Error Propagation of standard deviation?

Question

I have the following problem:

I do measurements with technical triplicates and biological duplicates. So I prepare the same sample two times and measure each one three times.

This gives me an average value of each duplicate with a standard deviation, calculated by the technical triplicates.

E.g. $$0,23 \pm 0,07\ \text{ and }\ 0,20 \pm 14.$$

Now I want to determine the total average and its error. And here I get confused: with the error propagation I can calculate the error for the average

$$\Delta x=\frac{1}{2}\sqrt{(Δx_1)^2+(Δx_2)^2}=0,16,$$

but there is also the standard deviation of the two values (~0,014).

I am wondering now, how I can combine both values to the error of the total average. Taking just the error propagation seems wrong to me, especially when the duplicate measurements differ greatly.

Answer 1

Your question comes about because you are actually trying to estimate two different quantities and so in general you would expect different answers:

1) The standard deviation of the repeated measurements tells you about the uncertainty just in making the measurement 2) The standard deviation between your replicates tells you about the combined uncertainty in redoing the experiment and in remeasuring the experiment. So if you are after the total error then this is the right quantity.

So from an uncertainty propagation approach this is pretty difficult to treat because you only have two replicates and that's just not enough to estimate a standard deviation reliably (see the reference in point 2 for more information). So you essentially have 3 choices:

1) The measurement uncertainties and the reproducibility are the same for each experiment and so you can estimate it from the entire data set. This is unsatisfying because it's unlikely to be correct.

2) You can use the standard deviation between your replicates but this is going to be very noisy because you only have two data points per experiment. If you want to turn this into a confidence interval you need to use a small sample size correction: http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Confidence_Intervals/BS704_Confidence_Intervals_print.html see section: Confidence Intervals for One Sample: Continuous Outcome for n<30 (because in your case n = 2)

3) You can partially pool and assume that the uncertainties are similar between different experiments. For example you can see about how to approach this here: https://jrnold.github.io/bayesian_notes/multilevel-models.html or https://cran.r-project.org/web/packages/bridgesampling/vignettes/bridgesampling_example_stan.html or https://mc-stan.org/rstanarm/articles/pooling.html or https://mc-stan.org/users/documentation/case-studies/radon.html. Unfortunately I wasn't able to find an example that was what you need (means individually estimated from the data but partial pooling on the uncertainty). But you should be able to use those examples to figure out how to do that.

Answer 2

The standard deviation of the two values 0,20 and 0,23, which you calculated to be 0,014, is not relevant: If these two values bring their own error, as it were, error propagation is the means of choice here. Calculating the error of the mean, as you did in your post, is correct, but it may make sense to use inverse-weighted variances instead:

An example from my metrology course was a situation where three different people measured the value of the same resistor and they end up with three different mean resistance values and their respective errors. Your case is different, i. e. the same person is measuring the value of two different samples and you introduce errors in both the preparation and the measurement. But I venture that if these samples are supposed to be identical (i. e. the preparation process and its experimental conditions were the same and you don't expect inherent fluctuations of your resulting data), this method should be acceptable, at least for a first-order approximation. Since the two values you provided are within each other's error margins (see last paragraph for a small comment on this), a very necessary condition for this method is fulfilled.

The formulas for $\hat{y}$ and $Var\left(\hat{y}\right)$ in the introduction of the wikipedia article about weighted means should give you the weighted mean of your two duplicates, and you can obtain the error of the weighted mean by calculating the square root of the resulting $Var\left(\hat{y}\right)$.

As Patrick already mentioned, you may have to apply a correction for small sample sizes, e. g. by multiplying the error of your weighted mean by a correction factor from the student-t value table. We didn't do this in our course (possibly because it was always the same resistor), but it would be reasonable in your case, since you compare different samples. Your number of statistical degrees of freedom should be 1 (since you only have two samples), and you'll have to choose a suitable two-sided confidence interval. These two values allow you to retrieve the correction factor in the table (e. g., 6.314 for a 95% confidence interval) and to calculate the corrected error margins.

Btw., there may still be an error in your numbers, since a result of 0,20 $\pm$ 14 should probably be discarded. Isn't this supposed to be 0,20 $\pm$ 0,14? Calculating the error with your figures yields 0,078 and not 0,16; you may have accidentally omitted the factor 1/2.