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Re: [matplotlib-devel] boxplot notch

From: Andrew S. <str...@as...> - 2009年12月19日 05:25:56
Pierre GM wrote:
> On Dec 18, 2009, at 10:34 PM, Andrew Straw wrote:
> 
>> Fernando Perez wrote:
>> 
>>> On Fri, Dec 18, 2009 at 2:28 PM, Andrew Straw <str...@as...> wrote:
>>>
>>> 
>>>> (This still leaves open the question of what the notches actually _are_...)
>>>>
>>>> 
>>> No idea. I'd still leave the code instead written as
>>>
>>> notch_max = med + (iq/2) * (pi/np.sqrt(row))
>>>
>>> 
>> Further searching turned this up: 
>> http://seismo.berkeley.edu/~kirchner/eps_120/Toolkits/Toolkit_01.pdf
>>
>> It says that
>>
>> median +/- 1.57 * (iq / sqrt(n)) is the median, plus or minus its standard error.
>>
>>
>> I can't find any further support for this notion, though.
>> 
>
>
> Looks like the std error of the median is (1.253*std error of the mean=1.253*std dev/sqrt(nb of obs)).
> The 1.57 looks like it's 1.253^2, but I wouldn't bet anything on it...
>
> 
Also, I think that formula is only for normally distributed data. Which, 
especially if you're using boxplots, medians, and quartiles, may not be 
a valid assumption.
Maybe we should at least raise a warning when someone uses notch=1. The 
current implementation seems dubious, at best, IMO.
-Andrew

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