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21

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 ...


15

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 ...


12

Here is how I interpret what happened: You used Excel to compute the coefficients of the Gaussian that best describe the data: mean $\mu$, standard deviation $\sigma$, and magnitude $A$ for a curve $$Y=Ae^{-(x-\mu)^2/2\sigma^2}$$ Then you evaluated that function at a number of X values. Since the X values are not symmetrical about the calculated mean, you ...


11

Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, ...


10

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 ...


10

Geant is a framework---which means that you use it to build applications that 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 ...


9

This is something that particle physicists are perfectly well aware of. For any given observed effect, there is always a nonzero probability that the observation will be a false positive that was caused by a random fluctuation. The name of the game is taking enough data that this probability is small enough. In general, the more data you take, the less ...


8

Half life is, by definition, the amount of time until half of an infinitely large sample would decay. That's precisely equivalent (according to the frequentist interpretation of probability, if that matters to you) to the time until an individual particle's probability of decay reaches one half. The half life is a theoretical quantity that doesn't depend on ...


8

You might like this 110-page paper by me and Alex Arkhipov, which is all about a quantum bosonic analogue of Galton's board (we even use the same graphic you did -- see Section 1.1!). In particular, we gave strong evidence that such a board (with an arbitrary configuration of "pegs," and with multiple entry points for the "balls") is exponentially hard even ...


8

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 ...


8

If there are enough data and the prior is not completely unreasonable, the frequentist and the Bayesian approach give essentially the same answer. This is related to the central limit theorem. If data are fairly scarce, the two approaches may differ a lot. In this case the Bayesian approach is far preferable but only if the prior reflects true prior ...


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 ...


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

No because none of them know the actual answer. The averaging process you describe only works if each estimate is of the exact answer plus noise. Otherwise it is known as the "Emperor's nose" problem. Nobody can see the Chinese emperor's face so they ask a million peasants how long his nose is, they average the results, and since they have such a large 'N' ...


7

Standard deviation adds uncertainties to the measured value: $23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of ${\rm m}$ in my case) and think, The value could be as low as $22.9\,{\rm m}$ or as high as $23.7\,{\rm m}$ without much thinking. Modifying this to being a percentage of the value would be confusing. Plus it would be ...


7

The sign of a good fit is that the residuals have the same distribution as your model for the errors. Usually the assumption that goes into fitting methods is that the errors are normally distributed. That is, given perfect inputs $x_i$, and an ideal relation $y_i = f(x_i)$, you will measure $y_i + \epsilon_i$, where $\epsilon_i$ are distributed normally. ...


6

The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" one used. Without a rigid mathematical definition of this quantity one often just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; ...


6

Yes, it is a statistical average in the sense that the measured half life will approach a single value of a true half life if you do lots of measurements. In other words, if you did the experiment many, many times you would find that on average you had 4 particles left after a half-life had passed. For any individual experiment, the results would vary. ...


6

I believe you are thinking of the Central Limit Theorem. The mean and variance of the averages of many measurements are better estimates of the precision of your measuring rule, but don't tell you anything about the accuracy of your measuring rule. Your measuring rule may be biased. The Central Limit Theorem is a part of mathematics. IMO you should also ...


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 ...


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

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 ...


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

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 ...


6

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 ...


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 ...


5

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 $$ ...


5

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} = ...


5

Statistics are used in physics to provide a conceptual link between the 'macroscopic view' and the 'microscopic view'. For example, when studying gases, we can examine the statistical distribution of particle velocities and energies to gain an understanding of the relationship between the macroscopically observable quantities (pressure, volume & ...



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