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Here is my problem. I have a large population of uncoated parts. The weight of the uncoated parts is normally distributed. I then apply a coating to the parts- the weight of the coating also being normally distributed. I take a sample of the parts before and after coating. I need to know the mean coating weight and the standard deviation of the coating and their confidence intervals.

I know how to calculate the mean coating weight, the standard deviation of the coating weight (square root of the difference in variances before and after coating), and confidence intervals for the mean. What I don't know is how to calculate the confidence interval for the standard deviation.

I know how to calculate the confidence interval for a sample standard deviation but not for a standard deviation calculated by taking the difference of two samples, each of which have their own inherent uncertainty.

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  • $\begingroup$ Could you clarify whether you are asking for a confidence interval for the standard deviation (which literally is what you write) or about how to use the standard deviations to compute a confidence interval for the difference of means (which is what the context would suggest you are interested in)? $\endgroup$
    – whuber ♦
    Commented Sep 13, 2015 at 17:20

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From the given data it's not obvious which distribution you're using. If you consider Gauss distribution, the strict way to account for risks is Propagation of Uncertainty $\sigma_y^2 = \mathbf{A\Sigma}_{xx}\mathbf{A}^T$.

If you have a variable $y$ which consist of other 'uncorrelated' variables $x$ where $\sigma_x^2$ is the variance of $x$, the error propagation for for the sum as for the difference is

$y = x_1 + x_2 + ... x_n$

$\sigma_{y}^2 = \sigma_{x_1}^2+\sigma_{x_2}^2+...+\sigma_{x_n}^2$.

Once the new standard deviation is computed, you can compute the confidence interval as usual.

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  • $\begingroup$ This is impractical since you hardly know $\Sigma_{x,x'}$ $\endgroup$
    – Henry.L
    Commented Sep 13, 2015 at 18:32
  • $\begingroup$ It does not matter if it's unpractical or not, this is the standard way to solve this type of problems. I can post you a book reference if you want. Another point is that for sure you know $\Sigma_{xx}$. The whole equation which I posted later is derived from this rule. If it's uncorrelated, you just have $\sigma_{x_i}^2$ on the diagonal. Just check the link from my OP. Section: Simplification. $\endgroup$
    – nali
    Commented Sep 13, 2015 at 22:41
  • $\begingroup$ @Henry.L, feel free to ask if you still miss some point. Especially the transition to statistical matrices might be not intuitive enough if it's not your every days work. $\endgroup$
    – nali
    Commented Sep 13, 2015 at 23:52

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