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I just read some topics that discuss how SPSS corrects the p value shown and it keeps 0.05 as significant when applying bonferroni correction.

The problem that I have is that I'm running a repeated measure ANOVA with several groups and when I compare the bonferroni correction table with the no correction table, it gives the exact same p value for both. Does this make sense? Am I doing something wrong? Thanks in advance

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For each of the analyses, it appears you are just comparing two groups. Thus, the Bonferroni correction, within each repeated time point, adjusts for only 1 pair of comparisons (thus, $\alpha/1$. You probably will need to select a different factor for comparison in SPSS if you want to compare the within comparisons (though I am not sure if SPSS even allows for this).

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  • $\begingroup$ Thanks Gregg. So what I have are two cycles (within factor) in 6 different groups (1min,15min,etc). I am comparing for each group if there is a difference between cycle 1 and cycle 2. So I thought bonferroni would correct for what would be multiple t tests. Any other pack that would allow this? Or maybe another test would be more suitable? $\endgroup$
    – Gonzalo Lerner
    Commented Apr 4, 2018 at 19:20
  • $\begingroup$ If the RQ is to determine if the cycles are different within each groups (I'm guessing a temporal grouping), and if there is no connection between those groups, then I would argue that you are conducting distinct experiments/analyses. Thus, I would not adjust the $P$-values. If, however, there is a possible connection between the groups that would suggest the analyses is first an omnibus test, with exploratory post hoc analyses to follow, then an adjustment would be appropriate. With the little info I have, I could suggest the Holms-Bonferroni adjustment, but only as a start. $\endgroup$
    – Gregg H
    Commented Apr 5, 2018 at 13:22
  • $\begingroup$ Thanks. The latest is correct. I want to determine if the cycles are different within each group but there actually is a connection (they have been exposed to different "Interference treatments"). I will try the Holms-Bonferroni then. $\endgroup$
    – Gonzalo Lerner
    Commented Apr 16, 2018 at 16:00

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