How Do You Calculate Uncertainty in Mean Difference?

AI Thread Summary
To calculate the uncertainty in the mean difference, one must first determine the mean difference, which is suggested to be 0.16. The uncertainty should not be calculated by simply adding the standard error of the means (SEM) but rather using the method of adding uncertainties in quadrature. This approach accounts for the propagation of errors when calculating the mean difference from two variables. In cases where correlations between uncertainties are not significant, the quadrature method is preferred for accuracy. Understanding these principles is essential for correctly reporting the mean difference with its associated uncertainty.
selsunblue
Messages
4
Reaction score
0

Homework Statement



321321.png


Homework Equations



Mean difference and SEM

The Attempt at a Solution



Would the mean difference be 0.16?. I'm not sure about the answer because they give that much working space, Do I have to find the uncertainty of the difference first and then my answer implement that into my answer to be 0.16 +- (uncertainty)?. How would I calculate the uncertainty of the mean difference? Do I just add the SEM's for both means?
 
Last edited:
Physics news on Phys.org
Would the mean difference be 0.16?
Sure. The large working space is a bit odd.
Do I have to find the uncertainty of the difference first and then my answer implement that into my answer to be 0.16 +- (uncertainty)?
That is part (ii).
How would I calculate the uncertainty of the mean difference? Do I just add the SEM's for both means?
No. Use the method (hopefully) described in section 1.5.
 
selsunblue said:
Do I just add the SEM's for both means?

Here are the basic rules for uncertainty:

Addition/Subtraction
$$(A ± ΔA) + (B ± ΔB) = (A+B) ± (ΔA+ΔB)$$
$$(A ± ΔA) - (B ± ΔB) = (A-B) ± (ΔA+ΔB)$$

When dividing and multiplying, changing to relative uncertainty, then converting back simplifies things.
where: $$ε = \frac{ΔA}{A}*100$$ 'relative uncertainty'

$$(A ± ΔA) / (B ± ΔB) = (A/B) ± (ε_A+ε_B)$$
$$(A ± ΔA)(B ± ΔB) = (AB) ± (ε_A+ε_B)$$

Then you can convert back to abs. uncertainty.
 
vanceEE said:
Here are the basic rules for uncertainty:

Addition/Subtraction
$$(A ± ΔA) + (B ± ΔB) = (A+B) ± (ΔA+ΔB)$$
$$(A ± ΔA) - (B ± ΔB) = (A-B) ± (ΔA+ΔB)$$
Are you sure? I believe the uncertainties should not simply be added, but added in quadrature. That is, if
x = au \pm bv,
then
\sigma_x^2 = a^2 \sigma_u^2 + b^2 \sigma_v^2.
This comes from the basic definition: If x is a function of measured variables u, v, \ldots,
x = f(u, v, \ldots),
then,
\sigma_x^2 \approx \sigma_u^2 \left ( \frac{\partial x}{\partial u}<br /> \right )^2 + \sigma_v^2 \left ( \frac{\partial x}{\partial v} \right )^2 + \ldots \;.

I did not check your other equations.
 
Yes, in general the uncertainties should be added in quadrature, if correlations are not important.

A linear addition is sometimes used as worst case estimate to include possible correlations between the uncertainties. We take the difference between measured heights here - there is no way the uncertainties could be correlated "the wrong way" (e. g. a systematic deviation towards larger men and smaller women at the same time).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top