VST and Kruskal-Wallis (using R)

1. The linear effects model is

y_ij = μ + τ_i + ε_ij

where μ is the overall mean, τ_i is the effect of estimation method i, and ε_ij is the random error.

ε_ij ~ N(0, σ²)

H0: μ1 = μ2 = μ3 = μ4 Ha: At least one population mean is different.

method <- factor(rep(1:4, each=6))

discharge <- c(.34, .12, 1.23, .70, 1.75, .12,
               .91, 2.94, 2.14, 2.36, 2.86, 4.55,
               6.31, 8.37, 9.75, 6.09, 9.82, 7.24,
               17.15, 11.82, 10.97, 17.20, 14.35, 16.82)

boxplot(discharge ~ method,
        xlab="Estimation Method",
        ylab="Peak Discharge")

qqnorm(discharge)
qqline(discharge)

The data do not appear to be normally distributed, and the variance does not appear to be constant.

model <- aov(discharge ~ method)
summary(model)

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

plot(model, which=1)

plot(model, which=2)

The residual plots indicate that the assumptions of normality and constant variance may not be satisfied.

library(MASS)

boxcox(model)

sqrt_discharge <- sqrt(discharge)

model_sqrt <- aov(sqrt_discharge ~ method)
summary(model_sqrt)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## method       3  32.69  10.898   81.17 2.27e-11 ***
## Residuals   20   2.69   0.134                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(model_sqrt, which=1)

plot(model_sqrt, which=2)

The Box-Cox plot suggests λ ≈ 0.5, so a square-root transformation was used. The ANOVA shows a significant difference among the four estimation methods (p < 0.05). Therefore, H₀ is rejected.

kruskal.test(discharge ~ method)

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

The Kruskal-Wallis test shows a significant difference among the four estimation methods (p < 0.05). Therefore, H₀ is rejected.