When considering the relationship among three or more variables, an interaction may arise. Interactions describe a situation in which the simultaneous influence of two variables on a third is not additive. Most commonly, interactions are considered in the context of regression analyses, but they may also be evaluated using two-way ANOVA.

A simple setting in which interactions can arise is a two-factor experiment analyzed using Analysis of Variance (ANOVA). Suppose we are interested in studying the effects of cocaine on sleep. We might design an experiment to simultaneously test whether both the use of cocaine and the duration of usage affect the number of hours a squirrel will sleep in a night. We might give half of the squirrels we test cocaine, and the other half a placebo substance (the substance variable). And we might vary the duration of usage by administering cocaine or placebo for one of two possible durations before test, 4 weeks or 12 weeks (the duration variable). We can then consider the average treatment response (e.g. number of hours slept) for each squirrel, as a function of the treatment combination that was administered (e.g. substance and duration). The following table shows one possible situation:

There are three null hypotheses to be tested:

• H₀₁: Both substance groups sleep for the same number of hours on average.

• H₀₂: Both treatment duration groups sleep for the same number of hours on average.

• H₀₃: The two factors are independent or there is no interaction effect.

We can start by computing the group means (for each cell, row, and column):

And by calculating the sum of squares (SS) for each group (cell):

• 4-Week Placebo Dataset = {7 8 6 7 6.5}, M = 7
  SS_{4-Wk Placebo} = (7-7)² + (8-7)² + (6-7)² + (7-7)² + (6.5-7)² = 2.25
  
  

  4-Week Cocaine = {5.5 3.5 4.5 6 5}, M = 4.9
  SS_{4-Wk Cocaine} = (5.5-4.9)² + (3.5-4.9)² + (4.5-4.9)² + (6-4.9)² + (5-4.9)² = 3.7
  
  

  12-Week Placebo = {8 10 13 9 8.5}, M = 9.7
  SS_{12-Wk Placebo} = (8-9.7)² + (10-9.7)² + (13-9.7)² + (9-9.7)² + (8.5-9.7)² = 15.8
  
  

  12-Week Cocaine = {5 4.5 4 6 4}, M = 4.7
  SS_{12-Wk Cocaine} = (5-4.7)² + (4.5-4.7)² + (4-4.7)² + (6-4.7)² + (4-4.7)² = 2.8
  
  

Now, we'll calculate the SS_B:

• SS_B = n [( M_{4-Wk Placebo} - M_Group )² + ( M_{4-Wk Cocaine} - M_Group)² + ( M_{12-Wk Placebo} - M_Group )² + ( M_{12-Wk Cocaine} - M_Group)²]
  

  • = 5 [(7 - 6.575 )² + (4.9 - 6.575)² + (9.7 - 6.575)² + (4.7 - 6.575)²]
    = 5 [0.180625 + 2.805625 + 9.765625 + 3.515625]
    = 5 [16.2675]
    = 81.3375
    
    

Now, we'll calculate the SS_W:

• SS_W = SS_{4-Wk Placebo} + SS_{4-Wk Cocaine} + SS_{12-Wk Placebo} + SS_{12-Wk Cocaine}
  

  • = 2.25 + 3.7 + 15.8 + 2.8
    = 24.55
    
    

  df_W = N - rc
  

  • = 20 - (2 * 2)
    = 16
    
    

  s_W² = SS_W / df_W
  

  • = 24.55 / 16
    = 1.534375
    
    

Now, we'll calculate the SS_R:

• SS_R = n [( M_Placebo - M_Group )² + ( M _Cocaine - M_Group)²]
  

  • = 10 [(8.35 - 6.575)² + (4.8 - 6.575)²]
    = 10 [3.150625 + 3.150625]
    = 10 [6.30125]
    = 63.0125
    
    

  • df_R = r - 1<BR>>

    • = 2-1
      = 1
      
      

    s_R² = SS_R / df_R
    

    • = 63.0125 / 1
      = 63.0125
      
      

Now, we'll calculate the SS_C:

• SS_C = n [( M_4-Week - M_Group )² + ( M _12-Week - M_Group)²]
  

  • = 10 [(5.95 - 6.575)² + (7.2 - 6.575)²]
    = 10 [0.390625 + 0.390625]
    = 10 [0.78125]
    = 7.8125
    
    

  • df_C = c - 1<BR>>

    • = 2-1
      = 1
      
      

    s_C² = SS_C / df_C
    

    • = 7.8125 / 1
      = 7.8125