1 / 2

Saturated Model

Saturated Model. Why, for a 3-d table, can’t we estimate a 3-way interaction? ln ( e ijk ) = u + u 1( i ) + u 2(j) + u 3(k) + u 12( ij ) + u 13( ik ) + u 23( jk ) + u 123( ijk ) the “ saturated model ”. Analogous to ANOVA w/o >1 replications per cell -- we’ve run out of df :

ranit
Télécharger la présentation

Saturated Model

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Saturated Model Why, for a 3-d table, can’t we estimate a 3-way interaction? ln(eijk) = u + u1(i) + u2(j) + u3(k) + u12(ij) + u13(ik) + u23(jk) + u123(ijk) the “saturated model”. Analogous to ANOVA w/o >1 replications per cell -- we’ve run out of df: For the 2x2x3 ex, fitting [ABC] would leave: 12 - 1 - 1 - 1 - 2 - 1 - 2 - 2 - 2 = 0 df 1(i)2(j)3(k)12(ij)13(ik)23(jk)123(ijk) In general for 3-d table, [ABC] leaves: IJK - 1 - (I-1) - (J-1) - (K-1) - (I-1)(J-1) - (I-1)(K-1) - (J-1)(K-1) - (I-1)(J-1)(K-1) =IJK - 1 - I+1 - J+1 - K+1 - IJ+I+J - 1 - IK+I+K - 1 - JK+J+K - 1 - IJK+IJ­+IK - I+JK - J - K+1 = 0 df.

  2. Saturated Model The saturated model will always (trivially) fit the data perfectly, because there are as many parameters to estimate as there are data points (i.e., 0 df). G2 would = 0 because all eijk = oijk, etc. In general, including the n-way interaction in hierarchial models for n-dimensional tables results in the saturated model.

More Related