Does anyone know how to calculate absolute uncertainty?

[nchin]

Does anyone know how to calculate absolute uncertainty??

A student repeatedly measures the time taken by an object to cover a certain distance and obtains the following data:

t1 = 2.432 s, t2 = 2.393 s, t3 = 2.402 s, t4 = 2.412 s, t5 = 2.424 s

The instrument uncertainty is 0.001 s

What should the student report as the absolute uncertainty in the measurement? HINT: The absolute uncertainty can be given by either the instrument uncertainty or the standard error (approximated by R/N) in the measurements

This is what I've done:

so since the instrument uncertainty is 0.001 s then.

t1 = (2.432 ± 0.001)s
t2 = (2.393 ± 0.001)s
t3 = (2.402 ± 0.001)s
t4 = (2.412 ± 0.001)s
t5 = (2.424 ± 0.001)s

help! what do i do? do i add the uncertainties together like 0.001 + 0.001...?

 

[haruspex]

Can you clarify what is meant by "standard error (approximated by R/N)"? In the textbook, or wherever this comes from, what are R and N?

 

[nchin]

Can you clarify what is meant by "standard error (approximated by R/N)"? In the textbook, or wherever this comes from, what are R and N?

I believe R stands for range and N is the number of values.

values"
t1 = 2.432 s, t2 = 2.393 s, t3 = 2.402 s, t4 = 2.412 s, t5 = 2.424 s


so range is highest - smallest. 2.432-2.393 = 0.039
and the number of values is 5

so 0.039/5 = .0078

hmm could this be the answer?

 

[nchin]

nvm i don't think this is the answer cause I am pretty sure i need to somehow include the instrument uncertainty in my calculations?

 

[haruspex]

I believe R stands for range and N is the number of values.

values"
t1 = 2.432 s, t2 = 2.393 s, t3 = 2.402 s, t4 = 2.412 s, t5 = 2.424 s


so range is highest - smallest. 2.432-2.393 = 0.039
and the number of values is 5

so 0.039/5 = .0078

hmm could this be the answer?

I considered that, but it doesn't make sense. Why would you divide by N? If you took 1000 measurements and they all fell in this range then R/N would become very small, yet the range of scatter has stayed the same.
An engineer might take the minimum reading - .001 to max reading + .001. But I can't square that with the hint. A statistician would treat this as the sum of two independent random variables, perhaps taking each to be normally distributed. But then you have to decide how many standard deviations you mean by 'absolute error'.

 

[nchin]

I considered that, but it doesn't make sense. Why would you divide by N? If you took 1000 measurements and they all fell in this range then R/N would become very small, yet the range of scatter has stayed the same.
An engineer might take the minimum reading - .001 to max reading + .001. But I can't square that with the hint. A statistician would treat this as the sum of two independent random variables, perhaps taking each to be normally distributed. But then you have to decide how many standard deviations you mean by 'absolute error'.

hey the answer was 0.039/5 = .0078

thanks !