Problem 5.9

library(GAD)
drill_speed <- c(rep(125,8),rep(200,8))
feed_rate <- c(rep(seq(0.015,0.060,0.015),4))
response <- c(2.70,2.45,2.60,2.75,2.78,2.49,2.72,2.86,2.83,2.85,2.86,2.94,2.86,
              2.80,2.87,2.88)
drill_speed <- as.fixed(drill_speed)
feed_rate <- as.fixed(feed_rate)
dat <- data.frame(drill_speed,feed_rate,response)
dat
model <- aov(response~drill_speed+feed_rate+drill_speed*feed_rate,data=dat)
summary(model)
##                       Df  Sum Sq Mean Sq F value   Pr(>F)    
## drill_speed            1 0.14822 0.14822  57.010 6.61e-05 ***
## feed_rate              3 0.09250 0.03083  11.859  0.00258 ** 
## drill_speed:feed_rate  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(drill_speed,feed_rate,response)

Model Equation: \(Y_{ijk} =\mu+\alpha_i+\beta_j+\alpha\beta_{ij}+\epsilon_{ijk}\) where, \(\epsilon_{ijk} ~ N(0,\sigma^2)\)

Hypothesis Test:

\(H_0: \alpha_i=0 \forall\) 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

Conclusions: From the summary of the model, we see that the p value (0.025567) of the interaction term is less than \(\alpha=0.05\) level so we reject the Null hypothesis \(H_0\) which indicates that the interaction term is significant.Also, the interaction plot shows that there is an interaction between these two factors.Because of this we would stop our testing and do not continue to test the hypothesis of the main effects.