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help > RE: Second level random effects analysis
Jun 9, 2015 01:06 AM | Alfonso Nieto-Castanon - Boston University
RE: Second level random effects analysis
Hi Kaitlin,
Yes, sorry about the confusion. The "2*..." in my previous post equation was meant to give you the total length of the CI errorbar around the effect-size mean-point (i.e. errorbars from x-CI/2 to x+CI/2), so if you are plotting the errorbars as (x-CI) to (x+CI) then yes, please remove the "2*" from the original equations (or divide by two the results). The formula corresponds to the 95% one-sided (or 90% two-sided) confidence intervals, which I believe are the same as those displayed by the 'display values' routine. If you want to use other CI percentiles simply change the .95 value there for the desired one-sided "1-alpha" level.
Hope this helps
Alfonso
Originally posted by Kaitlin Cassady:
Yes, sorry about the confusion. The "2*..." in my previous post equation was meant to give you the total length of the CI errorbar around the effect-size mean-point (i.e. errorbars from x-CI/2 to x+CI/2), so if you are plotting the errorbars as (x-CI) to (x+CI) then yes, please remove the "2*" from the original equations (or divide by two the results). The formula corresponds to the 95% one-sided (or 90% two-sided) confidence intervals, which I believe are the same as those displayed by the 'display values' routine. If you want to use other CI percentiles simply change the .95 value there for the desired one-sided "1-alpha" level.
Hope this helps
Alfonso
Originally posted by Kaitlin Cassady:
Hi Alfonso,
Thanks for the very helpful response. Just to clarify, In terms of the equation: 2*spm_invTcdf(.95,N-1)*std(x)/sqrt(N), which computes the CIs, does this compute 90% or 95% CIs? Also, when I input these CI values into a plot to produce error bars, the error bars appeared much larger than those that were produced with the CONN toolbox. Do I need to first divide these values by 2 if I want to display the error bars on both sides (both positive and negative) of the average contrast estimates?
Thanks!
Thanks for the very helpful response. Just to clarify, In terms of the equation: 2*spm_invTcdf(.95,N-1)*std(x)/sqrt(N), which computes the CIs, does this compute 90% or 95% CIs? Also, when I input these CI values into a plot to produce error bars, the error bars appeared much larger than those that were produced with the CONN toolbox. Do I need to first divide these values by 2 if I want to display the error bars on both sides (both positive and negative) of the average contrast estimates?
Thanks!
Threaded View
| Title | Author | Date |
|---|---|---|
| Kaitlin Cassady | May 30, 2015 | |
| Alfonso Nieto-Castanon | May 30, 2015 | |
| Kaitlin Cassady | Jun 4, 2015 | |
| Alfonso Nieto-Castanon | Jun 4, 2015 | |
| Kaitlin Cassady | Jun 5, 2015 | |
| Alfonso Nieto-Castanon | Jun 8, 2015 | |
| Kevin Mann | Jun 16, 2016 | |
| Kaitlin Cassady | Jun 8, 2015 | |
| Alfonso Nieto-Castanon | Jun 9, 2015 | |
| Kaitlin Cassady | Jun 9, 2015 | |