Problem 5.2

Problem 5.9

\[ H_0: \alpha_i=0 \ \forall_i \]

\[ H_a: \alpha_i \not=0 \ for \ some \ i \]

\[ H_0: \beta_j=0 \ \forall_j \]

\[ H_a: \beta_j \not=0 \ for \ some \ j \]

\[ H_0: \alpha\beta_ij=0 \ \forall_ij \]

\[ H_a: \alpha\beta_ij \not=0 \ for \ some \ ij \]

Feed_rate<-c(rep(0.015,4),rep(0.030,4),rep(0.045,4),rep(0.060,4))
Drill_speed<-c(125,125,200,200,125,125,200,200,125,125,200,200,125,125,200,200)
obs<-c(2.7,2.78,2.83,2.86,2.45,2.49,2.85,2.8,2.6,2.72,2.86,2.87,2.75,2.86,2.94,2.88)

data<-data.frame(Feed_rate,Drill_speed,obs)
data$Feed_rate <- as.factor(data$Feed_rate)
data$Drill_speed <- as.factor(data$Drill_speed)
model <- aov(obs~Feed_rate*Drill_speed,data = data)
summary(model)
##                       Df  Sum Sq Mean Sq F value   Pr(>F)    
## Feed_rate              3 0.09250 0.03083  11.859  0.00258 ** 
## Drill_speed            1 0.14822 0.14822  57.010 6.61e-05 ***
## Feed_rate:Drill_speed  3 0.04187 0.01396   5.369  0.02557 *  
## Residuals              8 0.02080 0.00260                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
interaction.plot(x.factor = data$Feed_rate,trace.factor = data$Drill_speed,response = data$obs)

The interactions P-value are lesser than \(\alpha\) = 0.05. Therefore, we can reject the Null Hypothesis, meaning that it is significant. However the interactions plot does not support that.