667 1.5255155 9.047818 0.0040668

6-3 7.7166667 3.9555155 11.477818 0.0000992

7-3 8.5300000 4.7688488 12.291151 0.0000325

5-4 2.5833333 -1.1778178 6.344484 0.2892414

6-4 5.0133333 1.2521822 8.774484 0.0063632

7-4 5.8266667 2.0655155 9.587818 0.0017029

6-5 2.4300000 -1.3311512 6.191151 0.3506392

7-5 3.2433333 -0.5178178 7.004484 0.1131016

7-6 0.8133333 -2.9478178 4.574484 0.9873690

2. V th

BIU NH NNG ENZYME N HM LNG VITAMIN C

BI 14:

>ngi=c(245,145,367,170,270,48)

> A=matrix( ng, nrow=2,ncol=3, byrow=T, dimnames = list(c("tang 1.5 - 3","tang 0.5-1"), c("td1", "td2", "td3")))

> chisq.test(A)

Pearson's Chi-squared test

data: A

X-squared = 249.9598, df = 2, p-value < 2.2e-16

Su tang can co lien quan den thuc don.

KT LUN : do gia tri p-value = 2.2e-16 < 0.05 nn c s khac nhau gia cc thc n 1,2,3 v s khc nhau ny co y nghia thng k, mc ngha 5%. Nhn thy thc n 3 c s tc ng mnh n cn nng nht(367 ngi tng 1.5- 3Kg/ thng trn tng s 415 ngi)nn ti chon thc n 3 giup bnh nhn lay lai cn nng ban u

BI 15:

>bai15 chisq.test(bai15)

X-squared = 97.153, df = 6, p-value < 2.2e-16

> # P-value < 2.2e-16 nn c s khc bit v cu trc sn phm khi s dng cc loi ph gia khc nhau c mc ngha anpha = 5%.

> barplot(bai15)

BU TH HIN KH NNG CI THIN CU TRC YAGOURT CA CC LOI PH GIA( LOI A,B,C,D).

BI 16:

Bng: nh hng ca nng enzyme ln hiu sut thu hi dch trit

Nng enzyme(%)

Hiu sut(%)

0

21.80a 1.67

0.05

25.60b 1.42

0.1

27.90bc 0.45

0.15

29.40c 1.07

0.2

30.90c 1.35

0.25

30.77c 0.83

0.3

30.87c 1.12

Ghi ch: k t a,b,c trn cng 1 ct ch s khc bit v hiu sut thu hi dch chit khi s dng cc nng enzyme khc nhau

Biu : nh hng ca nng enzyme ln hiu sut trch ly

Gi tr P-value = 1.062.10-6< cho thy hm lng phenol cc nng enzyme c s khc bit theo thng k vi . Hm lng polyphenol t cao nht(30.87% theo nng kh)khi s dng enzyme nng 0.3% v hm lng ny theo thng k khng c s khc bit khi s dng enzyme cc nng 0.2%, 0.25%, 0.15% . Tuy nhin li c s khc bit so vi khi s dng cc nng 0.1%, 0.05%. V vy trch ly polyphenol hiu qu chng ti ngh chn nng enzyme l 0.15%.

Ph lc:

> a a b lien lien

> attach(lien)

> analysis analysis

> anova(analysis)

Analysis of Variance Table

Response: b


a 6 211.011 35.169 24.833 1.062e-06 ***

Residuals 14 19.827 1.416

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

> res TukeyHSD(res)

diff lwr upr p adj

0.05-0 3.80000000 0.4821762 7.117824 0.0204195

0.1-0 6.06666667 2.7488429 9.384490 0.0003359

0.15-0 7.63333333 4.3155096 10.951157 0.0000276

0.2-0 9.10000000 5.7821762 12.417824 0.0000035

0.25-0 8.96666667 5.6488429 12.284490 0.0000042

0.3-0 9.06666667 5.7488429 12.384490 0.0000037

0.1-0.05 2.26666667 -1.0511571 5.584490 0.2944358

0.15-0.05 3.83333333 0.5155096 7.151157 0.0191830

0.2-0.05 5.30000000 1.9821762 8.617824 0.0012773

0.25-0.05 5.16666667 1.8488429 8.484490 0.0016225

0.3-0.05 5.26666667 1.9488429 8.584490 0.0013557

0.15-0.1 1.56666667 -1.7511571 4.884490 0.6776210

0.2-0.1 3.03333333 -0.2844904 6.351157 0.0835979

0.25-0.1 2.90000000 -0.4178238 6.217824 0.1057066

0.3-0.1 3.00000000 -0.3178238 6.317824 0.0886875

0.2-0.15 1.46666667 -1.8511571 4.784490 0.7355928

0.25-0.15 1.33333333 -1.9844904 4.651157 0.8072163

0.3-0.15 1.43333333 -1.8844904 4.751157 0.7542256

0.25-0.2 -0.13333333 -3.4511571 3.184490 0.9999992

0.3-0.2 -0.03333333 -3.3511571 3.284490 1.0000000

0.3-0.25 0.10000000 -3.2178238 3.417824 0.9999999

BI 17:

Qua php kim nh chiq.test ta thy p-value < 2.2e-16(bai17=matrix(c(170,40,180,50,150,60),nrow=2,ncol=3,

byrow=TRUE

>chisq.test(bai17)

# p-value < 2.2e-16

BI 18:

Mi quan h gi nng benzene v mt quang c biu th bngphng trnh

y= 1.7531x 0.1394. Nh vy khi nng benzene tng th mt quang thu c cng tng.

PH LC:

> am= c(0.2,0.5,1.0,1.5,2.0,2.5,3.0)

>cn= c(0.2,0.5,1.0,1.5,2.0,2.5,3.0)

> AM= c(0.2,0.37,0.64,0.93,1.22,1.50,1.80)

>thu= data.frame(cn,AM)

>lm(cn~AM)

lm(formula = cn ~ AM)

Coefficients:

(Intercept) AM

-0.1394 1.7531

2. Vth

th biu din mi quan h gia nng benzene v mt quang

BI 19:

CNG SUT SIU M

HM LNG VITAMIN C

0

50.53a +_ 0.85

150

96.57 b +_ 0.83

188

100.5 c +_ 0.82

225

103.3 d +_ 0.80

263

100.73c +_ 1.30

300

98.43b, c +_ 0.75

Ch thch: a,b,c l cc ch ci khc nhau th hin s khc bit c ngha khong tin cy 95% ca phng php kim nh anova mt yu t. GTTB= gi tr trung bnh,SD= lch chun

Nhn xt:

C s khc nhau v hm lng vitamin khi s dng cng sut siu m khc nhau mc ngha 5%.

Ti chn cng sut 225 v cng sut ny c s khc bit nht v c gi tr trung bnh hm lng vitamin C ln nht.

PH LC:

>vtmc=c(50.5,49.7,51.4,96.3,95.9,97.5,100.7,101.2,99.6,103.6,102.4,103.9,100.6,102.1,99.5,97.7,98.4,99.2)

>group= gl(6,3)

>group= as.factor(group)

>data= data.frame(group,vtmc)

> analysis=lm(vtmc~group)

>analysisanova(analysis)

# Pr(>F)= 2.57e-16

>res= aov(vtmc~group)

>TukeyHSD(res)

diff lwr upr p adj

2-1 46.0333333 43.5339177 48.53274899 0.0000000

3-1 49.9666667 47.4672510 52.46608233 0.0000000

4-1 52.7666667 50.2672510 55.26608233 0.0000000

5-1 50.2000000 47.7005843 52.69941566 0.0000000

6-1 47.9000000 45.4005843 50.39941566 0.0000000

3-2 3.9333333 1.4339177 6.43274899 0.0020483

4-2 6.7333333 4.2339177 9.23274899 0.0000122

5-2 4.1666667 1.6672510 6.66608233 0.0012540

6-2 1.8666667 -0.6327490 4.36608233 0.1959441

4-3 2.8000000 0.3005843 5.29941566 0.0253118

5-3 0.2333333 -2.2660823 2.73274899 0.9994785

6-3 -2.0666667 -4.5660823 0.43274899 0.1294844

5-4 -2.5666667 -5.0660823 -0.06725101 0.0429643

6-4 -4.8666667 -7.3660823 -2.36725101 0.0003088

6-5 -2.3000000 -4.7994157 0.19941566 0.0780319

2. V th

BIU NH HNG CNG SUT SIU M N HM LNG VITAMIN C

Bi 20:

Phng php

Hiu sut thu hi dch chit

B sung enzyme

80.11 b +- 1.67

Sng siu m

71.42 a +- 2.36

Kt hp

87.11 c +- 1.27

Ch thch: cc k t a,b,c trn cng 1 ct ch s khc bit hiu sut thu hi dch chit khi s dng cc phng php khc nhau.

Qua php kim nh anova ta thy p-value= 0.0001333 < 0.05 nn c s khc bit v ngh thng k gia cc nhm mc ngha 5%. Nhng t hu qu cao nht v hiu sut thu hi dch chit cao nht chng ti nghi s dng phng php kt hp sng siu m v s dng enzyme.

Biu th hin hiu sut thu hi enzyme qua cc phng php(ch thch:1-b sung enzyme;2-Sng siu m;3-Kt hp Enzyme v sng siu m)

PH LC:

>group=c(1,1,1,2,2,2,3,3,3)

>y=c(80.67,78.23,81.42,69.13,73.85,71.27,88.14,87.51,85.69)

>groupanalysisF)= 0.00013

>resTukeyHSD(res)

diff lwr upr p adj

2-1 -8.690000 -13.260802 -4.119198 0.0027037

3-1 7.006667 2.435865 11.577469 0.0079069

3-2 15.696667 11.125865 20.267469 0.0001059

BI 21:

> x nongdo doquang data21 plot(doquang~nongdo,pch=16,xlab="Nngnitrat (ml)",ylab="Mtoquang (Am)",sub="Miquanhgianngnitratvmtoquang")

> reg abline(reg)

> summary(reg)

> legend(5,0.5,c("y=0.196x\nRsquare=1"))

Tnhtonktqu:

- Mtoquangtrungbnhcamunc: (Am)

- Nngnitratcmutrongnc:

(ppm)

BI 22:

gi tr p-value = 0.02886 mdat=matrix(c(38, 92,462,708),nrow=2,ncol=2,byrow= TRUE)

>chisq.test(mdat)

X-squared = 4.7757, df = 1, p-value = 0.02886

Bi 23:

Loi tr

Mc yu thch

Tr rau m

6.29ab +- 0.24

Tr da

6.70c +- 0.21

Tr kh qua

6.40a +- 0.26

Ch thch: cc k t a,b,c trn cng 1 ct ch s khc bit mc yu thch ca ngi th i vi cc loi tr khc nhau.

Biu th hin mc yu thch i vi cc loi tr

Qua gi tr p-value=0.00212group=c(rep(1,10),rep(2,10),rep(3,10))

>y=c(6.2,6.2,6.5,6.2,6.4,6.5,6.5,6.2,5.8,6.2,6.6,6.3,6.7,6.6,6.7,7.0,6.5,6.9,6.4,6.9,6.4,6.4,6.5,6.0,6.6, 6.4,6.5,6.1,6.3,6.8)

>groupanalysis anova(analysis)


group 2 0.78867 0.39433 7.8 0.00212 **

>resTukeyHSD(res)

diff lwr upr p adj

2-1 0.39 0.1406846 0.63931542 0.0017111

3-1 0.13 -0.1193154 0.37931542 0.4113043

3-2 -0.26 -0.5093154 -0.01068458 0.0396760

BI 24:

> Am=c(0,0.375,0.73,1.127,1.43)

> nongdo=c(0,1,2,3,4)

> bai23=data.frame(nongdo,Am)

> plot(Am~nongdo,pch=16)

> abline(lm(Am~nongdo))

> lm(Am~nongdo)

lm(formula = Am ~ nongdo)

Coefficients:

(Intercept) nongdo

0.0100 0.3612

> # Phng trnh ng chun y = 0.3612x + 0.01

> Cx=(((0.875+0.89)/2)- 0.01)/0.3612

> y=hamluong.NO3=(Cx*10*100*100)/(1000*1000*3*10)

> y

[1] 0.008051864

KT LUN: Hm lng st c trong 2 mu thc n gia sc l 0.8%

BI 25:

Gi tr pvalue= 4.95e-05 < 0.05 nn c s khc bit gia A v B c ngha thng k mc anpha = 5%. Da vo mean of X (gi tr trung bnh ca A) v mean of Y (gi tr trung bnh ca B)ta thy gi tr trung bnh ca ging B ln hn ging A nn kh nng ca cho sa ca ging B tt. Nn ti chn ging B.

PH LC:>B=c(rep(205,7),rep(215,9),rep(225,8),rep(235,8),rep(245,14),rep(255,11),rep(265,6),rep(275,5))>A=c(rep(205,14),rep(215,21),rep(225,20),rep(235,13),rep(245,5),rep(255,4),rep(265,3),rep(275,2))[58] 265 265 265 265 265 265 275 275 275 275 275> var.test(A,B)

# pvalue=0.1219>0.05 nn s khc bit v phng sai gia A v B khng c ngha. Ta s dng T.test cng phng sai>t.test(A,B,var.equal=T)

t = -4.1809, df = 148, p-value = 4.95e-05

95 percent confidence interval:

-19.137101 -6.852856

sample estimates:

mean of x mean of y

225.9756 238.9706

BI 26:

Bng : Gi tr t l ny mm gia cc ging

Ging

T l ny mm

A

0.65

B

0.21

C

0.67

Ta thy p-value = 5.46e-12 < 0.05 nn c s khc nhau v ngha thng k gia cc ging la mc ngha anpha = 5% m theo bng ta thy t l ca i mch cao nht nn ta chn ging C sn xut malt

Biu th hin t l ny mm ca cc loi ging

BI 27:

GING NG LAI

NNG SUT

1

4.56 a +- 0.11

2

3.96 ab +- 0.18

3

5.28 ac +- 0.84

4

3.88 ab +- 0.497

Ch thch: cc k t a,b,c trn cng 1 ct th hin s khc bit nng sut gia cc ging ng lai khc nhau.

Qua gi tr p-value=0.001352giong=c(4.5,4.6,4.7,4.6,4.4,3.8,4.2,3.8,3.9,4.1,4.4,6.3,4.5,5.3,5.9,3.5,4.2,4.3,4.2,3.2)

>group=c(rep("giong1",5),rep("giong2",5),rep("giong3,5),rep("giong4",5))

> group=as.factor(group)

> analysis=lm(giong~group)

> anova(analysis)


group 3 6.312 2.10400 8.4583 0.001352

# P- value = 0.001352 (< 0.05) c s khc nhau gia cc nhm mc ngha thng k =5%

> av=aov(giong~group)

> TukeyHSD(av)

Fit: aov(formula = giong ~ group)

$group

diff lwr upr p adj

giong2-giong1 -0.60 -1.5024692 0.3024692 0.2659806

giong3-giong1 0.72 -0.1824692 1.6224692 0.1438289

giong4-giong1 -0.68 -1.5824692 0.2224692 0.1780807

giong3-giong2 1.32 0.4175308 2.2224692 0.0035176

giong4-giong2 -0.08 -0.9824692 0.8224692 0.9940433

giong4-giong3 -1.40 -2.3024692 -0.4975308 0.0021023

BI 28:

Phng trnh ng chun y= 0.1851 x +0.007

Hm lng ppm c trong 2 mu nc ln lt l 0.00339(ppm) , 0.00337(ppm)

> am= c(0,0.297,0.563,0.835,1.119)

>chuan=(0:4)

>cx= (chuan*15/10)

>th= data.frame(am,cx)

>lm(am~cx)

Coefficients:

(Intercept) cx

0.0076 0.1851

# phng trnh ng chun y= 0.1851x+0.0076

> m1= c((0.762-0.0076)/0.1851)

> m2= c((0.757-0.0076)/0.1851)

> Mn2= c((m2/1000)*(250/3)*(10/1000))

> Mn1= c((m1/1000)*(250/3)*(10/1000))

> Mn1

[1] 0.003396362

> Mn2

[1] 0.003373852

BI 29:

Nhn thy p-value th nht p-value=0.0306*< 0.05 , p-value th 2 bng 1.03e-05 *** < 0.05 nn ta c phng trnh y= 1.3966X.

Tnh ton:

%poli phenol=(Cx.Vmu)/(1000.Vm.100/p)

PH LC:

cm=c(1,2,3,4,5,6)*100/30

> am=c(0.087,0.273,0.407,0.544,0.694,0.902)

>ph=lm(cm~am)

>plot(cm~am,pch=16,sub="moi quan he giua nong do va mat do",xlab="matdo",ylab="nongdo")

>abline(ph)

>summary(ph)

>lengend(1,1,c("y=21.1972*am/nR-squared=0.9934

Bi 30:

x y bai30 summary(lm(y~x))

Residuals:

1 2 3 4 5

0.0136 -0.0148 -0.0032 -0.0036 0.0080

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -0.013600 0.009912 -1.372 0.264

x 6.810000 0.101160 67.319 7.22e-06 ***

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

Residual standard error: 0.0128 on 3 degrees of freedom

Multiple R-squared: 0.9993, Adjusted R-squared: 0.9991

F-statistic: 4532 on 1 and 3 DF, p-value: 7.223e-06

Phng trnh hi quy y=6.81x

Tm nng mu 2: 0.883 = 6.81 x

X = 0.1297

p dng cng thc pha long: 0.1297.10 = x.1

Nng ng ca mu 2 trong 100ml mu l 1.297 g/l

Khi lng ng c trong 100ml mu: 1.297 x 0.1 = 0.1297g

Hm lng l 1.297%

Khi lng cht kh trong 10g l 10.15%=1.5

Hm lng dg trong 1.5g cht kh: 0.1297/1.5

BI 31:

1. Dng kim nh ANOVA 1 yu t

- do gi tr p value=0.2552>0.05 ta thy Kh nng pht trin ca rau cu khng khc bit khi nh sng khc nhau mc nghi 5%

- Nu b qua nhit th ti s nui trng rau cu ch nh sng A2 v rau cu pht trin tt hn do c lng tng khi lng cao hn A1 (11.15 > 8.30).

PH LC:

> group group=as.factor(group)

> a=c(12.5,12.7,8.3,8.5,3.8,4,15.5,15.7,11.5,12,6,6.2)

> kq=data.frame(group,a)

> attach(kq)

> analysis=lm(a~group)

> anova(analysis)


group 1 24.368 24.367 1.4567 0.2552

Residuals 10 167.275 16.727

2. S dng ANOVA 2 yu t

Ch nhit

Lng tng khi lng(g)

B1

12.6a +- 0.14

B2

8.4b +- 0.14

B3

3.9c +- 0.14

B4

15.6d +- 0.14

B5

11.75e +- 0.35

B6

6.1f +- 0.14

Vi 2 gi tr p-value ca a =2.402e-07 b= gl(6,2,12)

>pt=c(12.5,12.7,8.3,8.5,3.8,4.0,15.5,15.7,11.5,12.0, 6.0,6.2)

> A=as.factor(a)

> A=as.factor(b)

>kq= data.frame(a,b,pt)

>attach(kq)

>twowayanova(twoway)


a 1 24.368 24.368 649.8 2.402e-07 ***

b 4 167.050 41.762 1113.7 9.725e-09 ***


> a=c(12.5,12.7,8.3,8.5,3.8,4,15.5,15.7,11.5,12,6,6.2)



> attach(kq)


> anova(analysis)


group 5 191.417 38.283 1020.9 1.061e-08 ***

Residuals 6 0.225 0.038

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

> res=aov(a~group)

> TukeyHSD(res)

diff lwr upr p adj

2-1 -4.20 -4.970694 -3.42930599 0.0000043

3-1 -8.70 -9.470694 -7.92930599 0.0000001

4-1 3.00 2.229306 3.77069401 0.0000368

5-1 -0.85 -1.620694 -0.07930599 0.0326881

6-1 -6.50 -7.270694 -5.72930599 0.0000007

3-2 -4.50 -5.270694 -3.72930599 0.0000032

4-2 7.20 6.429306 7.97069401 0.0000004

5-2 3.35 2.579306 4.12069401 0.0000180

6-2 -2.30 -3.070694 -1.52930599 0.0001747

4-3 11.70 10.929306 12.47069401 0.0000000

5-3 7.85 7.079306 8.62069401 0.0000002

6-3 2.20 1.429306 2.97069401 0.0002250

5-4 -3.85 -4.620694 -3.07930599 0.0000071

6-4 -9.50 -10.270694 -8.72930599 0.0000000

6-5 -5.65 -6.420694 -4.87930599 0.0000014

> by(a,group,mean)

group: 1

[1] 12.6

group: 2

[1] 8.4

group: 3

[1] 3.9

group: 4

[1] 15.6

group: 5

[1] 11.75

group: 6

[1] 6.1

BI 32:

1. . Nu b qua nh hng ca ging la ln nng sut

-Bng : Gi tr ca cc loi t

L t 1

31.55 +- 1.85

L t 2

31.63 +- 3.01

L t 3

33.93 +- 2.87

L t 4

33.23 +- 3.48

L t 5

33.90 +- 3.50

Qua php kim nh anova ta thy p-value=0.6551 > 0.05 nn ko c s khc bit c ngha thng k gia cc nhm mc ngha anpha =5% nn chn l t no cng c.

PH LC:


> a=c(32.8,33.3,30.8,29.3,34,30,34.3,28.2,34.3,36.3,35.3,29.8,35,36.8,32.3,28.8,36.5,34.5,35.8,28.8)


> attach(kq)


> anova(analysis)


group 4 22.357 5.5893 0.6201 0.6551

Residuals 15 135.212 9.0142

> res=aov(a~group)

> TukeyHSD(res)

diff lwr upr p adj

2-1 0.075 -6.48063 6.63063 0.9999996

3-1 2.375 -4.18063 8.93063 0.7944543

4-1 1.675 -4.88063 8.23063 0.9298920

5-1 2.350 -4.20563 8.90563 0.8004375

3-2 2.300 -4.25563 8.85563 0.8121928

4-2 1.600 -4.95563 8.15563 0.9398952

5-2 2.275 -4.28063 8.83063 0.8179607

4-3 -0.700 -7.25563 5.85563 0.9971524

5-3 -0.025 -6.58063 6.53063 1.0000000

5-4 0.675 -5.88063 7.23063 0.9975270

2 . Nu xt s nh hng ca ging la v khu vc canh tc ln nng sut thu hoch

Qua php kim nh anova ta thy p-value ca t =0.5985 > 0.05,p-value ca ging=0.2096 > 0.05 nn khng c s sai khc v nga thng k mc anpha=5% gia cc nhm nn chn nhm no cng c

PH LC:

>dat=gl(5,4,20)

>giong=gl(4,5,20)

>id=1:20

>score=c(32.8,33.3,30.8,29.3,34,30,34.3,28.2,34.3,36.3,35.3,29.8,35,36.8,32.3,28.8,36.5,34.5,35.8,28.8)

>dataattach(data)

>anova(lm(score~dat+giong) )


dat 4 22.357 5.5893 0.7134 0.5985

giong 3 41.191 13.7303 1.7524 0.2096

BI 33:

# s tc ng ring r ca ging v mi trng n kh nng cho sinh khi ti ca 6 chng nm men bnh m.

*GING:

GING

Nng sut cho sinh khi (g)

1

10.9 a+-4.49

2

13.6b+-4.19

3

9.89a+-3.25

4

11.8a b +-3.7

5

12.67b+-3.43

6

15.3c +-4.97

Ch thch: cc k t a,b,c trn cng 1 dng ch s khc bit kh nng cho sinh khi ti ca cc chng nm men khc nhau

P value=2.2e-16score=c(5.112,5.066,4.987,5.786,6.562,6.378,4.258,3.129,5.024,5.231,5.124,5.358,6.587,6.786,6.841,7.068,7.257,7.524,11.235,12.481,12.547,12.021,12.046,12.008,11.547,11.787,12.002,11.958,12.548,12.032,13.245,14.847,14.023,14.568,15.618,15.097,14.056,14.648,14.358,14.978,15.092,15.642,12.321,12.819,12.458,15.647,16.550,16.542,14.562,14.734,15.023,15.003,14.347,14.175,16.542,16.95,16.724,17.687,18.572,18.235,9.687,7.566,8.654,13.214,14.866,14.035,15.698,17.214,16.571,20.158,24.75,22.547,7.023,6.219,6.354,15.874,16.830,16.325,12.358,12.21,12.564,11.369,11.657,11.587,11.387,11.553,10.987,16.871,17.161,17.289)

> data=data.frame(moitruong,giong,id,score)

> attach(data)

> twoway=lm(score~moitruong + giong)

> anova(twoway)

# Qua trung bnh bnh phng(mean square),ti nhn thy nh hng ca mi trng c v quan trng hn l nh hng ca ging.Tuy nhin, c hai nh hng u c ngha thng k mc ngha anpha = 5% v tr s P rt thp cho hai yu t( moitruong:P-value= 2.2e-16;going: P-value= 5.21e-09).

>anova(

twowayF)

moitruong 4 1125.81 281.451 699.217 < 2.2e-16 giong 5 278.79 55.758 138.520 < 2.2e-16 moitruong:giong 20 330.90 16.545 41.103 < 2.2e-16

# S tng tc gia ging v mi trng

P-value=2.2e-16 res=aov(score~moitruong+giong)

> res=aov(score~moitruong+giong)

> TukeyHSD(res)

Fit: aov(formula = score ~ moitruong + giong)

$moitruong

diff lwr upr p adj

2-1 7.0851111 5.1252181 9.0450041 0.0000000

3-1 8.8265000 6.8666070 10.7863930 0.0000000

4-1 10.3106667 8.3507736 12.2705597 0.0000000

5-1 6.7522222 4.7923292 8.7121153 0.0000000

3-2 1.7413889 -0.2185041 3.7012819 0.1055540

4-2 3.2255556 1.2656625 5.1854486 0.0001532

5-2 -0.3328889 -2.2927819 1.6270041 0.9894838

4-3 1.4841667 -0.4757264 3.4440597 0.2245574

5-3 -2.0742778 -4.0341708 -0.1143847 0.0326370

5-4 -3.5584444 -5.5183375 -1.5985514 0.0000247

$giong

diff lwr upr p adj

2-1 2.6489333 0.4027711 4.8950956 0.0114153

3-1 -1.0612000 -3.3073622 1.1849622 0.7390421

4-1 0.8944000 -1.3517622 3.1405622 0.8530289

5-1 1.7170667 -0.5290956 3.9632289 0.2348854

6-1 4.3420667 2.0959044 6.5882289 0.0000036

3-2 -3.7101333 -5.9562956 -1.4639711 0.0000950

4-2 -1.7545333 -4.0006956 0.4916289 0.2140833

5-2 -0.9318667 -3.1780289 1.3142956 0.8300814

6-2 1.6931333 -0.5530289 3.9392956 0.2488588

4-3 1.9556000 -0.2905622 4.2017622 0.1243464

5-3 2.7782667 0.5321044 5.0244289 0.0067840

6-3 5.4032667 3.1571044 7.6494289 0.0000000

5-4 0.8226667 -1.4234956 3.0688289 0.8920120

6-4 3.4476667 1.2015044 5.6938289 0.0003424

6-5 2.6250000 0.3788378 4.8711622 0.0125404

BI 34:

PH LC:

>tl=gl(5,15,75)

>tg=gl(5,3,75)

>hs=c(40.41,42.50,39.50,45.58,45.21,46.67,47.27,47.36,48.79,49.44,49.53,50.53,49.74,50.17,51.17,37.38,37.47,38.47,49.56,49.65,50.56,50.91,51.00,52.00,53.75,53.55,54.55,54.54,54.63,55.63,45.91,46.00,47.00,54.24,54.33,55.33,59.49,59.58,60.58,60.65,60.74,61.74,61.04,61.13,62.13,52.04,52.13,53.13,62.26,62.35,63.35,68.17,68.26,69.26,74.44,74.53,75.53,76.36,76.45,77.45,59.86,59.95,60.95,68.52,68.61,69.61,73.68,73.77,74.77,78.85,78.94,79.94,79.47,79.56,80.56)

>dat= data.frame(tl,tg,hs)

>twowayanova(twoway)


tl 4 7277.9 1819.48 383.35 < 2.2e-16

tg 4 2916.5 729.11 153.62 < 2.2e-16

#c2 yu t t l(tl)v thi gian xay(tg)u c P-value< 2.2e-16,nn u c nh hng n hiu sut xay mc ngha 5%,tuy nhin nh hng ca yu t t l c v mnh hn thi gia xay(MeanSq ca tl=1819.48> MeanSq ca tg)

> anova(twowayF)

tl 4 7277.9 1819.48 3871.787 < 2.2e-16

tg 4 2916.5 729.11 1551.527 < 2.2e-16

tl:tg 16 289.8 18.11 38.536 < 2.2e-16

# tng tc cng gp gia tl:tg c nh hng n hiu sut xay (P-value< 2.2e-16) mc ngha 5% v tc ng cng gp ny thp hn so vi tc ng ring r ca tng yu t(MeanSq ca tl:tg=18.11).

# tc ng ring r ca tng yu t n hiu sut xay

>resTukeyHSD(res)

Fit: aov(formula = hs ~ tl + tg)

$tl (t l)

diff lwr upr p adj

2-1 2.652000 0.4209105 4.883089 0.0118786

3-1 9.734667 7.5035772 11.965756 0.0000000

4-1 20.122667 17.8915772 22.353756 0.0000000

5-1 25.544667 23.3135772 27.775756 0.0000000

3-2 7.082667 4.8515772 9.313756 0.0000000

4-2 17.470667 15.2395772 19.701756 0.0000000

5-2 22.892667 20.6615772 25.123756 0.0000000

4-3 10.388000 8.1569105 12.619089 0.0000000

5-3 15.810000 13.5789105 18.041089 0.0000000

5-4 5.422000 3.1909105 7.653089 0.0000000

$tg(thi gian)

diff lwr upr p adj

2-1 8.875333 6.644244 11.106423 0.0000000

3-1 12.812667 10.581577 15.043756 0.0000000

4-1 16.267333 14.036244 18.498423 0.0000000

5-1 17.155333 14.924244 19.386423 0.0000000

3-2 3.937333 1.706244 6.168423 0.0000522

4-2 7.392000 5.160911 9.623089 0.0000000

5-2 8.280000 6.048911 10.511089 0.0000000

4-3 3.454667 1.223577 5.685756 0.0004635

5-3 4.342667 2.111577 6.573756 0.0000076

5-4 0.888000 -1.343089 3.119089 0.7974094

11.110.9911.0511.0111.0211.0711.110.9811.0311.1211.2511.3111.2811.211.1811.21999999999999911.21999999999999911.1711.1911.209999999999999

mu thc phm

thi gian bn bt

1.16904519445001090.836660026534072340.81649658092772271.25830573921179671.16904519445001090.836660026534072340.81649658092772271.258305739211796718.16666666666666814.81916.25

Loi enzyme

Lng acid amin tng s (mg/kg)

6.896.637.14999999999999956.896.637.14999999999999950.50.300000000000000160.171.1463.7161.14

nng ph gia(%)

ng knh bnh (cm)

0.431194354587994542.88471175419000852.41997540325947740.431194354587994542.88471175419000852.41997540325947747.19571428571428664.21571428571428535.4785714285714304

VNG

Hm lng saponin (%)

0.900055553841002660.572741943054060561.30931279685183810.489931968066122990.739211291399960890.900055553841002660.572741943054060561.30931279685183810.489931968066122990.7392112913999608955708510011517.7221.06666666666666623.324.38333333333313424.986666666666629

thi gian

Hiu sut trch ly

0.894427190999912861.01980390271855591.01980390271855460.748331477354791440.894427190999912861.01980390271855591.01980390271855460.7483314773547914478.64.59999999999999964.8

Ging la

Nng sut (kg thc/n v din tch

2.03004926048606071.50706115779462961.33275404082423530.900388804905970510.719884249955038521.37711050149700931.16001436772711642.03004926048606071.50706115779462961.33275404082423530.900388804905970510.719884249955038521.37711050149700931.160014367727116400.050.10.150000000000000240.20.250.3000000000000003249.83000000000000578.12666666666665189.03666666666667691.74000000000002394.32333333333274196.75333333333328897.566666666666663

nong ch phm (%v/w)

Hm lng vitamin C (mg/g)

1.67000000000000191.420.451.071.350.830000000000000631.12000000000000011.67000000000000191.420.451.071.350.830000000000000631.120000000000000100.050.10.150000000000000240.20.250.3000000000000003221.825.627.929.430.930.7730.87

nng enzyme (%)

hm lng phenol (%)

0.20.511.522.530.20.370000000000000380.640000000000003010.931.221.51.8

Nng benzen (g/l)

Mt o quang (Am)

0.850490054811536610.832666399786447190.81853527718724950.793725393319380751.3051181300301240.750555349946513410.850490054811536610.832666399786447190.81853527718724950.793725393319380751.3051181300301240.75055534994651341015018822526330050.53333333333333196.566666666666663100.5103.3100.7333333333333198.433333333333309

cng sut (W)

Hm lng Vitamin C(mg/g)

1.66794284474461812.36341560740645921.27225521548541991.66794284474461812.36341560740645921.272255215485419980.10666666666666971.41666666666692787.113333333333259

phng php

Hiu sut thu hi dch chit (%)

0.240000000000000070.210000000000000080.260.240000000000000070.210000000000000080.26tr rau mtr datr kh qua6.296.76.4

loi tr

mc yu thch

ht ny mmABC815772ht khng ny mmABC194328

ging

t l(%)

0.114000000000000020.182000000000000190.838000000000000630.497000000000000390.114000000000000020.182000000000000190.838000000000000630.49700000000000039ging 1ging 2ging 3ging 44.55999999999999963.965.283.88

ging ng lai

nng sut

17

34 Bài tập xử lý số liệu trên phần mềm R

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