Solved – 2×2 design – one way anova or two way anova

anovaspss

So i did an experiment with 4 groups, each getting different advertisement communications tasty or healthy benefits:

group 1: control group, no health or taste benefits
group 2: only health benefits
group 3: only taste benefits
group 4: health & taste benefits

I have 4 different dependant variables: purchase intention, attitude, credibility and benefits all measured on 7 point likert scale

So I thought i had to do a 2 way anova because health and taste are the IV. But how do i test these? Because the interaction of them is already tested in group 4?

I work with SPSS

Best Answer

The two methods are statistically equivalent, but a two-way ANOVA will make it easier to test the specific effects of interest (i.e., an interaction between health benefits and taste and main effects of each). The model F-test in a two-way ANOVA will be equivalent to the F-test in the one-way ANOVA.

You can use a one-way ANOVA to test main and interaction effects. To test the main effect of health, you can test whether $\frac{\bar{x}_1+\bar{x}_3}{2}=\frac{\bar{x}_2+\bar{x}_4}{2}$. The t-test for this comparison will be equivalent to the F-test on the main effect of health benefits in a two-way ANOVA. For the interaction, you can test whether $\bar{x}_2-\bar{x}_1=\bar{x}_4-\bar{x}_3$. The t-test on that will be equivalent to the interaction F-test in a two-way ANOVA. A one-way ANOVA makes it easier to compare individual groups and other contrasts (e.g., is there a difference between group 1 and all other groups combined?) while a two-way ANOVA makes it easier to test interactions, main effects, and simple main effects.

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Solved – Post hoc test in a 2×3 mixed design ANOVA using SPSS

Answer edited to implement encouraging and constructive comment by @Ferdi

I would like to:

provide an answer with a full contained script
mention one can also test more general custom contrasts using the /TEST command
argue this is necessary in some cases (ie the EMMEANS COMPARE combination is not enough)

I assume to have a database with columns: depV, Group, F1, F2. I implement a 2x2x2 mixed design ANOVA where depV is the dependent variable, F1 and F2 are within subject factors and Group is a between subject factor. I further assume the F test has revealed that the interaction Group*F2 is significant. I therefore need to use post hoc t-tests to understand what drives the interaction.

MIXED depV BY Group F1 F2 
  /FIXED=Group F1 F2 Group*F1 Group*F2 F1*F2 Group*F1*F2 |  SSTYPE(3) 
  /METHOD=REML 
  /RANDOM=INTERCEPT | SUBJECT(Subject) COVTYPE(VC) 
  /EMMEANS=TABLES(Group*F2) COMPARE(Group) ADJ(Bonferroni)
  /TEST(0) = 'depV(F2=1)-depV(F2=0) differs between groups' 
    Group*F2 1/4 -1/4 -1/4 1/4 
    Group*F1*F2 1/8 -1/8 1/8 -1/8 -1/8 1/8 -1/8 1/8 
  /TEST(0) = 'depV(Group1, F2=1)-depV(Group2, F2=1)' Group 1 -1
    Group*F1 1/2 1/2 -1/2 -1/2 
    Group*F2 1 0 -1 0  
    Group*F1*F2 1/2 0 1/2 0 -1/2 0 -1/2 0 .

In particular the second t-test corresponds to the one performed by the EMMEANS command. The EMMEANS comparison could reveal for example that depV was bigger in Group 1 on the condition F2=1.

However the interaction could also be driven by something else, which is verified by the first test: the difference depV(F2=1)-depV(F2=0) differs between groups, and this is a contrast you cannot verify with the EMMEANS command (at least I did not find an easy way).

Now, in models with many factors it is a bit tricky to write down the /TEST line, the sequence of 1/2, 1/4 etc, called L matrix. Typically if you get the error message: "the L matrix is not estimable", you are forgetting some elements. One link that explains the receipt is this one: https://stats.idre.ucla.edu/spss/faq/how-can-i-test-contrasts-and-interaction-contrasts-in-a-mixed-model/

Solved – How to check for normality prior to using a Mixed Anova

Sphericity explained here is also assumed and, when violated, increases the Type I error rate substantially. Sphericity is almost always violated so the problem is not to test for it (although that can be done) but rather how to deal with it. Violations of normaility of residuals make less difference unless they are severe. A good argument against testing for normality is that the a priori probability that the distribution is exactly normal is essentially 0, so you know the null hypothesis is false before you do the test.

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Related Solutions

Solved – Post hoc test in a 2×3 mixed design ANOVA using SPSS

Solved – How to check for normality prior to using a Mixed Anova

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