Significant Digits for Uncertainty Values

[AllisonW4]

Homework Statement

Question on worksheet: What is the random uncertainty in this set of measurements? Use the instrumental uncertainty to limit the number of digits in your random uncertainty result.

Data for dropping a pencil from a height of one meter using a stopwatch that measures to the millisecond:
0.23 s 0.28 s 0.26 s 0.26 s 0.24 s 0.26 s 0.30 s

Homework Equations

We are given that σ = (high value - low value) / (number of measurements)^1/2
(I know that this formula is a really, really rough estimation - I think the teacher is just trying to keep it a bit simpler for us instead of making us find the standard deviation)

The Attempt at a Solution

σ = (high value - low value) / (number of measurements)^1/2
σ = (0.30 s - 0.23 s) / (7)^1/2
σ = 0.02646 s

This is where I am stuck - how many digits should I round this to? I believe that the instrumental uncertainty of our stopwatch was ±0.01 s because it measured to the nearest millisecond. In the question it says to "use the instrumental uncertainty to limit the number of digits in your random uncertainty result" - should I round the uncertainty to the nearest millisecond because the watch was only accurate to the millisecond? Or do something else? Our teacher also wrote on the worksheet to "use the instrumental uncertainty to determine the last significant digit"... I am not really clear on what she means by that.

 

[haruspex]

±0.01 s because it measured to the nearest millisecond.

0.01s is not a millisecond. Also, if it is to the nearest millisecond, what is the maximum error?

should I round the uncertainty to the nearest millisecond because the watch was only accurate to the millisecond?

That appears to be the instruction. Seems reasonable.

 

[AllisonW4]

0.01s is not a millisecond. Also, if it is to the nearest millisecond, what is the maximum error?

That appears to be the instruction. Seems reasonable.

Thank you! And oops I was being dumb, of course 0.01 is not a millisecond but one hundredth of a second. So would you agree that I should round it to ±0.03 s then?

 

[haruspex]

So would you agree that I should round it to ±0.03 s then?

No, you are still looking at the wrong decimal place.
Express 0.02646s in msec.

 

[alejandromeira]

Data for dropping a pencil from a height of one meter using a stopwatch that measures to the millisecond:
0.23 s 0.28 s 0.26 s 0.26 s 0.24 s 0.26 s 0.30 s

First of all: the units of measure you have in the data are seconds.

There is a problem in the statement. The statement says that the clock measures milliseconds. But the data is given with a precision of hundredths of a second.
If the chronometer measures milliseconds, why not appear 0.230s, 0.280s, etc? (add a zero at the end, and round to three decimal places)

If you use two decimals, rounding 0.03 is fine. But if you use 3 decimals then you have to round it with the next number.

I do not understand why in the numbers of the measures of time, there are only two decimal places. If the clock shows milliseconds, three should appear.