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My question is very practical and is about real physical value measurements and related uncertainty.

As I understand now most scientists work with Student's t-distribution when they are estimating error and confidence interval. Let me put an example to simplify the question.

Let's say I am measuring the diameter of a metal rod with calipers. My calipers can measure a minimum 0.1mm. I did 5 measurements (8.6, 8.7, 8.4, 8.3, 8.6), calculated the average (8.52) and standard deviation (0.164). Now if I want to apply Student's t-distribution I need degrees of freedom (4 = number of measurements - 1), I select a confidence interval 95% just because I want 95%. I can select any value. Now I am going here and selecting t-distribution value for my experiment which is 2.776.

enter image description here

Now I calculate error bound like this $$\Delta D=\frac{t(0.95, 4) \times S_n}{\sqrt{N}}=\frac{2.776 \times 0.164}{\sqrt{5}}=0.20$$ Finally, I can say that my real measured value $D$ with 95% probability is between 8.32mm and 8.72mm (average $8.52-\Delta D$ and $8.52+\Delta D$). I put everything in the table below: enter image description here

Now my questions:

  • If I read a scientific article and they publish some value like this here $6.67430(15) \times 10^{−11}$. How can I understand what is the real value with 95% confidence interval without knowing how many measurements N they did? As I understand from here section 7.2.2, which is supposed to be the world standard, this statement $6.67430(15) \times 10^{−11}$ means that the standard deviation is $0.00015 \times 10^{−11}$. But if I do know N, then it does not make much sense to know this value 15. If I would make 100 measurements in my example above, then $\Delta D$ would be 0.05 (4 times less) instead of 0.20.
  • In the same paper here section 7.2.4 they show an example but they do not divide by $\sqrt{N}$. Is it a typo or I do not understand anything?
  • Also here they are talking about 68.3% ("one sigma"), 95.4% ("two sigma"), or 99.7% ("three sigma"). How is it correlated with the example I did above?

Because this is practical question, I would prefer an answer with specific example with my numbers above.

P.S. I called it here standard deviation. Someone can argue that standard deviation is theoretical value, which we never know, etc. We can call it here standard uncertainty, sigma or whatever. Exact terminology does not matter here if we understand that it is calculated with Excel formula "STDEV.S" or like this $\sqrt{\frac{(D_i-D_{avg})^2}{N-1}}$. This is purely practical question.

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  • $\begingroup$ You are right: Usually we wish we knew the interval in which the true value of $D$ lies with a probability of 95%. However, this interval is called the 95% credible interval, and it differs from the 95% confidence interval. $\endgroup$
    – Semoi
    Commented Mar 2, 2023 at 21:59

2 Answers 2

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If I read a scientific article and they publish some value like this here 6.67430(15)×10−11. How can I understand what is the real value with 95% confidence interval without knowing how many measurements N they did?

This means that the value is $6.67430 \ 10^{-11}$ in the given units with a standard uncertainty of $0.00015 \ 10^{-11}$. To get a 95% confidence interval you would multiply the standard uncertainty by $1.96\approx 2$, but only if you can consider the distribution to be normal.

It is not necessary to know the number of measurements in order to determine the 95% confidence interval. The standard uncertainty represents the uncertainty for the whole experiment, including any repetitions. It is designed to have the above relationship with the confidence interval.

Essentially the number of measurements is already accounted for in the standard uncertainty itself.

Also here they are talking about 68.3% ("one sigma"), 95.4% ("two sigma"), or 99.7% ("three sigma"). How is it correlated with the example I did above?

The standard uncertainty is one sigma. So a result that is one times the standard uncertainty would be a one sigma result.

In the same paper here section 7.2.4 they show an example but they do not divide by $\sqrt{N}$. Is it a typo or I do not understand anything?

It is not a typo. There is no reason to divide by $\sqrt{N}$ there. The only time that the number of measurements would need to be used is if you were to repeat the whole experiment except use a different number of measurements. Then your anticipated standard uncertainty would differ from the previous experiment by the difference in $\sqrt{N}$.

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  • $\begingroup$ Thank you for your reply. Do I correctly understand that there are 2 approaches: 1. Normal distribution - 68.3% ("one sigma"), 95.4% ("two sigma"). 2. Student's t-distribution where this table above is used and divide by $\sqrt{N}$ $\endgroup$
    – Zlelik
    Commented Mar 1, 2023 at 21:29
  • $\begingroup$ Also, In wikipedia here en.wikipedia.org/wiki/… below the table they divide by $\sqrt{N}$. $\endgroup$
    – Zlelik
    Commented Mar 1, 2023 at 21:34
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You would understand this to indicate that $1.5\times 10^{-11}$ is an estimate of the standard deviation of the mean (often called the standard error).

The standard error would contain 68% of the probability distribution.

Assuming a normal distribution for the uncertainties, the 95% confidence interval would be about twice as big.

In your example it would correspond to the $\Delta D$ you would get if you had used $t(0.68, 4)$.

I do though understand your question and confusion. For small numbers of measurements where you are trying to estimate the standard deviation and the mean from the data, then there is no simple factor of 2 ratio between the 68% and 95% confidence intervals.

If that is so, then it is incumbent on someone reporting an error bar to say that the distribution is non-normal, otherwise a normal distribution would be the usual assumption.

As an aside, I am not so sure that using the t-distribution to estimate the confidence intervals is as common as you think. Most (in my field) would likely use the Bessel-corrected standard deviation of the population to estimate the standard error in the mean, which is at least a reasonable estimator of the variance, even if the standard error can be a little underestimated.

If you look carefully, these issues are discussed in section 4.2 of the document you have referenced. It notes there that the number of measurements should be large enough that the difference between the population and sample standard deviations is small, otherwise the t-distribution should be used to construct confidence intervals. The document also notes (section 4.2.6) that where confidence limits are estimated from repeated readings, then it is good practice to say how many readings were taken precisely for the reasons highlighted by your question. I would say as a rule-of-thumb, that you don't have to worry about this if the number of readings was $>5$ because it isn't likely to change the first significant figure of the standard error. Thus if you see no reference to the number of readings used to construct the mean and its standard error, you should (or are forced to) assume that either the error wasn't estimated from repeated readings or that the number of readings was large enough that any departure from a normal distribution can be neglected.

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