Uncertainties in fluid viscosity experimental trials

[physicsdude123]

Homework Statement

You are going to examine the viscosity of a fluid by dropping a ball bearing into a
cylinder of fluid and measuring the fall time. You will use a ruler to measure the
height, h, of the fluid column, and a stop watch to measure the fall time, t. The ruler
and stop watch have reading uncertainties of 0.5 cm and 0.05 s respectively. There
are four members in your group and you each do one measurement of each
quantity. Here are your results:
h1 = 36.0 cm t1 = 15.55 s
h2 = 36.0 cm t2 = 15.25 s
h3 = 36.0 cm t3 = 15.75 s
h4 = 36.0 cm t4 = 15.95 s
You would report your average value of h (in cm) and t (in seconds) as

Homework Equations

Don't know of any?

The Attempt at a Solution

I know to average the quantities, so for h=36.0 and t=15.63, but what I am unsure about
is how to do the uncertainties. Does the uncertainties stay the same as the measurement tool? So would they both be +/-0.5 or am I missing something?

Thanks!

 

[rude man]

You have 4 mesurements of h and four of t so the average of each of those two measurands would have their respective measurement errors reduced by a factor of 1/√4.

Since you will be determining avg. velocity = avg. h/avg. t you now need to combine the error on avg. h and the error on avg. t. What is the rule for error propagation for multiplication/division?

 

[haruspex]

You should treat the two kinds of measurement differently.
The height is fixed; the only uncertainty of it is in the measurement. For each measurer, the error range is ±0.5 cm. Even if they all work precisely, they will all generate the same error. The error in the ruler is not some random perturbation, it's just the limit of precision. Therefore the error in 36cm is still ±0.5 cm.
The time really does vary. You don't know the distribution for that, but you could take it to be normal. The error induced by the stopwatch, however, is uniform over the range ±0.05 s. You could approximate that with a normal distribution of the same variance. So the resulting time data come from the sum of two normal distributions, one with known mean and variance and the other with unknown parameters.

 

[rude man]

You should treat the two kinds of measurement differently.
The height is fixed; the only uncertainty of it is in the measurement. For each measurer, the error range is ±0.5 cm. Even if they all work precisely, they will all generate the same error. The error in the ruler is not some random perturbation, it's just the limit of precision. Therefore the error in 36cm is still ±0.5 cm.
.

Not so. h will vary randomly from reader to reader. I am neglecting the systematic error you're referring to, which implies an error in the ruler itself. I think the intent of this problem is to see how random errors propagate.

 

[haruspex]

Not so. h will vary randomly from reader to reader. I am neglecting the systematic error you're referring to, which implies an error in the ruler itself. I think the intent of this problem is to see how random errors propagate.

I admit the question could be more clearly worded, but I maintain that my interpretation is the more reasonable. I'm not saying there's an error in the ruler, only that the precision of reading is to the nearest 1cm. It's inevitable that that's how instrument readings work.
Under your interpretation, all the height readings being 36.0 would be somewhat unbelievable.
My guess is that recognising that the two sets of data should be treated differently is the whole point of the question.

The timings are more problematic. There's a bit of a mismatch between, on the one hand, the error being quoted as up to .05 s and, on the other, readings like 15.55s. With the obvious interpretation of the error range being "to the nearest .1 s", the readings should only have one decimal place. I think it possible the question is supposed to say the timing error is up to .025s.

The general statistical problem this raises is a very interesting one. We have a continuous r.v. X, and Y = [X], where the square brackets denote nearest integer. X and Y-X are clearly not independent. On the assumption that X is normal, I've been trying to derive the expectation and variance of Y, but it's messy.