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1. BO GIAO DUC VA DAO TAO
DAI HOC THAI NGUYfeN
T R U IN G DAI HOC NONG LAM THAI NGUYEN
TS. HOANG VAN PHU - ThS. D6 THINGOC OANH
Giaotrinh
PHl/ONG PHAP NGHIEN CL/U
TRONG TRONG TROT
■
(Gido trinh Cao hoc ngdnh Trong trot)
HOC Tnfei M-' •3■S' K
NHA XUAT BAN N 6N G NGHI$P
HA NOI - 2002

2. LCJlNOIDAU
Phuang phap dao tao "Lay ngudi hoc 1km trung tam" dang trd thanh xu th£ va duoc
ap dung r6ng rai tr6n the' gidfi va cf Viet Nam. Cung ca'p tai lieu, dac biet la giao tiinh cho
can b6 va sinh vien la dac biet ckn thiet de thuc hien xu hudmg nay, nhlm nang cao chat
lupng dao tao va nghien cun khoa hoc.
Mat khac vofi sur phat trien nhanh cua khoa hoc ky thuat, dac biet la su h6 tra cua tin
hoc ung dung, phucfng phap nghien curu n6ng nghi6p noi chung va nghien cun tr6ng trot noi
rieng co nhi^u d6i mai. V6i mong muon cung ca'p cho ngudi doc nhung kien thuc co ban
nha't, cap nhat thanh tmx khoa hoc mci v& phucfng phap thi nghiem trong nOng nghiep,
chung toi bien soan "Giao trinh phuang phap nghifen cuu trong tr6ng trot".
Giao trinh gom co ba phkn chinh:
Phan I: B6' tri thi nghidm - Chuang 1 va 2
Phan II: Phan tich s6' lieu - Chuang 3, 4 va 5
Phan III: Sir dung phan m6m SX 3.5 de phan tich ket qua trfin computer - Chuang 7
Tai lieu tham khao chinh trong qua trinh bi£n soan la "Statistical Procedures for
Agricultural Research" cua Gomez Kwanchai A. va Gomez Arturo A.; "Giao trinh phuang
phap thi nghiem d6ng ru6ng" 1976 va 1988 cua Pham Chi Thanh; va "Agricultural
Experimentation" cua Little Thomas M. va Hills Jackson F.
Nhan thurc duqc tinh phuc tap va da dang cua phucfng phap nghien curu v6i viec ap
dung thong ke sinh hoc trong phan tich thi nghiem, de ban doc d6 hieu chung t6i bien soan
nhung kien thurc ca ban cua phucfng phap. Nhung phln tinh toan qua phurc tap da co cac
phkn mem may tinh tra giup. Chung toi hy vong ring tai li6u se giup ich cho hoc tkp va
nghifen cun cua nghien cun sinh, sinh vi6n va ngucri lam c6ng tac nghien cun khoa hoc ndng
nghifep. Mac dau v6i su co gang cao trong qua trinh bien soan, nhung sach xult ban lkn dlu
chac chan con co phan han che'. Chung toi rat mong nhkn ducfc nhflng y kieh dong gop cua
ban doc de bo sung cho lan tai ban sau duac hoan chinh hcfn, gop phln phuc vu c6ng tac
hoc tkp va nghien cun cua ban doc.
Chung t6i xin chan thanh cam an Du an “Quan ly da't va nude nglm”, hop tac vod
Trudng Dai hoc Saskatchewan, Canada da tai tra m6t phkn cho bien soan va in In giao
trinh nay.
Cac tac gia
3

3. BANG CHtTVltfT TAT
Bang chu: vi£t tat nham giup cho vi£c tra curu va th6ng nh£t gifta c£c thu&t ngu: ti£ng
Vi6t va tieng Anh. Sap x£p theo a,b,c...
Viet tit Tieng Viet Tieng Anh
(i. Y hay X Trung binh tong thi, trung binh miu Treatment mean, sample mean
5, s 06 lech chuan tdng the, do lech chuin mau Standard deviation
ANOVA Phan tfch phifdng sai Analysis of variance
CF He so diiu chinh Correction factor
Cl Khoang tin c6y Confidence interval
CRD Ngau nhien hoan chinh Completed ramdomized design
CV He so bien dong Coefficient variance
df 06 td do Dgree offreedom
DMRT So sanh theo Dun can Duncan's multiple range test
G Tong I6n Grand total
LS 0 vuong La tinh Latin square
LSD Sai khac nho nhat co y nghTa Least significant difference
MS Trung binh tong binh phi/dng Mean square
ns Khong co y nghTa Non - significant
r He so tUdng quan Coefficient of correlation
R He so'tUdng quan dabien Multiple correlation ccoeficient
R2 He so ti/dng quan xac dinh Coeficience of determination
RCB Khdi ngau nhien ho&n chinh Completed randomized block design
SSE PhLfdng sai do nglu nhien Sum square of error
SX Phin mem Statistics 3.5 Statistics 3.5 software
TRT Cong there Treatment
TSS, <
?
, s2 Tong binh phddng, phifdng sai tdng the,
phUdng sai mau
Total Sum square, Variance
5

4. MUC LUC
Trang
l 6 i n 6 i DAU 3
BANG CHtTVlfiT TAT 5
Chuong 1: MOT SO KHAl NI$M
1.1 Nhftng khai nifim trong nghidn cuu khoa hoc ndng nghifep 11
1.2. PhSn tich su sai khac 12
1.3. Ba nguy&i tac ca ban cua thiet ke thi nghiem 13
1.4. Cac phuang phap nghien cuu ndng nghiep 15
1.4.1. Phuong phap thi nghiem trong phong 15
1.4.2. Thi nghiem ngoai d6ng ruong (On - station research) 15
1.4.3. Thi nghiem trong dieu kien san suat cua nong dM (On - farm research) 15
1.5 Mot so khai mem trong thdng kC 16
1.5.1 Thi nghiem 16
1.5.2 Bidn va tham s6 16
1.5.3 Cac tham s6 thdng k6 16
Chuong 2: B6 TRl THl NGHI$M
2.1. Nhan to thi nghifim 19
2.2. Cac kieu bd tri thi nghidm (Gomez, 1984) 19
2.2.1. Kieu ng&u nhi&n hoan toan (Completely Randomized Design - CRD) 19
2.2.2. KhO'i ngiu nhiSn hoan chinh (Radomized Complete Block Design) 22
2.2.3. Kieu 6 vuSng latinh (Latin Square Design - LS) 24
2.2.4. Kilu thi nghiem d phu - Split - Plot Design (Gomez, 1984) 26
2.3. Thu thap sd lieu 28
2.3.1. Cach IdymSu 28
2.3.2. Lay mSu trong thi nghifem d6ng rudng 30
2.3.3. Sd lidu thd va sd li6u tinh 31
2.4. Mot so quy tac ciin bi6t trong tinh toan 31
2.4.1. C onsdconghia 31
2.4.2. Phep tinh gin dting 31
7

5. Chuong 3: PHAN TICH BIEN DONG
3.1. Kilu thf nghi&n ngSu nhi6n ho^n toan (Completely Randomized Design - CRD) 32
3.2. Ki&i khdi ng&u nhifen hokn chinh (Radomized complete block design - RCB) 37
3.3. Kieu 6 vu6ng latinh (Latin Square Design - LS) 40
3.4. Thf nghiem hai nhan t6' v&Kilu thf nghiem 6 phu - SPLIT - PLOT DESIGN 44
3.4.1. Tuong tac gitta hai nhan t<5thf nghiem 44
3.4.2. Ki&i thf nghiem 16 phu - Split-Plot-Design 47
3.4.3. Thf nghi6m ba nhan t6 trd16n va cac ki£u thf nghi£m thfch hop 53
3.4.4. Ki£u thf nghi£m Strip - Split - Plot - Design 54
Chuong 4: SO SANH s 6 TRUNG BlNH
4.1. So sanh theo sai khac nho nh& - Lest Significant Diference Test - LSD 57
4.2 So sanh Duncan (Little M. va Hills, 1978) 61
Chuong 5: PHAN TfCH TUONG QUAN
5.1.Khaini£m 63
5.2. Quan h£ tuydn tfnh don bi£n (Simple Linear Regression) 65
5.3. Phan tfch he s6 tuong quan tuyen tfnh don gian (r) 69
5.4. Tuong quan tuyen tfnh da biSn 70
5.5. Tuong quan phi tuyS'n tfnh don bien 70
5.6. Tuong quan phi tuydn tfnh da biefn 72
5.7. Nhung sai l&n thucrng mac phai khi phto tfch tucmg quan 72
Chuong 6: TRINH BAY KfiT QUA NGHlfiN CUU
6.1 B6 cue va dinh dang bao cao khoa hoc 73
6.2. Cach trhih b&y s6 li£u 76
6.2.1. Phuong phap lap bi&i s6' li£u 76
6.2.2. Phuong pMp ve d6 thi 76
6.2.3 . Trrnh bay k£t qua phan tfch 76
Chuong 7: GlOl THIEU PHAN M*iM PHAN TlCH THONG Kfi
STATISTIX VERSION 3.5
7.1. Cai dat va khori d6ng 78
7.1.1 . Cai dat 78
7.1.2. Khdidbng 79
7.1.3. Sfr dung l£nh trong SX 79
8

6. 7.2. Phan rich phuctng sai-ANOVA v&so sanh s6 trung binh 83
7.2.1. Phan rich ANOVA kidu ngSu nhidn hoan toan CRD 83
7.2.2. Phan rich ANOVA va so sdnh sd trung binh cho Kidu RCB 88
7.2.3. Phan rich ANOVA, so sdnh trung binh kidu 6 vubng La tinh (LS) 92
7.2.4. Phan rich kidu Split-Plot-Design 96
7.2.5. T6m tat md hinh phan rich phtiong sai cua cdc kidu b<5trf thf nghidm
khac nhau 98
7.3. Phan rich tuong quan 98
7.3.1. Tuong quan tuydn tinh don gian 98
7.3.2. Quan hd tuydn tinh da bidn 100
7.3.3. Phan rich tuong quan phi tuydn tinh 102
TAI LlfiU THAM KHAO 107
PHULUC 108
Bang A: Bang s<5nglu nhifen 108
Bang B: Phan b6' tdn xuat tich luy z - Cumulative Normal Frequency Distribution
(During cong phan bri chu&i 0 - Z) 109
Bang C: Xac xua't phan bri't - Distribution of t probability 110
Bang D: Phan bri %
2- Chi-Square distribution 111
Bang E: Phan bri F (5%, va 1% in dam) - Point for the Distribution of F 112
Bang Fa: Tri sd Rp dung cho so sanh Duncan (5%) - Significant Studentized Factor (R)
to multiply by LSD for testing mean at various ranging (p) 5% level; n = degree
of freedom for error (Little M. Thomas va Hills F. Jackson, 1978). 116
Bang Fb: Tri sd Rp diuig cho so sanh Duncan (1% )- Significant Studentized Factor (R)
to multiply by LSD for testing mean at various ranging (p) 1% level; n = degree
of freedom for error (little M. Thomas va Hills F. Jackson, 1978). 117
Bang H: Gia tri hri sd tuong quan tuydn tinh don gian r - Simple Linear Correlation
Coeficients, r, at the 5% and 1% level of significance. 118
Bang K: Cac each sap xdp edng thtic trong kidu thf nghidm 6 vudng la tinh (LS) 119
9

7. Chirotigl
MOT SO KHAI NIEM
1.1 NHUNG KHAI NlfiM TRONG NGHlfcN CtJtJ KHOA HOC NONG NGHlftP
Nghidn cun khoa hoc Id mdt qud trinh tim hidu su that hay phat hidn cac quy ludt tu
nhidn. N6 duqc tidn hanh theo mdt phvtcttig phdp khoa hoc c6 tfnh hd thdng. Bat (Mu tit su
quan sdt su vat. Quan sat tit thuc td hoac tit cac s6 lidu, bao cdo da c6 san dd xac dinh van
dd cua thuc tidn cung nhu cac vSride khoa hoc doi h6i phai giai quySt. Tu: dd xac dinh muc
tieu nghien cun va hinh thanh cdc giA thidt dd giai thfch vdn dd. Nhung gia thidt se co gia tri
ndu no duqc kidm tra th6ng qua viec lam thf nghidm. Cudi cung la dua trdn nhung th6ng tin
thu duqc qua thf nghidm se phat hidn ban chat cua su that hay nhiing giai phap dd giai
quydt vdn dd. Dd ddnh gia dung va hidu ro y nghia cua kdt qua thf nghidm cdn thidt phai
phan tfch dd ddnh gid mdc dd tin cdy cua kdt qua thf nghidm.
Dd co nhung kdt ludn ddng tin cdy, thf nghidm cdn phai tudn theo phuong phdp dung.
Phucng phdp thf nghidm nay quy dinh cdch b6' tri thf nghidm vd cdch xft ly kdt qua thu
duqc. Cdc phuong phdp phan tfch thdng kd nhu: Phan tfch phuong sai (phan tfch sai khdc
hay bien ddng), so sdnh s6' trung binh, phan tfch ham h6i quy Id nhflng cdng cu hftu fch
giup cho phdn tfch va ddnh gid kdt qud thf nghidm. Lua chon phuong phdp tmh toan thd'ng
kd ndo phu thudc vdo muc dfch cua nha nghidn cun va cdch bd tri thf nghidm. Kdt qud thu
duqc sau khi qua xir ly thdng ke se cho kdt luan dung vd tin cdy.
Vi kdt qua thf nghidm thu duqc chi dua trdn trung binh mdu quan sat. Ndu chi dua
trdn kdt qua nay se c6 th i din tdi mac sai 1dm trong kdt luan. Muc dfch chfnh cua thdng ke
la dua ra nhflng co s6 khdch quan, kh6ng c6 su dinh hudng hay thidn vi cho vide phdn tfch
vdn de. Ddy chfnh la co sor cua m6n khoa hoc nay. Khoa hoc ndng nghidp sir dung phuong
phdp quy nap. Phdn tfch cdc quan sdt cu thd dd tir d6 rtit ra kdt luan khai quat.
Thf nghidm duqc dua trdn vide theo ddi anh hu&ng cua nhflng nhdn td thf nghidm
trong didu kidn cdc nhan td phi thf nghidm duqc kidm sodt. Nhdn td thf nghidm thay d6i
trong khi dd su anh huefng cua cdc nhdn td phi thf nghidm duqc gift 6 mfle tdi thidu. Vf du
nhu thf nghidm theo doi anh huong cua muc dam bdn ddn nang sudt lua thi cdc mflc dam
thay ddi edn hide phdh bdn khdc nhu ian vd kali duqc gid nguydn. Nhu vdy nang sudt thay
d^i la do anh hudng chfrih eda su thay ddi mdc dam bdn, trong khi dd anh hufrng do ldn vd
kali Id tdi thidu.
Tdm lai cd thd tdng kdt cdc bude nghidn cdu ndiighghidp nhu sau:
11

8. Set do l.l: Tien trinh nghien cucu thi nghiem
1.2. PHAN TICH SITSAIKHAC
S6' lieu thu ducfc sau khi lam thi nghiem c&i duoc phan tich d6 dua ra kSt lufn. Vf
du: Gia s i mudn bie't nang sudt cua gidng Ida mdi a c6 khde vdi gitfng Ida dia phuong b
hay khdng? Ngudi ta c6 thi chia mdt manh rudng ra lkn 2 6. Mdt 6 cSy gidng Ida a va 6
kia cdy gidng Ida b. Gidng n&o thu duorc nhilu hon cho k£t ludn la gidng c6 nang sudt cao
hon. Lieu k£t lu3n nhu vay c6 cMnh xdc khdng? E)£trd l£ti cdu h6i d6 edn xem xdt m6t s6
van dl sau:
12

9. Suesai khftc v6 nang suftt thu ducrc cua a vft b bdi hai nguyen nhan:
• Do khac nhau ve gidng: Gidng a khac gidng b - D6 la nhftn td thf nghiem
• Dinh duong: 6 deft eft'y ft khftc vdi d dftt eft'y b - D6 lft nhftn td phi thi nghiem
Neu efty gidng b d ca hai d trdn thi nang suftt thu duoc tit hai 6 cung khac nhau. Su khac
nhau nfty 1ft do anh hudng efia ydu td mdi tracing (dftt, nude, sftu bftnh...) khdng ki&n softt duoc.
Su khac nhau nfty n6i lftn su khac gifta hai 6 thl nghiem vft duoc goi 1ftsai sd'thi nghiem.
Nhu vfty su khftc nhau ve nang suftt gifta 2 6 thf nghiem 1ft do su khac nhau vb nhftn td
thi nghiem vft su khftc nhau gifta cftc d thf nghiem (sai sd thf nghiem). Vi vfty, chi can eft
vfto su khftc nhau de kbt luftn 1ft khdng chfnh xftc. Do dd eftn phfti sft dung phuong phftp
phftn tfch thdng ke dd phftn tfch su sai khftc. Phftn tfch thdng ke se tftch duoc su sai khac
(hay bie'n dong) do cftc nguyen nhan (hay nguSn) khftc nhau gfty ra.
Su khftc nhau co ban trong cftc kidu bd trf thf nghiem 1ft cftc d thf nghiem duoc phftn
theo nhom dua tren nhftn td thf nghiem. Trong mdt vfti kidu bd trf, cftc d thf nghiem con
duoc chia theo khdi hay dai. Viec phftn nhdm nfty tao ra su chat chS trong bd trf ngftu nhien
cftc cong thile thf nghiem vfto cftc d. Didu nfty eftn phfti duoc tfnh ddn khi phftn tfch cftc sd
lieu thf nghiem thu duoc. Phftn tfch ANOVA 1ft edng cu phftn tfch chfnh.
Phftn tfch sai khftc tdng (sai khftc toftn bd) ra cftc sai khftc do cftc nguyen nhftn
(ngudn) khftc nhau. Mdt trong nhfing nguyen nhftn chfnh din tdi sai khftc 1ft do nhftn td thf
nghiem.
Su sai khftc (trung binh tdng binh phuong) do cftc nguyen nhftn khftc nhau (nhftn td
thf nghiem, khdi, dai) duoc so sftnh vdi binh phuong efia sai sd thf nghiem dd co duoc gift
tri F thdng ke. Gift tri F duoc dung dd dftnh gift miic dd khftc nhau gifta cftc gift tri trung
binh efta mdt ngudn bibn ddng cu thd.
Phftn tfch sai khftc cho bidt mftc dd sai sd chudn cua cftc sd trung binh va su sai khftc
nay duoc tfnh toftn vft qua dd cd thd ude luong khoang tin efty chac chan cua su sai khftc do
ydu td thf nghiem gfty nen.
1.3. BA NGUYEN TAC CO BAN CUA THIEI KE THI NGHIEM
Thiet ke' thf nghiem dung se giam sai sd, va tang kha nang xftc dinh su khftc nhau do
yeu td thf nghiem tao ra dft su khftc nhau nay la nho. Dd kidm softt duoc sai sd khi bd trf thf
nghiem can phai theo ba nguyen tftc co ban sau:
Nhaclai
Nhac lai nghia 1ft thf nghiem phai duoc lap lai. Vf du mudn bidt mdt loai thude sftu
mdi cd hieu qua diet sftu khdng? Thude phai duoc phun thfr ft nhftt 1ft 2 lftn. Phai lftm
nhftc lai mdi ude luqng duoc sai sd efia thf nghidm. Sai sd duoc ude luong 1ft co sd cho
vide xftc dinh su khftc nhau hong sd lieu da quan sftt duoc cd thuc su khftc nhau vd mftt
thdng ke hay khdng.
13..

10. Vf du co hai gid'ng Ida khic nhau duoc cdy tren hai manh ddit khac nhau c6 chng mdt
dien tfch. Su khac nhau v6 nang sudt thu duoc cua hai gidng Ida cln phii xem cd thuc su Id
do gio'ng khac nhau khdng? hay chi do dinh duOng cua hai manh dd't khic nhau. Bcfi vi
cung mot gidng lua cay tren hai manh ddt khac nhau c6 cilng mdt dien tfch nang sudt thu
dUdc Cung khac nhau.
So trung binh cua cac Ian nhac lai ductc diing dd udc luqng anh hudng cua nhan td thf
nghiem nhu vay se dam bio su udc luqng chfnh xic hon.
N giu nhien
Nglu nhien co nghia Id su bd trf cia cic nhan td phi thf nghiem va cdng thdc thf
nghiem vao cac 6 thf nghiem hodn todn nglu nhien khdng co djnh hudng. Cac anh hudng
cua cac nhdn td khac (vf du Inh hudng tuong tac) cQng dttqc tfnh nhu Id ydu td nglu nhien.
Do su sai khac ludn xay ra nglu nhien vaikhich quan vi thd co scf c&a phdn tfch thdng
ke la dua tren ludt phdn bd chudtt (phdn bd ngin nhien). Bd tri ngiu nhien trong thf nghiem
n6ng nghiep se tranh nhung thidn vi hay chi quan do con ngucri gay nen. Day Id nguy&vtdc
corban de ap dung phdn tfch thdng ke. Phuong phip thdng ke con ydu cau eae sd lieu theo dot
la hodn todn do nglu nhien bdi vi nglu nhien 1dmcho cac gia thidt co gia tri khach quan.
Khoi
De so sanh chfnh xac, cac cdng thdc thf nghiem phai duoc bd trf if dieu kiftn cdng
gidng nhau cdng tdt. Vi ddt noi 1dm thf nghiem khdng hodn toan dong nhd't ndn no co the
duoc chia thanh cac khdi va cac cdng thuc thf nghiem deu duqc bd trf trong mdi khdi. Vf
du trong thf nghiem so sanh gidng cdy trdng, dd kiem soat sai sd cue bd, ngucfi ta chia
ruong ra lam nhidu khdi. Mdi khdi lai chia ra nhidu Id dd bd trf cac gidng. Nhu vdy khdi la
mot phdn hodn chinh cua thf nghiem. Su khic nhau (do didu kidn phi thf nghiem) gifla cac
d thf nghiem trong mdt khdi se nho hdn nhidu so vdi su khac nhau giua cic d thf nghiem if
cac khdi khac nhau.
Ky thudt khdi lam ting stt chfrih xac cho thf nghiem do’tich dttoc sai sd eda thf
nghiem ra khoi stt sai khic do nhdn td thf nghiem- Khdi edn cho phep so sinh su khic nhau
cua nhan td thf nghiem trong mdt khdi.
Mot thi nghiim tot can co nhvttig dac diem sau:
- Cac 6 thf nghiem khac nhau ve cac nhin td thf nghiem nhung khdng khac nhau theo
he thdng.
-Sai sd thf nghiem (sai sd rigid nhien) phii nh6
- Thiet ke' phai don gian nhd't de dat duoc dd chfrih Xac theo mdng dqi
- Phdn tfch theo phuong phap thdng ke thfch hop va khdng tao ra nhQng gia thie't
nhantao.
- Ket ludn phai cd gia tri rdng
Dd chfnh xic cua thf nghiem tang khi sai sd chuln cua trung binh cdng thttc
(Standard error of a treatirierit mean) giferi-Steel v i CS (1997) dua ra nidt sd giii phip <
$S
lam tang dd chfnh xac cua thf nghiem nhu sau:

11. - Tang kich thudc cua thi nghiem. Tuy nhidn kich thudc tang dd dua ddn su kem d6ng
nhlt cua thi nghiem.
- Hinh dang cua 6 thi nghiem: Theo Steel v l CS (1997) ndn bd tri khdi cua thi
nghidm cang gin vdi hinh vudng cang tdt, va cac 6 thi nghiem b6' tri theo hinh chu: nhat.
Neu thi nghiem bd tri trdn d6i ddc thi canh dai cua hinh chft nhat ndn b<5tri vudng gdc vdi
dufrng ddrig mdc (hay song song vdi hudng cua su thay d6i - dd ddc).
- Chon lua cdng thdc thi nghiem thich hop.
- Ky thuat thi nghiem ddng nhat.
1.4. CAC PHUGNG PHAP NGHlfiN CUlJ n 6N G NGHI$P
1.4.1. Phuong phap thi nghiem trong phong
La loai thi nghiem duqc bd tri trong didu kien nhan tao nhu nha kinh, trong chau, vai,
hay dia petri. Loai thi nghiem nay co dd chinh xlc cao. Uu didm 11 cd thd khdng chd cac
ydu td phi thi nghiem ndn dd pMt hidn cac nguyen nhan rieng re anh hudng va ban chit vln
dd. Do do ydu elu dd chinh xac thi nghiem cao nhat. Nhuqc didm 11 do dat trong didu kien
nhan tao do dd kdt luan rut ra chi co tac dung ly lufn, chu: khdng ap dung duqc vao san xult.
Y = f (V, M)
Trong do: Y = Nang suit; V = giong; M = cac bidn phap ky thuat
1.4.2. Thi nghiem ngoai dong ruong (On - station research)
Day la loai thi nghiem phd bidn trong cac co quan nghidn cuu khoa hoc ndng nghidp.
Cay trdng duqc trdng trong didu kidn tu nhidn, chiu anh hudng cua nhidu ydu td nhu khi
Mu, dlt dai. Loai thi nghidm nay cd uu didm la sit hon vdi didu kien tu nhidn do dd co thd
duoc set dung dd xay dung cac bidn phap ky thuat trong san suit. Tuy nhidn trpng loai thi
nghiem nay rridtsd ydu td cd thd bi khdng chd, hoac khac xa vdi didu kidn ndng dan. Vi
vay kha nang phd bidn ra san suit bi han chd.
Y = f (V, M, E)
Trong do: E = Mdi trudng d trai thi nghiem
1.4.3. Thi nghiem trong didu kien san suat cua ndng dan (On - farm research)
La loai thi nghiem duqc dat trong didu kien thuc td cua ndng dan, do ngudi dan quan
ly, theo doi va danh gia kdt qua. Vi trong didu kien thuc td san suit khd cd thd khdng chd,
do dd nhan td thi nghidm phai it v l don gian, ydu clu dd chinh xlc thlp hon. Cd uu didm
tidt kidm, kha nang phd bidn va ap dung cao, vi nd phit hop vdi didu kidn thuc td cua ndng
dan va do ho danh gia.
Y = f (V, M, E, S)
Trong dd: S = Ky nSng quin cua ndng din (Dd Kim Chung, 1999).
15

12. 1.5 M 0T Sd K H A I NI$M TRONG TH 6NG Kfc
1.5.1 Thi nghiem
Trong th<5ng kd, thuc hidn thi nghidm nham tao ra s6 lieu. Gieo d6ng tidn ia mdt
trong nhftng thx du cua thi nghiem, trong do c6 hai trudng hop xay ra la mat hinh va mat
chft. Gieo hdt xuc sac la thi nghiem c6 6 trudng hop xay ra. Phong vdn ndng dan b mdt
viing nao do de bidTt mdc thu nhap cua ho cung la du nghiem. Do luong nude mua b thuong
nguon sOng H6ng cung la mdt thi nghiem. Muc dich cua thi nghiem la d i ki6m tra gia
thuye't ve nguyen nhan thuc cua su vat hay hidn tuqng xay ra.
6 thi nghiem (experimental unit) la mdt ph£n hay don vi cua thi nghiem ma edng
thde thi nghiem duoc sap xep vao.
1.5.2 Bien va tham so
Bi& (Variable) la mdt dac didm hay tlnh chfit cdth^do ddm duoc cua thi nghiem. C6
hai loai bien: Bien ddc lap (Independent variable) va bi6i phu thude (Dependent variable).
Vf du trong thi nghiem so sanh gidng Ida, bien chieu cao cSy hay bien trong luong ngan
hat...ia bien ddc lap, trong khi do bien nang sudt ia bidh phu thude.
Tham sd (Parameter) la mdt dac didm cua qu&i th£ vi du nhu sd' trung bmh (p) cua
quan the hay phuong sai cua quin ihi (d2)... Cac dac di6m nay cua qu&i the chung ta khdng
the biet ma chi xac dinh chung thong qua ude luong.
1.5.3 Cac tham so thong ke
Tong the
Tat ca cac phln tit cua dd'i tuong ma ta nghien efiu daac goi la tdng the. Thi du
chung ta nghien cuu v6 mdc thu nhap b mdt vimg ndo dd thi tdng tbiTia tflt ca ede ndng hd
trong vCtng do. Trong mdt Id 10 tridu hat gidng hoa bao g6m hai loai: hoa vang va hoa do,
chung ta mudn bidt ty Id cua hai loai hat nay ia bao nhidu. Tdng thi trong trudng hop nay
la 10 tridu hat ndi trdn.
y y .
Trung binh tdng the: p = - ——
Yt = gia tri do ddm cua ca thi; N = sd' ca thi trong qu^n the
Mdu ngdu nhiin
Muc dfch cua chiing ta ia mudn bidt cac thdng tin v6 t6ng the. Tuy nhidn chung ta
khdng the quan s£t hdt tfit ca c£c ph&i tit ciia tdng the duoc vl han chd' v6 thdi gian, tai
chrnh'hoac cd nhidu trudng hop mSu bi hu^ hoai sau khi do ludng nhu trudng hop hat gidng
ndu trdn. Ndu la'y tat ck 10 tridu hat dem trdng di bidt bao nhidu cay hoa vang, bao nhidu
16

13. cay hoa do thi khOng cdn hat gidng dd dilng sau khi thii. Nghi&i cun rd cua cay thf nghifem
khOng thd nhd tdt ca dd do ddm.
VI cac f do trfin ndn cdc nhd nghifen cun phai lam th£ ndo dd chi la'y mOt sd phdn tit,
n, trong tdng thd dd quan sat ma vdn e6 thd bidt dude ede dde tmh ciia tdng thd. C6ng vific
nay dnqc goi ia "lay mdu”. Tap hop cac phdn tit ducqc lay ra dd quan sat dutqc goi Id "mdu”
va n la cS cha mdu.
De mdu dai di6n cho tdng thd, ta phai chon sao cho tat ca cac phdn tit cua mlu c6
cung m6t co h6i chon. Dd c6 dac tmh nay ta phai chon "ngdu nhien”cac ca thd trong tdng
thd nghifen cun. Do d6 m6i ca thd dnqc chon la m6t bidn ngdu nhidn c6 cung ham phan
phdi xac suat. Ngoai ra, su chon mdu con dam bao m6t dac tmh khac ntta ia su "doe lap”
gifla ede phdn tut trong mlu. Nghia ia kdtqua eta vide chon phdn tut nay kh6ng anh huctng
dfin vide chon ede phdn tut khde. M6t mdu thoa man dnqc hai didu kidn trSn duqc goi ia
"mdu ngdu nhien”. v
Trung binh mdu
TOmdu ngdu nhiSn cOn, trung binh mdu (Y ) dnqc dinh nghia nhu sau:
Y
n
Phitamg sal tdng the
i'hirong sai Id tdng ede sai khde cua ede cd thd so vdi gia tri tmng binh cua qudn thd.
N
Phurnig sai mdu
Tir mdu ngdu nhien cd n, phuerng sai mdu Id tdng binh phuctng cua sai khde gitta cd
the thd i vdi trung binh mdu. Sai khac ndy the hidn su bidn ddng cua qudn thd mdu so vdi
sd trung binh. Binh phuerng dd triet tidii gia tri dm, vi su khde nhau c6 thd Id dm hay duong.
Thudrng tdng binh phuong cua bidn d6ng ldn thi su sai khac gifta ede cd thd x c6 y nghia.
Vi du: c6 hai qudn thd mdu nhb sau:
Quin thi 1 Qudn thi 2
X
j £ X X
i £ X
1 2 6 3 12 3 3 4 3 2 12 3
X
, - X -2 - 1 3 0 Xi- X 0 1 0 - 1
(X; - X f 4 1 9 0 14 (x- X f 0 1 0 1 2
S2 = -x )2 = 14 s2= £ ( Xi - X )2 = 2
M N gc TmtNGii/t-Ni 17
jl!igri4QHfeLA*>
M irerH

14. Phuong sai do d i phln tan cua ting thi. G6 nhiiu truing hop Hai dim ding c6 s i
trang binh bang nhau nhung phuong sai khlc nhau thi hai dim d6ng d6 cung khlc nhau.
Vi du hai quin thi m lu trin duqc lly tit hai quin thi Ilia khlc nhau, vdi muc dfch chon
giing thi quin thi 1 se cho phep nha chon giing c i nhiiu co h ii chon ra nhiiu d6ng m ii
khlc nhau, nhung niu gilm dinh giing thi quin thi 2 c6 tinh 6n dinh tit hem.
Do lech chudn tong the d = -Jd*
D5 lich chuln II su sai khlc cua clc gil tri quan sit duorc (hay gil tri do dim) so veri
gil tri trung binh.
Do lech chudn mdu
U&c luong phuomg sai va d i lich chuln ting thi thing qua uic luong phuemg sai m lu
(s2) va d i lich chuln mlu.
( E X ) 2
s2=
Z Yi ~ Y>'
n - 1
a
n - 1
vas
He so bien dong CV (Cofficience of variance)
H i s i biin dong la gil tri so slnh d i lich chuln v6i gil tri trung binh
CV= = 100
Y
Gia tri t
La mot s i thing k i ding d l so slnh su sai khlc gifta trung binh m lu va trung binh
thuc cua ting thi trong m it don vi sai s i chuln (bod vi tuln theo lult phln b i chuln -
normal distribution).
t = ( Y - p ) / S ¥..
Gi&i han tin cay hay do tin cay CL (Confidence limits)
La khoang biin thiin cho phip cua gil tri trung binh m lu
CL = Y + 1-
Gia tri F
La ti s i gifta phuong sai m lu uic luong do nhln t i thi nghiim (s2) va phuong sai do
nglu nhien (s 2), duoc dung di xlc dinh su sai khlc c i f nghta gifta clc s i trung binh.
F
18

15. Chirong 2
BO TRf THf NGHIEM
2.1. NHAN l 6 THINGHIEM
Trong thf nghidm c6 hai ylu td: Yeu to thi nghiem va yeu to phi thi nghiem. Vf du thf
nghidm so sanh gidng Ma. D l so sanh dupe, cac gi<5ng Ilia phai dupe gieo cdy b nhflng dilu
ki|n (nhu dfit, nude, phan...) nhu nhau. Nhu v&y gidng Ma la ylu 16 thf nghitoi con cac
dilu kidn m6i trufrng Id yeu to phi thf nghidm hay c6n goi Id nln thf nghiem.
D l so sanh duoc, cac ylu td phi thf nghiem phai ddng dlu, vi vly mpt trong nhujig
nguyen tac quan trong cln tuan thu khi lam thf nghilm Id nguyen tac sai khac duy nhat.
D6 la cac yeu t<5phi thf nghilm (hay nln thf nghilm) phai duoc dam bao gidng nhau gitta
cac 6 thf nghilm, c6n ylu td thf nghiem thay dli. Nln chu ^ chon nln thf nghilm tien
tiln. Vf du nln phan b6n cho thf nghiem gidng phai dly du mdi the hiln duoc td't tiem
nang cua gidng.
De tiln cho so sanh, trong thf nghilm thuotng bd trf edng thiic dd'i chitng. Doi chiing
thuefng Id ky thuat phi high hoac tiln tiln. Vf du nhu trong thf nghiem gidng, gidng dupe
chon lam ddi chiing la gidng thuerng dung hoac gidng cd nang suat cao nhSt.
Thf nghiem chi co mdt nhan td thay d6i trong khi cac nhan td khac khdng thay ddi
dupe goi la thf nghilm mpt nhan td. Vf du nhu thf nghilm vl gidng Ma.
Thuc vat trong tu nhiln chiu tac ddng cua nhilu ydu td. Thf nghilm mdt nhan td
thufrng dupe coi la han chi vi phan ling cua thuc vat vdi mdt tac ddng nao do co the phu
thude nhilu vao miic dd tac ddng cua cac nhan td khac nhau. Vf du thf nghilm phan ling
cua cac gidng Ma khde nhau trln nln phan bon khac nhau la thf nghilm hai nhan td vi co
gidng va phan bdn thay d6i. Gidng chiu tham canh se cho nang suat cao trln nen phan cao,
ngupc lai gidng khdng chiu tham canh hay bi ldp, sau bdnh va nang suit khdng cao trln
nen tham canh cao. Vi vay ngudi ta cd thi lam thf nghilm cd nhilu nhan td (chung ta se
thao Man sau hon v l thf nghilm da nhan td trong Chuong 3, Muc Kilu thf nghilm Id phu -
Split - Plot - Design).
2.2. CAC KI^U B6 TRI T ffl NGHIEM (GOMEZ, 1984)
2.2.1. Kieu nghu nhien hoan toan (Completely Randomized Design - CRD)
Kieu nglu nhiln hoan toan la cac edng thiic thf nghilm cd thi dupe bd trf vao bat ky
d nao. Vf du thf nghilm cd 4 edng thiic (A,B,C va D) vdi 3 lln nhac lai. Tdng sd d thf
nghilm la 12. Thf nghilm dupe bd trf theo kilu ngSu nhiln hoan toan nhu so dd dudi day:
19

16. 1 D 2 C 3 A 4 A
5 A 6 B 7 A 8 A
9 B 10 D 11 D 12 B
ScfdS 2.1: Bo tri theo kieu ngau nhien hoan toan (CRD): t=4 (A,B,C,D); r=4
DSy la kilu W tri dcm gian va ft rang bulc nha't, ngoai c l m lt rang bulc la c6ng thdc
thi nghilm dlu c6 ca h li nhu nhau dl b l tri v&o bit k f I thi nghilm nao.
tHi dilm:
- P inh hoat c l thi dp dung vli bSit ky s i clng thdc va s i l&i nhac lai nao. Hem nfta
co thi cac clng thdc kMc nhau c6 s i l&i nhac lai khdc nhau
- Phan rich sai khac don gian k l ca vli s i l&i nhac lai khlng gilng nhau va khlng
phdc tap trong truing hop m lt hay thilu s i lilu
- Kilu thilt k l nay cho d l tu do cda sai s i lln nhlt
Nhuoc dilm:
- D l chxnh xac thlp nlu khu thi nghilm khlng ding dlu, va nhu vdy khl phat hiln
su sai kh^c do ylu t l thi nghilm gay nln.
Sddung:
- C6 d l chfnh xac cao nhlt nlu khu thi nghilm ding nh& vi vly thich hop vli
nhftng khu thi nghilm ding nhlt nhu thi nghilm trong phlng, chau vai.
- C l loi nlu khi m lt phln lln I thi nghilm khlng c l phan ting hoac bi m lt
- C l loi cho thi nghilm c l s i I till nghilm han chi bdi vi kilu nay c l d l tu do cua
sai s i lln nhdt.
C l nhilu each dl b l tri nglu nhiln cac clng thdc thi nghilm. Samday la each b l tri
nglu nhiln hoan toan theo Gomez (1984)
Vi du: Thi nghilm cl:
4 clng thdc (t=4; t: treatment) ia A, B, C, D
5 lln nhac lai (r=5; r: replication).
Bu&c 1: Xac dinh s i I thi nghilm: n = (r)(t) = (5)(4)=20.
Budc 2: Danh s i I thi nghilm td 1 - 20 nhu trong hinh vl.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
20

17. Buac 3: B6' tri c6ng thtic cho cac 6 thi nghifem b&ig phuong phap ng&u nhifen.
Ax: Chon 1 didm M't ky trong bang s6 ngSu nhidn bang cdch nham mat v&dat ngdn
tay vao 1 diem bdt ky (Phu luc A: Bang sd nglu nhidn).
14620 95430 12951 81953 17629
09724 85125 48477 42783 70473
56919 17803 95781 85069 61594
97310 78209 51263 52396 82680
07585 28040 26939 64531 70570
25950 85189 69374 37904 06759
82937 16405 81497 20863 94072
60819 27364 59081 72635 49180
59041 38475 03615 84093 49731
74208 69516 79530 47649 53046
39412 03642 87497 29735 14308
48480 50075 11804 24956 72182
95318 28749 49512 35408 21814
A2: Tit di&n bat ky' dd lay ra n = 20 s6 (vdi 3 chit s6) li&n tiSp nhau th trong bang.
Sap xdp cac s6 nay theo thd tu xudt hidn trong bang ngSu nhifen.
So ngau nhi£n Thfrtif S6 nglu nhi3n ThCrtir
937 1 918 11
149 2 772 12
908 3 243 13
361 4 494 , 14
953 5 704 15
749 6 549 16
180 7 957 17
951 8 157 18
018 9 571 19
427 10 226 20
A3: Xep hang n = 20 s<5theo thd tu tit nho ddn ldn hoac nguoc lai.
So ngau
nhien
Sdthljrtu X€p hang S6 nglu
nhien
So thirty Xep hang
937 1 17 918 11 16
149 2 2 772 12 14
908 3 15 243 13 6
361 4 7 494 14 9
953 5 19 704 15 12
749 6 13 549 16 10
180 7 . 4 057 17 20
951 8 18 157 18 3
018 9 1 571 19 11
427 10 8 226 20 5
21

18. A4: Chia n = 20 sb ra t = 4 nhom, m6i nh6m g6m r = 5 s6' theo thiS tu c&c s6 xuSt hi6n
trong bang ngSu nhibn.
Nhom Thtttiftrong nhom
1 17 2 15 7 19
2 13 4 18 1 8
3 16 14 6 9 12
4 10 20 3 11 5
A5: Gan t c6ng thftc cho n 6 thf nghi&n theo c£c nh6m da duoc chia b A4 theo so
cong thuc va thu: tu da xep hang trong tftng nh6m la s<5tint tu cua 6. Trong vf du tren nhom
s6' thu: nhat gan cho c6ng thftc A g6m cac 6 thu: 17, 2,15, 7, 19; nhom thft 2 gan cho c6ng
thurc B g6m cac 6 thiil 3, 4, 18, 1, 8; nhom thu: 3 g&n cho c6ng thftc C g6m cac 6 thu: 16, 14,
6,9, 12 va nhom thu: 5 gan cho c6ng thftc D g6m cac 6 thft 10, 20, 3, 11,5. Sau day la so d6
thf nghiem bo tri theo kieu nglu nhien hoan toan:
1 B 2 A 3 D 4 B
5 D 6 C 7 A 8 B
9 C 10 D 11 D 12 C
13 B 14 C 15 A 16 C
17 A 18 B 19 A 20 D
So do 2.2: Thi nghiem bo tri ngdu nhien hoan toon (CRD): t=4 (AM,CD); r=5
2.2.2. Khoi ngiiu nhien hoan chinh (Radomized Complete Block Design)
Kieu khoi ngSu nhien hoan chinh la chia khu thi nghifim th&nh khdi va yfeu ctiu stt co
mat cua cac cong thftc thf nghiem b tat ca cac khoi. Day la ki&i thi& k£ thf nghibm co ban
nhat va duoc ap dung r6ng rai nhat trong nghien cftu n6ng nghifip. RCBr& thfch hop v6i thf
nghiem co so 6 kh6ng lcm lam va bi£t duoc diSn bifih cua di&i ki&i mdi trucfng thf nghiem
nhu dat dai, anh sang... Do d6 dac diem cua kieu kh<5i ng&unhi£n hoan chinh la:
- Cac 6 thf nghifem duoc nhom theo kh<5i ma trong d6 su khdc nhau gifta cac 6 trong
m6t khoi la nho nhat.
- Cac c6ng thftc thf nghibm duoc bo tri nglu nhifen trong m6t khoi. Vcfi quy dinh
m6i c6ng three chi xuat hibn m6t lan trong m6t khdi.
Ifu diem
- Muc dfch cua khb'i la nham giam sai s6 thf nghifem bang each tach ngu6n (nguyen
nhan) bien d6ng do su kMc nhau gifta cac khtfi ra khoi bi£n d6ng do ngSu nhi&n kh6ng
duoc biet gifta cac 6 thf nghiem. Nhd v$y tang d6 chfnh x£c thf nghibm.
22

19. - Khi nh6m cac edng thtic vao mdt khdi, bien ddng trong khdi se dugc tdi thieu hoa
va bien dong gitta cdc khdi vdi nhau dugc toi da hod. Sau d6 tach bidn ddng giua cac khdi
ra khoi sai sd do nglu nhidn cua toan thi nghiem. Cach lam nay se tang kha nang phat hidn
su sai khac c6 y nghTa gida cac edng there thi nghiem vdi nhau.
- Co thd dp dung RCB vdi bdt ke sd edng there va sd ldn nhac lai nao vi sd ldn nhac
lai bang sd khdi.
Nhuorc diem
- Gap kho khan trong phdn tfch khi mdt sd lidu.
- Ndu bd tri khdi sai se gdy khd khan cho phdn tich.
- Kidu bd tri nay hidu qua kem hctn kidu bd tri khdc ndu ngudn bidn ddng ldn hem
mdt (chi kidm soat dugc mdt ngudn bidn ddng).
- Hieu qua cua kidu bd tri giam ndu sd edng thde tang vi kich thude cua khdi tang.
- Ndu khu thi nghidm ddng nhat thi kidu bd tri ngSu nhidn hoan toan (CRD) se cho
kdt qua chinh xac hem.
Svc dung
- Kidu bd tri nay loai dugc mdt ngudn bidn ddng (do khdi) tur sai sd thi nghidm va
nhd vay tang dd chinh xac.
- Cho phep su ude lugng khdng thidn vi cho cac gia tri trung binh cua khdi va thdm
thdng tin ve thi nghidm.
- Cho su chinh xac trong phdn ldn trudng hgp ma khdng edn phai cd nhQng thidt kd
phde tap.
Ggi y: Hinh dang va hudng khdi anh hudmg rdt ldn ddn y nghla cua bd tri theo khdi.
Nhin chung phai bd tri lam sao cho tdi da hoa bidn ddng gifia ede khdi vdi nhau. Do dd khi
bd tri sap xdp cac khdi theo dudmg cat ngang didrt bidn cua ddt, chd dd nude, dd ddc, hay
che dd anh sang v.v.
Cd mot sd ggi y nhu sau:
- Khi bidt dugc bidn ddng ddt dai theo hudng nhdt dinh thi bd tri khdi dai va hep,
hudng vudng goc vdi hudng bidn ddng ddt dai.
- Khi cd nhidu hudng bidn ddng thi chi quan tdm ddn hudng bidn ddng manh.
- Khi hudng bidn ddng ddt dai theo 2 hudng vudng gdc vdi nhau, hoac khdng bidt
hudng bidn ddng thi bd tri theo kidu d vudng latin, hoac bd tri khdi theo hinh vudng.
Vi du vdi thi nghidm cd sd edng there t = 6 va sd l&i nhac lai r = 4.
Bude 1:
Chia khu vuc thi nghidm ra r khdi nhu nhau. Vdi thi nghidm trdn chia khu thi nghidm
ra 4 khdi nhu sa d6 sau.
23

20. Hucfng thay d6i cua d6 mau mor cua ddt
06 mau m& thdp 06 mau mo cao
*-
Kh6i III
1 4
F A
2 5
D B
3 6
C E
Khoi I
1 4
C E
2 5
D B
3 6
F A
Khoi II
1 4
A C
2 5
E D
3 6
F B
Khd'i IV
1 4
E A
2 5
C F
3 6
D B
Set do 2.3: Bo tri theo khoi ngau nhien (RCB) voi t=6 (A, B, C, D, E, F); t=4
Mui ten chi huomg thay d6i cda ddt. Cac kh6i c6 hinh chu nhdt va vu6ng g6c v6i
hudng bien dong cua ddt.
Budc 2: B6 tri cac c6ng thitc cho tutng kh6'i m6L Chia kh6i 1 ra 16. t la s6 c6ng thiic
thi nghiem. Bo' trf cac cong thufc thf nghiem vdo edc 6 tmng khoi ngiu nhiSn (theo thu tuc
ngAu nhien giOi thieu a 2.1.1). Chu y bo tri ngdu nhifin cho tdng kh6i mot.
Vi du chon 6 s6 ngSu nhiSn va xdp thvr tir tit nho ddn Irtn nhtr sau:
So ngau nhi£n ThCftii (cong thttc) hang(vj trifrong khoi)
918 1 6
772 2 5
243 3 1
494 4 2
704 5 4
549 6 3
B6 tri cho 6 6 cua kh6i 1. Ti£p tuc b6 tri cac 6 cho ede khdi kbdc nhu tr6n.
2.2.3. Kieu 6 vuong latinh (Latin Square Design - LS)
Kieu 6 vu6ng la tinh chi thuc hipn dupe khi s6 c6ng thiic bang s6 ldn nhac lai. Khi do
khu thi nghi&m duoc chia theo hang va c6t va y6u cdu su c6 mat cua ede c6ng thue thi
nghiem d tat ca cac hang va cac c6t. 0ac dilm eda ki£u 6 vu6ng la tinh Id:
- Cac 6 thi nghidm ducfc nh6m theo hang dua tr6n su bi6n d6ng cua m6t y&u t6 va b6'
tri theo c6t theo su bi£n d6ng cua m6t y£u t6 khde. Nhu vdy ede 6 dupe b<5 tri theo hai
hudng d6c lap v6i nhau, theo hdng vd c6t. Ky thudt ndy c6 th£ tdeh sai s6 (bid'n d6ng) theo
hdng vd sai s6 theo c6t ra khdi sai s6 thi nghifim.
24

21. - Ydu cdu su c6 mat cua tat ca cac cOng thuc thf nghidm trong mdi hang va mdi cdt.
Mdi cdng thuc chi xudt hien mdt lSn trong m6i hang va m6i cdt. Do do sd nhac lai phai
bang sd cdng thuc.
Ifu diem
- Khac phuc duoc sir bidn ddng cua didu kidn thf nghidm theo hai hudng vudng goc
vdi nhau thSng qua ky thuat khdi. Nhd vay kidm sodt duoc hai ngudn bidn dOng.
Nhuac diem
- Yeu cau s6 6 thf nghidm phai la s6' binh phuong cua s<5cdng thuc thf nghidm. Vi
vay han cM sd c6ng thdc thf nghidm, tdi da la 10 hoac ft horn.
- Khi sd cdng thuc tang thi sai sd thf nghidm trdn 6 tang ldn
- Neu s6 cdng thuc nho thi dd tit do cua sai s6 dung dd so sanh qua nho va kh6 cho
vide ude luong dung sai sd thf nghidm.
- Neu sd cdng thuc ldn thi dien tfch thf nghidm se rdt ldn, thuc td rdt kho tim dien
tfch da't ddng deu tuong ddi rdng de lam thf nghidm.
- Phan tfch rat phu:c tap trong truemg hop mtft sd lidu.
Vf du thf nghidm cd t = 5; r = 5.
Bade 1: Chon so dd cua LS tut phu luc K. So d6 cho 5 x 5 tit phu luc K nhu sau:
A B C D E (hi)
B C D E A (h2)
C D E A B (h3)
D E A B C (h4)
E A B C D (h5)
Bade 2: Bd tri ngdu nhien cac cdng thuc thf nghidm theo chidu ngang va doc theo
phuong phap d 2.2.1
Vf du chon ngdu nhidn 5 sd 628, 846,475, 902, 452 va xdp hang tut nho ddn ldn.
So ngau nhi§n Thtrti/ X£p hang
628 <h)i 3
846 (h)2 4
475 (h) 3 2
902 (h) 4 5
452 (hi 5 1
Coi hang ia thd tu ctia ede hdng d so ddduoc chon. Nhu vdy hang thu: 1 d so dd duoc
chon duoc xdp thd 3, hang thti 2 duoc xdp thd 4. So dd mdi nhu sau:
25

22. cl c2 c3 c4 c5
E A B C D
C D E A B
A B C D E
B C D E A
D E A B C
Tuong tu chon 5 s6' d£ b6' tri ngSu nhiOn cho 5 cOt dia so d6 vita lam tr£n.
So ngau nhien ThtftiT X£p hang
845 (C) 1 4
785 (c)2 3
396 (c)3 1
856 (c)4 5
664 (c)5 2
Coi hang la thii tu cua cac cOt b so d6 mcfi. COt 1 nam b vi tri thur 4, cOt 2 nkn b vi tri
thd 3 va so d6 cudi cung nhu sau:
B D A E C
E B D C A
C E B A D
D A C B E
A C E D B
So do 2.4: Bo tri theo kieu 6 vuong la tinh vdi t- 5 (A.B.C.D.H): r=5
2.2.4. Kieu thi nghiem 6 phu - Split - Plot Design (Gomez, 1984)
Ki&i thi nghiOm 6 phu - Split - Plot Design thich hop dio thi nghiOm hai nhan tO':
Nhan to chinh va nhan td phu. Theo kieu nay thi dSu tiOn chia khu thi nghiem ra theo khoi
(s<5 khoi bang sd cOng thdc cua nhan tO' chinh). Sau d6 m6i khdi duoc chia ra thanh cac 6
nho (so 6 nho trong mOt khdi bang so c6ng thdc cua nhan td phu). Nhd vay:
- Anh hubhg cua nhan td chinh duoc uorc luong tit khdi, trong khi do anh hucmg cua
nhan tb phu va su tac dOng phOi hop gifta nhan t6' chinh v&nhan tO phu duoc udc luong til
cac 0 nho.
- Vi c6 hai nhan tO' thi nghiOm vdi sO cOng thftc khdc nhau nOn c6 hai sai s6 thi
nghiOm. Vi su d6ng ddu gifta cic khOi nh6 hon so vdi SUddng d£u gifta cic 0, do d6 sai s6
c£ia nhan tO phu (trong cdc 6 nh6) nho hon sai s6 dia nhan tO chinh (trong cic khdi).
26

23. Nhu vay nhan td chmh c6 dd chmh xac kem hem nhan td phu. Vi vay tuy vao muc
dfch cua thf nghiem dd chon nhan to chmh va nhan to phu. Vf du thf nghidm phan umg cua
cac gidng lua khac nhau vdi muc do phan bon khac nhau. Neu nha chon gidng quan tam
nhieu den giong thi chon giong la nhan to phu con mure phan bon la nhan td chmh. Tuy
nhien trong thuc td nhidu khi gap kho khan trong vide lua chon dd bd trf nhan td nao vao d
phu. Vf du trong thf nghiem so sanh gidng trong cac chd dd tudi nude khac nhau. Ndu ta
quan tam ddn nhan td chd dd tudi nude hem nhan td gidng thi se gap kho khan trong bd trf
Id phu la chd dd tudi nude vi ndoc dd dang th^m tha'u qua cac d. Do dd khi bd trf con phai
can cur vao dac tfnh cua timg nhan td dd co giai phap hidu qua giam chi phi, ddng thdi vSn
dam bao do chfnh xac cua thf nghiem.
Vu diem
- Cho phep svt dung co hidu qua khi mdt nhan td co nhidu edng there duqc phdi hop
vdi nhan td kia co ft edng thurc hom.
- Tang do chfnh xac khi so sanh mdt vai nhan td.
- Danh gia duoc anh hudng tuomg tac cua cac nhan td thf nghidm.
- Cho phep dua nhung edng thurc thd nghidm mdi vao mdt thf nghidm dang tidn hanh.
Nhttoc diem
- Phan tfch thdng kd phde tap vi nhung so sanh khde nhau co su bidn ddng sai sd
khac nhau.
- Do chfnh xac cua nhan td chfnh tha'p dan ddn sai khac 1dmtrd ndn khdng cd y nghia,
trong khi do su sai khac nho cua nhan td phu cung co thd cd y nghia vd mat thdng kd nhung
trong thuc td"khdng cd y nghia.
Svt dung
- Mdt nhan td thf nghidm cd rihidu edng tilde hem nhan td kia
- Dua mdt nhan td mdi vao thf nghidm dang duoc tiSh hanh
- Dung dd danh gia tuomg tac cua hai nhan td thf nghidm
Vf du thf nghidm cd
- 6 mure phan bdn khac nhau (nhan td chfnh A) vdi
- 4 loai gidng lua (nhan td phu B) va
- 3 ldn nhac lai.
Btfffc I: 1
Chia khu thf nghidm ra lam 3 khdi (tuomg duong vdi 3 ldn nhac lai),
Mdi khdi chia ra 6 d chfnh (6 mure phan).
Bude 2: Trdn cac ldn nhac lai (r=3) chia ra lam a = 6 d chfnh (ngau nhidn).
27

24. Bucfc 3: Chia cac ft chfnh (r)(a)=18 ra ldm b = 4 6 phu, bft tri ngdu nhiftn cho cac ft
phutrftncdc ft chfnh.
V4 N3 N1 NO N5 N2 N1 NO N5 N2 N4 N3 N N1 N4 N5 N N2
Nhlc lai I Nhdclaill Nhdclailll ,.
So dS 2.5: Bd'tri kieu 6 phu - Split - plot Design voi
• 4 giftng Ilia (V„ V2, V3, V4) dutdc coi lk nhdn tft phu va
• 6 mile phan b6n (N0, N1; N2, N3, N4, N5) duqc coi la nhdntft chfnh;
• 3 ldn nhac lai. ‘
Ngoai cac kiftu bft tri thTnghiftm thftng dung trftn ngUcri ta c6 thft dp dung nhiftu kiftu
khac nhu: Khfti ngdu nhiftn khftng hoan toan (Incomplete Randomize Block Design); Split -
Split - Plot Design, Trip - plot design... Tham khao thftm trong Gomez 1984; Statistical
Proceduresfor Agricultural Research.
2.3. THU 1HAP SO LIEU
2.3.1. Cach lay miu . s r ;
• Mot so van de ve Mdu
Mdu la mftt phdn cua Tong the. Thftng qua Mdu ngufti ta c6 thft uftc luqng vd hiftu
Tong the . Co 2 each uftc luqng tftng thft: Phuong phdp liftt kft va phuong phdp uftc luqng
thong qua mdu. Ly do ta phai ldy mdu la:
- Tong the c6 sft luqng ca thft rdt Ion, trong khi nhdn luc va tdi chlnh c6 han ta khftng
the nghien cun td't ca cac ca thft cua t6ng thft.
- Tong thft co the? biftn d6i theo thfti gian. Nftu dhng phuong phap liftt kft todn bft kftt
qua eft thft da lac hdu.
- Ly thuyet thftng ke da chung minh: Luat phdn bft cua Tftng thft tudntheo ludt phdn
bft chuan. Nftu mdu cua Tftng the Id dai diftn no cQng tudn theo ludt phdn bftchudn. Do dft
co the dung phdn tich thftng kft dft kiftm dinh dft chtnh xdc hay dft tin cdy cua mdu co dai
dien cho tftng thft hay khftng.,
• Ve do chlnh xdc cua phuong phap udeluqng qua mdu
Hai cdu hoi dat ra Id:
V4 V3 V3 V1 V2 V1
V2 V4 V2 V3 V3 V4
V1 V1 V4 V2 V4 V2
V3 V2 V1 V4 y t
c
o
>
V2 V1 V1 V2 V4 V3
V1 V4 V2 V3 V3 V2
V3 V2 V4 V1 V2 V1
V4 V3 V3 V4 V1 V4
V1 V4 V3 V1 V1 V3
V3 V1 V4 V2 V4 V2
V2 V2 V1 V4 V2 V4
V1 V3 V2 V3 V3 V1
28

25. - Phuomg phap liet ke day du cd dam bao dem lai chinh xdc khdng ?
Dem lai chinh xdc khi Tdng the c6 s6 lugng cd thfenho, ft bifen dOng.
Dd chinh xac giam khi Tdng th ec6 s6 lugng ca thfe ldn, bifen ddng nhifeu.
- Phuong phap udc lugng thong qua Mdu c6 dam bao dem lai chinh xac khdng?
Dem lai chinh xac hom khi Tdng the co s6 lugng ca thfe ldn hay bi^n ddng va lfey
mSu dft dung lugng va dai difen dung. Dd chinh xdc thdp hon khi 1% m&u It, thifen vi, khdng
dai difen.
C6 hai ngu6n gfey nfen sai sd' mSu Id do nglu nhifen vh do phvrong phdp ldy mdu gfey
nfen. Phuong phap khac phuc Id ldy mSu dai difen va tang s<5lutong mdu
* Nguyin tdc ldy mdu
- Gia dinh mdu phan dnh Tdng thfe. Do d6 cac cd thfe trong tdng the dfeu cd co hdi
gidng nhau dfe dugc chon lam mdu.
- Vay Mdu dieu tra = Mdu nglu nhien = Mdu xac suat va phai dugc lay nglu
nhifen.
- Phuong phap nay cung c<lp co s&logic tlnh todn xdc suat dd kidm dinh kfet qua.
* Dung lUffng mdu
Phu thude vao:
- Mdc dd bifen ddng efta mdu
- Dd chinh xac cfen co
- Difeu kifen nhdn luc, tai chinh
* Thietke lay mdu
+ Phuong phap lay mdu ngau nhien don gian:
- Mdi ca thfe trong Tdng thfe dfeu cd co hdi duoc chon nhu rihau.
- Chi duoc lam vdi Tdng thfe nho vh tucttig ddi ddng nhat.
+ Phuang phap lay mdu theonhom:
- Thu thap thOng tin vfe Tong th i
-Ldym duthii
- Phan tlch mdu thft
- Phan loai so bo Tong the theo (k) nhom
- Quy dinh ldy mdu hay xac dinh tifeu chudn cua mdu (nglu nhifen).
- Xac dinh dung lugng mdu (m) cua tirng nh6m.
Dung lugng mdu: n = (k) x(m)
29

26. + Phuong phap khd'i:
- Chia Ting thi ra ttcng kMi (r) theo tilu chuln.
- K l hoach chon mllu gifta cac khd'i la khdc nhau va ddc lap vdi khPi khac
- Trong mdt kh<5i phuong pMp chon m iu (ngSu nhite) la giPng nhau.
- Trong kilu My mSu nay, tfnh nglu nhite giam tir(r x t) xuPng r.
- ThuSn lpi cho vide thuc hipn My mlu, giam sai sP co giPi.
2.3.2. Lay mUu trong thf nghiem dong ruong
Muc dfch cua thf nghilm d6ng ruPng M thu hoach nang suat, nhung dl giai thfch vl
sir thay dpi nang suat gifta cac edng thdc thf nghilm, thi cSn phai 06 nhung sP lilu quan sat
ve cay tr6ng, dat va dilu kiln ngoai canh kMc dl tra Ida die cflu hoi:
- Thuc vat phan ung vdi dilu kiln ngoai canh va cac bite phdp ky thuat nhu th i nko?
Vi dunln phto bon khac nhau dSn din ty 16blnh khac nhau.
- Nguyen nhan cua su khac nhau vl nang suit gifta c&c edng thuc? Vf du nang suat
cac giPng lua khac nhau do sP nhanh de khac nhau hay trpngltfpng hat khac nhau.
NPi dung quan sat tuy thuPc v^o muc dfch, ylu cMqefta thf nghitei. Do khPng thi do
s *
«
dim dupe toan bp ca thi trong 6 thi nghilm nln phai chon m3u dai dite cho ca 6 thf
nghilm.
Vile My m&u cu thi thucmg co quy dinh cho tftng loai nghite eftu (nhu sinh truPng
cay, miic dp benh hai va dPi tupng nghiln eftu). Trong Mi lite nky khPng c6 tham vong
trinh bay cac phuong pMp My m iu cho tftng loai nghite edn khde nhau.
Sau day la mPt sP gpi y dl mSu dai diln va mang tfnh ngSu nhite c6 thi My mSu theo
cac each sau: .
- Lly ng&u nhiln bang each d&nh sP va gap tham
- Chon cay trung binh b nhung chP diln hinh eftathfnghitei
- Chon theo quy dinh may m6c theo dudng chte 5, 4* 3, 2 dilm, hoaw: theo mPt
khoang each nhlt dinh.
So do 2.6: Lay mdu theo dudng cheo
Tuy nhite tu^ loai cay trPng ma quy dinh cich M
ym iucu thi. Vf du b nghite eftu
dau d6 quy dinh My 10 cay trte 2 hang gifta, trft cay rrgoairta.
30

27. 2.3.3. So lieu thd va so Mu tinh
Sd M u thu duqc qua do ddm true tidp ngoki ddng rudng lk sd M u thd. Vi du bang sd
Mu sau.
Bang: Chiiu cao cky eba 2 gidng Ida (cm)
Nhac lai I
Cay so r 2 3 4 5 6 7 8 9 10 TB
Gitfng A 0.9 0.8 0.8 / 1.2 1.1 0.6 0.9 1 0.7 1.1
Gi6'ng B 1,2 1-1 1.3 0.9 0.6 0.7 0.9 1 0.8 1.0
Nhln vko bkng sd M u thd tr&n rkt khd nhkn dinh kdt qua. Gidng nko cao hon gidng
nao? V i dac tinh chiiu cao gidng nko 6n dinh hon gidng nko? D i tra lefi duqc cku hoi trin
ckn phai chuyin sd Mu thd thknh sd Mu tinh. Vi du nhu tinh chiiu cao trung binh. Cac sd
Mu co duqc sau khi tinh toan tut sd M u thd duqc goi lk sd M u tinh. Khi xb thdng kd se
dung sd Mu tinh.
2.4. MOT SO QUY TAG CAN Brffr TRONG TINH TOAN
2.4.1. Con so co nghia
- Vi du con sd chuih xkc tdil/10 lk con sd ngay sau dku (toil vi (,). Thi du 65,4
- Nhfing sd duqc goi lk ihfiih xdc Ik lAiiiig sd ckn thidt cho vide diin think khdng ki
sd 0 di dat dku don vi.
Vi du: Trong eke sd sau dky niu Iky dd chinh xkc lk 1/10 thi: Sd 65,4 c6 3 con sd c6
nghia (65,4); sd 4,530 cd 2 con sd cd nghia (4,5); sd 0,018 cd 2 con sd cd nghla (0,0)
2.4.2. Phep tinh gkn dung
Quy the: Trong cac phep nhkn, chia, khai can thi kit qua cudi cung khdng cd thi c6
nhiiu con sd cd nghla hon con sd cd nghia creon sd ckn tinh cd it nhtft.
Vi du: 73,245 x 4,52 = 331,04; kdt qua nky chi Iky din 2 con sd sau dku don vi lk cd
nghia vi sd 4,52 chi cd 2 con sd cd nghia.
1,684 : 0,02 = 84.20; kdt quk nky chi Iky ddh 2 con sd sau dku don vi cd nghia vi sd
0,02 chi Cd 2 con sd cd nghla. :
Quy tac: Trong eke phip edng, trir thi sd cd nghia (ki tit sau dku (,) eba kdt qua cudi
chng khdng thi cd nhiiu hon sd cd nghia (ki tir sau dku (,)d sd hanged it s6 c6 nghia nhkt.
Vi du: 3,16 + 2,7 = 5,9; kdt qua nky chi cd 1 sd cd nghia k i tCt sau dku (,) vl sd 2,7
chi cd 1 sd cd nghia k i tit sau dku (;).
83,12 - 72 = 11 vl sau dku (,) eba 72 khdng cd sd cd nghia.
31

28. Chuong 3
PHAN TfCH BIEN DONG
Sau khi lam thi nghidm, cdn xut ty thdng kd sd lidu thu duoc d l c6 co sd ehinh xdc
cho phan tich kdt qua thi nghidm. D l bidt duoe su sai khde cua cac sd lidu do ddm dude c6
f nghia hay khdng (hay ndi each khde ede yfZn td thi nghidm c6 dua ddn su khde nhau
khdng) edn phai phan tich bidn ddng (hay c6n goi la Phan tick ANOVA). Cach phan tich
bidn ddng phu thude vao each bd tri thi nghidm. Dudi day trinh bay each phan tich bidn
ddng cho mdt sd kilu bd tri thi nghidm thuemg gap.
3.1. KIEU THI NGHI$M NGAU NHl£N Ho AN TOAn (Completely Randomized
Design - CRD)
Vi du thi nghidm so sanh hidu qua cua ede loai thude trit rdy ndu anh hudng tdi nang
sudt cua lua, gdm:
- 7 edng thde (t = 7): Trong dd edng thde 7 la ddi chung.
- 4 lan nhac lai (r =4).
- Bd tri ngSu nhidn hodn todn CRD vdi tdng sd d thi nghidm:
n = (t) (r) = 7 x 4 = 28
C6 hai nguydn nhfin (hgudn) gfty ra bidn ddng trong n d thi nghidm:
- Ngudn bidn ddng do edng there tin nghidm vd
- Do nglu nhidn hay ndi each khde Id sai sd ngdu nhidn.
So sanh hai nguydn nhan (ngudn) ndy dd xac dinh su khde nhau Id thuc do ydu td thi
nghidm gay ndn hay Id do sai sd ngdu nhidn gay ndn. Ndu su khde nhau do ydu td thi
nghidm ldm hem su khde nhau do sai sd ngdu nhidn thi ta cd thi kit ludn rang su sai gitta
ede edng thiic thi nghidm Id dung.
Uu dilm cua kilu thi nghidm ndy la phan tich bidn ddng dem gian, thich hop vdi
trucmg hop cac edng thde cd sd ldn nhac lai khdng ddu nhau (chang han mdt sd lidu trong
mdt vdi d). .
Bude l : Nhom sd lidu theo edng there va tinh tdng cua edng there (T) vd tdng todn bd
(G - Grand Total) nhu d bang 1.
Bude 2: Thidt lap bang phdn tich bidn ddng - Phuong sai.
32

29. Bang phdn tick biin dong (Analysis of Variation - ANOVA)
Nguyen nhan
(nguon) bien
dong
Do tu do
(df)
Tong binh
phudng bi§'n
dOng (SS)
Trung binh t6ng
binh phuong bi£n
d6ng
(MS)
Fb
in
g
^
tin
h
5% 1%
Cong thOc
Ngiu nhien
Tong
SS: Sum of Square; MS: Mean Square
Bu&c 3: Tmh do tu do (degree of fifeedom - d.f) cfia tCtng ngudn bien ddng qua sd
cdng thdc t; sd lin nhic lai r:
Tong d.f
I
/*
■
S
4
—
>
/—
S
II
s
d.f cua cdng thtic: dfT = t -1
d.f cua sai sd ngiu nhidn: dfE = t(r- 1)
d.fEcon co thl tmh nhu sau: dfE =Tdf - dfT
Vdi vf du trdn:
Tdng d.f Tdf = (r)(t) -1 = (4X7) -1 = 27
d.f cua cdng there: dfT = t -1 = 7 - 1 = 6
d.f cua sai sd ngiu nhidn: dfE = t(r- 1) =7(4-1) =21
d.fEcon co the tmh nhu sau: d.fE=Tdf- dfT= 2 7 -6 =21
Bang 3.1: So sank hieu qua cua cac loai thuoc tritray nau
anh hitdng din nang mat lua (Gomez, 1984)
Cong thtfc
N9ng suat (Xj) Tong CT (T) Trung binh CT
I II III iv
1 2537 2069 ~ 2104 1797 8507 2127
2 3366 2591 2211 2544 10712 2678
3 2536 2459 2827 2385 10207 2552
4 2387 2453 1556 2116 8512 2128
5 1997 1679 1649 1859 7184 1796
6 ,'_v 1796 1704 1904 1320 6724 1681
Doi chtfng 1401 1516 1270 1077 5264 1316
Tong toan bo (G) 57110
T.binh toan bo 2040
Btfcfc 4: Tmh sd hifeu chinh va tdng binh phttong cfia bidn ddngcua toin bd.

30. So hieu chinh (correction factor) C.F =
G‘
n
ii
Tdng binh phuong bidn ddng cfia toan b6: TSS = ^ X 2 - CF
i=l
Z^
Tong binh phuong bidrt ddng do c6ng thvtc: SSj. =
_ i=l CF
T6ng binh phuong bidn ddng do sai sd ng5u nhidn: SS^ = TSS - SSj.
Theo dinh nghia t6ng binh phucmg bidn ddng duoc tmh nhu sau:
S (xi- x)1=S (x "- 2X!x +j?)=s XI2- xix +"X1=
2 x , 2 - 2 X £ x i + nX
- 2 X
Thay X = co
n
v n y
= 1 X ^ - 2
( E x , ) 2 , I X
n n
2 X
IX
n
Ky hieu £ se duoc dung cho "tdng cua", vi du nhu bidu thtic G = Xx + X2+ .. + Xtt
co the duoc vidt laG - ^T ^X j hay don gi&n I&G = £ X.
Vcfi vi du tren dung gia tri T v ^ G cua bang, tdng binh phuong bidn dOng t6ng thd
duoc tfnh nhu sau:
C F = G l = (5L n o ) l . = n 6 4 g 4 0 0 4
n (4X7)
Tong binh phuong bidn dCng cua toan bd
T S S = ^ X 2-C F =
i=l
TSS = [(2537)2 + (2069)2 +... + (1270)2 +(1077)2] -116.484.004 = 7.577.412
Tdng binh phuong bidn dong do cdng thdc
Z?
SSj. =
_ i=l CF =
_ (8507)2 +(10712)2 + ... + (5264)
116.484.004 = 5.587.174
SSe = 7.577.412 - 5.587.174 = 1.990.238
34
r

31. Budc 5: Tfnh trung binh cua binh phucfng bi£n d6ng (MS - Mean Square)) do cac
nguyen nhdn khde nhau ling v6i d6 tu do cua no.
Trung binh cua binh phuong bi£n ddng do cfing thiic:
MS]. = ^
t- 1
5.587.174
6
931.196
Trung binh cua binh phucfng bien d6ng do sai s6 ngSu nhifin:
M S ^
SSE
t ( r - l )
1.900.238
(7)(3)
94.773
Bu&c 6: TMi gia tri F (so sanh bi6'n d6ng do cdng thiic va bi£n dbng do ngSu nhi6n)
de kiem tra su sai khac co f nghia cua thf nghi&m
Tfnh
MSt
MSe
931.196
94.773
9,83
Bum 7: F ly thuyfit
Tra bang phu luc E vdfi f, = df cua c6ng thiic = dfT= (t -1)
f2= df cua sai sd ngdu nhi&n = dfE= t(r - 1).
V6i thf nghifem tr£n vdi f, = 6 va f2= 21;
F bang = 2,57 ofmiic 5% va = 3,81 ormiic 1%.
Bum 8: Lap bang phan tfch bi£n ddng (Bang ANOVA)v<Si cac gia tri tfnh duqc tii
budc 3 den 7.
Bu m 9: So sanh FT
£
llhvcii FW
ttg:
• F, > Fb Ket luan: Su khac nhau giua cac cong thde c6 y nghia. Hay noi m6t each
khac cac cong thiic khac nhau co anh huorng khac nhau hay su khac nhau la do yefu tef thf
nghiem gay nen. Co hai kha nang nhu sau:
- F, >Fb cf miic 1% k£t luan c6 su sai khac lorn giua cac c6ng thurc thf nghifem. Hay
noi mot each khac su sai khac gifia cac c6ng thiic thf nghiSm or miic d6 tin cay 99%. Va
duqc bieu thi "**" orgia tri cua F,
- Fb
0
1 > F, > Fb
0
S K6t luan: C6 su sai khde gifia ede c6ng thiic thf nghi&n. Hay n6i
m6t each khac su sai khac gifia ede c6ng thiic thf nghifim fymiic d6 tin cdy 95%. Va duqc
bi&i thi orgia tri cua F ,.
• F, < Fb0
5Ket ludn: Su khac nhau gifia ede cOng thiic khfing co y nghia. Duqc bieu
thi "ns" orgia tri cua F ,. (n.s = nonsignificant).
Hay noi m6t edeh khac ede c6ng thiic khde nhau kh6ng Id nguyen nhdn gdy n£n nang
sudt khde nhau.
35

32. Buac 10: Tinh trung binh t6ng va M s<5 bidn d6ng (coefficient of variation -cv,
thufrng goi la sai sd thi nghifem).
G
Trung binh t6ng
V6i vf du trdn
G =
n
cv =
_M
G
.100
G = G = 57110=2O4O
n 28
cv
G
E V94773
-.1 0 0 = — .100 =15,1 %
2040
cv chi ra d6 chinh xac cua thi nghidm, n6 chi ra sai s6 thi nghifem chi6n bao nhifeu %
s6 trung binh t6ng bidn dfing. Vdy cv cang nh6 thi nghifim cang chinh xac. Gid tri cua cv
co th£ ch^p nhdn ducfc la khac nhau gitta cac loai thi nghidm va cdc bidn theo doi.
Theo Gomez (1984), thuftng cv ducrc chap nhdn ddi vdi nang sudt cua Ilia cay:
- 6-8 % vdi thi nghidm gidng;
- 10-12% vdi thi nghidm phan bon;
- 13-15% vori thi nghidm v£ thudc trfit e6 vd trit sdu.
Thi nghidm vdi nhttng bb phan khac nhau cda cay cung ydu cdu cv khac nhau
- Vdi nang suat lua cv Id 10%
- Vdi s6 nhanh dd cua Ida cv la 20%.
Bdng phdn tichbiendong (ANOVA - Analysis of Variation) (CRD)a
cda nang sudt Ida d bdng 3.1
Nguon bie'n
dong
D6 tif do
(<#).
Tdng binh
phtfcmg (SS)
Trung binh tdng
b'mh phi/dng (MS)
F«b
^"b
in
g
5% 1%
Cong thuTc 6 5.587.174 931.196 9,83** 2,57 3,81
Ngau nhi^n 21 1.990.238 94.773
Tong 27 7.577.412
acv = 15,1%;b
**: c6 y nghla 61%;
Kdt luan: Ft>Fb0
1 vdy cd su sai khdc rat chic chan gida cdc cdhg thdc thi nghidm.
Hay n6i mdt cdch khdc, cdc cdng thdc ddng thudc sdu khdc nhau c6 M l hddng khdc nhau
d^ti hdng sudt rdt chdc chan - b mdc tin c&y 99%. Hay 99% khd ridng Id cdc cdng thdc
thudc khdc nhau cho ndng sudt khdc nhau.
36

33. 3.2. KIEU K H 6l NGAU NHlfeN HOAN CHINH (Radomized complete block
design - RCB)
Vi du thi nghiem anh hucmgcua lucmg hat gi<5ng gieo tdi nang su£t Ma. Thi nghiem
g6m:
- 6 cdng thiic (t=6), (6 lircfng hat gidng: 25, 50, 75,100, 125, 150 kg/ha).
- 4 l&i nhac lai (r=4),
- Bo tri theo kieu khd'i ng&u nhidn RCB.
Cac budc phan tich bi£n ddng nhu sau:
Bu&c 1: Nhom s6' Mu theo cdng thtic va Mn nhac lai. Tinh t6ng cua c6ng thiic (T),
cua nhac lai (R) va t6ng toan bd (G) nhu 6 bang 2 .
Bu&c 2: Thiet lap bang phan tich bi£n ddng.
Nguon bien dong df SS MS
^bing
5% 1%
• Nheic lai
• Cong thifc
• Nglu nhien
• Tong
Banx 3.2: Nang suat lua cua gidng IR8 v&i 6 lucmg hat gidng gieo trong th(nghiem
RCB, 4 Ian nhac lai (Gomez, 1984)
Cong thttc N3ng suat (kg/ha) (Xj) Tong cong Trung blnh cong
(kg hat/ha) l II III IV thOc (T) thufc
25 5.113 5.398 5.307 4.678 20.496 5.124
50 5.346 5.952 4.719 4.264 20.281 5.070
75 5.272 5.713 5.483 4.749 21.217 5.304
100 5.164 4.831 4.986 4.410 19.391 4.848
125 4.804 4.848 4.432 4.748 18.832 4.708
150 5.254 4.542 4.919 4.098 18.813 4.703
Tong nhic lai (R)
Tong I6n (G)
TB tong
30.953 31.284 29.846 26.947
119.030 4.960
Bu&c 3: Tinh dd tu do (df)
titng ngu6n bie'n ddng:
T6ng df:
df cda nhac lai
♦ ^
df cua cdng thtic
df cua sai s6' ngiu nhidn
qua t = so cdng thtic; r = sd l&i nhac lai; df = dd tu do cua
Tdf = rt -1
dfR= r -1
dfT= t -1
dfE= (r - l)(t - 1) = (rt - 1) - (r - 1) - (t - 1)
37

34. Vdi vf du trdn:
Tdng df
df cua nhac lai
df cua c6ng thdc
df c&a sai s6
'ngSunhidn dfE =(r - l)(t -1) = (it -1) - (r -1) - (t -1) = (3X5)=15
Budc 4: Trnh sd hidu chinh va t6ng binh phuang cua bidn ddng
Tdf = r t - 1= 24-1= 23
dfR = r - 1=4-1=3
dfT = t - 1 = 6 -1=5
Sd hieu chinh (correction factor) C.F =
G
n
Tdng binh phucfng bidn ddng cua toan bd:
TSS = ] > ] X? - CF i = sd' cdng thdc (t); j = sd nhac lai (r)
i=l i=l
Tong binh phucfng bidh ddng cua nhac lai:
_
_ J=
1
t
Tdng binh phuang bien dOng cua cfing thdc:
I t;2
CF
i=i
CF
SSj = J
Tdng binh phucfng bidn ddng cua sai sd ngSu nhidn:
Vdi vf du trdn:
CF =
SSg= TSS - SSr - SS^
G2 (119.030)2
rt (4)(6)
Tdng binh phucfng bidn ddng cua toan bd:
= 590.339.204
t r
t ss = S Z xF cf
i=l i=l
= [(5.113)2 +(5.398)2+... + (4.098)2] - 590.339.204 = 4.801.068
Tdng binh phucfng bien ddng cua nhac lai:
I * ?
SSr = — CF
^0-953)2+
i^L284)2t (29-846)2t g&gffl* _59o.339.204 =1.944i361
6

35. T6ng binh phuomg bidn d6ng cua c6ng thiic:
t t
_ i=l
SSj. =
(20.496)2 +... + (18.813)
CF
-590.339.204=1.198.331
T6ng binh phuong bi£n d6ng cua sai so ngiu nhidn:
SSe = TSS - SSr - SSr = 4.801.068n
- 1.944.361 - 1.198.331 = 1.658.376
Budc 5: Tinh trung binh cua binh phuong bi&i d6ng do cac nguyen nhan khac nhau
ling vdi d6 tu do cua n6.
Trung binh cua binh phuong bien d6ng do nhac lai:
MS> = ^ = 1944^ 6I =648.120
(r —1) 3
Trung binh cua binh phuong bi£n d6ng do cdng thdc:
wo SSr. 1.198.331
MS, = —+L- = ------------ = 239.666
( t - 1) 5
Trung binh cua binh phuong bien dbng do ngiu nhidn:
MSe =
ss, 1.658.376
= 110.558
(r - l)(t -1) 15
Budc 6: tmh gia tri cua F de kiem tra su sai khac
MSt _ 239.666 _
MSe ~ 110.558 ~ ’
Budc 7: so sanh Ftinhva Fb
4
llgnhu da gicfi thieu b budc 9 cua CRD
Tra bang E vdi
-f, =df, = 5;
-f2= dfE=15;
F0
5 = 2,90 va F0
1 = 4,56 .
Vay Ft < Fb: Sai khac gifia cac c6ng thtic thf nghi&m khdng chic chan.
Hay lucng hat giong gieo khdc nhau kh6ng lim anh hucng den nang suat lua (cong
thdc gieo 25 kg hat/ha cung co nang suit nhu cac c6ng thdc khac). Ket luan cac cong thdc
gieo luong hat gi6'ng khac nhau kh6ng anh hucng d£n nang suit. Can cii vao kft qua trfin,
39

36. nha nghien curu se khuye'n cao san suit nin ding cing thtfc 25 kg hat/ha se tiit kiim dugc
giong gieo trong khi nang suit khing dii).
Bu&c 8: Tfnh h i s6 biin ding hay sai so thf nghiim (cv)
JM SF Vl 10.558
cv = v- ^ E .100 = v ------ 100 =6,7%
G 4.960
Kit luan: Thf nghiim dam bao chfnh xac.
Buac 9: Lap bang phan tfch biin ding
Bang phdn tick biin dong cua nang sudt lua &bang 3.2“(Thi nghiim kieu CRB)
Nguon bien dcng df SS MS F h
r tinh
^bSng
5% 1%
Nhic lai 3 1.944.361 6 4 8 .1 2 0
C6ng thCfc 5 1.198.331 2 3 9 .6 6 6 2,17"® 2,90 4,56
Ngau nhien 15 1.658.376 1 1 0 .5 5 8
Tong 23 4.801.068
acv = 6,7%;bns = not significant = khong co y nghia;
3.3. KIEU 6 VUONG LATINH (Latin Square Design - LS)
Co 4 ngudn biin ding trong LS:
- Theo cit,
- Theo hang,
- Cing thdc,
- Va nglu nhiin.
Vf du: Thf nghiim so sanh nang suit
- 4 giing ngi: A, B, C (dii chtSng), va D
- B i trf theo kiiu LS (4x4).
Cac bucfc phan tfch biin ding nhu sau:
Buac l : Lap bang s i liiu theo hang, cit va cing thfic nhu b bang 3.
Buac 2: Tfnh ting cua hang (R), ting cua cit (C) va ting toan b i (G). Tfnh ting cua
cing thdc (T) va trung binh cita cing thtic nhu sau:
40

37. Bang 3.3a: K it quathi nghtem nang suit cua 3 giong ngd lai (A,B, vd D)
vd giong dd'i chitng (C), kieu LS
S6 thuftif hang NSng su£t hat (t£n/ha) (X,) Tdng h&ng
(R)
Cot 1 C6t 2 C6t 3 C6t 4
Hang 1 1.640(B) 1.210(D) 1.425(C) 1.345(A) 5,620
2 1.475(C) 1.185(A) 1.400(D) 1.290(B) 5,350
3 1.670(A) 0.710(C) 1.665(B) 1.180(D) 5,225
4 1.565(D) 1.290(B) 1.655(A) 0.660(C) 5,170
_ T o n g c6t (C) 6,350 4,395 6,145 4.475
TonglSn (G) 21,365
Tinh s6' trung binh cua ckc cdng thtfc:
C6ng thdc Tdng s6 Trung binh
A 5,855 1,464
B 5,885 • 1,471
C 4,270 1,068
D 5,355 1,339
Buac 3: Lap bang phto tich bi£n dbng
Ngudn bien dong df SS MS Ftfnh
F bang
5% 1%
• Hang
• Cot
• Cong thOc
• Nglu nhien
• Tdng
Budc 4: Tinh d6 tu do cua ngu6n biSn d6ng, coi t lk s<5c6ng thdc:
T6ng df (Tdf) = t2-1
df hang (dfR) = df c6t (dfc) = df c6ng thdc (dfx) = t -1
df sai s6' (dfE) = (t - l)(t - 2)
hay dfE= t6ng df - dfR- dfc- df,
Vcfi thi du tr£n
T d f= t2- 1 = 16-1 f 15
dfR= dfc = dfT = t - 1 = 4 - 1 = 3
dfE= (t - l)(t - 2) = (4 - 1)(4 - 2) = 6
hay dfE= Tdf - dfR- dfc- df, = 15 - 3 - 3 - 3 = 6
41

38. Buac J. Tinh s<5hiSu chinh v&t6n | binh phucmg cika bi£n d6ng:
= 28,528952
CP _ G2 (21,365)2
t 16
T6ng binh phucmg cda bi£n ddng cua toan b6
T S S = £ X 2 -G F =
T6ng bmh phucttig bi£n dbng cua h&ng
_ 2 X
ss.
T6ng binh phucmg bifih ddng cua c6t
- CF
s&
_ Z C -
CF
T6ng birth phucmg bi£n d6ng cha c6ng thtfc
Y ^ 2
SS, = — CF
T6ng biSn d6ng do ngSu nhifen
SSp = TSS- SS,- SSC- SS,
Vdi vf du trfin
T6ng binh phucmg cha bif&i dftng cuatoan bO
TSS = ^ X j 2 -C F = [(1,640)2 + (1,210)2 +.... + (0,660)2]-28,5289 = 1,4139
T6ng binh phucmg bi£n d6ng cda h&ng
SS,> - CF = R « 0 )2 + (5,3SO)1 H S,22Sy H SJTOy _ = ^
t 4
T6ng binh phucmg bi£n dOng cua c6t
2 . / a ~ ,n e 2 . ft - t >«r2
: : Z c . CT (6.350)2 +(4,395)2 +(6,l45)a +(4,47S)a
c t 4
28,528 = 0,827
Tdng binh phucmg bi£n d6ng cua c6ng thdc
s ^ 2 5 1 - c f = (5-855)* F y y ) L 28i528 = 0 ,4 2 6
t 4
T6ng bi£n dOng do ngSu nhifen
SSe=TSS - SSj - SSC- SS,=1,413 - 0,030 - 0,827 - 0,426 = 0,129

39. Budc 6: Tfnh trung blnh cda ting bmh phucfng c&a tfltng biin ding bing cieh chia
ting binh phuong cho d i tu do tuang ting.
Budc
MS,=
MSC=
MS, =
MSg =
ssr „ ° ’030
t - 1 3
ssc 0,827
t - 1 3
sst 0,426
t - 1 3
SSe
( t- l)(t - 2)
*
= 0,010
= 0,275
= 0,142
0,129
= (3)(2)"
7: Tuih F d i kiim tra hiiu qua cua thl nghiim
0,021
MSt
m se
0,142
0,021
Budc 8: So sanh F tlnh vtii F bing cfbang E vtii
f1=d f, = t - 1 v i f2= dfE= (t - l)(t - 2) va k€ft luan theo hudng din trong budc 9 phin
CRD.
Trong vi du nay: f,=3 va f2
=6 F0
5= 4,76 ; F0
1 = 9,78
Vi Fbos< F^n < Fbol kit luin thl nghiim c6 sai khic 6 mtic tin ciy 95%.
K it luin: Cac giing ngi khac nhau co nang suit khic nhau. Su sai khic nang suit cf
mtic tin c$y 95%.
Budc 9: Tlnh h i s i biin ding hay sai s i thl nghiim
c v = ^ 3 S x i o o = ^ ? 8 x l00. 11.0%
G 1,335
Budc 10: Lap bang phin tlch biin ding
Bang 3.3b: Phdn tick biin dong cua n&ngsuit hat &bdng 3.3“
(Ihl nghiim kiiu LS)
Nguon bi£n d6ng df SS MS
F
tfnhb
F bang
5% 1%
• Hang 3 0,030154 0,010051
• Cot 3 0,827342 0,275781
• Cong thOc 3 0,426842 0,142281 6,59* 4,76 9,78
• Ngau nhien 6 0,129585 0,021598
Tong 15 1,413923
*cv = 11,0%; b
*= c6 f nghla cfmtic 5%
Kit luin: 4 giing ngi c i nang suit khac nhau chic chin (timtic tin ciy 95%).
43

40. 3.4. THI NGHI$M HAI NHAN t 6 VA K lfu THI NGHI$M 6 PHU - SPLIT -
PLOT DESIGN
Trade khi di vk> phuong pMp phan tich phuong sai cua kidu thi nghifim nay chung ta
thao lu$n thdm vd thi nghifem nhidu nhan 16 nhu da d£ c$p trong chuong 2.
3.4.1. Tmmg tac giufa hai nh&n to thi nghiem
Trong ndng nghidp, cSy tr6ng chiu anh hudng eda nhidu nhan td. Fh&i ling cua cay
tr6ng vdi mdt nhan td nk> d6 la khde nhau d cac rmic dd khac nhau cua cac nhan td khac.
Hai nhan td dugc n6i la c6 tuong tac lin nhau anh hudng ddn cay tr6ng ndu anh hudng cua
nhan 16 nay bi anh hudng cua nhan td kia. Nghidn cun vf du dudi day d£ hidu ro anh hudng
tuong tac cua hai nhan td thi nghidm: Gidng (X v£ Y) va rmic bon dam I6n nang sudt Ma
(Gomez, 1984).
Gidng
N9ng suit lua (t/ha)
Kh&ng bon dam Bdn 60 kg/ha Trung binh
TrUdng hop khong co anh hudng tUdng tie giufa 2 nhan 1d
?
X 1.0 3.0 2.0
Y 2.0 4.0 3.0
Trung binh 1.5 3.5
TrUdng hop c6 anh hudng tuong tac giuTa 2 nhln td
X 1.0 1.0 1.0
Y 2.0 4.0 3.0
Trung binh 1.5 2.5
Phan tich:
+ Khi co tuong tac cua ca hai ydu td gidng vd dam
- Anh hudng cua gidng Iliakhi khdng bdn dam: 2.0 -1.0 = 1.0 t/ha (1)
- Anh hudng cua gidng Makhi cd bdn dam: 4.0 -1.0= 3.0 t/ha (2)
- Anh hudng bdn dam ddi vdi gidng X: 1.0 -1.0 = 0.0 t/ha (3)
- Anh hudng bdn dam ddi vdi gidng Y: 4.0 - 2.0 = 2.0 t/ha (4)
+ Khi khdng cd tuong tac cua hai ydu td gidng va dam
- Anh hudng cua gidng Ma khi khdng bdn dam: 2.0 -1.0 - 1.0 t/ha (5)
- Anh hudng cua gidng Ma khi co bdn dam : 4.0 - 3.0 = 1.0 t/ha (6)
- Anh hudng bdn dam ddi vdi gidng X: 3.0 -1.0 =2.0 t/ha (7)
- Anh hudng bdn dam ddi vdi gidng Y: 4.0 - 2.0 =2.0 t/ha (8)
44

41. Qua so sdnh nhttng truing hop trdn chung ta nhdn thd'y rang: Trong trudng hop khdng
cd anh huimg tuong tdc ctta hai nhdn td, khi luong bdn dam tang su thay d6i nang sudt Ida
ddi vdi hai gi<5ng Id gidng nhau (ddu Id thay ddi 2.0 t/ha); vd anh hucmg cua gidng trong
cung mOt ndn dam cOng la gidng nhau (ddu Id thay d6i 1.0 t/ha).
Nhung khi c6 su tuong tac gitta hai nhdn 16 ndy su thay d6i trong nang sudt ltta la
khac nhau. Vi du khi thay ddi luong dam bon tit 0kg ldn 60 kg/ha nang sudt gidng
lua X khdng thay ddi, nhung ddi vdi gidng Y nang sudt da tang tit 2.0 t/ha ldn 4.0 t/ha.
De xac dinh hai nhdn 16 c6 tuong tac vdi nhau ta c6 thd tfnh:
Tuong tac cua (Gidng x Dam) = 1/2 [(1) - (2)] = l/2[(4) - (3)]
= 1/2 [(6) - (5)] = l/2[(8) - (7)]
Vdy khi cd tuong tdc: 1/2 [3.0 - 1.0] = 1/2 [2.0 - 1.0] = 1.0 t/ha
va khi khdng cd tuong tac: 1/2 [1.0 - 1.0] = 1/2 [2.0 - 2.0] = 0.0 t/ha
Ta co thd dung d6 thi dd minh hoa tuong tac cua hai nhdn td. Ducmg bidu didn trdn dd
thi (a) cho thd'y khi khdng c6 anh hucmg tuong tac gitta gidng vd mttc b6n dam, nang sudt
lua cua hai gi<5ng tang gidng nhau khi d ndn phdn dam 60 kg^ia. Trong khi ndu c6 anh
hucmg tuong tdc, nang sudt cua cac gidng thay d6i la khdc nhau gitta cac ndn dam khac
nhau, nhu duoc bidu didn trdn d6 thi (b), (c), vd (d).
Trong thi nghidm ndng nghidp chung ta rdt cdn tun hidu anh hudng tuong tac nay
de c6 bien phap ky thudt dung. De tim hieu anh hucmg tuong tac cua cac nhdn td thi
nghiem chung ta sir dung thi nghidm nhdn td. Mdt thi nghidm ndu c6 ttt hai nhdn td trd
ldn duoc goi la thi nghiem nhdn td. C6 hai loai thi nghidm nhdn td: Hoan chinh vd khdng
hodn chinh.
Thi nghidm nhdn td hoan chinh la tdt ca cdc phdi hop cua cac nhdn td ddu duoc dua
vdo 1dm thi nghidm. Sd cdng thttc trong thi nghidm nhdn td tang ldn nhanh chdng cung vdi
su tang cua nhdn td tham gia va mttc thay d6i cua cdc nhdn td.
Sd cdng thttc thudc loai hoan chinh Id: (a x b x c x ... x n). 6 ddy a, b, c, n la mttc
thay ddi ctta cdc nhdn td tuong ttng. Vi du: Thi nghidm cd 3 nhdn td A, B, C c6 cdc mttc
thay d6i ctta cdc nhdn td la a = 2; b = 3; c = 5. Vdy sd cdng thttc la: 2 x 3 x 5 = 30. Ndu
trong trudng hop cac nhdn td ddu cd mttc thay ddi gidng nhau, sd cdng thttc ctta thi nghidm
ndy Id m“. 6 ddy n Id sd nhdn td, m Id mttc dd thay ddi ctta nhdn td.
Vi du: Thi nghidm hai nhdn td gidng vd hai mttc b6n phdn ddu c6 hai mttc thay ddi.
Sd cdng thttc thi nghidm Id 22 = 4. Thi nghidm c6 ba nhdn td vd ddu hai mttc thay ddi. Sd
cdng thttc Id 23= 8.
45

42. Vi sc> luang cdng thiic trong thi nghi&n nhan 16 tang nhanh ch6ng gay nhi&i kho
khan cho thidt lap thi nghidm hoan chinh. Do d6 ngudi ta chon loai bo m6t s6 c6ng thttc
khong can thiet. Thi nghi&m do goi la thi nghi&n nhan td khdng hoan chinh.
Do thi 3.1: Anh hudng cua tucfng tac giita nhan togidhg va mite bdn dam: (a) khdng tucfng tac;
(b,c, d) cd tucfng tac (Gomez, 1984)
46

43. 3.4.2. Kieu thi nghiem 16 phu - Split - Plot - Design
Thi nghifim m6t nhan t<5 c6 thd b6 tri theo c£c kilu khac nhau nhu CRD, RCB...
Nhung dtfi vdi thf nghifem c6 hai nhan t<5thi ki&i thi nghi&m phil hap la Kidu thi nghifem 16
phu - Split - Plot - Design. Ki&i thi nghifem n&y cho ph6p ta ki&n tra 3 loai anh hucfng:
- Anh hu6ng cua nhan t<5chinh
- Anh hucmg cha nhan t6'phu
- Anh hucfng cua tuang tic gifla 2 nhan t6
D6 chinh xac (hay n6i each khac la kha nang xic dinh sai khac) cua nhan t6' phu
tang I6n ga'p n l&i (6 day n la mdc thay d6i cha nhan t6 chinh) so vdi d6 chinh xic cua
nhan t6' chinh. Do d6 khi b6 tri thi nghifem, c6 gai $ ia: Nhan t6 nao kh6 kidm tra d6
sai khac ban thi n6n b6 tri nhan t6 ay la nhan t6 phu. Tuy nhi£n trong nhi£u truemg
hap vi do khac nhu di£u ki£n d6ng rudng hoac tinh chat c&a nhan tCf ma gai ^ tr£n
kh6 thuc hi£n.
Vi du: K6t qua thi nghidm hai nhan t6': vdi 6 mtic d6 b6n phan N khde nhau, tr£n 4
gi6'ng lua khac nhau vdi 3 Bin nhac lai) trinh bay b bang 4.
Gqi nhan t6 cua 6 chinh ia A - mdc b6n phan, nhan ttf efia 6 phu ia B - gidng lua.
Cac buofc phan tich su sai khac nhu sau:
Bu&c 1: Lap bang phan tich suesai khac
Bang phdn tich phuang sai - ANOVA
Ngu6n bien (Jong df SS MS ^tin
h
F bang
5% 1%
Nhlc lai r -1 =2
Nh§n t<
5chfnh (A) a -1 = 5
Ngiu nhien (a) (r- 1)(a -1) = 10
Nhiri ttf phu (B) b -1 =3
Tifdng tic (A x B) (a - 1)(b -1) = 15
Ngiu nhiin (b) a(r - 1)(b - 1) = 36
Tdng stf rab -1 = 71
47

44. Being 3.4: Nang sudt cua 4 gid'ng Ida v&i 6 mtic bdn dam v&i 3 Idnnhac lai
bo tri theo kieu Split - Plot Design (Gomez, 1984)
Gifing •* NSng suift (kg/ha)
Nhiclaii If III : -RSng
N0(0 kg N/ha)
V, (IR8) 4.430 4.478 3.850 12.758
V2(IR5) 3.944 5.314 3.660 12.918
V3(C4 - 63) 3.464 2.944 3.142 9.550
V4(Peta) 4.126 4.482 4.836 1 13.444
Tong 15.964 17.216 15.488
N1(60 kg N/ha)
v. 5.418 5.166 6.432 17.016
v 2 6.502 5.858 5.586 17.946 '
v 3 4.768 6.004 5.556 16.328
• V4 , 5.192 4.604 4.652 j 13.444
Tong 21.880 21.632 2.226
N2(90 kg N/ha)
v. 6.076 6.420 6.704 19.200
v2 6.008 6.127 6.642 18.777
v3 6.244 5.724 6.014 17.982
v4 4.546 5.744 4.146 14.448
Tong 22.874 24.015 23.506
N, (120 kg N/ha)
v. 6.462 7.056 6.680 20.198
v 2 7.139 6.982 6.564 20.685
v 3 5.792 5.880 6.370 18.042
v 4 2.774 5.036 3.638 11.448
Tong 22.167 24.954 23.252
•
• - -. |
N4(150 kg N/ha)
v. 7.290 7.848 7.552 22.690
v 2 7.682 6.594 6.576 , , 20.852
v3 7.080 6.662 6.320 20.062
v 4 1.414 1.960 2.766 6.140
Tong 23.466 23.064 23.214
N5(180 kgN/ ha)
v. 8.452 8.832 8.818 26.102
v 2 6.228 7.387 6.006 19.621
v 3 5.594 7.122 5.480 18.196
V4 2.248 1.380 2.014 5.642
Tdng 22.522 24.721 22.318

45. Bitoc 2: Lap 2 bang tdng nang suat
Bang thu nhat tong nang suat cua nhan ttf A
- So lan nhac lai x nhan to A
- Nhac lai,
- Nhan ttf A
- Toan bo.
Trong vi du nay la bang 3.5 - bang nang suat cua
- Nhac lai x muc phan dam (RA),
- Nhac lai (R),
- Cac rmic phan dam (A) va
- Toan b6 (G)
Bang 3.5: Nang suat theo mite phan dam &cac lan nhac lai ( Tinh tit bang 3.4)
MOc dam
Tong ndng suat (RA) Tong NS theo dam
(A) - nhan to chfnh
Nhac lai 1 Nhic lai II Nhic lai III
N0 15.964 17.218 15.488 48.670
N, 21.880 21.632 22.226 65.738
n2 22.874 24.015 23.506 70.395
n3 22.167 24.954 23.252 70.373
n4 23.466 23.064 23.214 69.744
n5 22.522 24.721 22.318 69.561
Tong cua nhic lai (R) 128.873 135.604 130.004
Tong toan bo (G) 394.481
Bang thu: hai la bang nang suat cua nhan t6' B
- Nhan to A x nhan to B
- Nhan to B.
Trong vi du nay la bang 3.6 bang nang suat theo
- Cac miic dam x cac giong khac nhau (AxB),
- Theo gidng (B).
Bang 3.6: Nang suat theo mite dam x giong (2 yen to nghien ettu) (tinh tit bang 3.4)
Mijfc dam T&ig n3nc suat (AB)
V, V2 V3 V4
N„ 12.758 12.918 9.550 13.444
N, 17.016 17.946 16.328 14.448
n2 19.200 ^ 18.777 17.982 14.436
n 3 20.198 20.685 18.042 11.448
n4 22.690 20,852 20.062 6.140
n 5 26.102 19.621 18.196 5.642
Tdng NS theo giong (B) 117.964 110.799 100.160 65.558
49

46. Buc/c 3: Trnh s6' hi&u chinh va tdng brnh phuong bidn ddng cua nhan t6' chrnh theo
cong thrtc sau:
C.F. = — = (394481) = 2.161.323.047
rab (3)(6)(4)
Tdng brnh phuong bidn ddng cua toan bd
TSS= £ X 2 - C-F.
T6ng bmh phuong bidn dOng do nhac lai
Y r:
SS^ = K
ab
C.F
Tdng brnh phuong bidn dong do nhan td A (dam)
.
SS,
rb
CF
Tdng bmh phuong bidn ddng do nglu nhidn A (dam)
_ Z (R A )2
SSe, - c f - s s r - ssa =
Trnh tdng bmh phuong bidn ddng cua khoi phu B (theo gidng):
Tong brnh phuong bidn dong cua B (gidng)
_ 2 > 2
SS„ = ------ CF
ra
Tdng brnh phuong cua AB (dam va gidng)
,2
SSAB' c f - ssb - ssa
Tong bmh phuong cua sai sd do B (gidng)
SSgb=T6ng SS - (tdng cua tat ca SS Jchac)
Vdi vr du trdn:
Tdng brnh phuong bidn dOng cua toan bd
TSS = £ X 2 -C .F .
= [(4.430)2 + ... + (2.014)2] - 2.161.323.047= 204.747.916
Tdng brnh phuong bidn ddng do nhac lai
SSR
= = (128.873)2 +(135.604)2 + (130004)2 323
ab ' (6)(4)
047 = 1.082.577
50

47. Tdng binh phucmg biSn d6ng do nMn t6 A (dam)
55 Z A* CT (48.670)2 +... + (69-561)2
A rb (3X4)
T6ng binh phucmg bi£n d6ng do ngSu nhi£n A (dam)
2.161.323.047 = 30.429.200
b
C F -S S r -S S a =
(15.964)2 +... + (22.318):
(4)
2.161.323.047 -1.082.577 - 30.429.200 = 1.419.678
Budc 4: tfnh t6ng binh phucfng biSn dCng cua kh<5i phu B (theo gi6ng):
T6ng binh phucmg bi6n dCmg cua B (gib'ng)
SSg = ^ ------ CF =
ra
(117.964)2 +... + (65.558)
(3)(6)
T6ng binh phucmg cua AB (dam va gib'ng)
S (A B ):
2.161.323.047=89.888.101
SSab ~ C F -S S b -SS,
= (12-758) + - + (5-642)— 2.161.323.047-89.888.101-30.429.200= 69.343.487
3
Tong binh phucmg cua sai s6' do B (gi6'ng)
SSj,b= T6ng SS - (t6ng cua tat ca SS khac)
= 204.747.961 - (1.082.577 + 30.429.200 + 1.419.687 + 89.888.101 + 69.343.487)
= 12.584.873
Budc 5: Tinh trung binh cua tdng binh phucmg (MS) cho titng ngu6n bi£n dbng bang
each chia t6ng binh phucmg cho d6 tu do tuemg ung.
MS cua nhac lai
SS„ 1.082.577
MSr =
_ R _
r - 1
= 541.228
MS cua nMn t6' A
MS* = = 3Q-429-200 = 6.085.840
a - 1
MS cua nghx nhifen c&a nhM 16 A
SSfia
( r - l) ( a - l)
1.419.678
10
= 141.968
51

48. MS cua nhan tb B
SS^ = 89.888.101 = 29 962 700
b -1 3
MS cua hai nhan t<5AB
M S ^ =
SSA.B
( a - l) ( b - l)
MS cua ngSu nhifen c&a nhan tb B
69.343.487
15
= 4.622.899
M S ^
SSEb 12.584.873
= 349.580
a(r - l)(b -1 ) 36
Buac 6: Tmh gia tri F cho timg nhan t6
' din ki&n tra, bang each chia MS cua nhan to
do cho MS ngiu nhien cua no:
6.085.840
MSa
F(A) = ------
MSEa 141.968
F(B)
MSb 29.962.700
F(AxB) =
MS*,
MSA.B
349.5^0
4.622.899
42,87
= 85,71
= 13,22
MSg, 349.580
Buac 1: Vbi nhftng nhan t(5co gia tri F tfnh ldn hem 1,
Tra F bang b bang E vbi
- fj = df cua MS tut s<5va
- f2= df cua MS cua m lu sb.
Vbi vf du tren F bang cua:
- F (A) vori df,la 5 va df2la 10 F0
5= 3.33;
- F (B) vbi dfjla 3 va df2la 36 F0
5= 2.86;
- F (AB) vdi dfjla 15 va df2la 36 F0
5= 1.96;
Buac 8: Tfnh he sb bien dbng cho khb'i chrnh va khbi phu.
F0
1 = 5.64
F0
1 = 4.38
F0
1 = 2.58
cv(A) = .100 = ^ 14L% 8 .100 =6.9%
G 5.479 •
cv(B) = f e l O O = ^349 58° .100 = 10,8%
G 5.479
52

49. Gia tri cua cv(A) chi dd chinh xac cua nhan td b khdi chinh. Gia tri cua cv(B) chi dd
chinh xac cua nhan 16 b khdi phu. Gia tri cua cv(B) thucmg nho hem cv(A) vi nhu da trinh
bay b phln trudsc nhan td c&a l^ ie h in h thuomg ydu clu dd chmh xac it horn so vdti nhan td
c&a khdi phu. =
Bude 9: Lap bang phan tich sai khac vdi cac gia tri da tinh duoc tut budre 3 ddn buctc
8. So sanh gia tri F finh vdfi F bang nhu hudmg din b CRD
Bang 3.7: Phan tick bien dong - ANOVA cua nang sua't Ida d bang 3.5 cho (hi nghiern
bo tri theo Men 6 phu - Split - plot designa
; Nguon bie'n dong df SS MS F ti'nhb
F bang
5% 1%
Nhlc lai 2 1.082.577 541.228
Dam (A) 5 30.429.200 6.085.840 42,87** 3,33 5,64 '
‘ Sai so (a) 10 1.419.678 141,968
Gidng (B) 3 89.888.101 29.962.700 85,71** 2,86 4,38
Ax B 15 69.343.487 4.622.899 13,22** 1,96 2,58
Sai so (b) 36 12.584.873 349.580
Tong so 71 204.747.916
a
cv(a)=6,9%,:cv(b)=10,8%;b
** = c6 f nghla bmtic 1%
Kdt luan:
- Cac mute dam khac nhau va giong khac nhau co anh hudng khde nhau t<5inang suit Ida.
- Cac mute bon N khac nhau co anh hudmg rdt khde nhau ddn nang suat (ct mute tin cay
99% ).
- Cac giong khac nhau co anh hudmg rdt khac nhau ddn nang suit (b mute tin cay
99%).
- Tac ddng phb'i hop cua cac giong vdi cac mute dam (anh hucrng tuemg tac) cd anh
hudmg ddn nang suat rdt chac chan (ct mute tin cay 99%).
3.4.3. Thi nghiern ba nhan to trot len va cac kieu thi nghiern thich hop
Ve mat nguyen tac, khi tang nhan td thi nghidm ta cd thd cai tien kieu thi nghidm
Split - Plot - Design bang each chia Id phu thanh cac mute thdp hon: Split - Split...Split - Plot
Vi du thi nghidm 3 nhan td: 2 Gidng (v=2), 3 loai thude true co (w=3) va 4 mute bdn
dam (n=4), 3 lln nhac lai. Bd tri theo kieu Split - Split - Plot - Design.
Cac Id chinh, Id phu clp 1 v i Id phu c'Sp 2 duoc bd tri nglu nhidn.
53

50. Sa do 3.1. Thi nghiem kieu Split - Split - Plot - Design
Bang ANOVA cua thi nghiem Split - Split - Plot - Design
Nguon bien dong Ddtifdo(df) ss MS F .
F
1bing
rtin
h
05 01
Nh&c lai
Nhint6 6 chfnh
- Giong (V)
- Thuoc trit co (W)
- Tifdng tac (VxW)
Nglu nhien (a)
Nhan to 6 phu (N)
Tifdng tac
- Tifdng tac: (NxV)
- Tifdng tac; (NxW)
- Tifdng tac:
(NxVxW)
Ngau nhi£n (b)
Tong bien dong
r - 1 = 2
vw- 1=5
v - 1=1
w - 1=2
(v-1Xw-1)=2
(r- 1)(vw -1)=10
n -1=3
(vw - 1)(n - 1)=15
(n - 1)(v-1)=3
(n - 1)(w -1)=6
(n- 1)(v- 1 )(w- 1)=6
vw(r- 1){n - 1)=36
rvwn -1=71
3.4.4. Kieu thi nghiem Strip - Split - Plot • Design
Vi du: Thf nghi&n nh&n ttf vdi 3 mdc phfln dam (Nj - N3), 6 gitfng Ida (Vj - V6), v&2
each thde gieo ca'y (xa Pj vd c&y P2) (Gomez, 1984).
54

51. Dac di&n cua kilu thf nghi6m nay la: Co 4 loai kich thucdc 16:
- L6 c6t (nhan t6 dam N)
- L6 hang (nh&n t6' gidng V)
- O Wong tac: co 3 loai: (N x V), (N x P), va (N x V x P)
- 6 phu (mat d6 cay P)
Tuomg ling vdi 4 loai kich thudc 16 nay co 4 d6 chmh xac khac nhau.
Dudi day la so d6 thf nghiem Strip - Split - Plot - Design. Cac 16, 6 dvtoc b6 tri ngau
nhi&n.
N, N3 N2
Set do 3.2. Thi nghiem kieu Strip - Split - Plot - Design
Bang ANOVA kieu tM nghiem kieu Strip - Split - Plot - Design
------------ ,
---------- ......... .
Nguon bi£n dong 06 tu do (df) SS MS F ^b4ng
■t»nh
05 01
Nh£c lai r- 1 = 2
Lo cot - Dam (A) 1 a - 1=2
Ngau nhien (a) (r * 1Xa -1) = 4
Lo hang - Giong (B) b - 1=5
Ngau nhien (b) (r- 1X b -1)=10
A x B (a - 1X b -1)=10
Ngau nhien (c) (r-1Xa-1Xb-1)=20
6 phu - kieu gieo cay (C) (C -1)=1
Ax C (a - 1)(c -1)=2
BxC (b - 1)(c - 1)=5
A x B x C (a-1Xb-1Xc-1)=10
Ngau nhien (d) ab(r-1Xc-1)=36
Tong bien dong rabc -1=107
55

52. Chiromg 4
SO SANH SO TRUNG BINH
So sknh cac s<5 trung binh cua cdng thftc thf nghidm vdi nhau dd phuc vu cho muc
dfch cua thf nghidm lk tim ra cdng thftc thf nghidm cd hidu qua nhkt. Vf du: So sknh 6 loai
thudc trftrky nku trongthf nghidm thft nghidm thudc sku (bang 1). Dd lkm duqc vide nay, ta
phki so sknh cac sd trung binh cua cac cdng thftc bang vdi nhau. Nhung vi sd trung binh
cua ckc cdng thftc duqc udc luqng thdng qua m iu do vky su sai khkc cua chung phai duqc
kilm chung theo phuong phkp thdng kd dd dam bko kdt lukn lk chung c6 thuc su khkc nhau
hay khdng.
Do muc dfch nghidn cftu rkt da dang do vky c6 nhidu ckch so sknh khac nhau Nhung
chftng duqc phkn loai thknh hai nh6m chfnh: So sank cap ddi (Pair Comparison) vk So
sanh nhom (Group Comparison). So sknh ckp ddi c6 hai phuong phkp: So sanh sai khac
nho nhdt co y nghia - Least Significant Diference Test (LSD), vk So sanh Duncan -
Duncan's Multiple Range Test (DMRT). So sanh nhdm c6 bdn phuong phkp: So sanh gida
cac nhom cdng thiic v&i nhau - Between - group Comparison; So sank gida cac cdng thiic
trong nhom - Within - group Comparison; So sank xu th i- Trend Comparison; va so sanh
nhan to - Factorial Comparison).
Trong cac nghidn cftu trdng trot thudng sft dung ckch so sknh ckp ddi do d6 trong tki
lidu nky chung tdi chi trinh bky ckc phuong phkp so sknh cap ddi. So sknh nhdm tham
khao trong Gomez, 1984.
Trong so sknh ckp ddi, hai phuong phkp LSD vk Duncan vd co ban lk gidng nhau: D6
la so sknh ckc sd trung binh cfta ckc cdng thftc bang ckch so sknh su sai khkc gifta chung
vdi gidi han cua su sai khkc c6 f nghia. Nhung trong thuc t€ khi sd luqng cdng thftc cua thf
, nghidm ldn, thi phuong phkp LSD se khdng thfch hop do su sai khkc gifta bkt eft ckc cap
cdng thftc lk khac nhau. Do dd khdng thd sft dung mdt gik tri gidi han nho nhkt cd $ nghia
(LSD) dd dung cho so sknh ckc cap ddi khkc nhau.
Ckc thf nghidm cd sd cdng thftc nh6 hon hoac bang 6, hay thf nghidm da duqc dinh
san ckc cap ckn so sknh, vf du nhu so sknh ckc cdng thftc trong thf nghidm vdi cdng thftc
ddi chftng thi ngudi ta sft dung phuong phkp LSD. Ndu muc dfch thf nghidm lk so sknh bkt
ckp ddi cdng thftc nko vdi nhau d thf nghidm cd sd cdng thftc ldn hon 6 thi ngudi ta sft
dung phuong phkp DMRT (hay dd goi lk phuong phkp Duncan).
Trudc khi thuc hidn so sknh ta ckn nghidn eftu kdt quk phkn tfch phuong sai
(ANOVA). Ndu F, > Fb cd nghia ring cd su sai khkc cd ^ nghia vk ta thuc hidn cdng vide
so sknh.
56

53. 4JL SO SA N H THEO SAIKHAC N H 6 NHAT - LEST SIGNIFICANT DIFERENCE
* T E S T -L S D
Sit dung phuong pMp LSD trong so sknh lk don gian nhkt va hay duac ditng d£ so
sfinh ckp. LSD Ik gik tri nhd nhkt chi ra gianh gidi gitta khac nhau c6 $ nghTa vk khac nhau
khdng c6 y nghta gitta ckc cap cia bkt ky cap cdng thtfc nko.
Khi so sknh gitta hai cdng thtfc d mtfc xac sudt “oc” nko dky vdi LSD* (Critical Value
for Comparison trong phkn mdm SX3), ndu su sai khkc cda chiing (D;j):
[Dj
j| > LSD* duac coi lk hai cdng thtfc i vk j khkc nhau chac chan d
rmtfc «;
jDjj] < LSD« thi duac coi lk hai cdng thtfc khdng khtfc nhau.
Cach ttuh:
Buffc 1: TMi su sai khac cua sd trung binh cua hai cdng thtfc thtf i vk j
d i i = x ; - x :
Trong dd X;vk Xj la trung binh cua cdng thtfc thtf i vkj.
Btfdc 2: Tmh LSD tai mtfc cd y nghta a
LSDa = (ta)(S5)
Sd 1a sai khkc chukn gitta ctfc s6 trung binh, vk ta (lk Critical T Value trong SX3) lk
gia tri t bang tra ttt phu luc C tai mtte cd ^ nghta a vdi dd tu do = n
Bu&c 3: So stfnh su khkc nhau gitta cac sd trung binh tftih duac &budc 1 vdi gia tri
cua LSD tinh duac d budc 2.
♦
Cach tmh sd
Khd'ihodn chink (RCB)
Vdi thi nghidm bd tri khdi hokn chinh nhu: Nglu nhi&n hokn torn, Khdi ngku nhidn
hokn chinh, Latin square chi cd mdt sai sd ngku nhidn vi vky sai sd chu&t Sd duac dttng cho
tat ca ctfc cap so sanh nhu nhau.
Trong dd: r = sd lkn nhac lai; s2= MSg cua sai sd ngSu nhidn.
Cdthdvidt S-
V i du: Ldy th£ nghidm CBD vdi 6 loai thudc sku va 1 ddi chtfng vk 4 lkn nhac lai cd
MSp = 94.773.
57

54. Mue dlch cua thf nghidm lk xac dinh trong 6 cOng thutc thi nghidm rihftng cOng tMc
nao tdt hon ddi chung. Vdi thi nghidm nay thich hop nhkt lk so sanh sd trung binh cua 6
c6ng thtic thf nghiem vdi sd trung binh cua dd'i chung.
Budc 1: Tmh su sai khke giuasd trung binh cua cOng tMc thi nghidm vdi ddi chutng,
kdt qua ghi d bang 4.1
dij = X; -Xd.c
Trong do X; = trung binh cua cOng tilde thut i ; Xd.c = trung binh cua cdng tMc ddi
chutng.
Budc 2: Tmh LSD tai mute cd y nghia a
LSDa = (ta) yj~~ = ( ^ ) ]] ^ f E-
Trong vf du nay, MSg = 94.773; dfE= 21; r = 4.
Tut bang C tra ta vdi dd tut do n = dfE= 21; duac = 2,080; ^ = 2,831.
Tmh LSD:
LSD 0
5= 2,080 J l& L IT Q =453 kg/ha
LSD.0
1 = 2,831 = 616 kgfca
Budc 3: Sosknh gik u i tuydt ddi c&a dy tmh d bode 1 vdi LSD tinh d budc 2. Ndu:
Ndu d;dc > LSD thi khke nhau gifia trung binh cdng tMc i vk ddi chiing lk dkng tin
cky (Ndu dj dc mang gik tri duong thi cdng thutc i ldn hon ddi chiing vk nguqc lai ndu didc
mang gia tri km thi cdng thutc i nho hon ddi chutng).
didc < LSD thi sai khac nhau khdng dkng tin cky.
De bieu thi mute sai khac cd y nghia, ta dung dku * dknh b dy.
>LSD 0
1 cd nghia su sai khke d mute tin cky 99% vk dknh 2 dku *
>LSD0
5cd nghia su sai khac d mute tin cky 95% vk dknh 1'dtfu *
Ndu
Ndu
dy
dy
Vdi vf du bang 1: dd_7= xi - X
dc = 2127 -1316 = 811 kg/ha > LSD0
1
Vky cdng tMc 1 cd nang sukt cao hon ddi chdng chac chan, d mute 99% vl vky dknh
dku d j.7 **
Kdt qua so sknh tidp cho thky tr£t cdng thutc 6 tkt ck eke cdng tilde khke ddu cho nkng
sukt cao hon ddi chdng dkng tin cky.

55. Bang 4.1: So sdnh trung binhndng sudft cua doichvmg vdi 6 cdng thitc
sii dung thuoc sdu khdc nhau
-- , 'U
, ..j "
Cdng thCte trung binh * kg/ha So vdi ddi chCfngbkg/ha X i —Xa.c
1 2.127 81T
2 2.678 1.362**
3 ■ : 2.552 ' 1.236**
4 2.128 812**
5 : 1.796 480*
6 1.681 365n
s
Boi chufng 1.316 -
a= Trung binh cho 4 lin nhac lai; b
** = Co y nghia bmtic 1%
* = Co dang tin c|y 6 mtic 5%; “ “Khdng cd f nghia - hay kh6ng ding tin c&y
Kdt ludn: Cac c6ng thic s i dung thu6
'c sdu 1, 2, 3, 4 vd 5 d£u cho nang sudt cao hcfn
han so vcri ddi ching. C6ng thic 6 kh6ng cho nang sudt khdc dtfi ching.
Split - Plot - Design
Vcri ki&i Split - Plot - Design c6 hai nhdn t<5vd hai h6 thdhg sai s6, do d6 c6 bdn ki&i
so sdnh cap.
• So sdnh hai s6' trung binh cua kh6'i chinh (Trung binh qua cac cftng thic cua khdi
phu (hay 16phu)).
• So sdnh hai s<5trung binh cua khd'i phu (Trung binh qua cdc c6ng thic cua khd'i chinh)
• So sdnh hai trung bihh cha khdl phu vdi cing m6t cdng thic 6&akhdi chinh
• So sdnh hai trung binh cua khdi chinh vdi cing hay khdc cdng thic cia kh<5i phu
(co nghia la cua bdt ky mdt cap cdng thic nao).
Bang 4.2: Tinh sai sd'chuan cua cac sd'trung binh cho bdh kieu so sdnh
cap cua Split - Plot - Design
Kieu so sanh sa
5
• So sanh 2 sd trung binh cCia khdi chfnh (Trung binh
qua c£c cdng thtfc cCia khdi phu)
V rb
• So s&nh 2 sdtrung binh cOa khdi phu (Trung binh /2iT
qua c£c cdng thifc cua khdi chfnh)
V ra
• Sd s£nh 2 trung binh cCiakhdi phu vdi cung 1 cdng (2e T
thdc cCia kho'i chinh
V r
* So sanh 2 trung binh cCia khdi chinh v6i cung ho3c 2 [(b -l)E b +E a]
kh£c cdng thifc cilia khdi phu
i rb
Ea= MS cda sai sd a, Eb= MS cua sai sd b, r = sd nhac lai, a =
*sd cdng thurc cia khdi chinh, b = sd
cdng thtic cua khdi phu.
59

56. Khi phan tich phucmg sai cho thf nghidm kidu Split - Plot - Design da ndu d phdn
trade cho thay c6 svt khac nhau ro rang gifta ede mdc dam vd ede gidng. Vi vdy vide so
sdnh trung binh nang sudt eda ede mdc dam (trung binh qua ede gi<5ng) hodc trung binh cua
ede gidng (trung blnh qua ede mute dam) khdng c6 f nghia. C6 $ nghia hon Id so sdnh ede
gidng vdi cung mdc dam hoac ede mdc dam vdi edng mdt gidng. Tuy nhidn chi c6 vide so
sdnh trung binh cua ede gidng vdi cuing mdc dam c6 f nghia thuc tiSn hem. Vi vdy 6 ddy
trinh bay so sdnh ndy.
Bude 1: Tinh sai sd chudn: vdi Eb= MS^, = 349.580
Sd
2Eb
r
2MSeb
r
2(349.580)
482,8 kg/ba
Bude 2: Tut bang C vdi n = df^, = 36 tra to5= 2,029; t^ = 2,722
Buac 3: Tinh LSD
LSD a = (ta)(s5)
LSD 0
5= (t 0
5)(s5) =(2,029)(482.8) =980 kg/ha
LSDo; = (t01)(s-d) = (2,722)(482.8) = 1.314 kg/ha
Bude 4: Lap bang nang sudt trung binh cua gi<5ng b ede mdc dam so sdnh s6 trung
binh cua hai gidng vdi cuing mdc dam nha b bang sau:
Bang 4.3: Nang sudt trung binh cua 4 gidng Ida vdi 6 note phdn dam
botri theo kiiu Split - Plot - Design (so lieu tit bang 4)
MlT
c dam
Trung binh kg/ha
IR8 IR5 C4 Peta
0 4.253 4.306 , 3.183 4.481
60 5.672 5.982 5.443 4.816
90 6.400 6.259 5.994 4.812
120 6.733 6.895 6.014 3.816
150 7.563 6.951 6.687 2.047
180 8.701 6.540 6.065 1.881
a= Trung binh cua 3 Lin nhac lai; LSD 0
5= 980kg/ha; LiSD^ = 1.314 kg/ha.
Mudn so sdnh hai gidng nao d6 vdi cimg mile dam, edn tfah sir khde nhau gitta chung
va so sdnh vdi LSD.
Vi du :
So sdnh hai gidng lua Peta va IR8 d mdc dam 0:
4.481 - 4.253 = 228 kg/ha < LSD0
5khde nhau khdng ddng tin cdy.
-Hay b mdc dam 0 gidng Peta vd ER.8 cd nang sudt nhu nhau.
60

57. So sanh C4 vdi IRS ormile dam 0
4.253 - 3.183 = 1.070 < LSD0
5khde nhau dang tin cdy
Hay IR8 c6 nang sudt hem C4 chac chan, dd tin cdy 95%
So sanh IRS voi Peta &mere dam 180
8701 - 1881 = 6.820 > LSD0
1 khac nhau rdt chac chan
Hay IR8 co nang sudt hern Peta rdt chac chan, dd tin cay 99%.
4.2 SO SANH DUNCAN (LITTLE M. VA H ILLS, 1978)
Vdi nhung thi nghidm co so c6ng there lcm va ydu cdu so sanh cac cap edng there bat
ky, dung LSD khdng thich hop. Trong tnicfng hop d6 dp dung so sanh Duncan (DMRT).
Trong so sanh Duncan edn phai tuih duoc set khac nhau nho nhdt Dp- cua tdt ca cac khoang
gifta cac so tnmg binh trong day xd'p hang.
Dp= R x LSD
Trong d6 Dpla set khac nhau nho nhdt cua mdt khoang edeh n&o dd
R Id gia tri bang tmh phu thude vao dd tu do cua ngdu nhidn va vi trf eda sd' trung
binh trong bang xdp hang.
VI du ve anh huomg cua lieu luomg chd't ddu qua dfih nang sudt ca chua. Co 4 edng
there nhu sau:
Gi5ng c§ chua
Lieu li/dng chfi <&u qua
0 3mg/100 m2
Dia phifcJng FS0 fs3
Lai MS0 m s 3
Bang 4.4: Nang sudt ca chua dat duqc sau khi svedung chdt ddu qud (kg/100 m2
)
Cong thCfc Khoi Cong thiirc
I il III IV Tong Trung binh
FSe 47 52 62 51 212 53
MS0 50 54 67 57 228 57
fs 3 57 53 69 57 236 59
m s 3 54 65 74 59 252 63
Tong khoi 208 224 272 224 928 58
Trung b]nh khoi 52 56 68 56
Bang 4.5: ANOVA - ket qud phdn tich biert dong nang sudt ca chua d bang tren
Nguon bi£n dong df SS MS F ti'nh
F bang
01 05
Tong 15 854
Khoi 3 576 192 24.69 3.86 6.99
Cong thCfc 3 208 69.3 8.91
Sai so 9 70 7.78
61

58. Bu&c 1: TmhLSD
LSDos = = 2 .2 6 2 ^^^— = 2.262^12(7^ 8) =4.46
Bu&c 2: Tinh gia tri Dpcho cac khoang gifta cac sd trung binh. Co bdn sd trung binh
vi vdy cac khoang gifta cac sd trang bmh la 2, 3 va 4 (hai edng thftc dung canh nhau thi
khoang gifta chung la 2 (2 mdc), hai edng thftc dftng each nhau qua mdt c6ng thftc thi
khoang each la 3 (3 moc). Nhir vay p co cac gia tri 2, 3, 4. Tra bang Fa crmftc 5% vdi dd tu
do n=dfE=9 co cac gia tri R nhu sau.
p 2 3 . 4
R 1.00 1.04 1.07
D = R x LSD 4.5 4.6 4.8
Bu&c 3: Xep cac sd trung binh theo sd tft nho ddn ldn vtt ki&n tra su sai khac
Cong thufe FS0 MS0 FS3 m s3
Trung binh 53 57 59 63
So sanh sd trung binh ldn nhat veri s<5trung binh nho nhSt dftng D4vi khoang gifta FS0
va MS3trong bang xep hang la 4
63 - 53 = 10 > 4,8 vi vay hai s<5trung binh nay khac nhau co <
/nghia
So sanh sd trung binh ldn nh& vdi sd trung binh be thft hai dung D3
63 - 57 = 6 > 4,6 vi v$y hai sd trung binh nay khac nhau chac chan
So sanh sd trung binh ldn nhdt vdi sd trung binh bd thft ba
63 - 59 = 4 < 4,5 VSy khdng cd su khftc nhau gifta hai sd trung binh
Tidp tuc so sanh sd trung binh ldn thft hai vdi c£c sd be hern
59 - 53 = 6 > 4,6 2 edng thftc khac nhau cd $ nghia
59 “57 = 2 < 4,52 edng thftc khdng khac nhau
So sanh sd trung binh ldn thft ba vdi sd be horn
57 - 53 = 4 < 4,52 edng thftc khdng khac nhau
Bu&c 4 : Lap bang kdt qua so sanh bang dudmg thang hoac chft cai
MS3 FS3 MS0 FS0 Hay MS3 FS3 MS0 FS0
63 59 57 53 63a 59ab 57bc 53c
Nhftng sd trung binh co cung dudng thang hay cung chft cai la khdng khac nhau d
mftc 5%.
Kdt luan vd thi nghidm tren: Dung lidu luqng chat ddu qua khac nhau co anh hudng
khac nhau ddn nang sua't eft chua, nhung khdng cd su khac nhau vd nang sudt gifta gid'ng
dia phuong va gidng lai.
62

59. Chirong 5
PHAN TfcH TUONG QUAN
5.1. KHAI NE&M
Trong tut nhien c6 nhilu mO'i quan hfe phttc tap anh hudng lin nhau. Nfeu ta bifet duoc
nhttng mO'i quan hfe nay se giup ta tifen doan duoc nhttng gi se xay ra va giup ta c6 quyfet
dinh hop ly, tifet kiem, va dem lai hieu qu& cao va nhanh chOng. Dang quan hfe gitta cac bifen
la rat khac nhau. MOt ham toan hoc bilu thi mOi quan hfe nay la mOt ham tuong quan.
Phan tfch tuong quan (Regression Analysis) chfnh la u6c luong mttc dO quan hfe
(tham so) gitta cac bifen, va xac dinh dO tin cfey cua cdc tham sfe nay co bilu thi dung mfei
quan hfe gitta cac bifen xay ra trong tu nhifen hay khfeng. Phan tfch hfe s<5 tuong quan
(Correlation Analysis - Coefficient of correlation -"r"or Coefficient of determination -"R2)
la danh gia ham ucrc luctng thOng qua tap hop kfet qua quan sdt trong thuc tfe co phan anh
dung quy luat cua su kifen hay khong.
Trong tr6ng trot, phan tfch tuong quan va hfe s<5tuong quan cho phfep xem xfet t6ng
hop cac quan hfe gitta c&c dac tfnh cua cay vcri nhau; quan hfe gitta phan ttng cua cay vci
nhan 16 thf nghifem va quan hfe gitta phan ttng cfia cfey vdimOi trucng trong thf nghifem.
Phan tfch tuong quan la phan tfch moi quan hfe anh hufefng cua mOt hoac nhifeu bifen dOc lap
Ifen mot bifen phu thuOc. Trong phan tfch nay c£n phai xac dinh no quan hfe gitta bifen phu
thuOc va bifen dOc lap phai la quan hfe co f nghla. Phan tfch hfe sOtuong quan cho bifet mttc
do phu thuOc gitta cac bifen nay. Ta co thi du dodn duoc gid tri cua bifen phu thuOc qua gia
tri bifen dOc lap, hay c6n goi la phan tfch hoi quy.
Co thi phan loai phan tfch tuong quan theo:
- SO
' bifen tham gia: Quan hfe don (Single) hay da bifen (Multiple)
- Loai quan hfe: Tuyfen tfnh (Linear) hay phi tuyfen tfnh (Curvelinear hay Non-linear).
Do do co bO'n loai quan hfe duoc bilu thi dudi ham ting quat dudi day:
- Quan hfe tuyfen tfnh don gian c6 mOt bifen dOc lfep
Y = a + bX
- Quan hfe tuyfen tfnh da bifen
Y = a + bX + cZ + cF +...
- Quan hfe phi tuyfen tfnh don bifen
Y = f(X)
- Quan hfe phi tuyfen tfnh da bifen
Y = f(X, Z, F...)
63

60. Y Y
Quart he phi tuyen ti'nh Khfing quan h6
Trong quan he phi tuy£n trnh, bifin Y c6 quan he phu thufic vdi c$c bi&i dfic lap c6
th£ la theo c£c hkn kMc nhau nhu ham mu, logarit, v.v. D i chon quan he nk) phh hop d l
ufic luong ham dfii vdi quan he hai bifin thl don gian, ta chi c&i dua die quan sfit len d6 thi,
nhin dang phan b6' cua cac di&n quan sat d£ quyfit dinh chon ufic hlpng loai quan he nao
tuyen tfnh hay phi tuyfih tfnh. Tuy nhifin n£u la ham da bi£h thl khfing th£ dung phuong
phip nay vi vfii ham da bi£n ta khfing th£ dua len dfi thi duqc (Tham khao Gomez, 1984).
Dufii day la mfit s<5dang quan he ham thuctag gap:
Ham tuyen tmh: Vf du quan he nang sua't ngfi (Y) v&mCrc bfin dam (X):
Y = a + bX
Ham Exponential: Vf du ham sinh trudrig thuc vat (Y) sau sfi ngay moc (X):
Y = aebX
Ham Asymptotic (Dang dac biet cua Exponential vdi b < 1). Vf du ham phan huy tan
du thuc vat (Y) theo thcri gian (X):
Y =
±a +(l/2)x
64

61. fflni Polynomials: Vf dufifen nangsudt thu hoach c&a dSQ xanh (Y) sau ngly thu
ho^cfaA btita(X ): -
Y = a + bX + cX2+ dX3+...
Thy theo sdm il duSlngbidu didn ham c6 dang kMc nhau:
$6 mu - Ten ham T^n (Judng bi^tfdiin
1 Tuyen tfnh Dudng th£ng
2 Bac 2 Parabol
3 BSc3 Parabol bSc 3
4 Bfc 4 Parabol bde 4
5 B§c5 Parabol bde 5
CM f khi phan tlch tuang quan, ham cang dan gian nhung phil hop vdi su phan bd
cua kdt qua quan s£t thi ia ham ndn dugc chon. Tranh chon ham qua phdc tap ft tinh kha
thi khi ap dung.
5.2. QUAN H $ T U Y fN TINH DON BIEN (SIMPLE LINEAR REGRESSION)
* /
M6
'i quan he gifta hai bidn la tuydn tinh ndu su thay ddi ia mdt hang sd trong pham vi
dugc xem xet. Duong bidu diln cfia quan he nay la mdt dudngthing.
Y = ct + pX
6 day Y la bidn phu thude, X ia bidn ddc lap, a - (intercept) v l p - (slope) la cac
tham sd (a ia didm giao gifta true tung vdi dudng thing bidu dilb mdi quan he Y v l X, va P
iahe sd quan he, hay chlnh la dd ddc cua dudng thing trdn).
- •
Trong phan tfch hdi qiiy hay tuang quan ta khdng did bidt dugc hai tham sd a va P
m l phli ude lugng chting thdng qua tham sd a va b tuang ting. Do dd sau khi ude lugng ta
phli kidm tra c£c tham sd v l him Vila dugc ude lugng cd dim bio tin ciy hay khdng.
Him hdi quy tuydn tinh don giln (hay edn goi 11 md hlnh hdi quy tuy6 i tinh don
bidn) cd dang sau:
Yj = a + bX + ei
Trong dd Yt iagil tri du doin efta bidn phu thude trong lfin quan sit thtS i; a iathdng
sd chi didm clt true tung; b iathdng sd diSn ti dd ddc efta dudng hdi quy; v i et ia gii tri
dao ddng nglu nhidn hay sai sd trong lln quan sit did i.
D£ dp dung tuang quart tuydn tinh m^t bidn sd efinphii c6 clc didu ki$n sau:
Chi cd mdt bidn ddc lap X Inh hudng tdi bidn phu thude Y
65

62. Quan h£ gifta X v ^Y da bifit hoac duqc gia dinh la theo ducmg thang
Sd li6u y&u cdu d l ap dung cho phdn tfch h6i quy dan gian la phai c6 n (n>2) cap s6
U eu ciaY v aX .
Vf du nghidn cun phan ting vdri mdc dam (N) trong thf nghi&n phan b6n c6 t mdc
dam, t cap gia tri cua X vh Y se la cua gia tri trung blnh cua Y va t$ 16 dam X. Minh hoa
phan rich tuang quan bang vi du quan h6 cda dam t<5i nang sudt 6 Bang 5.1.
Budc 1: Tihh gid tri trung blnh cua bi6n X ( X ) va bi£n Y (Y ), t6ng hi6u chinh cda
^ x 2, y2va xy ciia bi£n X v&Y.
- Y x
x = *=*—
n
n
2 V = 2 ](Xi-X)2
i=l
2 > 2 = Z ( Yi - Y )2
i=l
£ x y = j ;( X i -X X Y i - Y )
i=l
Trong d6 (X; va YJ la cap thtf i trong cdc gid tri cfia X va Y.
Trong vl du dudi day c6 n = 4 cSp gid tri c&anang sulfcY va mdc dam X.
Budc 2: lS6c luqng cdc h6 sd trong phuong trlnh dudng thing-tuong quan tuy£n tmh:
b
l]x y _ 249.475
23 x2 12.500
= 19,96
a = Y - b X = 5.870,75 - (19,96X75) = 4.374
Ta co phuong trlnh tuang quan tuy£n tinh (udc luqng) quan he gitta nang sua't IdaVa
mdc bon dam khac nhau:
Y = a + bX = 4.374 + 19,96X ;
Trong khoang 0 < X <150
C6 nghia la phuong trljih tr&n chi dting trong khoang mdc dam tit 0-150N. Thuc chat
do ia trong khoang cdc mdc N dd thfr nghi&n. NSu 6 ngodi khoang tr6n c£n phai lam thi
nghidm chd kh6ng th^ dilng ham tr6n d^ udc luqng.

63. Bang 5.1: Tinh cdc tham so trong tuang quart tuyert tinh mdt bie'n so trong quart he
gitia nang suat lua vd mute bdn dam trong thi nghiem phdn bon (Gomez, 1984)
Mifc dam
kg/ha (X)
N3ng suat
kg/ha(Y)
Sai kh£c so v6i trung binh Blnh phifdng cilia sai khac
(X) •(y)
X = (xt - jc
)
I
L
II
>
.
X2 y2
0 4230 -75 -1640,75 5625 2.692.061 123.056
50 5442 -25 -428,75 625 183.827 10.719
100 6661 25 790,25 625 624495 19.756
150 7150 75 1279,25 5625 1.636.481 95.944
Tdng
300 23483 0 0,00 12500 5.136.864 249.475
Trung blnh
75 5.870,75
Btidc 3: Lap d6 thi quan M Y va X
N3ng suat kg/ha
Do thi 5.1: Quan he tuyen tinh (tide Itiffng) gitia nang suat Y vd mtic phdn dam X
Cach ve:
Xac dinh cac diem quan sat va ve d6 thi bilu di§n duefng h6i quy duqc tide lttqng cr
budc 2.
- Danh da'u nhftng gia tri quan sat duqc. Trong vi du trfen c6 4 diem quan sat duqc d6
la cac gia tri cua X va Y.
- Dung dudng h6i quy ucrc luqng duqc ve gia tri.Y tuang ting vdi gia tri cua Xmin va
cac gia tri khac Xmax.
67

64. Ymin = a + bXmin; Ymax = a +bXmax
Trong vf du tren Xm
ill= 0 kg N/ha vk Xm
ax= 150 kgN/ba.
Y Here luong duoc tinh nhu sau:
Ynrin= 4.374 + 19,96 (0)= 4.374 kg
Ynuu[= 4.374 + 19,96 (150)= 7.368 kg/ha
Xkc dmh hai toa dd (X ^, Y „ J vk (Xm
ax, Y ^ vk kd dudng thang gitta hai toa dd trdn.
E)6 thi cua dirdng hdi quy tuydn tinh phai dat ydu cku sau:
- Duefng thang phki di qua hai didm X ^ va X ^
- Duefng thang phai di qua hai diem ( X ,Y ) trong d6 X va Y la gia tri trung binh
cua X va Y
- Dd ddc (slope) cua duefng thang la b
- Neu keo dai duemg thang phai gap true Y crgia tri dka a (intercept).
Bude 4: Kidm dinh y nghia cua tham sd P: Tinh tokn sai sd - chdnh ldch gitta duefng
ucrc lucrng va gia tri quan sat duoc (kdt qua theo thi nghidm) (s2yj)
S2
y.x
n - 2
5.136.864
(249.475)2
12,500
4 - 2
78.921
^ t b 19,96
Tinh gia tri tb.liull: tb.lm
b= ------ = ... ■--- = 7,94
R 7 ^ 9 2 T
v 12.500
So sanh tM
fahvdi tb.bingtrong bang Phu luc C.
vori dd tu do: (n - 2) = 4 - 2 = 2; n la sd cap so sknh- sd bidn (b md hinh tuydn tinh
don bien co sd bien lk 2).
Neu: tb_
l£
u
b> tb_
b4ttgco nghia la tham sd ude luong fi # 0. Hay co nghia la Y cd quan hd
phu thude vdi X.
Gia tri tbbing &mttc 5% hay mttc 1% chi mttc dd c6 f nghia etta tham sd b (hay noi
each khac la mttc dd tin cky etta he sd ucrc luong ft lk 95% hay 99%).
Trong vf du nay, tb_
b4ngcrmttc 5% = 4,303 va b mttc 1% = 9,925.
Vkytb(a=
oi)>tb
.riB
ll> tb(a=0
5)
Cd nghia la nang suat ltta cd phan ttng vdi mttc bdn dam trong khoang 0 - 150 kg/ha
la tin cky b mttc 95%.
Bude 5: Tinh tokn khoang tin cky (100-a)% (confidence interval - C.I) cho tham sd p.
68