Re: [MORPHMET] Doubts about Error Measurement
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 : > 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 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 : >> >> Liu Idárraga 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.) >>>> Individuo0,26857296 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 >>>> >>>&
Re: [MORPHMET] Doubts about Error Measurement
And respect to the total error, the sum of the % SSimage + % SSdig? Liu 2016-03-15 21:38 GMT-03:00 Carmelo Fruciano : > Liu Idárraga 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.) >> Individuo0,26857296 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 c.fruci...@unict.it > 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 > Groups "MORPHMET" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to morphmet+unsubscr...@morphometrics.org. > > -- *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 --- You received this message because you are subscribed to the Google Groups "MORPHMET" group. To unsubscribe from this group and stop receiving emails from it, send an email to morphmet+unsubscr...@morphometrics.org.
[MORPHMET] Doubts about Error Measurement
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.) Individuo0,26857296 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 -- *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 --- You received this message because you are subscribed to the Google Groups "MORPHMET" group. To unsubscribe from this group and stop receiving emails from it, send an email to morphmet+unsubscr...@morphometrics.org.
Re: MORPHMET - progress, but not there, yet.
Appreciated Dennis, I'll like to be in the new list, too. I big hug, Liu On 31/07/14 4:23, morphmet wrote: I have been making progress preparing the migration to the new system, but not quite there, yet. I hope to have this done by or over the weekend. I am writing this to let you know that when MORPHMET does come back up, you will receive an invitation to join the new list (actually just the old list in a new environment). You need to reply to this message when you receive it to continue to be a subscriber to the new (=old) MORPHMET. -dslice -- PK
