4 ms·

Really? > Do you take every observation: square it, average the total, then take the square root? Or do you remove the sign and calculate the average? Neither

Really? > Do you take every observation: square it, average the total, then take the square root? Or do you remove the sign and calculate the average? Neither of these is even an attempt to measure average daily temperature variation. (Assuming any reasonable definition...) If you're talking about variation from day to day, then looking at differences in max and min, or at given points in time, or on the same day in different years, would be some approaches. Averaging absolute values makes no sense whatsoever -- if the observations were 20 and -20 the result would be the same as if they were 20 and 20. And calculating the standard deviation of the observations is calculating something else again (it might be standard deviation, or it might just be something dumb). Neither of these are problems with standard deviation. It's sad if newspapers or their readers don't know what standard deviation means, but they're pretty much innumerate across the board so it's not clear whether further muddying terminology is going to help anyone.

enugu 10y ago
The averaging of absolute values in the article's context is happening around a mean of 0(averaging the variations). In general, one would average the absolute values of the difference from the mean. Yes, it would be silly to convert a -20 to 20 if the mean isn't 0. The advantage with STD is that it is easier to compute relative to MAD in a situation like a random walk. For instance when one has a +1, -1 equal probability one dimensional random walk, so the mean is 0, X = sum of X_i, E(X²) = n is straightforward, whereas computing E(|X|) is not so easy.
- Tloewald 10y ago
  That's as clear as mud. What does a -20 vs. a 20 mean? That it was colder at noon than at midnight? Assuming that there was some sensible definition (e.g. deviation from the past mean temperature, which is not at all what was stated) then "average deviation" has two obvious interpretations. Re-labeling "standard deviation" or "mean deviation" would not be helpful in this case, and the -20 and 20 values STILL don't help the case.
  - enugu 10y ago
    The article isn't clear but the deviation is either from the sample mean or the mean of the underlying random variable. You are trying to find the expected difference from the average and hence the absolute value(otherwise, the positive and negative deviations cancel out).
    - Tloewald 10y ago
      Average max, min, or average average?
scottlamb 10y ago
They used the word "change" in the paragraph before the one you quoted. I think by "observation" in this paragraph, they mean "x_i - M(x)" as in https://en.wikipedia.org/wiki/Average_absolute_deviation https://en.wikipedia.org/wiki/Average_absolute_deviation: the difference between the current value and some measure of central tendency (mean, median, mode). It's not explained well, but if you make this assumption, the article makes sense.

enugu 10y ago

The averaging of absolute values in the article's context is happening around a mean of 0(averaging the variations). In general, one would average the absolute values of the difference from the mean. Yes, it would be silly to convert a -20 to 20 if the mean isn't 0. The advantage with STD is that it is easier to compute relative to MAD in a situation like a random walk. For instance when one has a +1, -1 equal probability one dimensional random walk, so the mean is 0, X = sum of X_i, E(X²) = n is straightforward, whereas computing E(|X|) is not so easy.

Tloewald 10y ago
That's as clear as mud. What does a -20 vs. a 20 mean? That it was colder at noon than at midnight? Assuming that there was some sensible definition (e.g. deviation from the past mean temperature, which is not at all what was stated) then "average deviation" has two obvious interpretations. Re-labeling "standard deviation" or "mean deviation" would not be helpful in this case, and the -20 and 20 values STILL don't help the case.
- enugu 10y ago
  The article isn't clear but the deviation is either from the sample mean or the mean of the underlying random variable. You are trying to find the expected difference from the average and hence the absolute value(otherwise, the positive and negative deviations cancel out).
  - Tloewald 10y ago
    Average max, min, or average average?

scottlamb 10y ago

They used the word "change" in the paragraph before the one you quoted. I think by "observation" in this paragraph, they mean "x_i - M(x)" as in https://en.wikipedia.org/wiki/Average_absolute_deviation https://en.wikipedia.org/wiki/Average_absolute_deviation: the difference between the current value and some measure of central tendency (mean, median, mode). It's not explained well, but if you make this assumption, the article makes sense.