power.anova.test(groups=4,n=NULL,
                 between.var=var(c(18,19,19,20)),
                 within.var=3.5,
                 sig.level=0.05,power=0.80)
## 
##      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,
                 between.var=var(c(18,18.67,19.33,20)),
                 within.var=3.5,
                 sig.level=0.05,power=0.80)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 4
##               n = 18.21285
##     between.var = 0.7392667
##      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,
                 between.var=var(c(18,18,20,20)),
                 within.var=3.5,
                 sig.level=0.05,power=0.80)
## 
##      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
  1. The required sample sizes for the minimum, intermediate, and maximum variability cases are 21, 19, and 11 samples per fluid, respectively. Therefore, more samples are required when the variability among the fluid means is smaller.
Fluid1<-c(17.6,18.9,16.3,17.4,20.1,21.6)
Fluid2<-c(16.9,15.3,18.6,17.1,19.5,20.3)
Fluid3<-c(21.4,23.6,19.4,18.5,20.5,22.3)
Fluid4<-c(19.3,21.1,16.9,17.5,18.3,19.8)

dat<-data.frame(Fluid1,Fluid2,Fluid3,Fluid4)
dat
##   Fluid1 Fluid2 Fluid3 Fluid4
## 1   17.6   16.9   21.4   19.3
## 2   18.9   15.3   23.6   21.1
## 3   16.3   18.6   19.4   16.9
## 4   17.4   17.1   18.5   17.5
## 5   20.1   19.5   20.5   18.3
## 6   21.6   20.3   22.3   19.8
library(tidyr)
dat<-pivot_longer(dat,c(Fluid1,Fluid2,Fluid3,Fluid4))
dat
## # A tibble: 24 × 2
##    name   value
##    <chr>  <dbl>
##  1 Fluid1  17.6
##  2 Fluid2  16.9
##  3 Fluid3  21.4
##  4 Fluid4  19.3
##  5 Fluid1  18.9
##  6 Fluid2  15.3
##  7 Fluid3  23.6
##  8 Fluid4  21.1
##  9 Fluid1  16.3
## 10 Fluid2  18.6
## # ℹ 14 more rows
aov.model<-aov(value~name,data=dat)
summary(aov.model)
##             Df Sum Sq Mean Sq F value Pr(>F)  
## name         3  30.17   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

2-a. Null Hypothesis: The mean lives of all four insulating fluids are the same. Alternative Hypothesis: The mean lives of the four insulating fluids are not all the same. Since the p-value (0.0525) is less than α = 0.10, we reject the null hypothesis. There is sufficient evidence to conclude that at least one fluid has a different mean life.

plot(aov.model)

2-b. The residual plots do not show any obvious patterns, and the residuals are approximately normally distributed based on the Q-Q plot. The variability also appears reasonably constant across the groups. Therefore, the ANOVA model appears to be adequate.

TukeyHSD(aov.model, conf.level=0.90)
##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = value ~ name, data = dat)
## 
## $name
##                     diff        lwr       upr     p adj
## Fluid2-Fluid1 -0.7000000 -3.2670196 1.8670196 0.9080815
## Fluid3-Fluid1  2.3000000 -0.2670196 4.8670196 0.1593262
## Fluid4-Fluid1  0.1666667 -2.4003529 2.7336862 0.9985213
## Fluid3-Fluid2  3.0000000  0.4329804 5.5670196 0.0440578
## Fluid4-Fluid2  0.8666667 -1.7003529 3.4336862 0.8413288
## Fluid4-Fluid3 -2.1333333 -4.7003529 0.4336862 0.2090635
plot(TukeyHSD(aov.model, conf.level=0.90))

2-c. At a familywise error rate of 0.10, only Fluid 2 and Fluid 3 are significantly different (adjusted p-value = 0.0441). Fluid 3 has a significantly higher mean life than Fluid 2. All other pairwise comparisons are not significant.