4 ms·

People have this idea that when you take a measurement, you have so-and-so number of significant figures that are probably correct and the rest are just pure no

People have this idea that when you take a measurement, you have so-and-so number of significant figures that are probably correct and the rest are just pure noise. But that's not how measurement error works. In the real world, physical measurement errors are more-or-less normally distributed (we don't have to argue about the "more-or-less" part because my argument here holds for any distribution other than a uniform one). Let's say your measurement gives you a latitude of 45.73490534578° with a standard deviation of 0.00001° (that's about 11 meters). Those last few digits of your measurement are almost certain to be wrong. But does that make them meaningless? No! Because if your measurements are unbiased, then slightly more than half the time, 45.73490534578° is still going to be closer to the true value than 45.7349053457° is. By performing significant figure rounding, you aren't throwing very much information, you may not be throwing away any information you care about, but you are nonetheless throwing away information.

protonfish 7y ago
The problem is that these lat/lon figures are not displayed with precision information at all, so we can't even know the std dev. Sig figs are a simple and clear way to communicate the precision of a measurement, but if you want to be more statistically accurate, you can use parentheses to indicate the standard deviation. In your example (if I mess this up please correct me, but I think) it would be 45.73490534578(1000000)° Which again is silly because there is no way to know the standard deviation to that level of precision. A more reasonable number would be 45.734905(10)° Is it difficult to include precision when reporting measurements? No Is it sometimes valuable? Yes Is it really too much to ask for? No
- dfranke 7y ago
  I've never seen the convention you're using and I don't think I understand it. Conventions I've seen include: * Give an error bound like 45.73490534578° (±0.00002°) and indicate in prose that this is a 2σ bound. * Put non-significant figures in parenthesis, like 45.73490(534578)° (EDIT: possibly I've misinterpreted this one when I've seen it, see logfromblammo's reply) * Put a bar over the last significant figure, like 45.73490̄534578 (hopefully this one renders properly when I post this... (EDIT: nope))
  - JBorrow 7y ago
    The parentheses show something similar to your first, where this is the +- read from the right hand side of the number. So you could write yours: 45.73490534578(2000000)
  - logfromblammo 7y ago
    The value in the parentheses is the symmetric one-sigma bound. If I say the atomic weight of F is 18.998403163(6) g/mol... mean 18.998403163 g/mol std.dev. 0.000000006 g/mol If I say Planck time is 5.391245(60)e−44 s... mean 5.391245e−44 s std.dev. 0.000060e-44 s The standard rules for rounding imply that whenever a measurement is given to a certain number of significant figures, you're leaving out "0(5)" from the end. So 1.2345 is 1.23450(5) in parenthetical notation. Significant figures rules give you a close-enough propagation of error, but in order to be more exact, you need to combine absolute uncertainties when adding or subtracting, and combine relative uncertainties when multiplying or dividing.
  - busyant 7y ago
    You can see it done here: https://en.wikipedia.org/wiki/Standard_atomic_weight#List_of_atomic_weights https://en.wikipedia.org/wiki/Standard_atomic_weight#List_of...
    - logfromblammo 7y ago
      The brackets in the table mean that different sources of the element have different proportions of isotopes, so the mean atomic weights for specific deposits may cover a range that differs from the overall mean for every deposit of the element ever measured. I.e. if you measure carbon from the upper atmosphere, it's going to have more C-14 in it, from cosmic rays flipping protons in N-14 to neutrons. And if you measure carbon buried for thousands of years, it's going to have less C-14, from natural decay. If you look at https://en.wikipedia.org/wiki/List_of_physical_constants https://en.wikipedia.org/wiki/List_of_physical_constants you can see that the parentheses are omitted from defined constants, and included for measured constants.
  - gus_massa 7y ago
joshgel 7y ago
> if your measurements are unbiased This is a big IF which you can't actually know because its beyond your level of precision, by definition.
- dfranke 7y ago
  This is completely knowable, by taking repeated measurements of a reference object, one which was either checked by a more precise instrument or is definitionally correct (e.g. the old reference kilogram or the Greenwich meridian)
Dylan16807 7y ago
In the time it takes you to note down those extra digits, you could have improved your actual measurement by the same minuscule fraction of a bit ten times over.
mabbo 7y ago
The question isn't about mathematical rigor. It's about utility and distraction. In much of the world, phone numbers work even if you add extra numbers that aren't needed. So I could give you my phone number plus 10 extra digits that change nothing. The end result would be the same utility to you (you can contact me) but with an increased cost in recording, memorizing, chance of error. Using lat/lon to an unnecessary level of detail is the same thing. More digits are more chances to make mistakes, more cognitive load.
- lutorm 7y ago
  Exactly. Extraneous, useless, information is not no-value, it's negative-value, because of this.
kgwgk 7y ago
> slightly more than half the time, 45.73490534578° is still going to be closer to the true value than 45.7349053457° is. By performing significant figure rounding, That's not rounding, so I'll assume you meant "slightly more than half the time, 45.73490534578° is still going to be closer to the true value than 45.7349053458° is". And that may be true, but if 50.0001% of the time one is closer to the true value than the other that's essentially meaningless.

protonfish 7y ago

The problem is that these lat/lon figures are not displayed with precision information at all, so we can't even know the std dev. Sig figs are a simple and clear way to communicate the precision of a measurement, but if you want to be more statistically accurate, you can use parentheses to indicate the standard deviation. In your example (if I mess this up please correct me, but I think) it would be 45.73490534578(1000000)° Which again is silly because there is no way to know the standard deviation to that level of precision. A more reasonable number would be 45.734905(10)° Is it difficult to include precision when reporting measurements? No Is it sometimes valuable? Yes Is it really too much to ask for? No

dfranke 7y ago
I've never seen the convention you're using and I don't think I understand it. Conventions I've seen include: * Give an error bound like 45.73490534578° (±0.00002°) and indicate in prose that this is a 2σ bound. * Put non-significant figures in parenthesis, like 45.73490(534578)° (EDIT: possibly I've misinterpreted this one when I've seen it, see logfromblammo's reply) * Put a bar over the last significant figure, like 45.73490̄534578 (hopefully this one renders properly when I post this... (EDIT: nope))
- JBorrow 7y ago
  The parentheses show something similar to your first, where this is the +- read from the right hand side of the number. So you could write yours: 45.73490534578(2000000)
- logfromblammo 7y ago
  The value in the parentheses is the symmetric one-sigma bound. If I say the atomic weight of F is 18.998403163(6) g/mol... mean 18.998403163 g/mol std.dev. 0.000000006 g/mol If I say Planck time is 5.391245(60)e−44 s... mean 5.391245e−44 s std.dev. 0.000060e-44 s The standard rules for rounding imply that whenever a measurement is given to a certain number of significant figures, you're leaving out "0(5)" from the end. So 1.2345 is 1.23450(5) in parenthetical notation. Significant figures rules give you a close-enough propagation of error, but in order to be more exact, you need to combine absolute uncertainties when adding or subtracting, and combine relative uncertainties when multiplying or dividing.
- busyant 7y ago
  You can see it done here: https://en.wikipedia.org/wiki/Standard_atomic_weight#List_of_atomic_weights https://en.wikipedia.org/wiki/Standard_atomic_weight#List_of...
  - logfromblammo 7y ago
    The brackets in the table mean that different sources of the element have different proportions of isotopes, so the mean atomic weights for specific deposits may cover a range that differs from the overall mean for every deposit of the element ever measured. I.e. if you measure carbon from the upper atmosphere, it's going to have more C-14 in it, from cosmic rays flipping protons in N-14 to neutrons. And if you measure carbon buried for thousands of years, it's going to have less C-14, from natural decay. If you look at https://en.wikipedia.org/wiki/List_of_physical_constants https://en.wikipedia.org/wiki/List_of_physical_constants you can see that the parentheses are omitted from defined constants, and included for measured constants.
- gus_massa 7y ago

joshgel 7y ago

> if your measurements are unbiased This is a big IF which you can't actually know because its beyond your level of precision, by definition.

dfranke 7y ago
This is completely knowable, by taking repeated measurements of a reference object, one which was either checked by a more precise instrument or is definitionally correct (e.g. the old reference kilogram or the Greenwich meridian)

Dylan16807 7y ago

In the time it takes you to note down those extra digits, you could have improved your actual measurement by the same minuscule fraction of a bit ten times over.

mabbo 7y ago

The question isn't about mathematical rigor. It's about utility and distraction. In much of the world, phone numbers work even if you add extra numbers that aren't needed. So I could give you my phone number plus 10 extra digits that change nothing. The end result would be the same utility to you (you can contact me) but with an increased cost in recording, memorizing, chance of error. Using lat/lon to an unnecessary level of detail is the same thing. More digits are more chances to make mistakes, more cognitive load.

lutorm 7y ago
Exactly. Extraneous, useless, information is not no-value, it's negative-value, because of this.

kgwgk 7y ago

> slightly more than half the time, 45.73490534578° is still going to be closer to the true value than 45.7349053457° is. By performing significant figure rounding, That's not rounding, so I'll assume you meant "slightly more than half the time, 45.73490534578° is still going to be closer to the true value than 45.7349053458° is". And that may be true, but if 50.0001% of the time one is closer to the true value than the other that's essentially meaningless.