How Do Significant Figures Reflect Measurement Uncertainty?

[balboa]

When certain quantities are measured, the measured values are known only to within limits of the experimental uncertainty!

It's here where significant figures comes on scene.

Let us assume that the accuracy to which we can measure the length is +-0.1 cm. If we measured it to be 16.2 cm, there are three significant figures, but if we measure it to be 4.4 cm there are two. The uncertainty is same, but the number of significant figures is not. Why?

Also if we can measure something to be 0.00003 m and something else to be 0.03 m the uncertainty is much greater in the last measurement, but both numbers have only 1 significant figure. Can anyone explain this to me?

thank you for every reply!

 

[tiny-tim]

welcome to pf!

hi balboa! welcome to pf!

are we talking about input or output?

usually, we are given (or we measure) certain input numbers, from which we make a calculation to get an output number …

eg we measure m L and T and we calculate g …

we look at the significant figures in the input numbers, and use those to decide how many sig figs we should put in the output number …

in your example, if the input number was 16.2, that's 3 sig figs, so we probably use 3 sig figs in the output; if it was 4.4, we probably use 2 sig figs (and if it was 1998.2, we'd probably use 5 sig figs)

 

[balboa]

so as I understood the sig figs have nothing to do with how much we are certain about measures!

in my example those were input values, and I thought that sig figs have something with how certain those values are, but it seems they just show us how to calculate with them

thanks for your explanation and nice welcome to pf!

 

[BobG]

It does have to do with the uncertainty of the measurements.

If you have 16.2, then you really have some value between 16.16 and 16.25 inclusive. If you have 4.4, then you really have some value between 4.36 and 4.45 inclusive.

16.2 * 4.4 = 71.28

However, your value could really be as high as 72.3 or as low as 70.5. So your final answer should be 71, realizing that that could mean anywhere from 70 to 72. Obviously, an imperfect solution, but it at least better reflects the uncertainties than giving an answer that implies you have a better idea of the actual value than you really do.

For addition, you could (and do) get away with just keeping the smallest significant digit since:

16.2 + 4.4 = 20.6, but your real answer could be as high as 20.7 and as low as 20.5. Your final answer has an uncertainty of only plus or minus 0.1

If you use these results in further operations, your margin of error just keeps getting bigger, so it's best to just keep in mind that your last significant digit may not be very accurate. Then again, the more steps in your operations, the more likely it is that the highs cancel out the lows and the chances of your final answer being at the extreme edge of your error margin is pretty low.

 

[balboa]

It does have to do with the uncertainty of the measurements

thnx BobG, but what about this:

• 4.4 has two sig figs and 16.2 has three sig fig (different number of sig figs but same uncertainty <plus or minus 0.1>)
  
  
• both 0.03 and 0.0000003 have only 1 sig fig (same number of sig figs but different uncertainty)

This was my point when I was posting the thread.

 

[Vanadium 50]

There are two different concepts: absolute and relative uncertainty. 100 +/- 1 and 10 +/- 1 have the same absolute uncertainty, but different relative uncertainties: you know the first number to 1% of its magnitude and the second number to only 10%.

 

[Borek]

Don't pay too much attention to sig figs, they are faulty by design.

 

[balboa]

ok, I think I've got a point, thank you all for help :)