Analysis of Variance

Yuantao Hao

26th,Oct. 2009

Chapter5

Review:

1. The basic steps and logic of hypothesis testing;

2. One-sample t test;

3. Two-sample t test;

4. Paired –sample t test;

5. F test for homogeneity of variances;

6. Z test for the parameters of binomial distribution and

Poisson distribution when sample size is large

enough.

main steps:

(1) Set up the statistical hypotheses

(2) Select statistics and calculate its current value

(3) Determine the P-value

00 : H 01 : H

nS

Xt

/0

) statistic of aluecourrent v( ttPP

(4) Decision and conclusion

Comparing the P-value with the pre-assigned

small probability , if P ≤ , then reject ;

otherwise, not reject . Finally, issue the

conclusion incorporating with the background.

0H

0H

P value

P-value is defined as a probability of the event

that the current situation and even more

extreme situation towards appear in the

population. 0H

）8345.2( tPP

The P-value can also be thought of as the

probability of obtaining a test statistic as

extreme as or more extreme than the actual

test statistic obtained, given that the null

hypothesis is true.

）8345.2( tPP

Question:

1. One-sample;

2. Two-sample;

3. Paired –sample;

4. Three or more samples?

Analysis of variance (ANOVA) :

One-way ANOVA is used to test for

differences among two or more independent

groups.

Typically, however, the one-way ANOVA is

used to test for differences among at least

three groups, since the two-group case can

be covered by a t-test .

5.1 One-Way ANOVA for the Completely Random Design

The completely random design

For this design, there is only one treatment

factor with G (≥2) levels. The term level refers

to the possible status planned for the treatment

factor.

Example 5.1 Randomly assign 12

laboratory blood specimens (experiment

units) into three groups with 4 blood

specimens in each group.

How to assign the 12 units into three

groups randomly?

Table 11.1 Randomized grouping result Unit number 1 2 3 4 5 6 7 8 9 10 11 12 Random number 39 90 22 00 66 82 89 08 92 72 36 60 Rank (R) 5 11 3 1 7 9 10 2 12 8 4 6 Grouping result 2 3 1 1 2 3 3 1 3 2 1 2

Unit number 1 2 3 4 5 6 7 8 9 10 11 12

Random number 39 90 22 00 66 82 89 08 92 72 36 60

Rank (R) 5 11 3 1 7 9 10 2 12 8 4 6

Grouping result 2 3 1 1 2 3 3 1 3 2 1 2

Example 5.2 12 blood specimens are randomly assigned into three groups according to Table 2. Group 1 receives the treatment of anticoagulant （抗凝血剂） A; group 2 receives anticoagulant B; and group 3 receives anticoagulant C.

For each blood specimen, the erythrocyte sedimentation rate (ESR ，红细胞沉降率 ) after receiving the treatment is measured. The aim is to test whether the three mean ESRs are significantly different. The results are showed in Table 11.2.

Table 11.2 Erythrocyte sedimentation rate (ESR mm／h)

Anti-

coagulant ESR( ijX ) j

ijX j

ijX 2 in iX 2iS

A 17 16 16 15 64 1026 4 16.0 0.67

B 10 11 12 12 45 509 4 11.3 0.92

C 11 9 8 9 37 347 4 9.3 1.58

Total ij

ijX =146 ij

ijX 2 =1882 N =12 X =12.2 2cS =3.17

0H : 1 = 2 = 3 ,

1H : 1 , 2 and 3 are not all the same

BET

EEE SSMS /

BBB SSMS /

E

B

MS

MSF

G

i

n

jijT

i

XXSS1 1

2)(

G

iiiB XXnSS

1

2)(

G

i

n

jiijE

i

XXSS1 1

2)(

BET SSSSSS

1GB

1NT

GNE

TTT SSMS

If then reject H0

If then not reject H0

1,2,1FF

1,2,1FF

0H : 1 = 2 = 3 ,

1H : 1 , 2 and 3 are not all the same

Table 11.3 The table for one-way analysis of variance

Source DF SS MS F P

Between groups B = 1G BSS BMS = BSS / )1( G BMS / EMS

Within groups E = GN ESS = TSS - BSS EMS = ESS / )( GN

(Errors)

Total 1N TSS

Table 11.4 Table of one-way ANOVA for the effects of the anticoagulants

Source DF SS MS F P

Between the anticoagulants 2 96.17 48.09 45.37 <0.05

Errors 9 9.50 1.06

Total 11 105.67

Two assumptions on analysis of variance:

1. follows normal distribution , ;

2. homogeneity of variances, .

ijX ),( 2iiN Gi ,,2,1

222

21 G

Bartlett’s test :

222

21 G 0H

1H 222

21 ,,, G are not all equal

:

:

)ln()1(ln)1(

1 22

2ii

ci Sn

G

Sn

m

1G

)1(

1

1

1

)1(3

11

ii nnGm

Example 5.2 (cont.)

Test the homogeneity of variances for the

three populations in Table 2.

)14(

1

14

1

)13(3

11m

50.0

)58.1ln92.0ln67.0)(ln14(3

17.3ln)14(

148.1

12

213

20.10,2 4.61 P>0.10

Test for normality and transformations:

)log( aXY

XY

pY 1sin

5.2 Multiple comparisons

1. To examine whether a specified two means

are equal or not . LSD-t test.

2. To examine whether all the means of

comparison groups are equal or not . SNK-q

test.

LSD-t test

(least significant difference t test)

jiij XXd

)11( jiEdnnMSS

ij

ijdji StXX ,

H0 is rejected if：

SNK-q test

(Student-Newman-Keuls q test)

nMSS EX ij

nnnn G 21

21

1in

n NG N

All means should be sorted from the smallest to the

biggest to form contrasts.

Each contrast may contain a means, a=2,3,…,G.

In Example 11.2, the means of three groups is sorted as

9.3, 11.3 and 16.0; if 9.3 and 11.3 are selected to form a

contrast, a=2;

With the parameters a and, the critical value of

SNK-q test can be find out from Table 11 of Appendix 2.

, ,a vq

ijXaji SqXX ,,

H0 is rejected if：

For a =2, 0.01,2,9q =4.60, 0.01,2,9 Xq S =4.60×0.51=2.35.

For a =3, 0.01,3,9q =5.43, 0.01,3,9 Xq S =5.43×0.51=2.77.

Anticoagulants C B A

Means of ESR 9.3 11.3 16.0

Grouping (C, B) (A)

Figure11.1 The grouping of the effects of the three anticoagulants

5.3 Two-Way ANOVA for the Randomized Complete-Block

Design There are n blocks and each block contains G

experimental units to receive G treatments ran

domly. The total number of observations is N =

nG.

Example5.4 12 mice have been grouped into 4 blocks according to their birth litters and each block has 3 mice. Randomly assign 3 kinds of food to the 3 mice in each block.

Table 11.8 Random allocation of the randomized complete-block design

Block Unit No. Random No. (Rank) Unit No. (Treatment )

1 1 2 3 28 (1) 65 (3) 62 (2) 1 (A) 2 (C) 3 (B)

2 1 2 3 79 (3) 21 (2) 05 (1) 1 (C) 2 (B) 3 (A)

3 1 2 3 81 (2) 51 (1) 94 (3) 1 (B) 2 (A) 3 (C)

4 1 2 3 19 (1) 90 (3) 76 (2) 1 (A) 2 (C) 3 (B)

The advantage of this design comparing with

the completely random design is to reduce the

effect of the variation among the experimental

units if the difference among blocks was a

main source of variation.

The disadvantage is that all the sizes of

different blocks (or say, the numbers of

experimental units in different blocks) should

equal to the number of treatments, otherwise,

the statistical analysis will be difficult.

The observations of the randomized complete-block design

Treatment Block

1 2 … j … G

Total

1

2

i

n

X11 X12 … X1j … X1G

X21 X22 … X2j … X2G

… …

Xi1 Xi2 … Xij … XiG

… …

Xn1 Xn2 … Xnj … XnG

B1

B2

Bi

Bn

Total

Sum of squares

T1 T2 … Tj … TG

Q1 Q2 … Qj … QG

Example 5.5 The investigator used randomized block design to carry out the experiment to compare the anti-tumor effects of three anti-tumor drugs A, B, C on mice sarcoma （肉瘤） . 15 mice of the same race were selected and three anti-tumor drugs A, B, C randomly allocated into 3 mice within the same block.

With the observations of sarcoma’s weight, the experiment result sees Table 5. Please test if the effects of three anti-tumor drugs are different.

Table 11.11 The weight of mice sarcoma with different drugs (g)

Drugs Block

A B C Total (Bi)

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

Total (Ti) 3.07 2.17 1.57 6.81 Sum of Squares (Qi) 2.02 1.06 0.55 3.63

5

The table of analysis of variance for randomized complete-block design

Source DF SS MS F P

Treatment G-1 CTn

SSG

jjB

1

21

1 )1/(1! GSSMS BB EB MSMS /1

Block n-1 CBG

SSn

iiB

1

22

1 )1/(22 nSSMS BB EB MSMS /2

Error (G-1)(n-1) 21 BBTE SSSSSSSS )1)(1/( nGSSMS EE

Total Gn-1 CXSSn

i

G

jijT

1 1

2

2

1 1

1

n

i

G

jijX

nGC

Analysis of variance for the anti-tumor effects

Source DF SS MS F P Treatment 2 0.228 0.114 11.937 <0.004 Block 4 0.228 0.057 5.978 <0.016 Error 8 0.076 0.010 Total 14 3.624

0.532

Summary:

1. Basic logic of ANOVA;

2. ANOVA for completely random design data;

3. Multiple comparison;

4. ANOVA for randomized complete-block design

data.

THE END

THANKS ！