4
$\begingroup$

I have a set of $n$ measurements, $x_i$, representing distance to a point, and want to find the mean and the standard deviation of the distance, the problem is, I don't know how to take the precision of the measuring equipment into the equation.

The precision of the measuring equipment is given by the producer, say $\sigma_p$, so each measurement has a uncertainty: $x_i \pm \sigma_p$

When calculating the mean, I do this:

$\mu = \frac{1}{n} \sum_{i=1}^n (x_i \pm \sigma_p) = \frac{x_i}{n} \pm \sigma_p$

but I'm not sure if $\sigma_p$ should be in the equation at all or how to interpret it. Furthermore, I need to calculate the standard deviation, which usually is

$\sigma_x = \sqrt{\frac{1}{n-1} \sum_{i = 1}^n (x_i - \bar{x})^2}$,

but if I substitute my calculations I obtain

$\sigma_x = \sqrt{\frac{1}{n-1} \sum_{i = 1}^n (x_i \pm\sigma_p - (\bar{x}\pm \sigma_p))^2}$,

which I do not not how to manipulate further. Do the $+\pm\sigma_p$ and $-\pm\sigma_p$ cancel out or add up?

Should I rather just add the two together? That is

$\sigma_x^2 = \sigma_p^2 + \frac{1}{n-1}\sum_{i = 1}^n (x_i - \bar{x})^2$,

where the mean is calculated as normally,

$\bar{x} = \frac{1}{n} \sum x_i$

Any help is much appreciated!

Improve this question
$\endgroup$
5
  • $\begingroup$ The way I understand your setting, equipment error would increase the standard deviation but you don't need to add it, it's already there. It is in particular useful when you want to interpret a single measurement. $\endgroup$
    – Gala
    Commented Jun 19, 2013 at 13:57
  • $\begingroup$ Thank you for the reply @GaëlLaurans! Is there any litterature to refer to here? My professor (in statistics, not land surveying, which is where the problem arose) just said I should add the squared standard deviations,as I suggested at the end of the question. $\endgroup$
    – Gunnhild
    Commented Jun 19, 2013 at 14:20
  • 2
    $\begingroup$ The replicate measurements give you the Type A uncertainty whereas the specs from the manufacturer give the Type B uncertainty. The two are combined to give the combined standard uncertainty for the measurement. The Wikipedia entry on "Measurement Uncertainty" gives an overview of this process and gives references to the relevant ISO Standards. $\endgroup$
    – Tom
    Commented Jun 19, 2013 at 14:37
  • 1
    $\begingroup$ What exactly is the 'precision' in this context? Is it actually a standard deviation? A 'typical error'? Some bound on the bias? An error bound? Is it measured or is it arrived at in some other way? In any case, won't the observational noise from lack of instrument precision already be in the observations? Won't you be adding it in again? $\endgroup$
    – Glen_b
    Commented Jun 20, 2013 at 0:01
  • $\begingroup$ @Glen_b, I think what it is about at the end of the day is whether the equipment uncertainty contributes to the observed variance in the observation (Type A uncertainty), and I'm not sure how to evaluate that. $\endgroup$
    – Gunnhild
    Commented Jun 20, 2013 at 15:13

1 Answer 1

Reset to default
2
$\begingroup$

The primary assumption (which is often very reasonable) is that the measurement errors are independent, identically distributed, and conditionally independent on the true value of the variable of interest. With electronic devices, this is often the case.

So denote the 'true' distance variable $X_d$, the measurement error random variable $X_p$, and you observe a joint variable, $X_x=X_d+X_p$. You are asking for an estimate of $X_d$ which is corrupted with measurement error. So $\mu_x=\mu_d+\mu_p$ whose expected value is $E[X_x]=E[X_d + X_p] = E[X_d]+E[X_p]$ under independence. If your measurement error is zero-centered (that is, the measurement is unbiased) than $E[X_p]=0$, therefore $E[X_x]=E[X_d]$ and your best estimate is $\hat{X_d}=\hat{X_x}= \frac{1}{n}\sum{x_x^i}$. So basically, if you can assume zero centered, independent measurement errors, the best estimate of $\mu_d$ is the mean of your measurements.

Also, the variance of two independent random variables is the sum of the two individual variances. You can think of the variance of your data as representing $\hat{\sigma_x^2}=\frac{1}{n-1}\sum{(x_x^i-\bar{x})^2}$. Now we know $\sigma_x^2 = \sigma_p^2+\sigma_d^2 \to \sigma_d^2=\sigma_x^2-\sigma_p^2$. We have the estimator $\hat{\sigma_x^2}$ and we know $\sigma_p^2$, therefore our estimator $\hat{\sigma_d^2}=\hat{\sigma_x^2}-\sigma_p^2$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.