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.
If then reject H 0
If then not reject H 0
Two assumptions on analysis of variance:
follows normal distribution , ;
homogeneity of variances, .
Bartlett’s test : are not all equal : :
Example 5.2 (cont.)
Test the homogeneity of variances for the three populations in Table 2.
Test for normality and transformations:
5.2 Multiple comparisons
To examine whether a specified two means are equal or not . LSD- t test .
To examine whether all the means of comparison groups are equal or not . SNK- q test.
LSD- t test
(least significant difference t test)
H 0 is rejected if ：
SNK- q test
(Student-Newman-Keuls q test)
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.
H 0 is rejected if ：
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 randomly. 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.
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.
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.