1 Linear Effect Equation

\[ y_{ij} = \mu + \tau_{i} + e_{ij} \]

2 Data Box Plot

Based on the box plots, the assumption of normality may be questionable for some populations because there is potential skewness. The assumption of equal variance may also be violated because of the differences in the IQRs among these populations.

3 ANOVA (Parametric)

\(H_0 : \mu_1 = \mu_2 = \mu_3 = \mu_4\)

\(H_1 :\) at least one of \(\mu_i\) different

##             Df Sum Sq Mean Sq F value Pr(>F)    
## name         3  708.7   236.2   76.29  4e-11 ***
## Residuals   20   61.9     3.1                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Since the p-value is less than 0.05, there is enough evidence to reject \(H_0\). Therefore, at least one treatment mean is significantly different.

However, the model appears to be inadequate because the constant variance assumption is violated. The residual plots show that the variability of the residuals increases as the fitted values increase, especially for Method 4. The Q-Q plot also shows some deviations from normality at the tails.

4 ANOVA (Parametric) with Box Cox Transformation

\(H_0 : \mu_1 = \mu_2 = \mu_3 = \mu_4\)

\(H_1 :\) at least one of \(\mu_i\) different

## [1] "Optimal Lambda: 0.545454545454546"
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## name         3 149.66   49.89   83.69 1.71e-11 ***
## Residuals   20  11.92    0.60                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Since the p-value is less than 0.05, there is enough evidence to reject \(H_0\). Therefore, at least one treatment mean is significantly different.

5 Non-Parametric Kruskal-Wallace Test

\(H_0 :\) The four groups have the same distribution.

\(H_1 :\) At least one group has a different distribution.

## 
##  Kruskal-Wallis rank sum test
## 
## data:  value by name
## Kruskal-Wallis chi-squared = 21.156, df = 3, p-value = 9.771e-05

Since the p-value is less than 0.05, there is enough evidence to reject $H_0$. Therefore, at least one group has a significantly different distribution.

6 Executed Code

#data input
method_1 <- c(0.34, 0.12, 1.23, 0.70, 1.75 , 0.12)
method_2 <- c(0.91, 2.94, 2.14, 2.36, 2.86, 4.55)
method_3 <- c(6.31, 8.37, 9.75, 6.09, 9.82, 7.24)
method_4 <- c(17.15, 11.82, 10.97, 17.20, 14.35, 16.82)

boxplot(method_1,method_2,method_3,method_4)

#AOV raw data
library (tidyr)
dat <- data.frame(method_1, method_2, method_3, method_4)

dat<- pivot_longer(dat, cols = c("method_1", "method_2", "method_3", "method_4") )

aov.model <- aov(value ~ name, data = dat)
summary(aov.model)
plot(aov.model)

#aov boxcox 
library (MASS)

model <- lm(value ~ name, data = dat)
bc <- boxcox(model, lambda = seq(-2, 2, 0.1))

optimal_lambda <- bc$x[which.max(bc$y)]
print(paste("Optimal Lambda:", optimal_lambda))

dat_transformed <- dat
dat_transformed$value <- (dat_transformed$value^optimal_lambda - 1) / optimal_lambda

aov.model2 <- aov(value ~ name, data = dat_transformed)
summary(aov.model2)

#kruskal-wallace test
kruskal.test(value ~ name, data = dat)