Bioequivalence and Bioavailability Forum • Imbalance + Type III SS = Tricky for the sequence evaluation

Imbalance + Type III SS = Tricky for the sequence evaluation [🇷 for BE/BA]

posted by yjlee168 – Kaohsiung, Taiwan, 2009-09-08 00:23 (5762 d 13:51 ago) – Posting: # 4162
Views: 8,812

Dear Elmaestro,

I've tested your codes and got the following.

  Dependent Variable: Cmax                                                 



Type I SS

Analysis of Variance Table



Response: Cmax

          Df Sum Sq Mean Sq F value Pr(>F)

prd        1  37889   37889  0.8844 0.3656

drug       1  88706   88706  2.0705 0.1757

subj(seq) 13 720367   55413  1.2934 0.3313

Residuals 12 514123   42844               



Type III SS

Single term deletions



Model:

Cmax ~ prd + drug + subj + seq

          Df Sum of Sq     RSS     AIC F value  Pr(F)

<none>                  514123     307               

prd        1     37889  552013     307  0.8844 0.3656

drug       1     88706  602830     309  2.0705 0.1757

subj(seq) 12    719310 1233434     307  1.3991 0.2849

seq        0      1057  514123     307               



Tests of Hypothesis for SUBJECT(SEQUENCE) as an error term



Error: subj

          Df Sum Sq Mean Sq F value Pr(>F)

prd:drug   1   1057    1057  0.0176 0.8966

Residuals 12 719310   59943               



Error: Within

          Df Sum Sq Mean Sq F value Pr(>F)

prd        1  37889   37889  0.8844 0.3656

drug       1  88706   88706  2.0705 0.1757

Residuals 12 514123   42844

Find anything different from previous thread. Something can be wrong there... this is just a quick response.

❝ So here's my ugly proposal:

❝ Lm1=lm(lnAuc~Per+Trt+Subj+Seq) ## standard model, right?

❝ T3A=drop1(Lm1, test="F") ## this is our type III anova which gives a dumb

❝ Seq SS

❝ T3A[5,2] = anova(lm(lnAuc~Per+Trt))$Sum[3] -

❝ anova(lm(lnAuc~Per+Trt+Seq))$Sum[4]

❝ ## manually corrects the Seq SS accordingly to the text above

I translated your codes as follows.

cat("  Dependent Variable: Cmax                                                 \\n")

cat("\\n")

cat("Type I SS\\n")

Cmax<- lm(Cmax ~ prd + drug + subj + seq, data=TotalData)

BearAnova=anova(Cmax)

row.names(BearAnova)[3]="subj(seq)"

show(BearAnova)

cat("\\n")

cat("Type III SS\\n")

Cmax_drop1 <- drop1(Cmax, test="F")

row.names(Cmax_drop1)[4]="subj(seq)"

Cmax_drop1[5,2]= anova(lm(Cmax ~ prd + drug, data=TotalData))$Sum[3] - 

  anova(lm(Cmax ~ prd + drug + seq, data=TotalData))$Sum[4] 

show(Cmax_drop1)

cat("\\n")

Am I doing anything stupid here? Thanks.

—
All the best,
-- Yung-jin Lee
bear v2.9.2:- created by Hsin-ya Lee & Yung-jin Lee
Kaohsiung, Taiwan https://www.pkpd168.com/bear
Download link (updated) -> here

Complete thread:

Imbalance + Type III SS = Tricky for the sequence evaluation ElMaestro 2009-09-07 21:16
- Imbalance + Type III SS = Tricky for the sequence evaluationyjlee168 2009-09-07 22:23
  - Imbalance + Type III SS = Tricky for the sequence evaluation ElMaestro 2009-09-08 16:49
    - Imbalance + Type III SS = Tricky for the sequence evaluation yjlee168 2009-09-08 20:02
      - Sequential phenomena and some R code ElMaestro 2009-09-08 20:28
        
        Sequential phenomena and some R code yjlee168 2009-09-08 21:36
        
        Default in SAS ElMaestro 2009-09-08 21:43