Answers:
我认为将假设及其相应的检验清晰分开很重要。对于以下内容,我假设受试者之间采用均衡的CRF- 设计(相等的像元大小,柯克表示法:完全随机因子设计)。
是观察我在治疗 Ĵ因子的甲和治疗 ķ因子的乙与 1 ≤ 我≤ Ñ, 1 ≤ Ĵ ≤ p和 1 ≤ ķ ≤ q。该模型是 ÿ 我Ĵ ķ = μ Ĵ ķ + ε 我(Ĵ ķ ),
设计: 乙1个... 乙ķ ... 乙q 阿1 μ 11 ... μ 1 ķ ... μ 1个q μ 1 ... ... ... ... ... ... ... 甲Ĵ μ Ĵ 1 ... μ Ĵ ķ ... μ Ĵ q μ Ĵ 。... ... ... ... ... ... ... 甲p μ p 1 ... μ
是在小区中的预期值 Ĵ ķ, ε 我(Ĵ ķ )与人的测量关联的误差我在该小区中。的()表示法表示的索引 Ĵ ķ是固定的任何给定的人我因为该人仅在一个条件下观察。效果的一些定义:
(用于治疗的平均预期值Ĵ因子的甲)
(用于治疗的平均预期值ķ因子的乙)
(effect of treatment of factor , )
(effect of treatment of factor , )
(interaction effect for the combination of treatment of factor with treatment of factor ,
(conditional main effect for treatment of factor within fixed treatment of factor ,
(conditional main effect for treatment of factor within fixed treatment of factor ,
With these definitions, the model can also be written as:
This allows us to express the null hypothesis of no interaction in several equivalent ways:
(all individual interaction terms are , such that . This means that treatment effects of both factors - as defined above - are additive everywhere.)
(all conditional main effects for any treatment of factor are the same, and therefore equal . This is essentially Dason's answer.)
(all conditional main effects for any treatment of factor are the same, and therefore equal .)
: In a diagramm which shows the expected values with the levels of factor on the -axis and the levels of factor drawn as separate lines, the different lines are parallel.
An interaction tells us that the levels of factor A have different effects based on what level of factor B you're applying. So we can test this through a linear contrast. Let C = (A1B1 - A1B2) - (A2B1 - A2B2) where A1B1 stands for the mean of the group that received A1 and B1 and so on. So here we're looking at A1B1 - A1B2 which is the effect that factor B is having when we're applying A1. If there is no interaction this should be the same as the effect B is having when we apply A2: A2B1 - A2B2. If those are the same then their difference should be 0 so we could use the tests:
H_0 = \mu_{A1}=\mu_{A2}
或\mu_{A_1}
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