Chapter 4

analysis of variance

(ANOVA)

Section 1

the basic idea and

condition of application

Objective: deduce and compare several (or two ) population

means .

Method: analysis of Variance (ANOVA), ie F test for compari

ng several sample means .

Basic idea : according to the type of design, the sum of squa

res of deviation from means (SS) and degree of freedom (d

f) were divided into two or several sections . Except the cha

nce error, the variation of every section can be explained b

y a certain or some factors.

Condition of Application :

population : normal distribution and homogeneity

of variance.

Sample: independent and random

Types of design :

The ANOVA of completely random design;

The ANOVA of randomized block design;

The ANOVA of Latin square design;

The ANOVA of cross-over design ;

Table 4-1 the results of g groups

group Measure value Statistic

1level X11 X12 … X1j … 1nX1

n1 1X S1

2 level X21 X22 … X2j … 2nX2

n2 2X S2

…

…

…

…

…

…

…

…

…

…

g level Xg1 Xg2 … Xgj … ggnX ng gX Sg

total N S

The basic idea of ANOVA of completely random design

partition of variationsum of squares of deviations from m

ean ， SS ：

2 2

1 1 ,

2 ( 1)

ing N

iji j i j

SS X X X X

N S

1. total variation : the degree of variation of all variable values , the formula as follows

2 2

1 1 1

,

1

2 ,

i i

N

i

n ng g

ij iji j i j

ji j

SS X

C

X X

X

C

total

2 2

1 1 ,

( ) ( )ing N

ij iji j i j

X XC

N N amend factor:

1N total

2. between- group variation: the sum of squares of deviations from mean between groups means and grand mean show the effects of treatment and random error, the formula :

2

12

1 1

( )

( )

in

ijj

g g

i ii i i

X

SS n X X Cn

bg

1g bg

2

1 1

( )ing

ij ii j

SS X X

wg

g N g w

3.Within-group Variation : differences among values 3.Within-group Variation : differences among values

within each group .The formula as follows:within each group .The formula as follows:

b wgSS SS SS total g

total bg wg

the relation of three variation

SSMS

SSMS

bgbg

bg

wgwg

wg

mean square ， MS

Test statistic ：

•

If , were the estimated value of the random error , F value should be close to 1 .

• If were not equal , F value will be larger than 1.

1 2, , MS

FMS

bgbg wg

wg

1 2 g ,MS MSbg wg

2

1 2, , , g

Section 2

The ANOVA of Completely Random Design

All of objects were randomly distributed

to g groups (levels), and every group

give the different treatment. The effects

of treatment will be deduced by

comparing the groups means after

experimentation.

completely random design

Example 4-1 A doctor want to explore the cli

nic effect of a new medicine for reducing blood

fat, and selects 120 patients according to the s

ame standard. All of patients were divide into 4

groups by the completely random design. How

should he divide the groups?

The methods of dividing groups of completely random design 1. serial number: 120 patients was numbered from 1to

120 ( table 4-2 column 1) ；

2. choosing random figure: you can begin from the any

row or any column in the appendix 15(for example

beginning from the fifth row and seventh column),

and read three digit in turn as a random number to

write down the serial number, (table 4-2,column 2)

Table 4-2 the grouping result of completely random design

Serial number 1 2 3 4 5 6 7 8 9 10 … 119 120

Random number 260 873 373 204 056 930 160 905 886 958 … 220 634

rank 24 106 39 15 3 114 13 109 108 117 … 16 75

Grouping result A D B A A D A D D D … A C

3.edit serial number: edit serial number according to the number from

small to large (the same number according to early or late order)

(table 4-2,column 3)

4.define in advance: the serial numbers from 1-30 were defined the A

group; 31-60 were the B group; 61-90 were the C group; 91-120 were

the D group, (table 4-2,column 4)

（ 2 ） the choice of statistic methods

1. If the data accord with normal distribution an

d homogeneity of variance, one-way ANOVA

or independent t test was used (g=2) ；

2. If the data are not normal distribution or

heterogeneity of variance, the datum transfor

m or Wilcoxon rank sum test can be done.

decompose of variation

Example 4-2 A doctor wanted to explore the clinic e

ffect of a new medicine for reducing blood fat, and s

elected 120 patients according to the same standard.

He divided all of patients into 4 groups by the compl

etely random design. The low density lipoprotein wer

e measured after 6 weeks by double blind experimen

t, table 4-3. Is there difference among the population

means of low density lipoprotein of 4 groups ?

statistic group value

n

iX X 2X

3.53 4.59 4.34 2.66 3.59 3.13 2.64 2.56 3.50 3.25

3.30 4.04 3.53 3.56 3.85 4.07 3.52 3.93 4.19 2.96 Placebo group

1.37 3.93 2.33 2.98 4.00 3.55 2.96 4.3 4.16 2.59

30 3.43 102.91 367.85

New medicine

2.42 3.36 4.32 2.34 2.68 2.95 1.56 3.11 1.81 1.77

1.98 2.63 2.86 2.93 2.17 2.72 2.65 2.22 2.90 2.97 2.4g

2.36 2.56 2.52 2.27 2.98 3.72 2.80 3.57 4.02 2.31

30 2.72 81.46 233.00

2.86 2.28 2.39 2.28 2.48 2.28 3.21 2.23 2.32 2.68

2.66 2.32 2.61 3.64 2.58 3.65 2.66 3.68 2.65 3.02 4.8g

3.48 2.42 2.41 2.66 3.29 2.70 3.04 2.81 1.97 1.68

30 2.70 80.94 225.54

0.89 1.06 1.08 1.27 1.63 1.89 1.19 2.17 2.28 1.72

1.98 1.74 2.16 3.37 2.97 1.69 0.94 2.11 2.81 2.52 7.2g

1.31 2.51 1.88 1.41 3.19 1.92 2.47 1.02 2.10 3.71

30 1.97 58.99 132.13

Table 4-3 the low density lipoprotein value of 4 treatment groups (mmol/L)

三、 steps of analysis

H0：1 2 3 4 ，即4个试验组的总体均数相等

H1：4个试验组的总体均数不全相等

0.05

按表4-4中的公式计算各离均差平方和SS、自由度、均方MS和F值。 H0 ： ie. all of 4 population means are equal.

H1 ： not all of the population means are equal

1 2 3 4

0.05

2 . Calculate test statistic

1. State the hypotheses and test criteria

102.91 81.46 80.94 58.99 324.30ijX

2 367.85 233.00 225.54 132.13 958.52ijX 2(324.30) /120 876.42C

958.52 876.42=82.10SS total ，总=120－1=119 2 2 2 2(102.91) (81.46) (80.94) (58.99)

876.42 32.1630 30 30 30

SS bg

4 1 3 bg

82.10 32.16 49.94SS wg ， 120 4 116 wg

32.1610.72

3MS bg ， 49.94

0.43116

MS wg ，

10.7224.93

0.43F

variation source df SS MS F P total 119 82.10

Between-group 3 32.16 10.72 24.93 <0.01 Within-group 116 49.94 0.43

Table 4-5 the table of ANOVA of completely random design

list the ANOVA table

3. Calculate p value and deduce

according to a=0.05 level, reject , and accept ,

not all of 4 population means are equal; ie. differen

t dose medicines have different effects on ldl-c.

0H1H

attention ： if the result of ANOVA is to reject H0 , and accept H1, it d

oes not mean that all of population means have differ

ence each other. If analysing which groups have sign

ificant difference , we must compare among several

population means (section 6). When g=2, the ANOVA

of completely random design is equal to independent

t test, ie.

t F

Section 3

The ANOVA of randomized block design

randomized block design

Firstly, match the objects as the blocks according

to the non-treatment factor affecting the result

of experiment (such as sex, weight, age,

occupation , state of illness, course of disease et al) .

Secondly, the objects of each block were randomly

distributed to each treatment group or control group.

(1) grouping method of randomized block design ：

（ 2 ） characteristic of randomized block design • Random distribution was repeated many times for

objects of the blocks. The number of objects is

same in every treatment group.

• SS of the block variation was separated from SS of

the within-group variation of completely random

design; SS of within-group (sum of error square)

was decreased, and power of test was increased.

example 4-3 distribute 15 white mice of 5 blocks

to three treatment groups , how to do it ?

Grouping method: firstly, number the mice by the weight,

and match the 3 near weigh mice as a block (table 4-6).

Secondly, select 2 digit as one random number from any

row or any column in the random number table, for

example, from the 8th row and third column (table 4-6);

and rank the random number from small to large in every

block. The object of serial number in each block is 1,2,3

will accept A,B,C treatment respectively. (table 4-6)

Table 4-6 the distribution result of 5 blocks white mice

block 1 2 3 4 5

White mice 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Random

number 68 35 26 00 99 53 93 61 28 52 70 05 48 34 56

order 3 2 1 1 3 2 3 2 1 2 3 1 2 1 3

treatment C B A A C B C B A B C A B A C

Treatments (g level) Block number 1 2 3 … g

1 X11 X21 X31 … Xg1

2 X12 X22 X32 … Xg2

…

…

…

…

…

…

j X1j X2j X3j … Xgj

…

…

…

…

…

…

n 1nX 2nX 3nX … gnX

table 4-7 the result of random block design

partition of variation

(1)Total variation： SStotal.

(2) Treatment-group variation ： SStreatment.

(3) block-group variation： SSblock.

(4) Error variation： SSerror.

SS SS SS SS total treatment bl ock error

total treatment bl ock error

variaion df SS MS F

total N－1 2

1 1

g n

iji j

X C

treanment g－1 2

1 1

1( )

g n

iji jn

X C

SS

treatment

treatment

error

MS

MStreatment

Block n－1 2

1 1

1( )

gn

ijj ig

X C

SS

bl ock

bl ock

MS

MSbl ock

error

error （n－1）（g－1） SStotal－ SStreatment-－SSblock SS

error

error

table 4-8 the ANOVA of random block design

Steps of analysis

example 4-4

15 mice were divided into 5 blocks by th

e weight. there are 3 mice in every block

.the result showed in table 4-9. is there d

ifference among 3 treatment groups?

block A B C 1

g

iji

X

1 0.82 0.65 0.51 1.98

2 0.73 0.54 0.23 1.50

3 0.43 0.34 0.28 1.05

4 0.41 0.21 0.31 0.93

5 0.68 0.43 0.24 1.35

1

n

ijj

X 3.07 2.17 1.57 6.81 ( )ijX

iX 0.614 0.434 0.314 0.454 ( )X

2

1

n

ijj

X 2.0207 1.0587 0.5451 3.6245

2( )ijX

table 4-9 the variable values of different groups （ g ）

H0 ：

H1 ： not of all population means are equa

l

1 2 3

0.05

2

1 1

3.6245 3.0917=0.5328g n

iji j

SS X C

total

2 2 2 2

1 1

1 1( ) (3.07 2.17 1.57 ) 3.0917 0.2280

5

g n

iji j

SS X Cn

treatment

2 2

1 1

( ) / (6.81) /15 3.0917g n

iji j

C X N

2

1 1

2 2 2 2 2

1( )

1 (1.98 1.50 1.05 0.93 1.35 ) 3.0917 0.2282

3

gn

ijj i

SS X Cg

bl ock

Table 4-10 the ANOVA of example 4-4

variation df SS MS F P

total 14 0.5328

treatment 2 0.2280 0.1140 11.88 <0.01

block 4 0.2284 0.0571 5.95 <0.05

error 8 0.0764 0.0096

according to 1=2 、 2=8, check F value tabl

e:

At α=0.05 level ， reject H0， accept H1， not

all of population means are equal.

0.01(2,8

0.05(2,8) 0.01(2,8)

)

4.46, 8.65,

11.88 , 0.01

F F

F F P

。

section 6        

multiple comparison

can the above example be analyzed by t test ?Numbers of t test

a=0.05, the probability of non-type I error for one comparison :

1-0.05=0.95;

the probability of non-type I error for all of 6 times analysis :

=0.77;

the probability of type I error for 6 times analysis: 1-0.77=0.23

the probability of type I error will be increased

24 6C

60.95

Condition of application ：

when the result of ANOVA reject H0,

and accept H1, not all of population

means are equal. If wanting to know

the difference between any two group

means, we should do the multiple

comparison.

LSD-t test（ least significant differenc

e ）

The formula

LSD , i j

i j

X X

X Xt

S

error

1 1i jX X

i j

S MSn n

error

MS MS误差 组内

example 4-7example 4-7 for the example 4-2 d for the example 4-2 d

ataata ，， are there difference among the pare there difference among the p

opulation means of 2.4gopulation means of 2.4g 、、 4.8g4.8g 、、 7.2g 7.2g

and placebo groupand placebo group ？？

α=0.05

Comparing between 2.4g and placebo group ：

0 2.4g 0:H

1 2.4g 0:H

According to example 4-2， 2.4gX =2.72，

0X =3.43， 2.4gn = 0n =30，MSerror =0.43， error=116。

i jX XS =1 1

0.4330 30

=0.17

LSD-t =2.72 3.43

0.17

=－4.18

ν=116，t=4.18 check t value table，P<0.001。at

0.05 ，reject H0，accept H1，and there are

significant .

4.8g VS placebo group: LSD-t =-4.29

7.2g VS placebo: LSD-t =-8.59 。

Dunnett- t test

0

0

i

i

X X

X Xt

S

formulaformula ：：

Dunnett- , error

example 4-8 according to example

4-2, compare 3 population means of treatm

ent groups and placebo group,respectively?

H0： μi=μ0

H1： μi μ0

α=0.05

According to example 4-2， 2.4gX =2.72， 4.8gX =2.70， 7.2gX =1.97，

0X =3.43， in = 0n =30，MSerror =0.43， error=116.

2.4g

2.72 3.43

1 10.43

30 30

t

=－4.18

4.8g

2.70 3.43

1 10.43

30 30

t

=－4.29

7.2g

1.97 3.43

1 10.43

30 30

t

=－8.59

Dunnett-

Dunnett-

Dunnett-

ν =116、T=g－1=4－1=3 , check Dunnett-t

value table（two tail）， 0.01/2,116 0.01/2,120 =2.98t t 。

2.4g 0.01/2,116t t ，4.8g 0.01/2,116t t ，7.2g 0.01/2,116t t P<0.01。

at the level of 0.05 ，reject H0，accept

H1，there is significant difference。

三、 SNK-q test（ Student-Newman-Keuls ）

i j

i j

X X

X Xq

S

， ν=νerror

1 1

2i jX Xi j

MSS

n n

error

iX ， in and jX ， jn mean the group means and sample

numbers.

Example 4-9 according to 4-4,

compare the 3 group means by

SNK-q test

H0 ： μA=μB

H1 ： μA≠μB，

α=0.05

rank the 3 group means from small to

large and number them

mean 0.314 0.434 0.614

group C B A

number 1 2 3

group i jX X a q 0.05q 0.01q P

（1） （2） （3） （4） （5） （6） （7）

1，2 0.12 2 2.74 3.26 4.75 >0.05

1，3 1.30 3 6.85 4.04 5.64 <0.01

2，3 0.18 2 4.11 3.26 4.75 <0.05

Table 4-15 the comparing between two group means

Example 4-4 误差MS =0.0096， 8 误差 。numbe of

sample is 5， 0.0096 1 1

0.04382 5 5i jX XS

。

conclusion ：

there are significant difference

between A and B, A and C.