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I have a variable that changes over time. Let's call this variable "the error". I use a histogram to graph how many of intervals of error has occurred. For example

histogram

Here you can see that a particular error of 0.1 has occurred almost 1500 times and another of [0.1,0.2] has occurred a little less (like 1400 times) etc.

I have two questions:

  1. If I have another instance of this histogram for a different set of data, and the histogram is similar but with some slight difference in the values, how can I judge which case "is better"? (In general a case is better when the error is less)

  2. What other plots (or means?) can I use to judge whether set A (with its errors) or set B (with its own errors) is better?

Thanks

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  • $\begingroup$ Have you considered a QQ plot? $\endgroup$
    – whuber
    Commented Oct 1, 2023 at 14:14
  • $\begingroup$ Am I right that you are asking how you can decide whether one distribution by and large has lower values ("errors") in it than the other? This would require that you aggregate these distributions into single numbers that you compare. The most popular way for doing this would be to compute error means (or means of squared errors if these are statistical errors) and compare them, although there are alternatives. $\endgroup$ Commented Oct 1, 2023 at 14:34
  • $\begingroup$ @whuber I did not know what QQ plots were. So I am reading about them. How can they serve in this case? (The definition of qq plot says "we use them to determine to what extent the observed data points do or do not follow a given distribution" $\endgroup$ Commented Oct 10, 2023 at 10:22

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so in exploratory data analysis or maybe in making small multiples, you should change your frequencies to densities to standardize between y-axis of plots for different datasets. then ,visually, the better one is that has the least mass to the right. but it'd help if you use lines instead of bars so you can overlap them for easier visual comparison. a box plot is also an alternative approach to help you see which one is has errors closer to 0.

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  • $\begingroup$ Thanks. So my homework is 1. Change frequency to densities to be able to compare 2. use lines instead of bars 3. use boxplots. In this, the error closer to 0 judgement is based on the median? $\endgroup$ Commented Oct 2, 2023 at 3:06
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    $\begingroup$ I just read boxplots show the median not the mean. Would that represent a problem of misjudgment? $\endgroup$ Commented Oct 2, 2023 at 3:15
  • $\begingroup$ @KansaiRobot Well I wouldn't really go ahead and jump to say that mean is not a good representation because your data is not approxiamtely in the symmetric bell-curve shape so the mean is automatically not a good choice of central tendency. for kind of dataset, if you put a vertical line where the mean, median, and mode. you the arrangement would be like mode<median<mean. mean is not a particularly good choice because of the heavy right-tail $\endgroup$
    – Derf
    Commented Oct 2, 2023 at 3:51
  • $\begingroup$ I have received notification that the data I was given is incorrect and ignore the negative data and therefore this heavy right tail is not present. Should I edit this question or ask a new question with the correct data. I am thinking of asking a new one. $\endgroup$ Commented Oct 2, 2023 at 8:51

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