DF of source B can be calculated from the following equation
DF interaction : $=()() $ the only unknown value is \(\beta\) which comes out to 4.
SS_a can be calculated from \(MS_a*DF_a\), which comes out to .0002.
the remaining MS values can be solved by dividing the SS values from the respective DF values.
The F-stat can be solved by \(MS_(source)/MS_E\)
The P values can be calculated by using the following code;
pf(.00001,df1 =1,df2= 8,lower.tail = FALSE) # P value source A## [1] 0.9975543pf(2.271,df1 =4,df2= 8,lower.tail = FALSE) # P value source B## [1] 0.1503544pf(0.1424,df1 =3,df2= 8,lower.tail = FALSE) # P value source interaction effects## [1] 0.9316887The data frame below shows the full table
## Source DF SS MS Fstat Pvalue
## 1 A 1 0.0002 0.0002 0.00001 0.9975543
## 2 B 3 180.3780 45.0945 2.27100 0.1503544
## 3 Interaction 3 8.4790 2.8264 0.14240 0.9316887
## 4 Error 8 158.7970 19.8490 NA NA
## 5 Total 15 347.6530 NA NA NAThere are 4 levels for Factor B.
In general there are \(abn\) observations.
Since there are sixteen observations, and a, b are known
\(16=2*4*n\)
there are 2 replications
From the P and F values the results of this test were not significant at all
#5.9
x <- c(1,1,2,2)
Feedrate <- c(rep(1,4),rep(2,4),rep(3,4),rep(4,4))
DrillSpeed <- c(rep(x,times=4))
obs<- c(2.70,2.78,2.83 ,2.86,2.45,2.49,2.85,2.80,2.45,2.49,2.85,2.80,2.75,2.86,2.94,2.88)
Feedrate <- as.factor(Feedrate)
DrillSpeed <- as.factor(DrillSpeed)
dat <- data.frame(Feedrate,DrillSpeed,obs)
anov <- aov(obs~DrillSpeed*Feedrate)
summary(anov)## Df Sum Sq Mean Sq F value Pr(>F)
## DrillSpeed 1 0.2116 0.21160 108.51 6.25e-06 ***
## Feedrate 3 0.1345 0.04482 22.99 0.000275 ***
## DrillSpeed:Feedrate 3 0.0625 0.02083 10.68 0.003590 **
## Residuals 8 0.0156 0.00195
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1From the experimental design, the 2 factors and their interactions are statistically significant with p values under .05. It seems that drill speed has the greatest effects on the thrust force.