I am evaluating the accuracy of GPS watches, taking many readings over a known distance. I've been calculating standard deviation using the mean reading, but because I know what the reading should be, I could use that instead of the mean. Would this be a reasonable thing to do?
$\begingroup$
$\endgroup$
4
-
3$\begingroup$ Of what are you computing the standard deviation? The coordinates, or the distance to the known location, or something else? In most cases standard deviations are not relevant for assessing accuracy: they tell you about precision. $\endgroup$– whuber ♦Commented May 8, 2014 at 21:13
-
3$\begingroup$ It depends on what you want to measure. if you want to measure how consistent the GPS readings are, you might use a standard deviation. If you want to measure how close they are to the know answer, you might use the RMSE (from the known value). If you do that, note that your denominator should be n, not n-1. $\endgroup$– Glen_bCommented May 8, 2014 at 21:25
-
2$\begingroup$ A related question is discussed at stats.stackexchange.com/questions/65640. It explains the use of the RMS mentioned by @glen_b. $\endgroup$– whuber ♦Commented May 8, 2014 at 21:37
-
1$\begingroup$ Possible duplicate of How to calculate 2D standard deviation, with 0 mean, bounded by limits $\endgroup$– TimCommented Mar 21, 2016 at 12:20
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
Re:"Would this be a reasonable thing to do?"
As Wuber pointed out, it depends on what you are trying to measure. GPS signals for civilians are purposely degraded to prevent GPS signals being used for missile guidance against US targets. So the "location" of a particular position will vary with time on purpose and due to overall error in GPS.
When measuring the difference between two positions (at two different times) there will be about a SQRT(2) increase over the error of measuring one position.
-
4$\begingroup$ Department of Defense in the United States stopped degradation of GPS signals in 2000. $\endgroup$ Commented Nov 4, 2015 at 10:07
-
$\begingroup$ As MaxW pointed "true value" and "true mean of population" might be different. Anyway, if they are the same (that is, the GPS device has no systematic error or bias), it will be very reasonable to estimate standard deviation using true value, but then you should divide by n instead of n-1. $\endgroup$– PereCommented Jul 31, 2016 at 18:49
$\begingroup$
$\endgroup$
Maybe I'm oversimplifying this, but it sounds like just a mean squared error. If $\theta$ is the truth, then
$$ \text{E}(X_i - \theta)^2 = \text{E}(X_i - \bar{X} + \bar{X} - \theta)^2 \\ = \text{E}(X_i - \bar{X})^2 + \text{E}(\bar{X} - \theta)^2 $$
So on the variance scale, what you're computing is the ordinary variance plus an estimate of the bias squared.
answered Jan 23, 2017 at 14:34