Allyson,

MSE is the "pooled" estimate of the variance based on the "within"
sample variance.  If you have 2 groups with n1=n2, it is just the
average of the variances...MSE = (Var1 + Var2)/2.   If sample sizes
aren't equal, it's just
a weighted average ...MSE =(df1/dfT)*Var1 + (df2/dfT)*Var2 where
dfT=df1+df2.
If 3 groups with n1=n2=n3, MSE=(Var1+Var2+Var3)/3 or the weighted
version if not all equal.

Then, for the pairwise differences, you have...SE = sqrt(MSE/n +
MSE/n)
which is similar to the denominator of the usual t-test.

The assumption you have to make is that the variances are
equal...that's why your error bars are necessarily equal.

You are correct that it is the error term in the denominator of the
F-statistic that you should use for the estimated variance.  For your
repeated measures, that seems to be MS_Interaction so that SE =
sqrt(MS_Int/n + MS_int/n).
Again, if n is equal then SE will be the same for each pairwise
comparison.
You are assuming the variances are equal...look at the expected mean
squares to get an idea of what variance you are estimating here.

Warren May



[EMAIL PROTECTED] (Allyson) wrote in message 
news:<[EMAIL PROTECTED]>...
> Forgive me if this is a repost
> 
> I'm trying to generate error bars for a histogram of an ANOVA.  I
> would like to use the MSError for the effect.
> 
> Here's the ANOVA design:
> 1.2 Between groups (18 subjects in each)
> 2.3 repeated measures
> 
> I have been told that to derive the error bars for the effect one does
> the following:
> 
> errorbarsize=sqrt((MSE*2)/n)
> 
> Would I be correct in doing the following:
> sqrt((MSErrorInteraction*2)/(36*3))
> 
> Then the error bar would be the same size for all the values.  That
> doesn't sound right.  Am I clueless?  If so, please set me straight.
> 
> thanks,
> Allyson
.
.
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