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Hot answers tagged error-analysis

54

The second and the speed of light are precisely defined, and the metre is then specified as a function of $c$ and the second. So when you experimentally measure the speed of light you are effectively measuring the length of the metre i.e. the experimental error is the error in the measurement of the metre not the error in the speed of light or the second. ...

20

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 $$\frac{\sum_iw_ix_i}{\sum_i ... 20 Suppose you are analysing the weights of people in the UK to see what the distribution of weights looks like. Suppose also you can measure the weight to arbitrary precision, so that no two people's weights will be exactly the same. When you're finished you plot your data on a histogram, but the trouble is that because everyone has a different weight you get ... 17 To repeat Wikipedia: The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its value is exactly 299,792,458 metres per second, a figure that is exact because the length of the metre is defined from this constant and the international standard for time. In other words, it's exact ... 17 Simple error analysis assumes that the error of a function \Delta f(x) by a given error \Delta x of the input argument is approximately$$ \Delta f(x) \approx \frac{\text{d}f(x)}{\text{d}x}\cdot\Delta x $$The mathematical reasoning behind this is the Taylor series and the character of \frac{\text{d}f(x)}{\text{d}x} describing how the function f(x) ... 15 While appropriate in many important contexts, LeFitz's answer can fail in one important situation, and can lead you astray, for example, when plotting graphs in logarithmic scale. More specifically, LeFit'zs answer is only valid for situations where the error \Delta x of the argument x you're feeding to the logarithm is much smaller than x itself:$$ ...

15

There is no "one size fits all" answer to your question. First - the size of the smallest division on a meter ruler need not be one mm. I have a ruler that only goes down to half cm divisions, and I have one that gives half mm divisions. Second - a ruler may not be accurate to the nearest division. Wooden rulers in particular will grow and shrink with ...

13

This depends on the context. If you have 1500 of something and you counted them yourself and you are sure you have precisely 1500, then all four figures are significant. On the contrary, if you're guessing that you have 1500, implying a certainty of order 100, then only the leading two figures are significant. In scientific notation, one would write the ...

13

This apparent inconsistency comes about because the rule that you add the relative errors when multiplying or dividing the quantities only applies when those relative errors are uncorrelated. But the relative errors in $V$, $I$, and $R$ are correlated with each other. Correlation means, in this case, that if you calculate $V$ (with its uncertainty) from $I$ ...

10

The idea is just that if the uncertainties are small enough you can approximate the function by its Taylor series $$f(x_i + \delta_i) \approx f(x_i) + \sum_j \frac{\partial f(x_i)}{\partial x_j} \delta_j + \sum_{j,k} \frac{1}{2} \frac{\partial^2 f(x_i)}{\partial x_j \partial x_k} \delta_j \delta_k + \cdots.$$ If you neglect the second order terms the ...

10

It's not, at least not in the statistical sense. What you are doing is finding the (linearly approximated) change in $y$ obtained by changing the inputs by their standard deviations. This is okay as a rough approximation, but you can do better with almost no extra work. If you want the actual variance and standard deviation of $y$, the formula is different. ...

9

This problem is generally called propagation of error / uncertainty. You can google it and find a lot of info (I'd also recommend Taylor's "Introduction to Error Analysis"). Here's the gist of it, though. If you have independent measured quantities $x, y, z, \ldots$ with errors $\sigma_x, \sigma_y, \sigma_z, \ldots$, then the error on a function ...

9

I teach high school in the United States. I want to preface with that, because conventions common in one context are not necessarily universal. That being said, it's pretty standard teaching practice here (at least in every class I've ever taken, taught, or known a colleague to teach) to assume that trailing zeroes are not significant unless otherwise ...

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

When quoting results, there are a few good rules to follow: Avoid rounding errors in intermediate calculations. Write your error to 1 significant figure if your data set is smaller than $10^2$, 2 if it's smaller than $10^4$ etc. Write your estimate and its error with the same number of decimal places. Rules 1. and 3. are simple to understand. Rule 2. ...

8

You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated ...

8

Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right. But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the "errors" mean standard deviations as ...

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

CODATA is a group that compares and combines all the most accurate experimental measurements of fundamental constants to give recommendations for best-guess values that should be used. They periodically update their values as new experiments are done. You are seeing that some wikipedia pages use old (not-updated) CODATA recommendations, while others have ...

7

You should always find an answer that is a formula, and then only apply significant figures once you get to the one final step of substituting your numbers back into the problem in place of variables. Avoid multiple intermediate steps of substituting numbers at all costs. Not only will this save your pencil a lot of work, but it will also cause your ...

7

The typical gravitational acceleration on the surface of the Earth, $g \approx 9.8\: \mathrm{m/s^2}$, has uncertainty. That's one of the reasons why the $\approx$ symbol is used. The Earth's gravitational field varies a lot due to oceans, the thickness of the crust, mountains, non-uniform density in the crust and mantel, etc. A pair of satellites was ...

7

It's down the fact that different properties are used to calculate the ages and if you look at the margin of error in the calculations they're not incompatible. The lower limit for the age of HD140283 is: 14.46 - 0.8 = 13.68 billion years which is within the range for the age of the universe. Once better measurements of HD140283 are made it's age ...

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

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

Chaotic trajectories are perfectly deterministic, it's just that they demonstrate an extreme sensitivity to initial conditions. This is to say that if you start with the exact, precisely same initial conditions, you will get the exact, precisely same trajectories. But if you are even a tiny bit off from the initial conditions, the resulting trajectories will ...

7

Model two consecutive measurements as the real values plus some noise. Call the first measured temperature $T_1$ and the second $T_2$. Call the measured noises $\gamma_1$ and $\gamma_2$, and suppose that they are drawn from a distribution $\Gamma(\gamma)$ and are uncorrelated. The (approximation to the) derivative is \text{Derivative} \approx \frac{(T_2 ...

6

I think that the easiest way to understand this is in the formula for addition. If you consider your quantity $C$ which depends on $A$ and $B$ such that $C=A+B$, then the formula $\Delta C = \Delta A + \Delta B$ overestimates the error values. You can visualize this as a rectangle with $A$ on the x-axis and $B$ on the y-axis with the area enclosed being ...

6

As you can read in this Wikipedia article, it was decided recently to base all SI units on seven constants of nature. To be able to do so, these constants have to be set to absolute values. Therefore it was decided, that these constants are fixed without error margin at their commonly accepted values to derive all other SI units from those now fundamental ...

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

No, of course not. Yes, some people will overestimate and others will underestimate. Averaging would cancel out the bias to some extent, but there's no reason to expect it to cancel out the bias perfectly. We all have similar eyes and brains. We are all deceived by the same optical illusions, in the same way. We all have a shared cultural understanding of ...

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