# Tag Info

0

You don't say how the 0.05 ml error arises - and that is crucial. If it is a random error in estimating the volume, then I think the best estimate of the error is to calculate the mean and then calculate the uncertainty in that mean through standard error propagation, which assumes the measurements are independent of each other and that the error is ...

2

If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by  \bar{x} = \frac{1}{5} (x_1 + x_2 + ...

-1

Either options 1 or 3 would be valid options, but perhaps one is more suitable than the other, depending on the details. We can decide which one of option 1 and 2 is correct by understanding where the rules for propagating error come from. Mathematically, the rules for calculating the uncertainty in a derived or calculated value based on measured values ...

-2

Your uncertainity is 0.25ml here. To convince yourself this is the right value. Think about the maximum uncertainty that can be caused when you are adding the quantity. Suppose all the beakers have extra 0.05ml in the worst case scenario. Then you would get 0.25ml extra. Similarly if all the beakers had 0.05ml less, you would get 0.25ml less. So your ...

4

I like that this question is asking for some intuitive justification and calling for more understanding in error analysis. Too often, that's left out. From a teaching perspective, I think what's most important is that students come up with their own procedure, that's it's reasonable, and that they justify it. This encourages them to do the same sort of ...

1

OK, I'll take a stab at this. If you're looking for a real number to describe the effective width w of the statistical uncertainty in terms of a [0.5-w, 0.5+w] interval, then it should be obvious by now that no w will satisfy your requirements because you don't have a symmetric Gaussian distribution here. Assuming that you're talking about random events ...

1

There is no good reason to round intermediate calculations. "Round to one significant figure" means that an error of 0.16 and 0.24 would propagate the same way when they are different by 50%. Just don't believe the additional digits - but there is no reason to drop them.

1

The normal distribution has a very important special role in stochastics. One can prove mathematically that the distribution of the sum of many independent statistical processes is almost always a normal distribution. This is called the "Central Limit Theorem" in mathematics (there are actually several of these) and you can test it very easily yourself with ...

1

The gaussian normal distribution is a good guess when you don't know anything else, due to the Law of Large Numbers. If you know more about the process that generates the measurements, then you can choose another distribution. For example, if there is a lower bound on measurements, (for example: concentrations of a chemical compound in blood plasma cannot ...

0

Are you able to take multiple measurements to be able to estimate the correlations/covariance involved? It's not really clear how exactly your channel counts enter the formula, but the "dirty solution" works everytime: Estimate the covariance matrix OR make a lot of observations of correlated counts Based on the estimated covariance matrix, randomly ...

Top 50 recent answers are included