library(pwr)
sqrt(1/4.5)
## [1] 0.4714045
pwr.anova.test(k=4,n=NULL,f=0.471,sig.level = 0.05,power =0.80)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 13.30502
##               f = 0.471
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group
# Ans to the ques 1.(a).
# 14 samples of each fluid we will need to be collected 

sqrt(.5^2/4.5)
## [1] 0.2357023
pwr.anova.test(k=4,n=NULL,f=0.2357,sig.level = 0.05,power =0.80)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 50.05016
##               f = 0.2357
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group
# Ans to the ques 1.(b).
# 51 samples of each fluid we will need to be collected 


type1 <- c(17.6,    18.9,   16.3,   17.4,   20.1,   21.6)
type2   <- c(16.9,  15.3,   18.6,   17.1,   19.5,   20.3)
type3   <- c(21.4,  23.6,   19.4,   18.5,   20.5,   22.3)
type4 <- c(19.3,    21.1,   16.9,   17.5,   18.3,   19.8)

dat <- data.frame(type1,type2,type3,type4)
f <- (1)^2/(2.0447)^2

pwr.anova.test(k=4,n=6,f,sig.level = .1,power =NULL)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 6
##               f = 0.2391888
##       sig.level = 0.1
##           power = 0.213476
## 
## NOTE: n is number in each group
# Ans to the ques 2.(a).
# The power will be 0.213476 for a hypothesis test with an alpha = 0.10 level 
# of significance 

library(tidyr)
fluidsample<-pivot_longer(data = dat, c(type1,type2,type3,type4))
aov.model<-aov(value~name, data = fluidsample)
summary(aov.model)
##             Df Sum Sq Mean Sq F value Pr(>F)  
## name         3  30.16   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Ans to the ques 2.(b).
# Since p value = 0.0525 is less than alpha = 0.1, so we reject null hypothesis.

plot(aov.model)

## hat values (leverages) are all = 0.1666667
##  and there are no factor predictors; no plot no. 5

# Ans to the ques 2.(c).
# From the plot, it seems like Normality assumption is satisfied.
# The model is adequate.


TukeyHSD(aov.model, conf.level = 0.9)
##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = value ~ name, data = fluidsample)
## 
## $name
##                   diff        lwr       upr     p adj
## type2-type1 -0.7000000 -3.2670196 1.8670196 0.9080815
## type3-type1  2.3000000 -0.2670196 4.8670196 0.1593262
## type4-type1  0.1666667 -2.4003529 2.7336862 0.9985213
## type3-type2  3.0000000  0.4329804 5.5670196 0.0440578
## type4-type2  0.8666667 -1.7003529 3.4336862 0.8413288
## type4-type3 -2.1333333 -4.7003529 0.4336862 0.2090635
plot(TukeyHSD(aov.model, conf.level = 0.9))

# Ans to the ques 2.(d).
# Type 2 and Type 3 fluid significantly differs

Source Code

library(pwr)
sqrt(1/4.5)
pwr.anova.test(k=4,n=NULL,f=0.471,sig.level = 0.05,power =0.80)
sqrt(.5^2/4.5)
pwr.anova.test(k=4,n=NULL,f=0.2357,sig.level = 0.05,power =0.80)
type1 <- c(17.6,    18.9,   16.3,   17.4,   20.1,   21.6)
type2   <- c(16.9,  15.3,   18.6,   17.1,   19.5,   20.3)
type3   <- c(21.4,  23.6,   19.4,   18.5,   20.5,   22.3)
type4 <- c(19.3,    21.1,   16.9,   17.5,   18.3,   19.8)
dat <- data.frame(type1,type2,type3,type4)
f <- (1)^2/(2.0447)^2
pwr.anova.test(k=4,n=6,f,sig.level = .1,power =NULL)
library(tidyr)
fluidsample<-pivot_longer(data = dat, c(type1,type2,type3,type4))
aov.model<-aov(value~name, data = fluidsample)
summary(aov.model)
plot(aov.model)
TukeyHSD(aov.model, conf.level = 0.9)
plot(TukeyHSD(aov.model, conf.level = 0.9))