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Chapter 4 analysis of variance (ANOVA). Section 1 the basic idea and condition of application.

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Presentation on theme: "Chapter 4 analysis of variance (ANOVA). Section 1 the basic idea and condition of application."— Presentation transcript:

1 Chapter 4 analysis of variance (ANOVA)

2 Section 1 the basic idea and condition of application

3 Objective: deduce and compare several (or two ) population means. Method: analysis of Variance (ANOVA), ie F test for comparing several sample means. Basic idea : according to the type of design, the sum of squares of deviation from means (SS) and degree of freedom (df) were divided into two or several sections. Except the chance error, the variation of every section can be explained by a certain or some factors.

4 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 ;

5 The basic idea of ANOVA of completely random design

6 partition of variation  sum of squares of deviations from mean , SS :

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

8 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 :

9 3.Within-group Variation : differences among values within each group.The formula as follows:

10 the relation of three variation

11 mean square , MS

12 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.

13 Section 2 The ANOVA of Completely Random Design

14 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

15 Example 4-1 A doctor want to explore the clinic effect of a new medicine for reducing blood fat, and selects 120 patients according to the same standard. All of patients were divide into 4 groups by the completely random design. How should he divide the groups?

16 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)

17 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)

18 ( 2 ) the choice of statistic methods 1. If the data accord with normal distribution and 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 transform or Wilcoxon rank sum test can be done.

19 decompose of variation

20 Example 4-2 A doctor wanted to explore the clinic effect of a new medicine for reducing blood fat, and selected 120 patients according to the same standard. He divided all of patients into 4 groups by the completely random design. The low density lipoprotein were measured after 6 weeks by double blind experiment, table 4-3. Is there difference among the population means of low density lipoprotein of 4 groups ?

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

22 三、 steps of analysis H 0 : ie. all of 4 population means are equal. H 1 : not all of the population means are equal 2. Calculate test statistic 1. State the hypotheses and test criteria

23

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

25 3. Calculate p value and deduce according to a=0.05 level, reject, and accept, not all of 4 population means are equal; ie. different dose medicines have different effects on ldl-c.

26 attention : if the result of ANOVA is to reject H0, and accept H1, it does not mean that all of population means have difference each other. If analysing which groups have significant 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.

27 Section 3 The ANOVA of randomized block design

28 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 :

29 ( 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.

30 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)

31

32 table 4-7 the result of random block design

33 partition of variation (1)Total variation : SS total. (2) Treatment-group variation : SS treatment. (3) block-group variation : SS block. (4) Error variation : SS error.

34 table 4-8 the ANOVA of random block design

35 Steps of analysis example 4-4 15 mice were divided into 5 blocks by the weight. there are 3 mice in every block.the result showed in table 4-9. is there difference among 3 treatment groups?

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

37 H 0 : H 1 : not of all population means are equal

38

39

40 according to 1 =2 、 2 =8, check F value table: At α=0.05 level , reject H 0 , accept H 1 , not all of population means are equal.

41 section 6 multiple comparison

42 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

43 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.

44 LSD-t test ( least significant difference )

45 The formula

46 example 4-7 for the example 4-2 data , are there difference among the population means of 2.4g 、 4.8g 、 7.2g and placebo group ? example 4-7 for the example 4-2 data , are there difference among the population means of 2.4g 、 4.8g 、 7.2g and placebo group ?

47 α=0.05 Comparing between 2.4g and placebo group :

48

49 4.8g VS placebo group: LSD-t =-4.29 7.2g VS placebo: LSD-t =-8.59 。

50 Dunnett- t test

51 formula : Dunnett-

52 example 4-8 according to example 4-2, compare 3 population means of treatment groups and placebo group,respectively? H 0 : μ i =μ 0 H 1 : μ i μ 0 α=0.05

53 Dunnett-

54

55 三、 SNK-q test ( Student-Newman-Keuls )

56

57 Example 4-9 according to 4-4, compare the 3 group means by SNK-q test H 0 : μ A =μ B H 1 : μ A ≠ μ B , α=0.05

58 rank the 3 group means from small to large and number them

59 Table 4-15 the comparing between two group means

60 conclusion : there are significant difference between A and B, A and C.


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