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Say we use a ruler to make a measurement of the width of a block of wood. We get some value like 4.2 +/- 0.1 cm, where the error is our estimated error of our ruler's precision. Now we have four new individuals measure the same block of wood, and two of them obtain 4.5 +/- 0.1cm and two obtain 4.2 +/- 0.1 cm. If we take the mean of these five measurement we obtain 4.32 +/- 0.16 cm, where I'm ignoring significant figures in this example and the uncertainty is the standard deviation of the samples. Now, in stating the total uncertainty for the width of the block of wood, should I combine the 0.16 in quadrature with the 0.1, or should I propagate the error of each measurement through the standard deviation formula and obtain an uncertainty on my uncertainty, like 0.16 +/- alpha, where alpha is the propagated uncertainty value? Or perhaps I should ignore the standard deviation of the measurements and use the classic 0.1/sqrt(5)---but this is silly given the spread of our measurements. This seems like a simple question but I can't seem to find two sources that agree on the matter (har har).

Thanks for any help.

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  • $\begingroup$ How do you know the +/- 0.1 cm value? Is it yet another estimate from the individuals estimating the width of the block of wood? $\endgroup$ – John Oct 18 '11 at 18:42
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If you assume all the noises are Gaussian (and especially that +/- 0.1 really isn't... but anyway), then I think the 0.16 is already an estimate of the combination of the two noises. So I would report 4.32 +- 0.16/$\sqrt{n}$, where n is the number of measurements. A derivation:

So we're trying to measure the width $\mu$ of a block of wood. Suppose every measurement has some Gaussian error (perhaps because different people are measuring differently). We can write these as $X_{1},\ldots,X_{n}\sim\mathcal{N}(\mu,\sigma^{2})$. But the ruler itself has some imprecision, which for simplificity we'll model as Gaussian as well. So our final observations are $Y_{1},\ldots,Y_{n}$, distributed as $Y_{i}\sim\mathcal{N}(X_{i},\rho^{2})$. Suppose $\rho^{2}$ is known (e.g. the precision of the ruler). Then our goal is estimate $\mu$, and $\sigma^{2}$ is a nuisance parameter. This is easy because the $X$'s and $Y$'s are jointly normal. Write $(X,Y)$ for a generic pair of these variables. Their joint distribution is $$ \begin{pmatrix}X\\ Y \end{pmatrix}\sim\mathcal{N}\left(\begin{pmatrix}\mu\\ \mu \end{pmatrix},\begin{pmatrix}\sigma^{2} & \sigma^{2}\\ \sigma^{2} & \rho^{2}+\sigma^{2} \end{pmatrix}\right) $$ So $Y\sim N(\mu,\rho^{2}+\sigma^{2})$.

Let $\bar{Y}=(Y_{1}+\cdots+Y_{n})/n$ be the our estimator for $\mu$. Then $\bar{Y}\sim\mathcal{N}(\mu,[\rho^{2}+\sigma^{2}]/n)$. Since we don't know $\sigma^{2}$, I'd recommend plugging in your favorite sample estimate for the variance of $Y$, call it $\hat{\sigma}_{Y}^{2}$, and plugging that in for $\rho^{2}+\sigma^{2}$. So you end up with $\bar{Y}\pm\hat{\sigma}_{Y}/\sqrt{n}$.

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