FA9_Group6

Question 1

min <- c(18,19,19,20)
inter <- c(18,18.66667,19.33333,20)
max <- c(18,18,20,20)

power.anova.test(groups = 4, between.var = var(min), within.var = 3.5, sig.level = 0.05, power = 0.8)

## 
##      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, between.var = var(inter), within.var = 3.5, sig.level = 0.05, power = 0.8)

## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 18.1787
##     between.var = 0.7407393
##      within.var = 3.5
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

power.anova.test(groups = 4, between.var = var(max), within.var = 3.5,sig.level = 0.05, power = 0.8)

## 
##      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

Answer: The sample value for minimum is 21, for intermediate is 19 and maximum is 11.

Question 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)

##             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

1. Answer: It fail to reject the hypothesis.

plot(aov.model)

2. Answer: The model shows constant variance, normally distributed and the model seems adequate.

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))

3. Answer: None of the fluids values differs significantly, the third fluids has a little difference than the three other, but not significantly.

Complete Code

min <- c(18,19,19,20)
inter <- c(18,18.66667,19.33333,20)
max <- c(18,18,20,20)

power.anova.test(groups = 4, between.var = var(min), within.var = 3.5, sig.level = 0.05, power = 0.8)

power.anova.test(groups = 4, between.var = var(inter), within.var = 3.5, sig.level = 0.05, power = 0.8)

power.anova.test(groups = 4, between.var = var(max), within.var = 3.5,sig.level = 0.05, power = 0.8)

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)

plot(aov.model)

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