# Random uncertainies too small to be of use?

I'm writing an assignment report on a lab experiment (measuring switch on volts of LEDs to find $E$ vs $f$ and then calculate the Planck Constant). The digital voltage readings are to three significant figures and two decimal places (eg $1.34~\rm V$). So I can quote scale readings as being $\pm 0.01~\rm V$ - all good.

But when I do the random uncertainty $\left(\rm largest\ reading-smallest\ reading \over number\ of\ readings\right)$, I think I have a problem. In most cases the readings are very close (e.g. $1.34, 1.34, 1.34, 1.34$ & $1.35$ which gives $\frac{0.01}{5} = 0.002$). Am I correct that I can't quote this to three decimal places because the readings are to only two? If so then I need to round down to zero! Only one mark for the random uncertainty calculations but I want to be accurate (and not drop the mark by just writing down the numbers I am coming up with).

Seems like the voltage readings would need to be done to more significant figures to pick up whatever is really going on. So I should I state that the random uncertainty appears small (too small to be allowed to mention, except for ONE led where I have a value of $\pm 0.014$ so can say "$\pm 0.01$") and say in my Results Assessment section that the experiment would be better done with a more accurate voltmeter? Odd thing is the result was quite close to real value of 'h' (admittedly the best fit curve was tried a couple of times!)

(I initially gave a wrong answer; now I've corrected it.)

Never heard of this way to estimate errors. What you should really do is to calculate the standard deviation $s$ of your sample and then write your best estimate as

$$\mu \pm \frac{s}{\sqrt N}$$

where $N$ is your sample size. If the sample is

$$\{1.34,1.34,1.34,1.34,1.35\}$$

You will get

$$\mu=1.342$$ $$s=0.004$$

You can therefore report the result as

$$1.342 \pm \frac{0.004}{\sqrt 5} = 1.342 \pm 0.002$$

Notice that the estimated error is identical to what you estimated: however this is only a coincidence!

If however you have a stable reading, i.e. if your sample is

$$\{1.34,1.34,1.34,1.34,1.34\}$$

then $s$ would be $0$, which cannot be correct. In this case, since your instrument only give you a number up to the second decimal place, the best estimate you could give would actually be

$$1.340 \pm 0.005$$

This is not a random error, but really a limitation in the resolution of your instrument.

Notice the apparently paradoxical result that a stable reading results in a more "uncertain" result.

• Agreed - but the assignment instructions SPECIFICALLY state to calculate "Random uncertainty" as described. It's only 1 mark out of 20 but my feeling is that if I follow as instructed the 0.002 figure this yields isn't valid as it implies the reading are accurate to 3 dec places NOT the allowable 2. The limitations of the voltmeter used is indeed the issue. FYI - this is in Scottish SQA exam assignment. Napier, Clerk Maxwell and Lord Kelvin must be spinning in their graves! Mar 2, 2018 at 17:16