Homework Chapter 11

11.2.4

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

  1. There were 4 groups.

  2. 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

  1. There were 5 groups

  2. There were 54 observations.

11.4.1

  1. We would reject the null hypothesis.The first groups seems to have much higher values.
library(rpsychi)
## Warning: package 'rpsychi' was built under R version 3.6.3
## Loading required package: gtools
## Warning: package 'gtools' was built under R version 3.6.3
m<-c(9.81,6.28,5.97)
s<-c(3.62,2.88,3.19)
n<-c(18,16,8)
d<-data.frame(m,s,n)
with(d,ind.oneway.second(m,s,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. 39+2=41 total degrees of freedom.

  2. F = 6.352

  3. a=.05 pvalue=0.004 pvalue<a

We have strong evidence to reject the null.

  1. Spooled= sqrt(10.73)=3.28

11.4.5

  1. It looks like the null hypothesis would be rejected. North and west may have some outliers, which means that they would be different.

  2. H0: μ(North)=μ(East)=μ(South)=μ(West)=μ(open)

mean<-c(9.81, 6.28, 5.97)
sd<-c(3.62,2.88, 3.19)
n<-c(18,16,8)
d<-data.frame(mean,sd,n)
with(d,ind.oneway.second(mean,sd,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

Reject the null?

11.4.7

(a)There were 3 groups.

  1. We would fail to reject the null. There is no evidence to suggest a difference in the mean yield of the different varieties of oats.

(c)Spooled= sqrt(95.5342)=9.77

11.4.8

  1. I would say plot 2 due to the one groups which values are much higher than the other

(b)I would say plot 1 due to having fewer outliers (more clumped together)