# How to write measured value with error

• I
During a labexercise we measured a value to be 26.3 mT. The teslameter was said to have an accuracy at +- 0.05%. This means we measured 26.3 +- 0.0132mT. However we are supposed to write a measured value with it's error in the last digit. But the error is smaller than the smallest digit we got from the measurement. Does this mean I shold not write 26.3 +- 0.0132mT and just write 26.3 mT without any estimation of the error?

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Not sure, If I can be right, since all the significant digits in 0.0132 are uncertain, and in 26.3, all digits are certain, How about writing off a value with all certain digits and one uncertain digit at end? Like 26.31? That last digit '1' shows uncertainty in your measurement denoting error.

It is better you get this verified with someone before I misguided you.

Thank you, that is what I would like to do. But the Teslameter only gave me the first decimal digit so I wouldn't know what to write after 26.3. I don't think I could just add zero when I didn't measure it.

After referring my physics textbook regarding taking errors, You can write the measured value with error as 26.3 ± 0.01 as in 0.0132, you can consider 32 as insignificant and round them to 30 getting, 0.013 and then to 0.01. Now you can't neglect 0.01. Simply writing 26.3 will say kind of no error.

hilbert2
Gold Member
When I was doing undergraduate lab work, we had a "15 unit rule" which said that ##3.77\pm 0.24## is rounded to ##3.8\pm 0.3##, while ##5.171\pm 0.124## is rounded to ##5.17\pm 0.13##. Not everyone uses the same rule, though.

AlphaLearner
When I was doing undergraduate lab work, we had a "15 unit rule" which said that ##3.77\pm 0.24## is rounded to ##3.8\pm 0.3##, while ##5.171\pm 0.124## is rounded to ##5.17\pm 0.13##. Not everyone uses the same rule, though.
Couldn't even apply any single rule regarding way you rounded off digits. Can you explain the logic behind your rounded figures hilbert2.

hilbert2
Gold Member
If the first nonzero digits in the error are less than "15", the error is rounded to two significant figures, otherwise it's rounded to one significant figure. And an error is always rounded upwards.

If the first nonzero digits in the error are less than "15", the error is rounded to two significant figures, otherwise it's rounded to one significant figure. And an error is always rounded upwards.
Following you, should the measure come as 26 ± 0.14?

hilbert2
Gold Member
The result should have the same precision as the error. Here the result 26.3 mT doesn't have as good precision as the error rounded to two or one significant figures, which implies that the actual error is larger than OP thinks. I don't really get why the measuring equipment wouldn't show as many figures as the magnitude of error allows.

CWatters
Homework Helper
Gold Member
If the meter can only display 3 digits the the meter could have measured any value between 26.25 to 26.35 and it would display 26.3.

Then the meter could be inaccurate by +/- 0.0132 so the actual value could be anywhere from 26.25-0.0132=26.2368 to 26.35+0.0132=26.3632

I think.

If the meter can only display 3 digits the the meter could have measured any value between 26.25 to 26.35 and it would display 26.3.

Then the meter could be inaccurate by +/- 0.0132 so the actual value could be anywhere from 26.25-0.0132=26.2368 to 26.35+0.0132=26.3632

I think.
When you have two readings one was 26.2368 and other was 26.3632, how about taking the closest value to true value by simply taking their mean? Last digit will tell error. Even probability of getting a true value from it is pretty less due to too many uncertain digits in error.

Tom.G
During a labexercise we measured a value to be 26.3 mT. The teslameter was said to have an accuracy at +- 0.05%. This means we measured 26.3 +- 0.0132mT. However we are supposed to write a measured value with it's error in the last digit. But the error is smaller than the smallest digit we got from the measurement. Does this mean I shold not write 26.3 +- 0.0132mT and just write 26.3 mT without any estimation of the error?
With any digital readout the accuracy is never better that ±1 least-significant-digit (LSD). That's because you have no more information being given. The specifications of a device with a digital readout are given as ± a percent of either reading or full scale, ± a number of counts. If the percent error works out to less than LSD/2 then the LSD error dominates and the percent error is ignored.

pixel and AlphaLearner
CWatters
Homework Helper
Gold Member
When you have two readings one was 26.2368 and other was 26.3632, how about taking the closest value to true value by simply taking their mean? Last digit will tell error. Even probability of getting a true value from it is pretty less due to too many uncertain digits in error.

CWatters
Homework Helper
Gold Member
With any digital readout the accuracy is never better that ±1 least-significant-digit (LSD). That's because you have no more information being given. The specifications of a device with a digital readout are given as ± a percent of either reading or full scale, ± a number of counts. If the percent error works out to less than LSD/2 then the LSD error dominates and the percent error is ignored.
+1

Although I don't think you can ignore the percent error unless it's much less than LSD/2.

It appears that there is an assumption made in this thread (and seemingly everywhere else on the Internet) that digital instruments round their values before displaying. But I don't think that is correct.

For example, if I measure the time it takes for a ball to roll down a ramp with a timer that only measures to the whole seconds, then a typical result could read (say) 45 seconds. However, the digital display isn't going to display "45" until the measurement reaches at least 45.000000000001 seconds and will retain the "45" until the measurement reaches "45.999999999999" seconds.

It would seem to me that a digital device displaying "45" should be expressed as 45.5 \pm 0.5 seconds, not 44\pm 1 second nor 45 \pm 0.5 seconds.

Do I have this right?