How to round in error analysis

  • #1
Hi, to start with my questions I will show you what I have done so far.

(23.56+/-0.05) km/h x (56.3+/-0.4) h

So I ended up with (1326.428+/-12.234) km
But I know the real answer is (1330+/-10) km
What I dont understand is how I would round to that answer. I do not know what I have to look for to see that I have to round it.
 

Answers and Replies

  • #2
Quantum Defect
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Hi, to start with my questions I will show you what I have done so far.

(23.56+/-0.05) km/h x (56.3+/-0.4) h

So I ended up with (1326.428+/-12.234) km
But I know the real answer is (1330+/-10) km
What I dont understand is how I would round to that answer. I do not know what I have to look for to see that I have to round it.
Basically, all of the extra digits in the error that you have are extraneous. The 0.234 past the decimal is essentially meaningless.

The error will be the square root of the sum of the squares of the individual relative errors.

speed: 0.2% error; time: 0.7% error ==> resultant error = sqrt (0.002^2 + 0.007^2) = 0.0074 --> 0.74%
distance = 1326.4 km (0.74% of this is the error) -- error = 9.8 km ==> which has been rounded to 10 (one sig fig)
So, the answer is 1330 +/- 10 km (you round the answer to have the same precision as the precision in the error)
 
  • #3
haruspex
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The error will be the square root of the sum of the squares of the individual relative errors.
Not necessarily. [soap box alert]
To an engineer, the range of error in the answer is all values consistent with the given inputs. This makes Syed's original answer correct, except for some overstatement of precision. 1326.4+/-12.2 would be reasonable.
In scientific circles, it is customary to do as you say and take a more statistical approach. Sadly, there are serious flaws with the way that is usually done.
The basis of it is that the error range is interpreted as some (unstated) number of standard deviations of an approximately normal distribution. The calculation you mention then obtains the same number of standard deviations of the result. But in many, if not most, practical situations the original error is clearly not normally distributed. A classic example is rounding a reading to a number of digits. If my lab scales show a weight of 0.120N, in a digital display, that's a uniform distribution from 0.1195 to 0.1205. The range +/- 0.0005 then represents some calculable number of s.d. But after performing the calculation that combines this weight with other uniformly distributed data, the distribution is no longer uniform. Thus, it may be appropriate to adjust the computed error if the +/- expression of it is to have a consistent interpretation.
sqrt (0.002^2 + 0.007^2) = 0.0074
0.00728
 

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