Hi I'm reading the book "OpenIntro Statistics" about ANOVA and in section 5.5.4 when explaining the conditions for applying ANOVA under the voice 'Independence' it states:

Independence. [...] For processes and experiments, carefully consider whether the data may be independent (e.g. no pairing). [...]

I'm thinking about an experiment such as benchmarking n databases operation time, for example the writing time of some datas. The experiment would have the following structure:

  1. generate m random data to insert,
  2. insert data to all n databases and measure writing time for each of the m data inserted,
  3. apply ANOVA.

After the expiriment I imagine to have a dataframe that looks something like the following:

data_id | database_id | time
  ...   |     ...     | ...

where there are any possible combinations of (data_id, database_id). But AFAIK these are paired data as I'm using the same subject (the data being inserted) in different conditions (the different databases), is this correct? Does this violate the ANOVA preconditions?

The alternative is to use different randomly generated datas for each database.

  • 1
    $\begingroup$ Yes. You have a repeated measures ANOVA design. $\endgroup$
    – amoeba
    Feb 18, 2016 at 11:18
  • $\begingroup$ @amoeba thanks for the response! Didn't know about repeated measures ANOVA. And in the case that I use different randomly generated datas for each database does it become a simple ANOVA? $\endgroup$ Feb 18, 2016 at 17:31
  • 1
    $\begingroup$ Yes. But I would suggest to use the same datasets and repeated measures ANOVA, as it is probably a more sensitive test. $\endgroup$
    – amoeba
    Feb 18, 2016 at 21:33
  • 1
    $\begingroup$ A more appropriate term should be "pseudo-replication" when the measurements are actually correlated. However, as @amoeba pointed out in this case you have a rmANOVA. I agreed with the last comment left by amoeba, since there seems to be a natural sphericity in your experiment, and that is why you should choose rmANOVA. $\endgroup$
    – Henry.L
    Feb 18, 2016 at 21:35


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