0
$\begingroup$

I have run a 1 way basic ANOVA on 4 categorical variables. Afterwards, I ran multiple comparisons on all 4 groups (6 comparisons) & adjusted using the Tukey-Kramer method. This was all done in Matlab.

Now I want to plot a boxplot showing the errorbars of each group adjusted for multiple comparisons - i.e. if the error bars of two groups dont overlap then those groups are significantly different. I know Matlab is able to do this, but I haven't been able to find any resource on HOW this adjustment is performed.

Can anyone point me towards a formula/theory of how to do this by hand?

Thanks!

Improve this question
$\endgroup$
4
  • $\begingroup$ By "boxplot", you presumably meant "barplot", right? The Tukey test uses the sampling distribution of differences. Any 2 means that differ by more than a specified amount are significantly different. I suppose you could simply have error bars whose length in each direction is the required difference. Then, when the error bars on one mean don't overlap another mean, those means differ. However, the error bars wouldn't have the usual interpretation & I suspect it would lead to misinterpretations. $\endgroup$
    – gung - Reinstate Monica
    Apr 28, 2016 at 2:01
  • $\begingroup$ Hi @gong, I actually meant boxplot, since I could make my notches that size, but the errorbar on a barplot would give the same result. Also, this is just for myself to visualize for data exploration, so misinterpretation isnt a concern. Finally- would the required difference be the same for all groups. Ie would would the length of the bars depend on which groups are being compared (in which case this is invalid), or same for all? $\endgroup$
    – DankMasterDan
    Apr 28, 2016 at 18:23
  • $\begingroup$ I would do separate boxplots & barplots, given your situation. I expect combining them will create plots that are just too busy. Re: Tukey's test, I should look it up, but I'm pretty sure it's the same for all groups. I don't quite remember how it's done if the n's differ or if you aren't assuming homogeneity of variance. $\endgroup$
    – gung - Reinstate Monica
    Apr 28, 2016 at 18:26
  • $\begingroup$ Thanks! In my case the n's do differ, but I (at least so far) do assume homogeneity of variance $\endgroup$
    – DankMasterDan
    Apr 28, 2016 at 18:37

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.