Thank you very much Carmelo for your help.

Finally I think that understand the equation. It is some basic statistical
concepts relate to the variance components calculation. Now, I think that
it is better way to report the error: as the statistical books show it.
In my case (I'm sorry but is in Spanish):

Modelo general:

*Yijk =  **m** + **a**i** + **b**j(i)** + **e**ijk*


Y para el caso de la sumas de los cuadrados:

SCT = SCA + SCB (A) + SCE

 Donde:

SCA = suma de cuadrados entre individuos.

SCB(A) = suma de cuadrados para las imágenes anidadas en los individuos.

SCE = suma de cuadrados para las digitalizaciones anidadas en los
individuos y las imágenes.

El valor F correspondió a:

F factor A (individuo): división del cuadrado medio de A (CMA) por el
cuadrado medio de B o la imagen (CMA/CMB)

F factor B (imagen): división del cuadrado medio de B por el cuadrado medio
del error o digitalización (CME).

Los cuadrados medios fueron obtenidos por la división de la suma de
cuadrados de cada efecto por sus grados de libertad (g. l.).

g. l. factor A =  (a-1) (k – 4)*[1]* <#_ftn1>

g. l. factor B = a (b - 1) (k – 4)

g. l. factor E = ab(r-1) (k – 4)

g. l. total = abr (k – 4) - 1

Donde:

a = número de individuos digitalizados

b = número de imágenes repetidas (= 2)

r = número de digitalizaciones de cada imagen (= 2)

k = número de coordenadas = # puntos anatómicos - # semilandmarks (ya que
cada semilandmark sólo aporta un grado de libertad.

Las esperanzas de los cuadrados fueron:

*E(MCA)* = *s**2  **+ r**s**2**b* *+ br**s**2**a*

*E(MCB(A))* = *s**2  **+ r**s**2**b*

*E(MCE)* = *s**2  *

Y las estimas de los componentes de varianza:

S2*a*  = [CMA ind – CMB(A)imag] / ( b * r)

S2*b*  = [CMB(A)imag – CME dig] / r

S2dig = CME

Estos efectos representaron las diferencias reales entre individuos, el
error metodológico e instrumental y el error personal, respectivamente.

Esta estimación de las varianzas corresponde a la teoría clásica
estadística de modelos anidados con efectos aleatorios, la cual es retomada
por Bailey y Birnes (1990).

------------------------------

[1] <#_ftnref1> Los cuatro grados de libertad que se le restan a k en todos
los cálculos corresponden a los que se pierden durante el ajuste de
Procrustes.


Again, thanks a lot for the attention and help.


Liu




2016-03-17 1:57 GMT-03:00 Carmelo Fruciano <[email protected]>:

> Hi Liu,
> I'm not sure I understand your question.
> Those are the percentages of the total sum of squares due to each of those
> terms and, yes, as you can see from the numbers, they all add up to 100%.
>
> Of course, you can quantify only what you measure, and you cannot exclude
> that there are other sources of error than the ones you tested (I say this
> because of your wording "total error").
>
> I hope this helps,
> Best,
> Carmelo
>
>
>
>
> Liu Idárraga <[email protected]> ha scritto:
>
> And respect to the total error, the sum of the % SSimage + % SSdig?
>>
>> Liu
>>
>> 2016-03-15 21:38 GMT-03:00 Carmelo Fruciano <[email protected]>:
>>
>> Liu Idárraga <[email protected]> ha scritto:
>>>
>>>
>>> Appreciated colleagues.
>>>
>>>> I have some doubts about the correct calculation of error in my data. I
>>>> had
>>>> understood that the error corresponde to the proportion that represent
>>>> the
>>>> SS of each effect (image / digitalization) in the SStotal; but I found
>>>> in
>>>> the classical paper of Bayley & Byrnes 1990. Syst. Zool. 39(2):126 that
>>>> the
>>>> way to calculate the percent of the measurement error is:
>>>>
>>>> % ME = 100%  s2 within / (s2 within - s2 among)
>>>>
>>>> s2 within = MS within
>>>> but
>>>> s2 among = (MS among -MS within)/# replicates
>>>>
>>>> ​I have the followed results for 71 ind in 2D, and 44 coord (66 - 4 -
>>>> 18 semi landmarks)
>>>>
>>>>
>>>> Procrustes ANOVA para la forma: con g.l. corregidos
>>>> Efecto SS     SS (%) MS df F p (param.)
>>>> Individuo    0,26857296 92,84 8,59709E-05 3124 25,97 <.0001
>>>> Imagen 0,01048868 3,63 3,31082E-06 3168 2,05 <.0001
>>>> Digitalización 0,01021206 3,53 1,61175E-06 6336
>>>> Total 0,28927370 100,00
>>>>
>>>> ​My questions are:
>>>> ​1) Is correct to say that 3,63 and 3,53 are the errors of image and
>>>> digitalization?
>>>> 2)  Is the total error equal to 3,63 + 3,53 or I should calculate it
>>>> with
>>>> the equations of Bayley & Byrnes 1990?
>>>>
>>>> In advance, thanks a lot for the help.
>>>>
>>>> Liu
>>>>
>>>>
>>> Dear Liu,
>>> Perhaps the best way to report this would be to say that 3.63% of the
>>> total sum of squares is accounted for by/due to a certain term
>>> (digitization, preparation)?
>>> Best,
>>> Carmelo
>>>
>>>
>>>
>>> --
>>> Carmelo Fruciano
>>> Postdoctoral Fellow - Queensland University of Technology - Brisbane,
>>> Australia
>>> Honorary Fellow - University of Catania - Catania, Italy
>>> e-mail [email protected]
>>> http://www.fruciano.it/research/
>>>
>>> --
>>> MORPHMET may be accessed via its webpage at http://www.morphometrics.org
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>>>
>>>
>>>
>>
>> --
>> *Liu Idárraga*
>> Estudiante doctoral
>> Facultad de Ciencias Naturales y Museo
>> Universidad de La Plata - Argentina
>>
>> --
>> MORPHMET may be accessed via its webpage at http://www.morphometrics.org
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>>
>
>
> --
> Carmelo Fruciano
> Postdoctoral Fellow - Queensland University of Technology - Brisbane,
> Australia
> Honorary Fellow - University of Catania - Catania, Italy
> e-mail [email protected]
> http://www.fruciano.it/research/
>
> --
> MORPHMET may be accessed via its webpage at http://www.morphometrics.org
> ---You received this message because you are subscribed to the Google
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> email to [email protected].
>
>


-- 
*Liu Idárraga*
Estudiante doctoral
Facultad de Ciencias Naturales y Museo
Universidad de La Plata - Argentina

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