在双向方差分析中交互作用的零假设是什么?


20

假设我们有两个因子(A和B),每个因子有两个级别(A1,A2和B1,B2)和一个响应变量(y)。

在执行类型的双向ANOVA时:

y~A+B+A*B

我们正在测试三个原假设:

  1. 因子A的均值没有差异
  2. 因子B的均值没有差异
  3. 因子A和B之间没有相互作用

写下后,很容易提出前两个假设(对于1来说是)H0:μA1=μA2

但是假设3应该如何表述呢?

编辑:以及如何将其制定为两个以上级别的情况?

谢谢。


3
我没有名声让我的编辑,但我想你想H0=μA1=μA2(或μA1,如果你想有一个双标)哎呀,它已经自动TEX-指明分数是: H_0 = \mu_{A1}=\mu_{A2}或\mu_{A_1}]
— Ben Bolker 2010年

1
Oups,没看到你正在使用大写字母来表示因子的名字和他们的水平-修复它(以下@Ben符号)。
— chl 2010年

Answers:


18

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)。pq

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 ),YijkijAkB1≤i≤n1≤j≤p1≤k≤qYijk=μjk+ϵi(jk),ϵi(jk)∼N(0,σϵ2)

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 B1…Bk…Bq A1μ11…μ1k…μ1qμ1.…………………Ajμj1…μjk…μjqμj.…………………Apμp1…μpk…μpqμp. μ.1…μ.k…μ.qμ

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:μjkjkϵi(jk)i()jki

(用于治疗的平均预期值Ĵ因子的甲)μj.=1q∑k=1qμjkjA

(用于治疗的平均预期值ķ因子的乙)μ.k=1p∑j=1pμjkkB

αj=μj.−μ (effect of treatment j of factor A, ∑j=1pαj=0)

βk=μ.k−μ (effect of treatment k of factor B, ∑k=1qβk=0)

(αβ)jk=μjk−(μ+αj+βk)=μjk−μj.−μ.k+μ
(interaction effect for the combination of treatment j of factor A with treatment k of factor B, ∑j=1p(αβ)jk=0∧∑k=1q(αβ)jk=0)

αj(k)=μjk−μ.k
(conditional main effect for treatment j of factor A within fixed treatment k of factor B, ∑j=1pαj(k)=0∧1q∑k=1qαj(k)=αj∀j,k)

βk(j)=μjk−μj.
(conditional main effect for treatment k of factor B within fixed treatment j of factor A, ∑k=1qβk(j)=0∧1p∑j=1pβk(j)=βk∀j,k)

With these definitions, the model can also be written as: Yijk=μ+αj+βk+(αβ)jk+ϵi(jk)

This allows us to express the null hypothesis of no interaction in several equivalent ways:

  1. H0I:∑j∑k(αβ)jk2=0
    (all individual interaction terms are 0, such that μjk=μ+αj+βk∀j,k. This means that treatment effects of both factors - as defined above - are additive everywhere.)

  2. H0I:αj(k)−αj(k′)=0∀j∧∀k,k′(k≠k′)
    (all conditional main effects for any treatment j of factor A are the same, and therefore equal αj. This is essentially Dason's answer.)

  3. H0I:βk(j)−βk(j′)=0∀j,j′∧∀k(j≠j′)
    (all conditional main effects for any treatment k of factor B are the same, and therefore equal βk.)

  4. H0I: In a diagramm which shows the expected values μjk with the levels of factor A on the x-axis and the levels of factor B drawn as separate lines, the q different lines are parallel.


1
A really impressive answer Caracal - thank you.
— Tal Galili

9

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:

H0:C=0vs.HA:C≠0.


1
Thanks Dason, that helped. Also, after reading your reply, it suddenly became clear to me that I am not fully sure how this generalizes in case we are having more factors. Could you advise? Thanks again. Tal
— Tal Galili

2
You can test multiple contrasts simultaneously. So for example if A had three levels and B had 2 we could use the two contrasts: C1 = (A1B1 - A2B1) - (A2B1 - A2B2) and C2 = (A2B1 - A2B2) - (A3B1 - A3B2) and use a 2 degree of freedom test to simultaneously test if C1 = C2 = 0. It's also interesting to note that C2 could equally have been (A1B1 - A1B2) - (A3B1 - A3B2) and we would come up with the same thing.
— Dason

Hi @Dason: you seem to have multiple accounts. Could you please complete the form at stats.stackexchange.com/contact and request that they be merged? That will simplify your use of this site (and give you the combined net reputation of both accounts).
— whuber
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