HW-Module 4
Juan Carlos Soto
2024-10-11
Problem 3.23
# Input the data manually
Fluid_Type <- factor(rep(1:4, each = 6))
Life <- c(17.6, 18.9, 16.3, 17.4, 20.1, 21.6, # Fluid Type 1
16.9, 15.3, 18.6, 17.1, 19.5, 20.3, # Fluid Type 2
21.4, 23.6, 19.4, 18.5, 20.5, 22.3, # Fluid Type 3
19.3, 21.1, 16.9, 17.5, 18.3, 19.8) # Fluid Type 4
# Combine into a data frame
data <- data.frame(Fluid_Type, Life)
# Display the first few rows of data
head(data)## Fluid_Type Life
## 1 1 17.6
## 2 1 18.9
## 3 1 16.3
## 4 1 17.4
## 5 1 20.1
## 6 1 21.6Perform ANOVA
# Perform ANOVA
model <- aov(Life ~ Fluid_Type, data = data)
# Display the summary of the ANOVA
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## Fluid_Type 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 ' ' 1Answer for part (a)
While the result is borderline, with a p-value of 0.0525, we would fail to reject the null hypothesis at the 0.05 significance level. This means that, based on this analysis, we cannot say with confidence that the fluid types differ in terms of life expectancy.
# Check p-value
if (summary(model)[[1]][["Pr(>F)"]][1] < 0.05) {
cat("There is a significant difference between the fluids.")
} else {
cat("There is no significant difference between the fluids.")
}## There is no significant difference between the fluids.Part (b)
# Calculate the mean life for each fluid type
mean_life <- aggregate(Life ~ Fluid_Type, data, mean)
# Display the mean life for each fluid type
mean_life## Fluid_Type Life
## 1 1 18.65000
## 2 2 17.95000
## 3 3 20.95000
## 4 4 18.81667# Identify the fluid with the longest average life
best_fluid <- mean_life[which.max(mean_life$Life), ]
cat("The fluid with the longest life is Fluid Type", best_fluid$Fluid_Type, "with an average life of", best_fluid$Life, "hours.")## The fluid with the longest life is Fluid Type 3 with an average life of 20.95 hours.part (c)
# Perform Kruskal-Wallis test
kruskal_test <- kruskal.test(Life ~ Fluid_Type, data = data)
# Display the result of the test
kruskal_test##
## Kruskal-Wallis rank sum test
##
## data: Life by Fluid_Type
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015interpretation
Since the the p-value >0.05, we fail to reject the null hypothesis and conclude that there is no significant between fluid types.
# Extract p-value
p_value <- kruskal_test$p.value
if (p_value < 0.05) {
cat("There is a significant difference between the fluid types (p-value =", p_value, ").\n")
} else {
cat("There is no significant difference between the fluid types (p-value =", p_value, ").\n")
}## There is no significant difference between the fluid types (p-value = 0.1014857 ).Problem 3.28
# Input the data
Material <- factor(rep(1:5, each = 4))
Failure_Time <- c(110, 157, 194, 178, # Material 1
1, 2, 4, 18, # Material 2
880, 1256, 5276, 4355, # Material 3
495, 7040, 5307, 10050, # Material 4
7, 5, 29, 2) # Material 5
# Combine into a data frame
data <- data.frame(Material, Failure_Time)
# Display the first few rows of data
head(data)## Material Failure_Time
## 1 1 110
## 2 1 157
## 3 1 194
## 4 1 178
## 5 2 1
## 6 2 2perform ANOVA
# Perform ANOVA
model <- aov(Failure_Time ~ Material, data = data)
# Display the ANOVA summary
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## Material 4 103191489 25797872 6.191 0.00379 **
## Residuals 15 62505657 4167044
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# Extract p-value
p_value <- summary(model)[[1]][["Pr(>F)"]][1]
if (p_value < 0.05) {
cat("There is a significant difference between the materials (p-value =", p_value, ").\n")
} else {
cat("There is no significant difference between the materials (p-value =", p_value, ").\n")
}## There is a significant difference between the materials (p-value = 0.003785956 ).part (b): Plot residuals vs predicted and Q-Q plot
# Extract residuals and fitted values
residuals <- model$residuals
fitted_values <- model$fitted.values
# Residuals vs Fitted Plot
plot(fitted_values, residuals,
xlab = "Fitted Values", ylab = "Residuals",
main = "Residuals vs Fitted Values")
abline(h = 0, col = "red")
# Q-Q Plot
qqnorm(residuals)
qqline(residuals)
Part (c)
# Perform Kruskal-Wallis test
kruskal_test <- kruskal.test(Failure_Time ~ Material, data = data)
# Display the result of the test
kruskal_test##
## Kruskal-Wallis rank sum test
##
## data: Failure_Time by Material
## Kruskal-Wallis chi-squared = 16.873, df = 4, p-value = 0.002046# Interpretation
if (kruskal_test$p.value < 0.05) {
cat("Based on the Kruskal-Wallis test, there is a significant difference between the materials (p-value =", kruskal_test$p.value, ").\n")
} else {
cat("Based on the Kruskal-Wallis test, there is no significant difference between the materials (p-value =", kruskal_test$p.value, ").\n")
}## Based on the Kruskal-Wallis test, there is a significant difference between the materials (p-value = 0.002045539 ).###Problem 3.29 # part (a)
# Input the data
Method <- factor(rep(1:3, each = 5))
Count <- c(31, 10, 21, 4, 1, # Method 1
62, 40, 24, 30, 35, # Method 2
53, 27, 120, 97, 68) # Method 3
# Combine into a data frame
data <- data.frame(Method, Count)
# Display the first few rows of data
head(data)## Method Count
## 1 1 31
## 2 1 10
## 3 1 21
## 4 1 4
## 5 1 1
## 6 2 62Performa ANOVA
# Perform ANOVA
model <- aov(Count ~ Method, data = data)
# Display the ANOVA summary
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## Method 2 8964 4482 7.914 0.00643 **
## Residuals 12 6796 566
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# Extract p-value
p_value <- summary(model)[[1]][["Pr(>F)"]][1]
if (p_value < 0.05) {
cat("There is a significant difference between the methods (p-value =", p_value, ").\n")
} else {
cat("There is no significant difference between the methods (p-value =", p_value, ").\n")
}## There is a significant difference between the methods (p-value = 0.006430238 ).#part (b) Residual Analysis
# Extract residuals and fitted values
residuals <- model$residuals
fitted_values <- model$fitted.values
# Residuals vs Fitted Plot
plot(fitted_values, residuals,
xlab = "Fitted Values", ylab = "Residuals",
main = "Residuals vs Fitted Values")
abline(h = 0, col = "blue")
# Q-Q Plot
qqnorm(residuals)
qqline(residuals)
# part (C)
# Perform Kruskal-Wallis test
kruskal_test <- kruskal.test(Count ~ Method, data = data)
# Display the result of the test
kruskal_test##
## Kruskal-Wallis rank sum test
##
## data: Count by Method
## Kruskal-Wallis chi-squared = 8.54, df = 2, p-value = 0.01398# Interpretation
if (kruskal_test$p.value < 0.05) {
cat("Based on the Kruskal-Wallis test, there is a significant difference between the methods (p-value =", kruskal_test$p.value, ").\n")
} else {
cat("Based on the Kruskal-Wallis test, there is no significant difference between the methods (p-value =", kruskal_test$p.value, ").\n")
}## Based on the Kruskal-Wallis test, there is a significant difference between the methods (p-value = 0.01398178 ).Problem 3.51
# Input the data
Fluid_Type <- factor(rep(1:4, each = 6))
Life <- c(17.6, 18.9, 16.3, 17.4, 20.1, 21.6, # Fluid Type 1
16.9, 15.3, 18.6, 17.1, 19.5, 20.3, # Fluid Type 2
21.4, 23.6, 19.4, 18.5, 20.5, 22.3, # Fluid Type 3
19.3, 21.1, 16.9, 17.5, 18.3, 19.8) # Fluid Type 4
# Combine into a data frame
data <- data.frame(Fluid_Type, Life)
# Display the first few rows of data
head(data)## Fluid_Type Life
## 1 1 17.6
## 2 1 18.9
## 3 1 16.3
## 4 1 17.4
## 5 1 20.1
## 6 1 21.6# Perform the Kruskal-Wallis test
kruskal_test <- kruskal.test(Life ~ Fluid_Type, data = data)
# Display the result of the test
kruskal_test##
## Kruskal-Wallis rank sum test
##
## data: Life by Fluid_Type
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015# Extract the p-value
p_value <- kruskal_test$p.value
if (p_value < 0.05) {
cat("There is a significant difference between the fluid types (p-value =", p_value, ").\n")
} else {
cat("There is no significant difference between the fluid types (p-value =", p_value, ").\n")
}## There is no significant difference between the fluid types (p-value = 0.1014857 ).# Perform the ANOVA test for comparison
anova_model <- aov(Life ~ Fluid_Type, data = data)
# Display the ANOVA summary
summary(anova_model)## Df Sum Sq Mean Sq F value Pr(>F)
## Fluid_Type 3 30.16 10.05 3.047 0.0525 .
## Residuals 20 65.99 3.30
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Problem 3.52
# Input the data
Fluid_Type <- factor(rep(1:4, each = 6))
Life <- c(17.6, 18.9, 16.3, 17.4, 20.1, 21.6, # Fluid Type 1
16.9, 15.3, 18.6, 17.1, 19.5, 20.3, # Fluid Type 2
21.4, 23.6, 19.4, 18.5, 20.5, 22.3, # Fluid Type 3
19.3, 21.1, 16.9, 17.5, 18.3, 19.8) # Fluid Type 4
# Combine into a data frame
data <- data.frame(Fluid_Type, Life)
# Display the first few rows of data
head(data)## Fluid_Type Life
## 1 1 17.6
## 2 1 18.9
## 3 1 16.3
## 4 1 17.4
## 5 1 20.1
## 6 1 21.6# Perform the Kruskal-Wallis test
kruskal_test <- kruskal.test(Life ~ Fluid_Type, data = data)
# Display the result of the test
kruskal_test##
## Kruskal-Wallis rank sum test
##
## data: Life by Fluid_Type
## Kruskal-Wallis chi-squared = 6.2177, df = 3, p-value = 0.1015# Extract p-value from Kruskal-Wallis test
kruskal_p_value <- kruskal_test$p.value
if (kruskal_p_value < 0.05) {
cat("There is a significant difference between the fluid types based on the Kruskal-Wallis test (p-value =", kruskal_p_value, ").\n")
} else {
cat("There is no significant difference between the fluid types based on the Kruskal-Wallis test (p-value =", kruskal_p_value, ").\n")
}## There is no significant difference between the fluid types based on the Kruskal-Wallis test (p-value = 0.1014857 ).#Perform ANOVA (for Comparison)
# Perform ANOVA
anova_model <- aov(Life ~ Fluid_Type, data = data)
# Display the ANOVA summary
summary(anova_model)## Df Sum Sq Mean Sq F value Pr(>F)
## Fluid_Type 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# Extract p-value from ANOVA
anova_p_value <- summary(anova_model)[[1]][["Pr(>F)"]][1]
if (anova_p_value < 0.05) {
cat("There is a significant difference between the fluid types based on ANOVA (p-value =", anova_p_value, ").\n")
} else {
cat("There is no significant difference between the fluid types based on ANOVA (p-value =", anova_p_value, ").\n")
}## There is no significant difference between the fluid types based on ANOVA (p-value = 0.05246316 ).Compare results
if (kruskal_p_value < 0.05 & anova_p_value < 0.05) {
cat("Both the Kruskal-Wallis test and ANOVA conclude that there is a significant difference between the fluid types.\n")
} else if (kruskal_p_value >= 0.05 & anova_p_value >= 0.05) {
cat("Both the Kruskal-Wallis test and ANOVA conclude that there is no significant difference between the fluid types.\n")
} else {
cat("The results from the Kruskal-Wallis test and ANOVA differ, suggesting that the assumptions of ANOVA may not be met.\n")
}## Both the Kruskal-Wallis test and ANOVA conclude that there is no significant difference between the fluid types.Problem 4.3
Input data
# Data: Tensile Strength of Cloth by Chemical and Bolt
Chemical <- factor(rep(1:4, each = 5))
Bolt <- factor(rep(1:5, times = 4))
Strength <- c(73, 68, 74, 71, 67, # Chemical 1
73, 67, 75, 72, 70, # Chemical 2
75, 68, 78, 73, 68, # Chemical 3
73, 71, 75, 75, 69) # Chemical 4
# Combine into a data frame
data <- data.frame(Chemical, Bolt, Strength)
# Display the first few rows of data
head(data)## Chemical Bolt Strength
## 1 1 1 73
## 2 1 2 68
## 3 1 3 74
## 4 1 4 71
## 5 1 5 67
## 6 2 1 73Perform Two-way Anova(randomized block design)
# Perform two-way ANOVA
model <- aov(Strength ~ Chemical + Bolt, data = data)
# Display the ANOVA summary
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## Chemical 3 12.95 4.32 2.376 0.121
## Bolt 4 157.00 39.25 21.606 2.06e-05 ***
## Residuals 12 21.80 1.82
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Interpretations
# Extract the p-value for Chemical factor
p_value_chemical <- summary(model)[[1]][["Pr(>F)"]][1]
if (p_value_chemical < 0.05) {
cat("There is a significant effect of the chemicals on the tensile strength (p-value =", p_value_chemical, ").\n")
} else {
cat("There is no significant effect of the chemicals on the tensile strength (p-value =", p_value_chemical, ").\n")
}## There is no significant effect of the chemicals on the tensile strength (p-value = 0.1211445 ).Check assumptions of ANOVA
# Extract residuals and fitted values
residuals <- model$residuals
fitted_values <- model$fitted.values
# Residuals vs Fitted Values Plot
plot(fitted_values, residuals,
xlab = "Fitted Values", ylab = "Residuals",
main = "Residuals vs Fitted Values")
abline(h = 0, col = "red")
#Normal Q-Q plot
# Q-Q Plot for normality of residuals
qqnorm(residuals)
qqline(residuals)
# Conclusion The p-value for the Chemical factor is 0.1211, which is
greater than the significant level of 0.05. Therefore, we fail to reject
the null hypothesis.Meaning there is no difference in the tensile
strength between the four chemicals.
Problem 4.16
Input data
# Data: Tensile Strength of Cloth by Chemical and Bolt
Chemical <- factor(rep(1:4, each = 5))
Bolt <- factor(rep(1:5, times = 4))
Strength <- c(73, 68, 74, 71, 67, # Chemical 1
73, 67, 75, 72, 70, # Chemical 2
75, 68, 78, 73, 68, # Chemical 3
73, 71, 75, 75, 69) # Chemical 4
# Combine into a data frame
data <- data.frame(Chemical, Bolt, Strength)
# Display the first few rows of data
head(data)## Chemical Bolt Strength
## 1 1 1 73
## 2 1 2 68
## 3 1 3 74
## 4 1 4 71
## 5 1 5 67
## 6 2 1 73Estimate the Overall Mean
# Calculate the overall mean
mu <- mean(data$Strength)
mu## [1] 71.75Estimate the chemical effects
# Calculate the mean response for each chemical
chemical_means <- aggregate(Strength ~ Chemical, data, mean)
# Estimate the chemical effects (tau_i)
tau <- chemical_means$Strength - mu
tau## [1] -1.15 -0.35 0.65 0.85Estimate the Bolt Effects
# Calculate the mean response for each bolt
bolt_means <- aggregate(Strength ~ Bolt, data, mean)
# Estimate the bolt effects (beta_j)
beta <- bolt_means$Strength - mu
beta## [1] 1.75 -3.25 3.75 1.00 -3.25Display the results
Chemical Effects
chemical_means$Chemical## [1] 1 2 3 4
## Levels: 1 2 3 4tau## [1] -1.15 -0.35 0.65 0.85Bolt Effects
bolt_means$Bolt## [1] 1 2 3 4 5
## Levels: 1 2 3 4 5beta## [1] 1.75 -3.25 3.75 1.00 -3.25Problem 4.22
Input data
# Input the data
Batch <- factor(rep(1:5, each = 5))
Day <- factor(rep(1:5, 5))
Ingredient <- factor(c('A', 'B', 'D', 'C', 'E', # Batch 1
'C', 'E', 'A', 'D', 'B', # Batch 2
'B', 'A', 'C', 'E', 'D', # Batch 3
'D', 'C', 'E', 'B', 'A', # Batch 4
'E', 'D', 'B', 'A', 'C')) # Batch 5
Reaction_Time <- c(8, 7, 1, 7, 3, # Batch 1
11, 2, 7, 3, 8, # Batch 2
4, 9, 10, 1, 5, # Batch 3
6, 8, 6, 6, 10, # Batch 4
4, 2, 3, 8, 8) # Batch 5
# Combine into a data frame
data <- data.frame(Batch, Day, Ingredient, Reaction_Time)
# Display the first few rows of data
head(data)## Batch Day Ingredient Reaction_Time
## 1 1 1 A 8
## 2 1 2 B 7
## 3 1 3 D 1
## 4 1 4 C 7
## 5 1 5 E 3
## 6 2 1 C 11Perform Two-way ANOVA(Latin Square Design)
# Perform two-way ANOVA (Latin square design)
model <- aov(Reaction_Time ~ Ingredient + Batch + Day, data = data)
# Display the ANOVA summary
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## Ingredient 4 141.44 35.36 11.309 0.000488 ***
## Batch 4 15.44 3.86 1.235 0.347618
## Day 4 12.24 3.06 0.979 0.455014
## Residuals 12 37.52 3.13
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Interpretation of ANOVA
# Extract p-value for Ingredient
p_value_ingredient <- summary(model)[[1]][["Pr(>F)"]][1]
if (p_value_ingredient < 0.05) {
cat("There is a significant difference in reaction times due to the ingredients (p-value =", p_value_ingredient, ").\n")
} else {
cat("There is no significant difference in reaction times due to the ingredients (p-value =", p_value_ingredient, ").\n")
}## There is a significant difference in reaction times due to the ingredients (p-value = 0.0004876512 ).Analyze Residuals (Check ANOVA assumptions)
# Extract residuals and fitted values
residuals <- model$residuals
fitted_values <- model$fitted.values
# Residuals vs Fitted Values Plot
plot(fitted_values, residuals,
xlab = "Fitted Values", ylab = "Residuals",
main = "Residuals vs Fitted Values")
abline(h = 0, col = "green")
Normal Q-Q plot
# Q-Q Plot for normality of residuals
qqnorm(residuals)
qqline(residuals)
## Conclusion
if (p_value_ingredient < 0.05) {
cat("We conclude that the different ingredients significantly affect the reaction time.\n")
} else {
cat("We conclude that the different ingredients do not significantly affect the reaction time.\n")
}## We conclude that the different ingredients significantly affect the reaction time.cat("Check the residual plots above to confirm if the ANOVA assumptions (normality and equal variance) are satisfied.")## Check the residual plots above to confirm if the ANOVA assumptions (normality and equal variance) are satisfied.