# Tag Info

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To begin with ask yourself two questions: Has the device been re-calibrated in the middle of data taking? Is the calibration known to drift over time similar to the length of the data taking? If both of the answers are "no" then you can reasonably assume that the calibration effect is the same on each and every data point. So the mean is off by that ...

1

I think what you have described is not the correct method to estimate a 68% confidence interval in one parameter of interest. The error is in freezing the other parameter when minimising chi-squared. A better procedure to evaluate the uncertainty in parameter 1, is to evaluate the minimum chi-squared for a set of values of parameter 1, whilst allowing ...

1

In the case of a 2-parameter fit, the 68% confidence space is typically an ellipse, not necessarily aligned with the axes. If you want to figure out the size of the ellipse, you should find the orientation of the ellipse, not just the intercepts of the ellipse with a horizontal and vertical line through its center. Example: fit $y=a+b x$ for a large set of ...

24

Use the second derivative (or third, or whatever). The reason we use that formula is that $$df \approx \frac{df}{dx} dx$$ is the first order Taylor approximation to df. If the first order term vanishes, you should include higher terms: $$df \approx \frac{df}{dx} dx+\frac{1}{2}\frac{d^2f}{dx^2} dx^2+...$$ In your case, with $f=x^2$, and $x=0$, we'd ...

9

This is a situation where naive error propagation breaks down. Those methods (i.e. giving uncertainty for $f(\mathbf{x})$ for some values $\mathbf{x} \pm \Delta \mathbf{x}$) are based on linear approximation, which fails for $f(x) = x^2$ near $x = 0$. If you're not too worried about statistics issues, you can use the 'min-max' technique: your error bars on ...

1

Assume the values are normally distributed without correlation (i.e. $\Delta(x+y) = \sqrt{\Delta x^2 + \Delta y^2}$), then \begin{align} \Delta h &= \Delta ( x \sin \theta ) \\ &= \sqrt{ (\Delta x \cdot \sin \theta)^2 + (x \cdot \Delta (\sin \theta))^2 } \tag{product rule} \\ &= \sqrt{ (\Delta x \cdot \sin \theta)^2 + (x\cos \theta \cdot \Delta ...

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The measurements are repeated $N$ times for the same physical conditions; there are thus $N$ points for this single datum, which will be reported as the mean of the sample , along with the measured standard deviation. The interval is conventional, based on common usage. This makes it easier to interpret the meaning of the reported measurements. ...

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For some detailed analysis of the limits for electron microscopy, see: Viewpoint: What Are the Resolution Limits in Electron Microscopes? This is a brief review of the technology, and they summarize the recent improvements in resolution with: "The authors estimate that the resulting resolution limit is in the range $0.50–0.8Å$, which is consistent with the ...

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If your curve-fitting program allows it, instead of fitting $y=a+bt+ct^2+dt^3$ you could try fitting $y=d(t-a)(t-b)(t-c)$. The output parameters $a$, $b$, $c$ will then be the required roots and the errors (or rather variances and/or co-variances) will be included in the statistics.

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Basically what you are doing for $y=\frac{AB}{C}$ is adding up all the percentage errors, which is wrong. Take log function on both side, so you get $\log(y)=\log(A)+\log(B)-\log(C)$ so for percentage error becomes: $\frac{dy}{y} = \frac{dA}{A} + \frac{dB}{B} - \frac{dC}{C}$ so if you actually follow your step, you get: $%error= 3 + 2 - 7/3 = 2.6%$ so 2.6% ...

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This seems to be addition in quadrature of multiple independent uncertainties in a measurement. In particular, if you have a measurement which depends on two quantities $a$ and $b$ whose uncertainties $\delta a$ and $\delta b$ are completely independent and uncorrelated, then their uncertainties will often be combined as \delta(a+b)=\sqrt{\delta ...

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