I assume SD is standard deviation, and also in unit [ml], I also assume that the error has no sign, i.e. it is the absolute value and cannot become negative.

If this is not so, I do not understand your problem. If these assumptions hold, how do you know, you have a Gauss distribution?

When I get such large SD, I find something like a mixture of two Gauss distributions. Did you use graphical methods, like plots of the empirical distribution function or the empirical hazard function?

Dear Dr. Günter Becker, I will try to clarify how the data are collected in order to facilitate the understanding about the problem:

-> A volume test is conducted using a material measure (a “bucket”) with a scale that varies from - 200 ml to + 200 ml, where “0” represent 20 liters. If the petro pump dispenses, for example 19.9 l, the error for this device is = (19.9 – 20.0) l = - 0.1 l.

-> These tests are conducted periodically in order to check the market, and it is expected that the values (errors) are closer to “0” as possible, what means that the consumers are not paying for a volume of fuel they are not consuming.

 -> So, for the average we have negative and positive errors leading to an average close to “0”, and a big standard deviation (SD) since the scale is not from “0” to positive, but from negative to positive values.

In order to obtain a macroeconomic indicator for the fuel market, I want to present the result as: average +- uncertainty.

I assume that an indicator for the measuring uncertainty in this case is the SD, however I have doubts if it is the best to be used in the present case.

I hope that I made the problem clear in order to have contributions!

Regards!

If you start with the assumption that the variations are statistically independent, then the SD of (your estimate of the average) will be the SD of the individual values divided by the square root of the number of samples.

In your example, you have 78,804 samples. The square root of that is about 280.7. We divide 45.99 by 280.7, and you get about 0.160.

So, you could report your estimate of the average as 7.99 plus/minus 0.160.

With this information, Gauss distribution appears reasonable. I am not quite sure, whether a negative error means the customer pays more than he gets and a positive one means, he pays less. (Or is it the other way round? Than the argument below can be reversed)

So, you could put it this way:

• On an average, a customer pays 0.04 % less than what he gets.
• Despite the variation of the measuring error, less than 44 % of the customers pay for more than they get..

It depends a bit on what your client requires. Although there are many ways to communicate statistics, I think to stress that the deviation is very small is fair in this case.

John Seymour is right. If your question is: How accurate is my estimate of the average, take his approach.

Incidentally, the results I have do not make any assumptions about whether the distribution is Gaussian. The "square root of n" effect on SD is based on the assumption of statistical independence, and not on the distribution.

If you wanted to make inferences about what percentage of customers wind up paying a certain amount more for the petrol, then you would need to make assumptions about the distributions.

Dear Dr. Günter Becker and John Seymour,

I usually use this approach (ISO GUM) - where uncertainty = SD/(N)^1/2 – to one instrument tested several times.

My doubt is if this approach suits for the present case, where several instruments are tested once.  

According to the comments, I understand that the assumption that the variations are statistically independent is what matter to apply the ISO GUM approach, right?

Thanks for contributions again!

Rgds.

Yes, Bruno. That's correct.

Thanks again!

Hi,

I think you must pay attention to the real question you want to answer : do you want to know the "measurement" uncertainty (i.e. the uncertainty on each measurement of your database), the confidence interval of the mean volume delivered by the 78,804 pumps, or the mean error one customer can meet when using one pump ?

For the first question, you can estimate the measurement uncertainty due to the method you used to produce each of the 78,804 measurements, including instruments used to determine the bias of each pump (GUM approach).

The two other questions can be answered by using classical statistics on your database, but, strictly speaking, it's not really the "GUM".

Best regards

You have tested 78,804 pumps for the accuracy to dispense 20 L. Your test device is designed to measure in mL and is scaled +/- 200 mL at 20,000 mL. The average value of the 78,804 measurements was 20,007.99 mL +/- 45.99 mL. The SD is small and not larger than the average.

Your evaluation of the pumps is an evaluation of the population of pumps. The SD is a population metric. Dividing by the square root of the number of pumps returns the standard error, which does not apply to the population distribution. The SE shows how well the average is known, but tells you nothing about the population distribution.

You could describe your results in round terms by stating that 95% of the pumps dispensed 20,000 mL within +/- 50 mL or 20 L +/- 0.05 L.

What was the range or how bad were the worst pumps? What percentage of the pumps were off 100 mL or more?

Dear Dr. Joseph L Alvarez,

Sorry for my delayed message, but I had to remade the evaluations in order to contribute to the discussion. I have talked to some statistics but you were the first to point that the average is actually 20007.99 mL, and I thank you for this out of the box point of view.

I summed 20 L (20000mL) to the results, as you suggested. The results are AVR = 20007.99 mL and SD=45.99mL. The results were already expected, once I am only moving the curve forward.   

However, I am attaching a value to the measuring error, for example: for a pump selling 20L (for 1L = 1$), and an error = 0.1 mL, the consumer is paying 20 $ but receiving 19.9 mL, then a loss of $ 0.1 for the consumer. So, as I am interest in the losses, I used the error as a distribution (error = instrument – standard) and not the instrument as you suggested.

I am only moving the curve backward, diminishing the values in order to have the error distribution. So the average is around 0, and the SD “seems” to be bigger than the average, but not, as you showed.  Am I right?

The worsts pumps limits delivered 15000 mL (error = -5000 mL) and 22000 mL (error = 2000mL) for a 20000 mL test. The percentage of pumps within SD is around 80% and within 2xSD is around 99.2%.

Thanks for the contribution!

Dear Dr. Bertrand Blanquart ,

I am actually interest in aswer the second question. In average, what the mean error one consumer can meeting the market.

Rgds, and Thanks!

You are correct. The SD is small. 

The error bounds 80% at one SD and 99.2% at two SD shows that the distribution is not normal, but reasonably close. The range from -5000 mL to 2000 mL is not surprising, but unfortunate. One should think such errors would be obvious.

Good study.

Dr. Joseph L Alvarez,,

Thanks for contributions. These extreme values are mostly related to frauds. I am conducting a secondary study in order to treat this case, once these errors are no statistical, but intentional.