# Which 'error' to choose random or absolute?

WIn the experiment, to measure the distance I used a ruler (the least digit is 0.5 cm)

For each idep. variable step I repeated the experiment 10 times, and then calculated the standard deviation from the set.The deviation for each set turned out to be less than the absolute error.

Presenting the result for each set, do I use the absolute error (the least digit) or the deviation?

• What do you mean by "absolute error"? – David Z Aug 23 '15 at 19:53
• The least digit of the ruler divided by 2. In my case 0.5 mm – Алексей Наумов Aug 23 '15 at 19:56
• it is worth noting that "half of the finest division on the scale" is a rule of thumb which is often a bit conservative (i.e. overestimates the real reading error). If you would be willing to write down an value between millimeters (like $2.25\,\mathrm{cm}$ in the event that it fell right between two marks), then the rule of thumb probably overestimates your real reading error, and you might use $0.3$ times the smallest scale division instead. I've also occasionally seen scales where I felt the finest division were optimistic and used a bigger reading error. – dmckee Aug 23 '15 at 19:56
• Ok, but in general. Should I stick to the "half of the finest division on the scale" (or 1/3 or 1/5) or use the deviation from the repeated set? Or, in other words, how do I decide between the smallest scale division divided by something and the standard deviation? – Алексей Наумов Aug 23 '15 at 20:02
• Related (possible duplicate, though currently without an answer): physics.stackexchange.com/questions/73785/… – David Z Aug 23 '15 at 20:13

Ad-hoc estimates of individual measurement errors, like the $0.5\ \mathrm{cm}$ in your example, don't play any role in maximum likelihood estimation. So if you have multiple measurements, it's better to forget about the individual estimated errors and just use the standard deviation.
As you describe it, each measurement is guaranteed to be within $\pm 0.5$ mm. If you measure the same object $10$ times and are consistent, the standard deviation of the measurements will be $0$. That clearly does not reflect reality-there is no random error, but there is a systematic error that could be as large as $\pm 0.5$ mm.
If you have ten different objects, measure each one, and add up the lengths, the error in the sum is guaranteed to be less than $\pm 5$ mm. It is very unlikely that all the individual errors are at the same end of the interval. A practical approach is to regard $0.5$ mm as the error of each measurement, then root sum square them to get $\pm 0.5 \sqrt {10}\approx 1.5$ mm as the error. It could certainly be worse than this. Even more aggressively, you can use the variance of your distribution being $\frac 1{12}$ mm$^2$, so the variance of the sum is $\frac {10}{12}$ mm$^2$ and the error of the sum is a little less than $\pm 1$ mm.