1 QUESTION 1

Model Equation: \[ Y_i = \mu + \tau_i + \epsilon_i \]

Null Hypothesis ($H_0$): $\mu_A = \mu_B$

Alternative Hypothesis ($H_a$): $\mu_A \neq \mu_B$

1.1 Analysis and Results:

# Loading Data
lifespans <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/lifespans.csv")

# Boxplot 
boxplot(lifespans$Lifespan ~ lifespans$Supplier, 
        main = "Boxplot of Lifespans by Supplier",
        xlab = "Supplier",
        ylab = "Lifespan (hours)",
        col = c("blue", "red"))

# two-sample t-test with pooled variances
t_test_result <- t.test(Lifespan ~ Supplier, data = lifespans, var.equal = TRUE)
t_test_result

## 
##  Two Sample t-test
## 
## data:  Lifespan by Supplier
## t = 2.6682, df = 18, p-value = 0.01567
## alternative hypothesis: true difference in means between group A and group B is not equal to 0
## 95 percent confidence interval:
##   2.411193 20.269776
## sample estimates:
## mean in group A mean in group B 
##        504.4806        493.1401

1.1.1 conclusion:

The p-value<0.05.
we fail to reject $H_0$
There is no significant differnce in the mean lifespan between Supplier A and Supplier B.

2 QUESTION 2

2.1 Analysis and Results:

# defining data
Before <- c(35, 40, 32, 38, 36, 42, 39, 41)
After <- c(30, 38, 31, 36, 34, 40, 37, 39)

# Paired t-test
t_test_result <- t.test(Before, After, paired = TRUE)
t_test_result

## 
##  Paired t-test
## 
## data:  Before and After
## t = 5.4628, df = 7, p-value = 0.0009431
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  1.276065 3.223935
## sample estimates:
## mean difference 
##            2.25

2.1.1 conclusion:

The P-value <0.05.
we fail to reject $H_0$
The training program significantly reduces tasks completion time

3 QUESTION 3

Model equation : \[ Y_{ij} = \mu + \tau_i + \epsilon_{ij} \] Null Hypothesis ($H_0$): All material means are equal. Alternative Hypothesis ($H_a$): At least one material mean differs.

3.1 Analysis and Results:

# Load the dataset from the csv file
tensile_strength <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/tensile_strength.csv")

# Performing ANOVA
anova_result <- aov(Strength ~ Material, data = tensile_strength)
summary(anova_result)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## Material     2  972.2   486.1   3.039 0.0855 .
## Residuals   12 1919.3   159.9                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

#Anova Result significant: Performing Tukey's post-hoc test if significant
if (summary(anova_result)[[1]]["Material", "Pr(>F)"] < 0.05) {
  tukey_result <- TukeyHSD(anova_result)
  tukey_result
}

3.1.1 Conclusion:

The ANOVA p-value <0.05.
we reject $H_0$
Atleast one material differs significantly in mean tensile strenght.

4 QUESTION 4

Model Equation : \[ Y_{ijk} = \mu + \tau_i + \beta_j + (\tau\beta)_{ij} + \epsilon_{ijk} \]

4.1 Analysis and Results:

# Loading the yield data
yield_data <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/yield_data.csv")
yield_data$Temperature <- as.factor(yield_data$Temperature)
yield_data$Pressure <- as.factor(yield_data$Pressure)

# Performing two-way ANOVA
anova_result <- aov(Yield ~ Temperature * Pressure, data = yield_data)
summary(anova_result)

##                      Df Sum Sq Mean Sq F value   Pr(>F)    
## Temperature           1  433.8   433.8  26.637 0.000863 ***
## Pressure              1   95.9    95.9   5.889 0.041412 *  
## Temperature:Pressure  1    1.0     1.0   0.060 0.811915    
## Residuals             8  130.3    16.3                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

# Extracting and displaying main effects and interaction effects
summary(anova_result)

##                      Df Sum Sq Mean Sq F value   Pr(>F)    
## Temperature           1  433.8   433.8  26.637 0.000863 ***
## Pressure              1   95.9    95.9   5.889 0.041412 *  
## Temperature:Pressure  1    1.0     1.0   0.060 0.811915    
## Residuals             8  130.3    16.3                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

4.1.1 conclusion:

P-Value <0.05 for both temperature and pressure.
we reject $H_o$ for all factors. both Temperature and pressure as well as their interaction significantly affect yield.

# Combined R Code from Questions 1-4
# Question 1
lifespans <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/lifespans.csv")
boxplot(lifespans$Lifespan ~ lifespans$Supplier, main = "Boxplot of Lifespans by Supplier", xlab = "Supplier", ylab = "Lifespan (hours)", col = c("blue", "red"))
t_test_result <- t.test(Lifespan ~ Supplier, data = lifespans, var.equal = TRUE)
print(t_test_result)

# Question 2
Before <- c(35, 40, 32, 38, 36, 42, 39, 41)
After <- c(30, 38, 31, 36, 34, 40, 37, 39)
t_test_result <- t.test(Before, After, paired = TRUE)
print(t_test_result)

# Question 3
tensile_strength <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/tensile_strength.csv")
anova_result <- aov(Strength ~ Material, data = tensile_strength)
summary(anova_result)
if (summary(anova_result)[[1]]["Material", "Pr(>F)"] < 0.05) {
  tukey_result <- TukeyHSD(anova_result)
  print(tukey_result)
}

# Question 4
yield_data <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/yield_data.csv")
yield_data$Temperature <- as.factor(yield_data$Temperature)
yield_data$Pressure <- as.factor(yield_data$Pressure)
anova_result <- aov(Yield ~ Temperature * Pressure, data = yield_data)
summary(anova_result)

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Okeoghene Jonathan Benjamin

12/6/2024