##11.2.4a

df<-c(3,12,15)
SS<-c(135,337,472)
MS<-c(45,28.08, NA)

table<- data.frame(df,SS,MS); table
##   df  SS    MS
## 1  3 135 45.00
## 2 12 337 28.08
## 3 15 472    NA

##11.2.4.b-c 4 groups 16 observations ##11.2.5

df<-c(4,49,53)
SS<-c(159,964,1123)
MS<-c(39.75,19.67, NA)

table<- data.frame(df, SS, MS); table
##   df   SS    MS
## 1  4  159 39.75
## 2 49  964 19.67
## 3 53 1123    NA

##11.2.5b-c 5 groups 54 observations ##11.4.1

#install.packages("rpsychi")
library(rpsychi)
## Loading required package: gtools
d.mean<-c(9.81, 6.28, 5.97)
d.sd<-c(3.62,2.88, 3.19)
d.n<-c(18,16,8)
d<-data.frame(d.mean,d.sd,d.n)
with(d,ind.oneway.second(d.mean,d.sd,d.n))
## $anova.table
##                 SS df     MS     F
## Between (A) 136.29  2 68.148 6.352
## Within      418.42 39 10.729      
## Total       554.72 41             
## 
## $omnibus.es
##       etasq etasq.lower etasq.upper 
##       0.246       0.032       0.423 
## 
## $raw.contrasts
##     mean.diff  lower upper   std
## 1-2      3.53  1.254 5.806 1.125
## 1-3      3.84  1.025 6.655 1.392
## 2-3      0.31 -2.559 3.179 1.418
## 
## $standardized.contrasts
##        es  lower upper   std
## 1-2 1.078  0.383 1.773 0.344
## 1-3 1.172  0.313 2.032 0.425
## 2-3 0.095 -0.781 0.970 0.433
## 
## $power
##  small medium  large 
##  0.076  0.271  0.608
  1. Base on the is graphical display I would say that the null hypothesis is going to be rejected. b 41 total degrees of freedom c F=6.352 d The P-velue for alpha =0.05 of 0.004 means that we reject the null and accept that alternative that at least one of the groupsis signifcantly different. e 10.72 A Little unsure about e ##11.4.5
  2. Based on the dotplots, I would say accept the null if we are just talking about sides of the building, but to reject the null if also comparing with open areas.
  3. Ho:mu(N)=mu(E)=mu(S)=mu(W); Ha: Stem lengths are not equal among all four directions
  4. No, p-value provided by R
#install.packages("rpsychi")
library(rpsychi)
d.mean<-c(41.4,43.8, 46.5, 43.2)
d.sd<-c(9.3,6.1,6.6,10.4)
d.n<-c(13,13,13,13)
d<-data.frame(d.mean,d.sd,d.n)
with(d,ind.oneway.second(d.mean,d.sd,d.n))
## $anova.table
##                 SS df     MS     F
## Between (A)  174.0  3 58.013 0.843
## Within      3305.0 48 68.855      
## Total       3479.1 51             
## 
## $omnibus.es
##       etasq etasq.lower etasq.upper 
##       0.050       0.000       0.157 
## 
## $raw.contrasts
##     mean.diff   lower upper   std
## 1-2      -2.4  -8.944 4.144 3.255
## 1-3      -5.1 -11.644 1.444 3.255
## 1-4      -1.8  -8.344 4.744 3.255
## 2-3      -2.7  -9.244 3.844 3.255
## 2-4       0.6  -5.944 7.144 3.255
## 3-4       3.3  -3.244 9.844 3.255
## 
## $standardized.contrasts
##         es  lower upper   std
## 1-2 -0.289 -1.078 0.499 0.392
## 1-3 -0.615 -1.403 0.174 0.392
## 1-4 -0.217 -1.006 0.572 0.392
## 2-3 -0.325 -1.114 0.463 0.392
## 2-4  0.072 -0.716 0.861 0.392
## 3-4  0.398 -0.391 1.186 0.392
## 
## $power
##  small medium  large 
##  0.076  0.279  0.640

##11.4.7 a) 3 groups b) Fail to reject the null. All oat yields (bu/acres) were equal/the same/not statistically different. c) The s pooled = 95.5342 ##11.4.8 a) Plot II because a much larger mu difference between groups b) Plot I because equivalent mu difference between groups with less variability within groups than II.