Relative Potancy 95% CI

[#1]

Hi there, 

I am fitting a curve and applying a sigmoidal 4PL Rodbard model and testing parallelism. I have to determine the relative potency using the inflection points (IC50). So I calculate relative potency by dividing inflection point of the standard by the inflection point of the sample  (0.6479693 / 0.6477584 = 1.0003).

How would I create a 95% confidence interval for the relative potency?

[#2]

Maybe try looking under the Relative Potencies outline bar, which should be below the graph under the Parallel Fit Parameter Estimates.  The relative potencies are calculated and the standard error is included.  (at least it is in JMP15). 

It may be that there isn't conventional method of calculating the confidence interval, so the JMP developers only included the standard error? In that case, if you happen to have JMP Pro, you can bootstrap the relative potency and use the bootstrap confidence limits.

[#3]

You can read the background information about ratio estimators here, to go with @Byron_JMP's answer.

[#4]

The problem, even with the solution I proposed is that the interval around the relative potency is likely non-symmetric. So, a simple +/- the t-quantile*SE is going to overestimate on the low side and underestimate on the high side.

[#5]

Thank you @Byron_JMP and @Mark_Bailey ,

When I look at the relative potencies tab below the graph, it is not using the inflection points to generate the relative potency. For the bioassay I'm using I need to generate the RP using the relative potency. 

I understand what you are saying about the over/under estimation. I will read more on the link.