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In a physics text book I read the following:

$$e/m=1.758820150(44) ×10^{11} \mathrm{C/kg} $$ In this expression, $(44)$ indicates the likely uncertainty in the last two digits, $50$.

How should I understand this uncertainty? Does it mean $\pm 44$ on the last two digits?

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2 Answers 2

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The digits in parentheses are the uncertainty, to the precision of the same number of least significant digits. (The meaning of the uncertainty is context-dependent but generally represents a standard deviation, or a 95% confidence interval.) So: $$e/m=1.758\,820\,1\color{blue}{50}\,\color{magenta}{(44)}×10^{11} \ \mathrm{C/kg}=\left(1.758\,820\,1\color{blue}{50}×10^{11} \pm 0.000\,000\,0\color{magenta}{44}×10^{11}\right) \ \mathrm{C/kg}$$

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The uncertainty is equal to the standard deviation $\sigma = 0.000000044×10^{11} C/kg = 4400 C/kg$. It means you are

  • approx. 68% confident that the true value $m_e$ lies within the interval $[\hat m_e-\sigma, \hat m_e+\sigma]$, where $\hat m_e$ is the measure value.
  • and approx. $95\%$ confident that the true value $m_e$ lies within the interval $[\hat m_e-2\sigma, \hat m_e+2\sigma]$.
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  • $\begingroup$ Excuse me, but do you have any sources that use this kind of method and/or this notation? Sorry, it's just that I've only taken one statistics course and that was quite some time ago; although I know confidence intervals relatively well, I've never seen the math you've used before. $\endgroup$
    – Mailbox
    Feb 11, 2023 at 3:31

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