DOE - HW - Week 10

Answer 5.4

FeedRate <- c(rep("0.20",12), rep("0.25",12), rep("0.30",12))
Depth <- c("0.15","0.18","0.20","0.25")
Cut <- c(rep(Depth,9))
Data <- 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)
dat5.4 <- data.frame(FeedRate,Cut,Data)
dat5.4$FeedRate <- as.fixed(dat5.4$FeedRate)
dat5.4$Cut <-  as.fixed(dat5.4$Cut)

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\alpha\beta_{ij} = 0\)

Alternative Hypothesis: \(\alpha\beta_{ij} \neq 0\)

Main Effect:

Null Hypothesis: \(\alpha_i = 0\) For all i

Alternative Hypothesis: \(\alpha_i \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\)

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

dat.model <- aov(Data~FeedRate+Cut+FeedRate*Cut, data=dat5.4)
GAD::gad(dat.model)

## Analysis of Variance Table
## 
## Response: Data
##              Df  Sum Sq Mean Sq F value    Pr(>F)    
## FeedRate      2 3160.50 1580.25 55.0184 1.086e-09 ***
## Cut           3 2125.11  708.37 24.6628 1.652e-07 ***
## FeedRate:Cut  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

(a.) Therefore we see that the interaction p value = 0.01797 < alpha = 0.05, we reject the null hypothesis and claim that there is an interaction between the two factors.

And for depth of cut, p = 1.652e-07 and Also, for feed rate, p = 1.086e-09 < 0.05, both also have an effect.

interaction.plot(x.factor=FeedRate, trace.factor=Cut, response=Data)

Above is the interaction plot.

plot(dat.model)

(b.) The plots seem normally distributed, also the plot for residuals vs fitted shows that varaince are constant.

mean(dat5.4$Data[1:12])

## [1] 81.58333

var(dat5.4$Data[1:12])

## [1] 205.5379

mean(dat5.4$Data[13:24])

## [1] 97.58333

var(dat5.4$Data[13:24])

## [1] 64.08333

mean(dat5.4$Data[25:36])

## [1] 103.8333

var(dat5.4$Data[25:36])

## [1] 36.87879

(c.) Point Estimates:

For Feed Rate 0.20 = The Mean is 81.58333 & Variance is 205.5379

For Feed Rate 0.25 = The Mean is 97.58333 & Variance is 64.08333

For Feed Rate 0.30 = The Mean is 103.8333 & Variance is 36.87879

(d.) P values:

For Interaction of Factors p = 1.086e-09

For Depth of Cut main effect p =1.652e-07

For Feed Rate main effect p =0.01797

Answer 5.34

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\gamma_k\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\alpha\beta_{ij} = 0\)

Alternative Hypothesis: \(\alpha\beta_{ij} \neq 0\)

Main Effect:

Null Hypothesis: \(\alpha_i = 0\) For all i

Alternative Hypothesis: \(\alpha_i \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\)

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

FeedRate <- c(rep("0.20",12), rep("0.25",12), rep("0.30",12))
Depth <- c("0.15","0.18","0.20","0.25")
Cut <- c(rep(Depth,9))
Depthn <- c(rep("1",4),rep("2",4),rep("3",4))
Block <- c(rep(Depthn,3))
Data <- 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)
datn <- data.frame(FeedRate, Cut, Block, Data)
datn$FeedRate <- as.fixed(dat5.4$FeedRate)
datn$Block <- as.factor(datn$Block)
datn$Cut <-  as.fixed(dat5.4$Cut)

dat.modeln <- aov(Data~FeedRate+Cut+Block+FeedRate*Cut, data=datn)
summary(dat.modeln)

##              Df Sum Sq Mean Sq F value   Pr(>F)    
## FeedRate      2 3160.5  1580.2  68.346 3.64e-10 ***
## Cut           3 2125.1   708.4  30.637 4.89e-08 ***
## Block         2  180.7    90.3   3.907  0.03532 *  
## FeedRate:Cut  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

Therefore we see that the interaction p value = 0.00726 < alpha = 0.05, we reject the null hypothesis and claim that there is an interaction between the two factors.

And for depth of cut, p = 4.89e-08 and Also, for feed rate, p = 3.64e-10 < 0.05, both also have an effect.

The P value of the Block is = 0.03532 < 0.05, hence we conclude that it has an effect, but in both the cases we do reject the null hypothesis.

interaction.plot(x.factor=FeedRate, trace.factor=Cut, response=Data)

Above is the interaction plot.

Answer 13.5

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\sigma^2_{\alpha\beta} = 0\)

Alternative Hypothesis: \(\sigma^2_{\alpha\beta} \neq 0\)

Main Effect:

Null Hypothesis: \(\sigma^2_\alpha = 0\)

Alternative Hypothesis: \(\sigma^2_\alpha \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\) For all j

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

FurnacePosition <- c(rep(1,9),rep(2,9))
pos <- c("800","825","850")
Temperatures <- c(rep(pos,6))
Furnacepos <- c(570, 1063, 565, 565, 1080, 510, 583, 1043, 590, 528, 988, 526, 547, 1026, 538, 521, 1004, 532)
Dat13.5 <- data.frame(FurnacePosition,Temperatures,Furnacepos)
Dat13.5$Temperatures <- as.fixed(Dat13.5$Temperatures)
Dat13.5$FurnacePosition <- as.random(Dat13.5$FurnacePosition)

dat.model13.5 <- aov(Furnacepos~FurnacePosition+Temperatures+FurnacePosition*Temperatures, data = Dat13.5)
GAD::gad(dat.model13.5)

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

Therefore we see that the interaction p value = 0.4271101 > alpha = 0.05, we fail to reject the null hypothesis and claim that there is no interaction between the two factors.

And for Temperature, p = 0.0008647 and Also, for Furnace Position, p = 0.0017624 < 0.05, both also have an effect.

Answer 13.6

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\sigma^2_{\alpha\beta} = 0\)

Alternative Hypothesis: \(\sigma^2_{\alpha\beta} \neq 0\)

Main Effect:

Null Hypothesis: \(\sigma^2_\alpha = 0\)

Alternative Hypothesis: \(\sigma^2_\alpha \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\) For all j

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

PartNo <- 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))
Measurements <- c("1","1","1","2","2","2")
Operators <- c(rep(Measurements, 10))
Datas <- 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)
Dat13.6 <- data.frame(PartNo,Operators,Datas)
Dat13.6$Operators <- as.fixed(Dat13.6$Operators)
Dat13.6$PartNo <- as.random(Dat13.6$PartNo)

dat.model13.6 <-  aov(Datas~PartNo+Operators+PartNo*Operators, data = Dat13.6)
GAD::gad(dat.model13.6)

## Analysis of Variance Table
## 
## Response: Datas
##                  Df Sum Sq Mean Sq F value    Pr(>F)    
## PartNo            9 99.017 11.0019  7.3346 3.216e-06 ***
## Operators         1  0.417  0.4167  0.6923    0.4269    
## PartNo:Operators  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

Therefore we see that the interaction p value = 0.9270 > alpha = 0.05, we fail to reject the null hypothesis and claim that there is no interaction between the two factors.

DOE - HW - Week 10

Tajammul Siddiq Mohammed

11/7/2021

Answer 5.4

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\alpha\beta_{ij} = 0\)

Alternative Hypothesis: \(\alpha\beta_{ij} \neq 0\)

Main Effect:

Null Hypothesis: \(\alpha_i = 0\) For all i

Alternative Hypothesis: \(\alpha_i \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\)

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

(a.) Therefore we see that the interaction p value = 0.01797 < alpha = 0.05, we reject the null hypothesis and claim that there is an interaction between the two factors.

And for depth of cut, p = 1.652e-07 and Also, for feed rate, p = 1.086e-09 < 0.05, both also have an effect.

Above is the interaction plot.

(b.) The plots seem normally distributed, also the plot for residuals vs fitted shows that varaince are constant.

(c.) Point Estimates:

For Feed Rate 0.20 = The Mean is 81.58333 & Variance is 205.5379

For Feed Rate 0.25 = The Mean is 97.58333 & Variance is 64.08333

For Feed Rate 0.30 = The Mean is 103.8333 & Variance is 36.87879

(d.) P values:

For Interaction of Factors p = 1.086e-09

For Depth of Cut main effect p =1.652e-07

For Feed Rate main effect p =0.01797

Answer 5.34

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\gamma_k\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\alpha\beta_{ij} = 0\)

Alternative Hypothesis: \(\alpha\beta_{ij} \neq 0\)

Main Effect:

Null Hypothesis: \(\alpha_i = 0\) For all i

Alternative Hypothesis: \(\alpha_i \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\)

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

Therefore we see that the interaction p value = 0.00726 < alpha = 0.05, we reject the null hypothesis and claim that there is an interaction between the two factors.

And for depth of cut, p = 4.89e-08 and Also, for feed rate, p = 3.64e-10 < 0.05, both also have an effect.

The P value of the Block is = 0.03532 < 0.05, hence we conclude that it has an effect, but in both the cases we do reject the null hypothesis.

Above is the interaction plot.

Answer 13.5

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\sigma^2_{\alpha\beta} = 0\)

Alternative Hypothesis: \(\sigma^2_{\alpha\beta} \neq 0\)

Main Effect:

Null Hypothesis: \(\sigma^2_\alpha = 0\)

Alternative Hypothesis: \(\sigma^2_\alpha \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\) For all j

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

Therefore we see that the interaction p value = 0.4271101 > alpha = 0.05, we fail to reject the null hypothesis and claim that there is no interaction between the two factors.

And for Temperature, p = 0.0008647 and Also, for Furnace Position, p = 0.0017624 < 0.05, both also have an effect.

Answer 13.6

Model Equation:

\(Y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{ij}\) + \(\epsilon_{ijk}\)

Stating Hypothesis:

Interaction:

Null Hypothesis: \(\sigma^2_{\alpha\beta} = 0\)

Alternative Hypothesis: \(\sigma^2_{\alpha\beta} \neq 0\)

Main Effect:

Null Hypothesis: \(\sigma^2_\alpha = 0\)

Alternative Hypothesis: \(\sigma^2_\alpha \neq 0\) for some i

Null Hypothesis: \(\beta_j = 0\) For all j

Alternative Hypothesis: \(\beta_j \neq 0\) for some j

Therefore we see that the interaction p value = 0.9270 > alpha = 0.05, we fail to reject the null hypothesis and claim that there is no interaction between the two factors.

And for Operators, p = 0.4269 > 0.05, it does not have an effect.

And also, for PartNo, p = 3.216e-06 < 0.05, it has an effect.