I've done some non-linear regressions, and trying to determine if parameters from different curves are significant from one another or not. I have the best-fit values for a parameter for both curves, the standard error, and the 95% confidence intervals.

1. If the confidence intervals of two values overlap, but do not reach the value itself, are they considered significantly different? (eg. is 10 with C.I. 7-13 different from 6 with C.I. 4-8?)

2. Or should I just do a t-test with the best fit value, standard error and N as the number of replicates to which I fit the curve?

Thanks a lot.

I've done some non-linear regressions, and trying to determine if parameters from different curves are significant from one another or not. I have the best-fit values for a parameter for both curves, the standard error, and the 95% confidence intervals.

1. If the confidence intervals of two values overlap, but do not reach the value itself, are they considered significantly different? (eg. is 10 with C.I. 7-13 different from 6 with C.I. 4-8?)

2. Or should I just do a t-test with the best fit value, standard error and N as the number of replicates to which I fit the curve?

Thanks a lot.

Assuming the regression model is the same, it is best to do this using a z-test (although in practise, using a t-test won't make much difference)

The significance of the difference between parameter estimates bi and bj a standardized z-test is computed as:

z(bi ,bj) = [(bi - bj)]/[SE (bi - bj)]

where bi and bj are the parameter estimates. The estimated standard error for the difference between bi and bj is simply the square root of VAR(bi - bj). VAR(bi - bj) is given by the sum of the variances of bi and bj minus twice their covariance:

VAR(bi - bj) = VAR(bi) + VAR(bj) - 2COV(bi ,bj)

The variances themselves can be calculated from the standard error as follows:

VAR(bi) = SE(bi)^2*n, VAR(bj) = SE(bj)^2*n

The significance can then be calculated using a z-distribution table (an online one is here (http://faculty.vassar.edu/lowry/tabs.html#z))

thanks for your help.