Question 5.4

An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut. He selects three feed rates and four depths of cut.

Reading Data

Feed <- c(rep(1,12), rep(2,12), rep(3,12))
Depth <- rep(seq(1,4),9)
Response <- c(74, 79, 82, 99, 64, 68, 88, 104, 60, 73, 92, 96, 92, 98, 99, 
              104, 86, 104, 108, 110, 88, 88, 95, 99, 99, 104, 108, 114, 
              98, 99, 110, 111, 102, 95, 99, 107)
Data <- data.frame(Feed, Depth, Response)

Part A

Analyze the data and draw conclusions. Use a = 0.05.

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_{i}\) + \(\beta_{j}\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Analysis

library(GAD)
Feed <- as.fixed(Feed)
Depth <- as.fixed(Depth)
Model <- aov(Response~Depth+Feed+Depth*Feed)
GAD::gad(Model)

## Analysis of Variance Table
## 
## Response: Response
##            Df  Sum Sq Mean Sq F value    Pr(>F)    
## Depth       3 2125.11  708.37 24.6628 1.652e-07 ***
## Feed        2 3160.50 1580.25 55.0184 1.086e-09 ***
## Depth:Feed  6  557.06   92.84  3.2324   0.01797 *  
## Residual   24  689.33   28.72                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Answer: At a significance level of 0.05, starting testing our hypotheses backwards we can see that the interaction between the two factors is significant. Also the main effects of both the factors are significant since p-values are very small.

Part B

Prepare appropriate residual plots and comment on the model’s adequacy.

plot(Model)

Answer: If we look at “Normal Q-Q” and “Residuals vs Factor Levels” plots, we can see that they satisfy the assumptions of normality and constant variance. Data is not perfect but the model can be reasonably considered adequate.

Part C

Obtain point estimates of the mean surface finish at each feed rate.

mean(Data$Response[1:12])

## [1] 81.58333

var(Data$Response[1:12])

## [1] 205.5379

mean(Data$Response[13:24])

## [1] 97.58333

var(Data$Response[13:24])

## [1] 64.08333

mean(Data$Response[25:36])

## [1] 103.8333

var(Data$Response[25:36])

## [1] 36.87879

Point estimates of mean surface finish at each feedrate alongwith the variance values have been calculated above.

Part D

Find the P-values for the tests in part (a).

P-value of Depth of Cut & Feedrate Interaction: 0.01797

P-value of Depth of Cut : 1.652e-07

P-value of Feedrate : 1.086e-09

Question 5.34

Reconsider the experiment in Problem 5.4. Suppose that this experiment had been conducted in three blocks, with each replicate a block. Assume that the observations in the data table are given in order, that is, the first observation in each cell comes from the first replicate, and so on. Reanalyze the data as a factorial experiment in blocks and estimate the variance component for blocks. Does it appear that blocking was useful in this experiment?

Reading Data

Feed <- c(rep(1,12), rep(2,12), rep(3,12))
Depth <- rep(seq(1,4),9)
Block <- c(rep(1,4), rep(2,4), rep(3,4), rep(1,4), rep(2,4), rep(3,4), rep(1,4), rep(2,4), rep(3,4))
Response <- c(74, 79, 82, 99, 64, 68, 88, 104, 60, 73, 92, 96, 92, 98, 99, 
              104, 86, 104, 108, 110, 88, 88, 95, 99, 99, 104, 108, 114, 
              98, 99, 110, 111, 102, 95, 99, 107)
Data <- data.frame(Depth, Feed, Block, Response)

Analysis

library(GAD)
Feed <- as.fixed(Feed)
Depth <- as.fixed(Depth)
Block <- as.fixed(Block)
Model <- aov(Response~Depth+Feed+Block+Depth*Feed)
summary(Model)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Depth        3 2125.1   708.4  30.637 4.89e-08 ***
## Feed         2 3160.5  1580.2  68.346 3.64e-10 ***
## Block        2  180.7    90.3   3.907  0.03532 *  
## Depth:Feed   6  557.1    92.8   4.015  0.00726 ** 
## Residuals   22  508.7    23.1                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Looking at the p-value of interaction between the two factors we conclude that there is an interaction between the factors. Also if we look at the p-values of the main effects of both the factors we conclude that both feedrate and depth of cut have a significant effect.

Variance component of Blocks:

VarBlock <- c((90.3 - 23.1)/(3*4))
VarBlock

## [1] 5.6

Variance component of Blocks: (MS of Blocks - MSE)/(i*j) = 5.6

If we look the p-value of blocks “0.03532”, we can see that it is less than the 0.05 level of significance and we can comprehend that blocking was effective in this scenario and it has a signficant effect.

Question 13.5

Suppose that in Problem 5.13 the furnace positions were randomly selected, resulting in a mixed model experiment. Reanalyze the data from this experiment under this new assumption. Estimate the appropriate model components using the ANOVA method.

Position <- c(rep(1,9), rep(2,9))
Temperature <- rep(seq(1,3),6)
Response <- c(570, 1063, 565, 565, 1080, 510, 583, 1043, 590, 528, 988, 526, 547, 1026,
              538, 521, 1004, 532)
Data <- data.frame(Position, Temperature, Response)

Analysis

library(GAD)
Position <- as.random(Position)
Temperature <- as.fixed(Temperature)
Model <- aov(Response~Position+Temperature+Position*Temperature)
GAD::gad(Model)

## Analysis of Variance Table
## 
## Response: Response
##                      Df Sum Sq Mean Sq  F value    Pr(>F)    
## Position              1   7160    7160   15.998 0.0017624 ** 
## Temperature           2 945342  472671 1155.518 0.0008647 ***
## Position:Temperature  2    818     409    0.914 0.4271101    
## Residual             12   5371     448                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Answer: From the results of the model, we comprehend that interaction between the two factors is not significant while the main effects of both the effects are significant.

Question 13.6

Reanalyze the measurement systems experiment in Problem 13.1, assuming that operators are a fixed factor.Estimate the appropriate model components using the ANOVA method.

Reading Data

Part <- c(rep(1,6), rep(2,6), rep(3,6), rep(4,6), rep(5,6), rep(6,6), rep(7,6), rep(8,6), rep(9,6), rep(10,6))
Operator <- c(rep(1,3), rep(2,3), rep(1,3), rep(2,3), rep(1,3), rep(2,3), rep(1,3),
              rep(2,3), rep(1,3), rep(2,3), rep(1,3), rep(2,3), rep(1,3), rep(2,3),
              rep(1,3), rep(2,3), rep(1,3), rep(2,3), rep(1,3), rep(2,3))

Response <- c(50, 49, 50, 50, 48, 51, 52, 52, 51, 51, 51, 51, 53, 50, 50, 54, 52, 51,
              49, 51, 50, 48, 50, 51, 48, 49, 48, 48, 49, 48, 52, 50, 50, 52, 50, 50,
              51, 51, 51, 51, 50, 50, 52, 50, 49, 53, 48, 50, 50, 51, 50, 51, 48, 49,
              47, 46, 49, 46, 47, 48)

Data <- data.frame(Part, Operator, Response)

Analysis

library(GAD)
Part <- as.random(Part)
Operator <- as.fixed(Operator)
Model <- aov(Response~Part+Operator+Part*Operator)
GAD::gad(Model)

## Analysis of Variance Table
## 
## Response: Response
##               Df Sum Sq Mean Sq F value    Pr(>F)    
## Part           9 99.017 11.0019  7.3346 3.216e-06 ***
## Operator       1  0.417  0.4167  0.6923    0.4269    
## Part:Operator  9  5.417  0.6019  0.4012    0.9270    
## Residual      40 60.000  1.5000                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

HW 10

Muneeb

11/1/2021

Question 5.4

An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut. He selects three feed rates and four depths of cut.

Reading Data

Part A

Analyze the data and draw conclusions. Use a = 0.05.

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_{i}\) + \(\beta_{j}\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Analysis

Answer: At a significance level of 0.05, starting testing our hypotheses backwards we can see that the interaction between the two factors is significant. Also the main effects of both the factors are significant since p-values are very small.

Part B

Prepare appropriate residual plots and comment on the model’s adequacy.

Answer: If we look at “Normal Q-Q” and “Residuals vs Factor Levels” plots, we can see that they satisfy the assumptions of normality and constant variance. Data is not perfect but the model can be reasonably considered adequate.

Part C

Obtain point estimates of the mean surface finish at each feed rate.

Point estimates of mean surface finish at each feedrate alongwith the variance values have been calculated above.

Part D

Find the P-values for the tests in part (a).

P-value of Depth of Cut & Feedrate Interaction: 0.01797

P-value of Depth of Cut : 1.652e-07

P-value of Feedrate : 1.086e-09

Question 5.34

Reading Data

Analysis

Looking at the p-value of interaction between the two factors we conclude that there is an interaction between the factors. Also if we look at the p-values of the main effects of both the factors we conclude that both feedrate and depth of cut have a significant effect.

Variance component of Blocks:

Variance component of Blocks: (MS of Blocks - MSE)/(i*j) = 5.6

If we look the p-value of blocks “0.03532”, we can see that it is less than the 0.05 level of significance and we can comprehend that blocking was effective in this scenario and it has a signficant effect.

Question 13.5

Suppose that in Problem 5.13 the furnace positions were randomly selected, resulting in a mixed model experiment. Reanalyze the data from this experiment under this new assumption. Estimate the appropriate model components using the ANOVA method.

Analysis

Answer: From the results of the model, we comprehend that interaction between the two factors is not significant while the main effects of both the effects are significant.

Question 13.6

Reanalyze the measurement systems experiment in Problem 13.1, assuming that operators are a fixed factor.Estimate the appropriate model components using the ANOVA method.

Reading Data

Analysis

Looking at the p-value of interaction, we can see that interaction between the factors is not significant.

From the p-values of main effects, we conclude that main effect of Operator is not significant while the main effect of Part is significant.