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You pretty much have the right ideas here. The MC is necessary because the functional dependence of the detector response and the underlying physics is too complex to calculate in any other way. And therein lies the actual analysis problem: it's not enough to simulate the physics based on the standard model (or some other hypothetical interaction). In order ...

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The weighted mean of three values is given by $$\bar{x} = \frac{ \sum_{i=1}^{3} x_i \alpha_i}{\sum_{i=1}^{3} \alpha_i},$$ where here $\alpha_i$ represents the weight that you give to each measurement. If you wished to just find the weighted mean from your data as you presented it originally, then the weight $\alpha_i = 1/\sigma_i^{2}$. If you then wish to ...

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The way that I would approach this is to calculate $\chi^2$ with $$\chi^2 = \sum_{i=1}^N\left({p - x_i \over \sigma_i}\right)^2$$ where $p$ is the best value for the fit and $N$ is the number of measurements $x_i$ with error $\sigma_i$. Now with a computer or Excel sheet I would plot $\chi^2$ as a function of $p$. As you know the value of $p$ for the ...

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You may be surprised to learn that you have done the right thing and have got the right answer. It's just that you've used non-standard units. Because you've put the distance in as millimetres and the time as milliseconds the value you calculate for $g/2$ is in units of millimetres per millisecond squared. You need to multiply it by a thousand to convert it ...

1

Let's start out by understanding what's going on when we take logaithms. We have the relationship $$y=kx^n$$ There's no obvious method for working out either $k$ or $n$ from the graph of $y$ against $x$; instead, we plot $\log(y)$ against $\log(x)$. Why do we do that? Well, if we take logarithms of both sides of the above equation, we get: $$... 2 First method: The principle You can always transform one distribution to another through a transformation of the independent variable, so we get if p:\mathbb{R}^N\to(\mathbb{R}^+\bigcup\{0\})^N is a probability density function of N variables and we transform the independent variables \mathbb{R}^N\to\mathbb{R}^N by some differentiable transformation ... 7 The sign of a good fit is that the residuals have the same distribution as your model for the errors. Usually the assumption that goes into fitting methods is that the errors are normally distributed. That is, given perfect inputs x_i, and an ideal relation y_i = f(x_i), you will measure y_i + \epsilon_i, where \epsilon_i are distributed normally. ... 3 Using the linearization procedure has become so common in experimental chemical kinetics that practitioners have taken to using it to define the activation energy for a reaction. That is the activation energy is defined to be (-R) times the slope of a plot of ln(k) vs. (1/Tâ€‰) Thus, you should go with the linearized version because it is the common ... 2 Within the specification I can glean from the question - here is what I would do. (i) Find the best fit Gaussian - which I am assuming is what you have done. (ii) Your best fit should return a chi-squared value You should compare the chi-squared value with critical values of the chi-squared distribution for the appropriate number of degrees of freedom of ... 12 Here is how I interpret what happened: You used Excel to compute the coefficients of the Gaussian that best describe the data: mean \mu, standard deviation \sigma, and magnitude A for a curve$$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 ...

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A gaussian fit is symmetrical by definition, because it is a gaussian. Your orange fit doesnt look like a gaussian, it is not even smooth. I do no think excel had a gaussian fit function (but I dont use excell so cannot tell for sure. You can use other software such as matlab, or likely free ones on the web. Or, just use that data to calculate the parameters ...

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