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Re: st: Average of Sample vesus Sub-Samples
From
Erasmo Giambona <[email protected]>
To
[email protected]
Subject
Re: st: Average of Sample vesus Sub-Samples
Date
Fri, 19 Mar 2010 17:11:45 +0100
Thank you all very much!
Erasmo
On Fri, Mar 19, 2010 at 5:05 PM, Tim Scharks <[email protected]> wrote:
> Ah, I see That's what I get for faulty inference and reading too fast...thanks!
>
> On Fri, Mar 19, 2010 at 8:59 AM, Jeph Herrin <[email protected]> wrote:
>> The original question was not whether these two
>> numbers could differ, but whether the average of the
>> larger group could be greater than the mean of the
>> two sub averages. In your case they differ, but the
>> average of the larger group is smaller.
>>
>> I think you have to have negative numbers to reverse
>> the inequality.
>>
>> cheers,
>> Jeph
>>
>>
>> Tim Scharks wrote:
>>>
>>> but what about
>>>
>>> A = 1, 1, 1
>>> B = 3
>>> m(A) = 1
>>> m(B) = 3
>>> (m(A)+m(B))/2=2
>>> mean (A+B) =1.5
>>>
>>> uh oh...
>>>
>>> The problem is not negative numbers--it is your failure to weight the
>>> subsample means according to their relative size:
>>>
>>> (m(A)*3+m(B)*1)/4 =1.5
>>> m(A+B) = 1.5
>>>
>>>
>>> On Fri, Mar 19, 2010 at 6:49 AM, Jeph Herrin <[email protected]> wrote:
>>>>
>>>> Yes, if they are negative.
>>>>
>>>>
>>>> A = -1,-1,-1
>>>> B = -3
>>>> then
>>>> mean(A) = -1
>>>> mean(B) = -3
>>>> (mean(A)+mean(B))/2 = -2
>>>> mean(A+B)= -6/4 = -1.5
>>>>
>>>> -1.5 > -2
>>>>
>>>> So the average of all 4 numbers is larger than
>>>> the average of the means of the two subsamples.
>>>>
>>>>
>>>> hth,
>>>> Jeph
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> Erasmo Giambona wrote:
>>>>>
>>>>> Dear Statalisters,
>>>>>
>>>>> Is it possible that the arithmetic average of a sample is larger than
>>>>> the averages of two sub-samples containing overall all the
>>>>> observations of the full samples?
>>>>>
>>>>> Any thoughts would be appreciated,
>>>>>
>>>>> Erasmo
>>>>> *
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>
> *
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