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Question: r/m calculation with CI95%! #119
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When using the Yes if two 95%CI do not overlap then it suggests that there is a significant difference at the p=0.05 level. When using embranch option things are a bit different since there will be parameter estimated for each branches. |
@xavierdidelot thank you! But in principle the calculations are correct either way? By calculating as I did and also as you suggested? |
The intervals you get using my method based on the output of emsim are more statistically valid than the ones calculated using the method you described, but otherwise yes what you described is a correct application of what I had described in other issues. |
Thank you again |
Hello @xavierdidelot,
I have read the issues #99 and #114 regarding CI95% calculations for r/m. Do I understand it correctly:
CFML has been ran with following parameters:
First to get overall CI95% I have to use following input like in issue #99.
Second I calculate r/m as follows:
r/m = 1/1/delta * R/theta * nu
therefore ther/m = 1/0.000530892 * 0.297706 * 0.0157936 = ~8.86
Third I calculate CI95% for each parameter using the formula in R
qgamma(c(0.025,0.975),shape=a_post,rate=b_post)
:R/theta CI95% is calculated
qgamma(c(0.025,0.975),shape=2162.84,rate=7261.02)
1/delta CI95% is calculated
qgamma(c(0.025,0.975),shape=2007.28,rate=3780950)
nu CI95% is calculated
qgamma(c(0.025,0.975),shape=59768,rate=3784310)
Summary of results of CI95% calculations:
Fourth I will calculate 0.025:
r/m(0.025) = 1/0.0005079198 * 0.2854475 * 0.01566726 = ~8.81
Fifth I will calculate 0.975:
r/m(0.975) = 1/0.0005543673 * 0.3105534 * 0.01592050 = 8.92
Finally the r/m(CI95%) is 8.86[8.81-8.92]
So here are my question:
-embranch true
parameter and because of the values being factors as stated in issue How to get the parameter estimates and 95% confidence intervals from result? #87.Thank you in advance
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