how to calculate relative potency in parallel line model?Hi,
I fitted the parallel line model LogY=Sample+Log_Dose in fit curve.
The parallelism test and model parameters are fine. They are exactly same as what I got in R or PLA software. But the relative potencies are totally different. Any place to find the technical details about its calculation? Thank you!
Advanced Statistical ModelingBasic Data Analysis and ModelingRe: how to calculate relative potency in parallel line model?A small test data table, the different codes used and an overview of the results will facilitate the investigation.
Re: how to calculate relative potency in parallel line model? Predictive and Specialized Modeling > Fit Curve > Model Fit Options > Test Parallelism does contain some information
Re: how to calculate relative potency in parallel line model?I used the example from PLA. The RP from R is 2.170981, which is the same as what I got from PLA software. But JMP's result is 5.9..
The estimated parameters are the same, so I think both are using the same model, just the RP from JMP is too large.
data <- read_excel("PLA Example_parallel line.xlsx", sheet = "Sheet1")
data$Log_Dose = log(data$Dose)
data$LogY = log(data$Y)
fit <- lm(LogY ~ Sample+Log_Dose , data = data) #parallel line model: common slope
summary(fit)
coefficients <- coef(fit)
db <- coefficients["SampleT"] #
beta <- coefficients["Log_Dose"]
LogRP <- db / beta
RP<-exp(logRP)
In JMP, I fitted the model in specialized modeling-> fit curve, polynomials->Fit linear, and test parallelism
Re: how to calculate relative potency in parallel line model?Hi @acton ,
The reason you are seeing the differences is that the estimates in the parallel line test in JMP is not taking the difference from the 'baseline' of S, instead showing each term, you can get the same values as your screen shot (which is using dummy coding for the sample groups):
- Intercept = 5.71347 (for S)
- SampleT = 0.70423 → T’s intercept = 5.71347 + 0.70423 = 6.4177 (matches JMP table)
- SampleU = 0.51262 → U’s intercept = 6.2261
- SampleV = 0.61602 → V’s intercept = 6.3295
The Std Error estimates differ because JMP takes it from the covariance matrix, whereas your estimates from the screenshot look to be the raw estimates.
Thanks,
Ben
Re: how to calculate relative potency in parallel line model?Hi Ben,
I am ok with JMP's intercept and slope estimates but the relative potencies seem not correct. I clicked test parallelism and then got the relative potency part. I am not sure if they are relative potencies for parallel line model. I cannot find JMP help file for this analysis.
thanks,
acton
