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## New answers tagged error-analysis

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1) This varies by textbook. A common format you'll see is h=6.62606957(29)×10−34 (from Wikipedia: Planck constant). The digits in parentheses indicate they are uncertain. Hence, you'd expect that h is known to at least 0.00000001/6.6260957 (pretty well known.) Other references will explicitly state what the error bars are, or may simply cite the sources. ...

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Working with significant digits is very prone to error, because it can be misleading. It is much better to work with explicit errors. So to rewrite your example with explicit errors: $\left(1.0 \pm 0.5\right) \times 10^3 + 1.00\pm0.05$ We now add the errors quadratically (assuming they are uncorrelated): $\sqrt{(0.5\times ... 0$1000$has an ambiguous number of significant digits. It could have 1, it could have 4. I think it is generally assumed that it has four unless otherwise stated. (For example, I've seen people put a line over the last significant digit, like this:$1\overline000$.) This is why scientific notation is useful. If you are saying that$1000$has$1$... 0 1000 has 4 significant digits as mentioned before, stating that the measured value is between 999.5 and 1000.5. 1.0 has 2 significant digits, staying it is measured within an accuracy of 0.05. Adding the numbers gives you a result within an accuracy of 0.5, so noting the result with 1 decimal digit is nonsense. If your measured value of 1000 indeed has only ... 4 The last digits in 1000 are absolutely significant, they state that you have not 1200, not even 1001 but exactly 1000. In scientific notation, you would write this as$1.000 \times 10^3$. Compare this to$1\times10^3$where you have just one significant digit. Update: consider the example from the question$1\times10^3+1.0$. The first term could be anything ... 0$k$is just the mean of$k_1$and$k_2$No, the best value of k is calculated using a weighted mean, weighting by the reciprocals of the squares of the respective individual uncertainty values. An accurate measurement must contribute more to the best value than an inaccurate measurement. I thought that I would need to square-sum the errors ... 2 NIST has its own more sophisticated guidelines for reporting uncertainty of measurments. http://physics.nist.gov/Pubs/guidelines/TN1297/tn1297s.pdf There is nothing wrong with reporting two digits in an uncertainity, and many peer reviewed journal articles do. This is especially true when the first digit of the uncertainity is 1. If you report +/-1, 1 ... 0 When adding and subtracting, you can only go to the lowest number of decimal places. That is to say, we are dealing with precision and not significant figures when adding/subtracting numbers. If you have two measuring devices and one is accurate to 0.1mm and the other to 1mm, then you cannot definitively state the combined measure to 0.1mm, you can only ... 0 First, can it be written as the following? $$N_A=6.022\,141\,29\times10^{23}\pm0.000\,000\,27\times10^{23} {\rm mol}^{-1}$$ Yes. A cleaner way to write it is $$N_A=6.022\,141\,29(27)\times10^{23}{\rm mol}^{-1}$$ where the$(27)$indicated the uncertainty in the last$N$digits (where$N$is the number of digits inside the parenthesis). If I have a ... 1 ok, here goes... Direct from "Data Reduction and Error Analysis for the Physical Sciences," Bevington, McGraw-Hill, Chapter Four, when$ z = f(x,y) $then since$ \sigma_z^2 = \Sigma(x_j-<x>)^2 $and$\Delta z = \Delta x *\frac{\delta z}{\delta x} + \Delta y * \frac{\delta z}{\delta y}$, some equation hacking leads to$ \sigma_z^2 = \sigma_x^2 * ...

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