DOE - HW - Week 6

knitr::opts_chunk$set(echo = TRUE)
library(tidyr)
library(agricolae)
library(MASS)

Answer 3.23

Entering Data

FT1 <- c(17.6, 18.9, 16.3, 17.4, 20.1, 21.6)
FT2 <- c(16.9, 15.3, 18.6, 17.1, 19.5, 20.3)
FT3 <- c(21.4, 23.6, 19.4, 18.5, 20.5, 22.3)
FT4 <- c(19.3, 21.1, 16.9, 17.5, 18.3, 19.8)

FT <- cbind.data.frame(FT1, FT2, FT3, FT4)
FTs <- pivot_longer (data = FT, c(FT1, FT2, FT3, FT4))

FTs$name <- as.factor(FTs$name)
FTs$value <- as.numeric(FTs$value)

3.23 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)
Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

dat1 <- aov(value~name, data = FTs)
summary(dat1)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## name         3  30.16   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the above ANOVA test we see that the P-value = 0.0525 which is > alpha = 0.05. Hence we fail to reject the Null Hypothesis. This indicates that the mean life for all fluid types does not differ.

3.23 (b).

LSD.test(dat1,"name", alpha = 0.05, console = TRUE)

## 
## Study: dat1 ~ "name"
## 
## LSD t Test for value 
## 
## Mean Square Error:  3.299667 
## 
## name,  means and individual ( 95 %) CI
## 
##        value      std r      LCL      UCL  Min  Max
## FT1 18.65000 1.952178 6 17.10309 20.19691 16.3 21.6
## FT2 17.95000 1.854454 6 16.40309 19.49691 15.3 20.3
## FT3 20.95000 1.879096 6 19.40309 22.49691 18.5 23.6
## FT4 18.81667 1.554885 6 17.26975 20.36358 16.9 21.1
## 
## Alpha: 0.05 ; DF Error: 20
## Critical Value of t: 2.085963 
## 
## least Significant Difference: 2.187666 
## 
## Treatments with the same letter are not significantly different.
## 
##        value groups
## FT3 20.95000      a
## FT4 18.81667     ab
## FT1 18.65000      b
## FT2 17.95000      b

As fluid type 3 has the higest average life, it should be selected if the objective is long life.

3.23 (c).

plot(dat1)

As shown in the above plots, we can conclude that the data is normally distributed. Also, the residual vs fitted value plot shows that the variance is constant.

Answer 3.28

Entering Data

M1 <- c(110, 157, 194, 178)
M2 <- c(1, 2, 4, 18)
M3 <- c(880, 1256, 5276, 4355)
M4 <- c(495, 7040, 5307, 10050)
M5 <- c(7, 5, 29, 2)
MFT <- cbind.data.frame(M1, M2, M3, M4, M5)
MFTs <- pivot_longer (data = MFT, c(M1, M2, M3, M4, M5))

MFTs$name <- as.factor(MFTs$name)
MFTs$value <- as.numeric(MFTs$value)

3.28 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)
Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

dat2 <- aov(value~name, data = MFTs)
summary(dat2)

##             Df    Sum Sq  Mean Sq F value  Pr(>F)   
## name         4 103191489 25797872   6.191 0.00379 **
## Residuals   15  62505657  4167044                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the above ANOVA test we see that the P-value = 0.00379 which is < alpha = 0.05. Hence we reject the Null Hypothesis. This indicates that atleast one of the mean differs.

3.28 (b).

plot(dat2)

As per the above plots, the data follows a S-shape and is not normally distributed. Also, the assumption of constant variance is not satisfied either as there is a significant difference in the data widths between populations. Hence, we need to transform the data using boxcox.

3.28 (c).

boxcox(dat2)

The value of lambda lies very close to zero. Hence we need to do a log transformation of the data.

MFTs$value <- log(MFTs$value)
dat2<-aov(value~name, data = MFTs)
summary(dat2)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## name         4 165.06   41.26   37.66 1.18e-07 ***
## Residuals   15  16.44    1.10                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(dat2)

Therefore, now we can see that the data is normally distributed and the residual vs fitted value plot shows that the variance is constant.From the ANOVA test we see that the P-value = 1.18e-07 which is < alpha = 0.05. Hence we reject the Null Hypothesis and conclude that atleast one of the mean differs.

Answer 3.29

Entering Data

Me1 <- c(31, 10, 21, 4, 1)
Me2 <- c(62, 40, 24, 30, 35)
Me3 <- c(53, 27, 120, 97, 68)
MeC <- cbind.data.frame(Me1, Me2, Me3)
MeCs <- pivot_longer (data = MeC, c(Me1, Me2, Me3))

MeCs$name <- as.factor(MeCs$name)
MeCs$value <- as.numeric(MeCs$value)

3.29 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)
Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

dat3 <- aov(value~name, data = MeCs)
summary(dat3)

##             Df Sum Sq Mean Sq F value  Pr(>F)   
## name         2   8964    4482   7.914 0.00643 **
## Residuals   12   6796     566                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the above ANOVA test we see that the P-value = 0.00643 which is < alpha = 0.05. Hence we reject the Null Hypothesis and state that at least one of the methods has a different effect on the mean particle count.

3.29 (b).

plot(dat3)

As per the above plots, the data is not normally distributed. Also, the assumption of constant variance is not satisfied either as there is a significant difference in the data widths between populations. Hence, we need to transform the data using boxcox.

3.29 (c).

boxcox(dat3)

The value of lambda is approximately = 0.4

MeCs$value <- (MeCs$value)^0.4
dat3a <- aov(value~name, data = MeCs)
summary(dat3a)

##             Df Sum Sq Mean Sq F value  Pr(>F)   
## name         2  21.21  10.605   9.881 0.00291 **
## Residuals   12  12.88   1.073                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(dat3a)

Therefore, now we can see that the data is normally distributed and the residual vs fitted value plot shows that the variance is constant.From the ANOVA test we see that the P-value = 0.00291 which is < alpha = 0.05. Hence we reject the Null Hypothesis and conclude that atleast one of the mean differs.

Answer 3.51

kruskal.test(value~name, data = FTs)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  value by name
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015

We can see that P-value = 0.1015 is > than alpha = 0.05 hence we accept the Null Hypothesis and conclude that fluid does not differ. Also, the p-value obtained in the Kruskal-Wallis test is larger as compared to the one from the ANOVA test as it does not assume normality. Overall, our conclusion is same in both the methods.

DOE - HW - Week 6

Tajammul Siddiq Mohammed

10/10/2021

Answer 3.23

Entering Data

3.23 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)

Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

From the above ANOVA test we see that the P-value = 0.0525 which is > alpha = 0.05. Hence we fail to reject the Null Hypothesis. This indicates that the mean life for all fluid types does not differ.

3.23 (b).

As fluid type 3 has the higest average life, it should be selected if the objective is long life.

3.23 (c).

As shown in the above plots, we can conclude that the data is normally distributed. Also, the residual vs fitted value plot shows that the variance is constant.

Answer 3.28

Entering Data

3.28 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)

Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

From the above ANOVA test we see that the P-value = 0.00379 which is < alpha = 0.05. Hence we reject the Null Hypothesis. This indicates that atleast one of the mean differs.

3.28 (b).

As per the above plots, the data follows a S-shape and is not normally distributed. Also, the assumption of constant variance is not satisfied either as there is a significant difference in the data widths between populations. Hence, we need to transform the data using boxcox.

3.28 (c).

The value of lambda lies very close to zero. Hence we need to do a log transformation of the data.

Answer 3.29

Entering Data

3.29 (a).

Stating the Hypothesis:

Null Hypothesis: \(H_o: \mu_1= \mu_2= \mu_3= \mu_4 = \mu_i\)

Alternative Hypothesis: \(H_a\): At least one of the \(\mu_i\) differs

From the above ANOVA test we see that the P-value = 0.00643 which is < alpha = 0.05. Hence we reject the Null Hypothesis and state that at least one of the methods has a different effect on the mean particle count.

3.29 (b).

As per the above plots, the data is not normally distributed. Also, the assumption of constant variance is not satisfied either as there is a significant difference in the data widths between populations. Hence, we need to transform the data using boxcox.

3.29 (c).

The value of lambda is approximately = 0.4

Answer 3.51