Problem 5.2

PART A:

Fill in the blanks in the ANOVA table. You can use bounds on the P-values.

Finding the P values

pf(0.000010075,1,8,lower.tail = FALSE)
## [1] 0.9975452
pf(3.029,3,8,lower.tail = FALSE)
## [1] 0.09335136
pf(0.1424,3,8,lower.tail = FALSE)
## [1] 0.9316887
Source DOF SS MS F P
A 1 0.0002 0.0002 1.00757e-5 0.9975
B 3 180.378 60.126 3.029 0.0933
Interaction 3 8.479 2.8263 0.1424 0.9317
Error 8 158.797 19.8496
Total 15 347.653

PART B:

How many levels were used for factor B?

Since the Degree of Freedom of factor B is 3 as calculated in part A.

The no of levels will be (DOF + 1).

Therefore, no of levels of factor B = 4

PART C:

How many replicates of the experiment were performed?

There are k = 2 replicates per set.

Because the replicates are the runs with the same levels or factor combination.

PART D:

Model Equation

\(Y_{ijk} = \mu + \alpha _{i} + \beta _{j} + \alpha \beta _{ij} + \epsilon _{ijk}\)

Hypothesis

Null Hypothesis: \(H_{0}\) : \(\alpha _{i}\) = 0 for all ij

Alternate Hypothesis : \(H_{0}\) : \(\alpha _{i} \neq 0\) for some ij

If we assume that the conditions of anova are met i.e. constant variance and normality.

Then for alpha = 0.1 level of significance, for factor A we cannot conclude that it is significant because the p value is greater than 0.1 (alpha). But for factor B we can say that it is significant because the p value is less than 0.1(alpha).

Problem 5.9

A mechanical engineer is studying the thrust force developed by a drill press. He suspects that the drilling speed and the feed rate of the material are the most important factors. He selects four feed rates and uses a high and low drill speed chosen to represent the extreme operating conditions. He obtains the following results. Analyze the data and draw conclusions. Use alpha = 0.05.

Entering the data

DS<-c(125,125,125,125,125,125,125,125,200,200,200,200,200,200,200,200)
DS<-as.factor(DS)
FR<-c(0.015,0.015,0.03,0.03,0.045,0.045,0.06,0.06,0.015,0.015,0.03,0.03,0.045,0.045,0.06,0.06)
FR<-as.factor(FR)
TF<-c(2.7,2.78,2.45,2.49,2.6,2.72,2.75,2.86,2.83,2.86,2.85,2.8,2.86,2.87,2.94,2.88)
dat1<-data.frame(DS,FR,TF)

Performing Analysis

Model Equation

\(Y_{ijk} = \mu + \alpha _{i} + \beta _{j} + \alpha \beta _{ij} + \epsilon _{ijk}\)

Hypothesis

Null Hypothesis: \(H_{0}\) : \(\alpha _{i}\) = 0 for all ij

Alternate Hypothesis : \(H_{0}\) : \(\alpha _{i} \neq 0\) for some ij

model<-lm(TF~DS+FR+DS:FR,data=dat1)
anova(model)
## Analysis of Variance Table
## 
## Response: TF
##           Df   Sum Sq  Mean Sq F value    Pr(>F)    
## DS         1 0.148225 0.148225 57.0096 6.605e-05 ***
## FR         3 0.092500 0.030833 11.8590  0.002582 ** 
## DS:FR      3 0.041875 0.013958  5.3686  0.025567 *  
## Residuals  8 0.020800 0.002600                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Plotting Graph

interaction.plot(x.factor = dat1$FR, trace.factor = dat1$DS, response = dat1$TF, type = "b")

Conclusion:

From the analysis we see that the drill speed, feed rate and their interaction all have a p value less than alpha specified (0.05), hence we reject the null hypothesis and state that they all are significant.