Hot answers tagged statistics
16
No, one doesn't need to measure the material for years - or even millions or billions of years. It's enough to watch it for a few minutes (for time $t$) and count the number of atoms $\Delta N$ (convention: a positive number) that have decayed. The lifetime $T$ is calculated from
$$ \exp(-t/T) = \frac{N - \Delta N}{N}$$
where $N$ is the total number of atoms ...
9
Because physicists learn the math and do it themselves. Why do you need a special expert class of people nowadays?
EDIT: Deconstructing statistics
In response to comments that "statisiticians go through years of study", I would like to say why I think all this studying is counterproductive. The theory of statistics (when it isn't about statistical ...
8
I agree with @Ron Maimon that these ETS questions are problematic. But this is (i think) the reasoning they go with. Unlike @Mike's assumption you should not take the normal average, but as stated in the question the weighted average. A weighted average assigns to each measurement $x_i$ a weight $w_i$ and the average is then
$$\frac{\sum_iw_ix_i}{\sum_i ...
7
The formula you've specified
$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$
is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression
$$ k = k_1 + k_2.$$
Generally, to obtain experimental error of a dependent quantity (and the expression stated in your question), you start with ...
7
The fact that the quantities $\langle p^2\rangle$ and $\langle p\rangle^2$ are different is not something specific to quantum mechanics, but exists in any context where one can define an average value. Let's take a simple example and suppose that $p$ is a balanced binary random variable which can take the values $+1$ and $-1$. Since the variable is ...
6
Typically one knows the functional form of something when they are interested in the FWHM. In this case, you can use least-squares fitting to fit the function to the data and extract the parameters. A very common functional form of a resonance is the Lorentzian:
$$f(A,\gamma,x_0;x) = A\frac{1}{1 + \gamma^2\left(x-x_0\right)^2}$$
where $1/\gamma$ gives the ...
6
Errors in particle physics are of two kinds. Statistical, and systematic. Statistical is the usual standard deviation of gausian distributions, sqrt(n)/N for 1 sigma. It is the systematics that take a lot of effort, and often are not taken well into account.
Systematic errors come from
1) the background to the signal expected. The background is calculated ...
6
The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept $\beta$, your friend takes into account that the true zero of the mass parameter $m$ may be shifted from what you think it is. So an affine model $T^2=Cm+\beta$ is ...
6
The ultimate answer is the JCGM 100:2008 guide followed by most of the metrology institutes around the world. The specific chapter on combining uncertainties is Chapter 5.
Specifically, for a two-variable function $f(t_1, t_2)$ of two random variables, Eq. (16) of Section 5.2.2. gives $$\Delta f^2= \left (\frac{\partial f}{\partial t_1} \right )^2 \Delta ...
6
Well, the difference between the two expressions is exactly $(\Delta p)^2$, i.e. the squared uncertainty (variance) of the momentum, as the very question correctly says.
To be sure, the real question is why it's not zero. It's not zero. Just write the function $\psi$ in the momentum representation. Then there is a probabilistic distribution
$$ ...
6
Geant is a framework---which means that you use it to build applications the simulate the detector and physics you are interested in. The simulation can include all of physics and the complete detector including electronics and trigger (i.e. you can write your simulation so that it output a data file that looks just like the one you are going to get from the ...
6
think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result.
When the functions are arbitrary, the average of the product quantifies statistical ...
5
If you're talking about deriving the value of Planck's constant, then no, that is not possible. The value is simply a consequence of our chosen unit system.
If you're talking about deriving the fact that something analogous to Planck's constant has to exist at all, then I believe the answer is still no. To some extent that is also a consequence of our unit ...
5
I don't think there's any way you're going to do this in six months.
I'll give a calculation below, but first an order of magnitude estimate. If you've detected a total of $N_{\rm events}$ events, your measurement of a modulation will have an error of order $N_{\rm events}^{-1/2}$ --
-- these things always do! -- so the number of events required is going ...
5
Maybe for the same reason that experimental physics groups do not have a theoretician as a group member.
One could think of experimental groups as ruled by "control freaks", they need as members experimetnalists who have mastered enough theory to set up the experiments and enough theory to interpret the results. Within this "theory" one could count ...
5
I think you're exercising an incorrect picture of statistics here - mixing the inputs and outputs. You are recording the result of a measurement, and the spread of these measurement values (we'll say they're normally distributed) is theoretically a consequence of all of the variation from all different sources.
That is, every time you do it, the length of ...
4
I believe those are two different measures. The detection significance is the likelihood that a signal is real; the error is a statement about how precise your measurement is. It's the old accuracy-versus-precision distinction. The detection significance is measured by comparing how strong your signal is to the background noise of the image.
If you want the ...
4
For statistical analysis Gnumeric works very well, as it has passed a lot of statistical test. This report explains why it is a much better choice than Excel.
Of course there is also R, which is the largest free statistical package and is used in a lot of research areas.
Personally I think going the plain python route is also not a bad idea, as there is ...
4
Look, Dr. Zaslavsky is completely correct. But. The great mathematician Jean Leray once, after being asked to think about Maslov's work on asymptotic methods to approximate the solutions of partial differential equations which were generalisations of the WKB method, decided, in the 70's, to write an entire book titled
Lagrangian Analysis and Quantum ...
4
That depends entirely on what you consider to be "expected range of values."
When you see a value like $3.43\pm 0.04$ (I will omit units for brevity), in many cases, it actually represents a normal probability distribution with a mean of $3.43$ and a standard deviation of $0.04$. If the $3.43\pm 0.04$ is the result of an experiment, for example, then the ...
4
Yes, the only sensible formula for the total error is the sum in quadrature,
$$ \Delta X_{\rm total} = \sqrt { \Delta X_{\rm syst}^2 + \Delta X_{\rm stat}^2 } $$
The key assumption behind the validity of the formula is that the two sources of error are independent i.e. uncorrelated.
$$ \langle \Delta X_{\rm syst} \Delta X_{\rm stat} \rangle = 0$$
Because of ...
4
You're measuring the quantity $X$ and you got results $+1,0,-1$ and perhaps $+1,-1$ again. Assuming that your systematic error is zero, these numbers are randomly generated around the right value you want to know.
That's why you want to estimate the right value as the average of the results you obtained. That's
$$ \overline{X}= \frac{(-1)+0+(+1)}{3} = ...
4
On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges).
However, in statistical mechanics one ofte studies effective theories where there are additional means of distinguishing particles. Two important examples:
In modeling molecular fluids, two atoms on the same molecule are ...
4
The solution is to realise that that function is merely a Gaussian. In fact Each component of the velocity vector has a normal distribution with mean =0 and st-dev $\sqrt {kT/m}$.
All that is left to do at that point is to get the Gaussian CDF (well known) and sample from it, making sure to plug in our mas and temperature.
$$CDF(x)=\frac{1}{2}\times \left[ ...
3
Here's the general derivation of the commonly used, and often (but not always) valid, uncertainty propagation formula for independent small Gaussian errors.
$\newcommand{\bbv}[1]{\mathbf{#1}}$
Consider a quantity $y$, calculated from measured quantities $\bbv{x}$
$$
y + \Delta{y} = f(\bbv{x}+\Delta\bbv{x}) = ...
3
Yes, it is simple to prove using moment generating functions. And yes, the mathematics is very closely related to that of quantum field theory.
You compute $G(j) = <exp(\sum j_i x_i)>$ where each $j_i$ is a "source" for the corresponding $x_i$. This is easily shown to be something like $G(j) = exp(\sum j_i \mu_{ij}^{-1} j_j)$ To get expectation ...
3
One of the characteristics of physics research is the regular use of advanced methods and techniques from other fields, such as mathematics, computer science, probability theory and even biology. Physicists therefore often need to dive deep into complex topics in other fields.
None of these topics is trivial and the level of understanding of physicists in ...
3
In a number of areas in experimental physics you simply do not need a detailed statistical view on your data.
At the research institutes I have visited there were a great number of experimental and theoretical physicists but basically nobody that does anything that would need a deep understanding of statistical methods. While in some disciplines like ...
3
No, I would say it is wrong to immediately conclude that there is no scale just because the variance diverges. Only functions of $x$ of the form $x^n$, a power law, have a chance to be considered scale-free; none of these functions may be considered a probability distribution because the integral diverges. Any other function – and therefore any normalizable ...
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