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 >>> 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 "MO
Re: [MORPHMET] Doubts about Error Measurement
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 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. -- 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.
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.
Re: [MORPHMET] Doubts about Error Measurement
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.
[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.
