Question 1

## Loading required package: carData
power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,19,19,20)), within.var = 3.5)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 20.08368
##     between.var = 0.6666667
##      within.var = 3.5
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group
power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,18.66667,19.3333,20)), within.var=3.5)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 18.17886
##     between.var = 0.7407326
##      within.var = 3.5
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group
power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,18,20,20)), within.var = 3.5)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 10.56952
##     between.var = 1.333333
##      within.var = 3.5
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

Comment: The N value required for the Minimum Variance is 21. Intermediate Variance is 19 and Maximum is 11.

Question 2

aov.model <- aov(value~name, data=df)
summary(aov.model)
##             Df Sum Sq Mean Sq F value Pr(>F)
## name         3  19.83   6.611   1.373   0.28
## Residuals   20  96.33   4.816

Comment: Fail to reject the hypothesis at an alpha of 0.1

plot(aov.model)

Comment: #The Model appears adequate. There may be one outlier but nothing to cause major concern.

TukeyHSD(aov.model)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = value ~ name, data = df)
## 
## $name
##                 diff       lwr      upr     p adj
## dat2-dat1  0.9666667 -2.579757 4.513090 0.8700640
## dat3-dat1  2.3000000 -1.246424 5.846424 0.2956275
## dat4-dat1  0.1666667 -3.379757 3.713090 0.9991586
## dat3-dat2  1.3333333 -2.213090 4.879757 0.7214052
## dat4-dat2 -0.8000000 -4.346424 2.746424 0.9206966
## dat4-dat3 -2.1333333 -5.679757 1.413090 0.3577999
plot(TukeyHSD(aov.model, conf.level = 0.9))

Comment: None of the values differ significantly so none of them will be rejected

#Complete Code

library(car)
library(agricolae)
library(tidyr)
library(pwr)

?power.anova.test

power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,19,19,20)), within.var = 3.5)
#20.1
power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,18.66667,19.3333,20)), within.var=3.5)
#18.2
power.anova.test(groups = 4,n = NULL, sig.level = 0.05, power = 0.8, between.var=var(c(18,18,20,20)), within.var = 3.5)
#10.6




#2
dat1 <- c(17.6,18.9,16.3,17.4,20.1,21.6)
dat2 <- c(16.9,25.3,18.6,17.1,19.5,20.3)
dat3 <- c(21.4,23.6,19.4,18.5,20.5,22.3)
dat4 <- c(19.3,21.1,16.9,17.5,18.3,19.8)

df <- data.frame(dat1,dat2,dat3,dat4)
df <- pivot_longer(df, c(dat1,dat2,dat3,dat4))
df$name <- as.factor(df$name)


aov.model <- aov(value~name, data=df)
summary(aov.model)
#Fail to reject the hypothesis at an alpha of 0.1

plot(aov.model)
#The Model appears adequate. There may be one outlier but nothing to cause major concern.

TukeyHSD(aov.model)
plot(TukeyHSD(aov.model, conf.level = 0.9))