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I have a list of values each with a corresponding error. I can determine the average/mean of the list, but I'm not sure how to best calculate the error over the entire population.

e.g.

107 +/- 0.2
120 +/- 0.1
184 +/- 0.4
44  +/- 0.3

Average is simply 113.75, but what is the error?

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  • $\begingroup$ I've clarified by question a little by removing "average error" as I don't think that's what I'm after exactly. What I need to work out is the error of the whole dataset when averaged. $\endgroup$
    – Carl
    May 20, 2012 at 5:03

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If errors can be either negative or positive it does not make sense to average them because positive errors cancel negative error. That is why statisticians use variance (average squared error) or the mean absolute error to summarize the magnitude of the errors.

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  • $\begingroup$ I understand how to get a standard mean error (stdev / sqrt(number-of-values), but I'm not sure I understand the average squared error you mentioned. How would I calculate that given my example? $\endgroup$
    – Carl
    May 19, 2012 at 15:25
  • $\begingroup$ A standard error is the standard deviation of a mean. It has no relation to mean absolute error or mean square error. The model error at a fitted point is the difference between the observed value and its corresponding fitted points. Square all these errors, then average them and you have the mean square error. Least squares regression minimizes the mean square error. That makes it sensitive to outliers because large errors are penalized a lot because the square of the error is the term in the average. $\endgroup$ May 19, 2012 at 16:18
  • $\begingroup$ So the line is forced to move closer to the outlier at the expense of the other observations. Absolute eror is the absolute value of the error so with average absolute error we take the absolute value of the error and average it. Using that criteria the penalty for a large error is big by not nearly as big as for least squares. So it is less influenced by outliers and doesn't sacrifice the accuracy at the other points as much. So we say it is robust against outliers. $\endgroup$ May 19, 2012 at 16:22

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