Multifactor ANOVA

See pag 954-955 in Neter et al, 1996. Read the data Ch23pr12 into a data.frame and execute lm for ANOVA analysis.

Ch23pr12=read.table("c://temp//ch23pr12.txt")

Reading directly from the Web-server.

Ch23pr12<-read.table("http://www.biw.kuleuven.be/vakken/statisticsbyR/datasetsTXT/CH23PR12.txt")

names(Ch23pr12)<-c('minutes','gender','sequence','experience','replic')
attach(Ch23pr12)
gender=factor(gender)
sequence=factor(sequence)
experience=factor(experience)
ResLM1=lm(minutes~gender*sequence*experience)

Check residuals:

 layout(matrix(c(1,2,3,4),2,2))
 plot(ResLM1)

What is your answer for 23.12 a and b?

Check anova table

anova(ResLM1)
 

Answer 23.13 c to f on page 955.

Main effects study by differences (problem 23.14 a )

 mui..=tapply(minutes,gender,mean);mui..
#                1              2
# 1155.9333 966.1333
 mu.j.=tapply(minutes,sequence,mean);mu.j.
#             1           2            3
# 1044.15 1101.40 1037.55

mu..k=tapply(minutes,experience,mean);mu..k
#              1            2
# 1140.867 981.200

Calculate the Bonferroni 90% family condifence interval.

 dfMSE=ResLM1$df.residual;dfMSE
#   [1] 48
 alfa=0.1
 ncomp=5
 tscore=qt(1-alfa/2/ncomp,dfMSE);tscore
#  [1] 2.406581

MSE=sum(ResLM1$residuals^2)/dfMSE;MSE
# [1] 858.0417
 

Now all elements are available to answer the question.

Confidence interval on μ₂₃₁ (problem 23.14 b)

muijk=tapply(minutes,gender:sequence:experience,mean);muijk
    1:1:1    1:1:2    1:2:1      1:2:2     1:3:1   1:3:2        2:1:1     2:1:2    2:2:1     2:2:2     2:3:1     2:3:2
1218.6 1051.0 1274.2  1122.4 1218.2   1051.2 1036.4    870.6 1077.4   931.6 1020.4    860.4
Average for i=2; j=3 and k=1

23 April, 2003 by Guido Wyseure