Question 5.4

(a) Analyze the data and draw conclusions. Use alpha 0.05.

Model Equation yijk=μ+αi+βj+αβij+ϵijk

library(GAD)

## Warning: package 'GAD' was built under R version 4.2.2

## Loading required package: matrixStats

## Loading required package: R.methodsS3

## R.methodsS3 v1.8.2 (2022-06-13 22:00:14 UTC) successfully loaded. See ?R.methodsS3 for help.

Feed_rate <- c(rep(1,12),rep(2,12),rep(3,12))
depth <- rep(seq(1,4),9)
Feed_rate <- as.fixed(Feed_rate)
depth <- as.fixed(depth)
observation <- 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)
Test1 <- aov(observation~Feed_rate+depth+Feed_rate*depth)
summary(Test1)

##                 Df Sum Sq Mean Sq F value   Pr(>F)    
## Feed_rate        2 3160.5  1580.2  55.018 1.09e-09 ***
## depth            3 2125.1   708.4  24.663 1.65e-07 ***
## Feed_rate:depth  6  557.1    92.8   3.232    0.018 *  
## Residuals       24  689.3    28.7                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

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

plot(fitted(Test1),residuals(Test1))

plot(Test1)

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

Feed_rate1 <- c(74,79,82,99,64,68,88,104,60,73,92,96)
Feed_rate2 <- c(92,98,99,104,86,104,108,110,88,88,95,99)
Feed_rate3 <- c(99,104,108,114,98,99,110,111,102,95,99,107)
Feed_rate1_mean <- mean(Feed_rate1)
Feed_rate2_mean <- mean(Feed_rate2)
Feed_rate3_mean <- mean(Feed_rate3)
Feed_rate1_mean

## [1] 81.58333

Feed_rate2_mean

## [1] 97.58333

Feed_rate3_mean

## [1] 103.8333

a) The p-value of interaction is less then level of signifance (0.05) Therefore we reject the Null Hypothesis.

b) The Residual plots we can conclude that our variance is not equal

C) point estimate of the mean surface finish at each feed rate

d) p-values

Feed rate = 1.09e-09

Depth = 1.65e-07

Interaction = 0.01797

Question 5.34

library(GAD)

observation <- c(74,79,82,99,92,98,99,104,99,104,108,114,64,68,88,104,86,104,108,110,98,99,110,111,60,73,92,96,88,88,95,99,102,95,99,107)

block <- c(rep(1,12),rep(2,12),rep(3,12))
block

##  [1] 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3

Feed_rate <- c(rep(0.20,4),rep(0.25,4),rep(0.30,4))
Feed_rate

##  [1] 0.20 0.20 0.20 0.20 0.25 0.25 0.25 0.25 0.30 0.30 0.30 0.30

Feed_rate<- rep(Feed_rate,3)
Feed_rate

##  [1] 0.20 0.20 0.20 0.20 0.25 0.25 0.25 0.25 0.30 0.30 0.30 0.30 0.20 0.20 0.20
## [16] 0.20 0.25 0.25 0.25 0.25 0.30 0.30 0.30 0.30 0.20 0.20 0.20 0.20 0.25 0.25
## [31] 0.25 0.25 0.30 0.30 0.30 0.30

depth<- rep(seq(1,4),9)
depth

##  [1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

block <- as.fixed(block)
Feed_rate<- as.fixed(Feed_rate)
depth <- as.fixed(depth)
Test2<- lm(observation~block+Feed_rate*depth)
gad(Test2)

## Analysis of Variance Table
## 
## Response: observation
##                 Df  Sum Sq Mean Sq F value    Pr(>F)    
## block            2  180.67   90.33  3.9069  0.035322 *  
## Feed_rate        2 3160.50 1580.25 68.3463 3.635e-10 ***
## depth            3 2125.11  708.37 30.6373 4.893e-08 ***
## Feed_rate:depth  6  557.06   92.84  4.0155  0.007258 ** 
## Residual        22  508.67   23.12                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(Test2)

block_1 <- c(74,79,82,99,92,98,99,104,99,104,108,114)
block_2 <- c(64,68,88,104,86,104,108,110,98,99,110,111)
block_3 <- c(60,73,92,96,88,88,95,99,102,95,99,107)
block_1 <- var(block_1)
block_2 <- var(block_2)
block_3 <- var(block_3)
block_1

## [1] 146.5455

block_2

## [1] 261.2424

block_3

## [1] 169.6061

The p value of the 0.007258 < level of significance = 0.05

we have sufficient evidence to reject the null hypothesis

Question 13.5

position <- c(rep(1,9),rep(2,9))
temperature <- rep(c("800","825","850"),6)
observation <- c(570,1063,565,565,1080,510,583,1043,590,528,988,526,547,1026,538,521,1004,532)
data<-data.frame(position,temperature,observation)

library(GAD)


position <- as.random(position)
temperature <- as.fixed(temperature)
Test3 <- aov(observation~position+temperature+position*temperature)
GAD::gad(Test3)

## Analysis of Variance Table
## 
## Response: observation
##                      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

The p value for position is 0.00176 < 0.05 level of signifance

The temperature p value 0.0008647< 0.05 level of signifance

Question 13.6

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))
Operators<-rep(c(rep(1,3), rep(2,3)), 10)
observation<-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)

data1<-data.frame(partno,Operators, observation)

library(GAD)

partno <-as.random(partno)
Operators<-as.fixed(Operators)
Test4<-aov(observation~partno+Operators+partno*Operators)
gad(Test4)

## Analysis of Variance Table
## 
## Response: observation
##                  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

The p value for the Part Number 3.216e-06 < 0.05 level of signifance so the Part Number is significant effect.

The p value for the Operators of 0.4269 > 0.05 level of signifance so the Operator does not have a significant effect.

Homework 10

Rishidhar, Hemanth

2022-11-06

Question 5.4

(a) Analyze the data and draw conclusions. Use alpha 0.05.

Model Equation yijk=μ+αi+βj+αβij+ϵijk

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

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

a) The p-value of interaction is less then level of signifance (0.05) Therefore we reject the Null Hypothesis.

b) The Residual plots we can conclude that our variance is not equal

C) point estimate of the mean surface finish at each feed rate

d) p-values

Feed rate = 1.09e-09

Depth = 1.65e-07

Interaction = 0.01797

Question 5.34

The p value of the 0.007258 < level of significance = 0.05

we have sufficient evidence to reject the null hypothesis

Question 13.5

The p value for position is 0.00176 < 0.05 level of signifance

The temperature p value 0.0008647< 0.05 level of signifance

Question 13.6

The p value for the Part Number 3.216e-06 < 0.05 level of signifance so the Part Number is significant effect.

The p value for the Operators of 0.4269 > 0.05 level of signifance so the Operator does not have a significant effect.