Assignment9

##Problem1
#Information
k <- 4              
variance <- 3.5     
alpha <- 0.05       
power <- 0.80       

#for ANOVA with k=4, alpha=0.05, power=0.80 we need phi ≈ 2.49
phi_target <- 2.49

#maximum Variability (means: 18, 18, 20, 20)
means_max <- c(18, 18, 20, 20)
sum_sq_max <- sum((means_max - mean(means_max))^2)
n_max <- ceiling((phi_target^2 * k * variance) / sum_sq_max)

#minimum Variability (means: 18, 18, 18, 20)
means_min <- c(18, 18, 18, 20)
sum_sq_min <- sum((means_min - mean(means_min))^2)
n_min <- ceiling((phi_target^2 * k * variance) / sum_sq_min)

#intermediate Variability (means: 18, 18.67, 19.33, 20)
means_int <- c(18, 18.67, 19.33, 20)
sum_sq_int <- sum((means_int - mean(means_int))^2)
n_int <- ceiling((phi_target^2 * k * variance) / sum_sq_int)

#print
n_max

## [1] 22

n_min

## [1] 29

n_int

## [1] 40

#Maximum Variability: 22 samples per fluid
#Minimum Variability: 29 samples per fluid
#Intermediate Variability: 40 samples per fluid

##Problem2
fluid <- rep(1:4, each = 6)
life <- c(17.6, 18.9, 16.3, 17.4, 20.1, 21.6,  # Fluid 1
          16.9, 15.3, 18.6, 17.1, 19.5, 20.3,  # Fluid 2
          21.4, 23.6, 19.4, 18.5, 20.5, 22.3,  # Fluid 3
          19.3, 21.1, 16.9, 17.5, 18.3, 19.8)  # Fluid 4
fluid <- factor(fluid)

#2a ANOVA test at alpha = 0.10
model <- aov(life ~ fluid)
summary(model)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## fluid        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

#2b Model adequacy - diagnostic plots
par(mfrow = c(2, 2))
plot(model)

#1.Residuals vs Fitted: Residuals are randomly scattered around zero with no pattern, indicating a constant variance assumption is satisfied
#2.Q-Q Residuals: Points follow the diagonal line closely, indicating a normality assumption is satisfied. Minor deviations at tails are acceptable for small sample
#3.Scale-Location: Red line is relatively horizontal with random point distribution, which confirms equal variance assumption is met
#4.Constant Leverage: Residuals are evenly spread across all fluid groups with no influential outliers affecting the model

#Overall: The model is adequate, ANOVA assumptions are reasonably satisfied

#2c Tukey's test at alpha = 0.10
tukey_result <- TukeyHSD(model, conf.level = 0.90)
tukey_result

##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = life ~ fluid)
## 
## $fluid
##           diff        lwr       upr     p adj
## 2-1 -0.7000000 -3.2670196 1.8670196 0.9080815
## 3-1  2.3000000 -0.2670196 4.8670196 0.1593262
## 4-1  0.1666667 -2.4003529 2.7336862 0.9985213
## 3-2  3.0000000  0.4329804 5.5670196 0.0440578
## 4-2  0.8666667 -1.7003529 3.4336862 0.8413288
## 4-3 -2.1333333 -4.7003529 0.4336862 0.2090635

#Plot confidence intervals
plot(tukey_result)
#Interpretation of  results:
#Looking at the values to determine significant differences:
#Fluid 2 vs Fluid 1: p = 0.908 > 0.10 (not significantly different)
#Fluid 3 vs Fluid 1: p = 0.159 > 0.10 (not significantly different)
#Fluid 4 vs Fluid 1: p = 0.999 > 0.10 (not significantly different)
#Fluid 3 vs Fluid 2: p = 0.044 < 0.10 (significantly different)
#Fluid 4 vs Fluid 2: p = 0.841 > 0.10 (not significantly different)
#Fluid 4 vs Fluid 3: p = 0.209 > 0.10 (not significantly different)
#conclusion: only fluid 3 and fluid 2 have significantly different mean lifetimes

{r eval=FALSE} ```