# Propagation of standard error of fit through to the average and standard deviation of replicates

I am using an equation to fit measured data. I have measured multiple replicates for each condition, each of which is then individually fitted to this equation.

Once the data are fitted I obtain a constant and associated standard error for each fit (this should represent the magnitude of the average distance each point is from the fitted line if my understanding is correct?).

The measurement error is negligible (tiny) given the magnitude of the fitted error and the differences between replicates.

So I have fitted constants with standard errors $$C1\pm\epsilon1$$, $$C2\pm\epsilon2$$ & $$C3\pm\epsilon3$$ and I would like to take the average of these and calculate their standard deviation.

If we consider $$C1 = 38.4 \pm 0.06$$ $$C1 = 42.5 \pm 0.08$$ and $$C1 = 41.2 \pm 0.07$$ then the mean and SD, ignoring the SE, are $$40.77 \pm 2.12$$. We can see that the SE is very small given this variation but intuitively we would still expect a slight increase in the error.

As I understand it, we can calculate the SD from the SE by multiplying by the square root of the sample size. In this case, the sample size would be the number of data points along the curve to be fitted, let's say $$240$$ points. Now the small SE of fitting becomes much larger.

$$SD = \frac{SE}{\sqrt n}$$

$$0.06$$ becomes $$0.93$$ while $$0.08$$ and $$0.07$$ become $$1.24$$ and $$1.08$$ respectively.

Would it still be correct to then plug these new standard deviations into the traditional formula for error propagation for multiplication/division (i.e. is the derivation of a SD from a SE of fit in this manner compatible with typical error propagation afterwards?):

SD of the mean $$= \sqrt{(\frac{0.93}{38.4})^2 + (\frac{1.24}{42.5})^2 + (\frac{1.08}{41.4})^2} = 0.046$$

However, this error is much smaller than expected and fails to take into account the variance in samples I think. Is this extra error a value than can be linearly added to the larger error for sample variation? Or is there a better formula/approach to correctly propagate this type of error when averaging fitted constants?

A similar question has suggested pooling all the original data ("Average" standard errors of multiple samples) but I do not see how this could appropriately be applied to curve fitting.

This question (Average of Average Quantities and Their Associated Error) seemed like it might be what I'm after but applying the suggested formula yielded a similarly small final SD:

SD of the Mean $$=\sqrt{\frac{1}{N^2} \sum_i^N{Var(X)}} = \sqrt{\frac{1}{9} * \left( 0.93^2 + 1.24^2 + 1.08^2 \right)} = 0.63$$

While larger than the previous attempt of 0.046, it is still much smaller than the 2.12 calculated without the corresponding SE.

How should one go about propagating the error associated with fitted data through to their averages when performed in replicates?