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I have collected data on the stress relaxation ($\sigma$) of 3 chicken tendons over time ($t$), and the following nonlinear regression model was fit through each individual tendon: $$\sigma = A + B e^{-\frac{t}{\tau}} + C e^{-\frac{t}{\eta}}$$

For each tendon, I have an estimate of each parameter ($A$, $B$, $C$, $\tau$, $\eta$) and their confidence intervals. In other words, I have 3 sets of each parameter:

Tendon 1: $A_1$, $B_1$, $C_1$, $\tau_1$, $\eta_1$

Tendon 2: $A_2$, $B_2$, $C_2$, $\tau_2$, $\eta_2$

Tendon 3: $A_3$, $B_3$, $C_3$, $\tau_3$, $\eta_3$

Now, I want to use the 3 sets of parameters from each tendon to estimate the true chicken tendon parameters, but I am not sure how to do this. I am very skeptical of simply using the arithmetic mean (e.g. $\tau_{true} = \frac{1}{3}(\tau_1+\tau_2+\tau_3)$) because the variance within each individual tendon model is not the same for the 3, and I feel like the parameter associated with the tendon with less variance should be weighted more heavily.

How would I do this?

Note: I am not able to combine the 3 data sets into 1 to fit only 1 nonlinear regression (although I wish I could do that). I had to fit 3 different regressions and work with the resulting parameters for each tendon.

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Here is a path to explore that could assist in answering your question.

Consider investigating the use of a model for alternative parameters estimates or for variance determinations based on a natural log transformation. For example, specify separate models for each of the three types of tendons as follows:

$${Ln(\sigma_k) = ln(B) + 1/{\tau}*t + \epsilon_k}$$

The employment of the log transformation may also assist in making the errors (${\epsilon}$) more normal and even improve the homogeneity of the error variance in compliance with least-squares regression theory.

Further, the regression can supply a variance estimate for the derived regression coefficients.

As such, to answer your question the latter variance estimates may be of assistance in formulating a weighted estimate for ${\tau}$ (say by weighting inversely by respective variance estimates).

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