Required Packages

library(ggplot2)
library(reshape2)
library(car)
library(ggpubr)

One sample t-test

Manual Calculation

set.seed(123)
sample_data <- rnorm(30, mean = 50, sd = 10)
mean(sample_data)
[1] 49.52896
sd(sample_data)
[1] 9.810307

Hypothesis:
\(H_0: \mu = \mu_0\)
\(H_1: \mu \neq \mu_0\)

# mu_0 = 55
SE <- sd(sample_data)/sqrt(length(sample_data))
t_cal <- (mean(sample_data) - 55)/SE
print(t_cal)
[1] -3.054553
t_crit <- qt(0.05/2, df = length(sample_data)-1, lower.tail = TRUE)
t_crit
[1] -2.04523
t_cal <= t_crit  # decision: reject null
[1] TRUE
pt(t_cal, df = 29, lower.tail = TRUE) + pt(-t_cal, df = 29, lower.tail = FALSE)
[1] 0.0047971

Calculating confidence interval:

mean(sample_data) + c(-1, 1) * abs(t_crit) * SE
[1] 45.86573 53.19219

Using Function

Generate sample data:

set.seed(123)
sample_data <- rnorm(30, mean = 50, sd = 10)

Testing normality using Shapiro-Wilk test:

shapiro.test(sample_data)

    Shapiro-Wilk normality test

data:  sample_data
W = 0.97894, p-value = 0.7966

\(H_0:\) Data follows normal distribution.
\(H_1:\) Data does not follow normal distribution.

Since the p-value is greater than the level of significance (\(\alpha\) = 0.05), we do not have enough statistical evidence to reject the null hypothesis.

One-sample t-test:

Hypothesis:
\(H_0: \mu = 50\)
\(H_1: \mu \neq 50\)

Using function, perform the two tailed t-test:

t.test(sample_data, mu = 53, conf.level = 0.95)

    One Sample t-test

data:  sample_data
t = -1.9379, df = 29, p-value = 0.06242
alternative hypothesis: true mean is not equal to 53
95 percent confidence interval:
 45.86573 53.19219
sample estimates:
mean of x 
 49.52896

If p-value < 0.05, then reject null. Decision is not rejected.

t.test(sample_data, mu = 53, conf.level = 0.99)

    One Sample t-test

data:  sample_data
t = -1.9379, df = 29, p-value = 0.06242
alternative hypothesis: true mean is not equal to 53
99 percent confidence interval:
 44.59198 54.46595
sample estimates:
mean of x 
 49.52896

If p-value < 0.01, then reject null. Decision is not rejected.

t.test(sample_data, mu = 53, conf.level = 0.90)

    One Sample t-test

data:  sample_data
t = -1.9379, df = 29, p-value = 0.06242
alternative hypothesis: true mean is not equal to 53
90 percent confidence interval:
 46.48564 52.57228
sample estimates:
mean of x 
 49.52896

If p-value < 0.10, then reject null. Decision is rejected.

Hypothesis:
\(H_0: \mu <= 40\)
\(H_1: \mu > 40\)

t.test(sample_data, mu = 40, alternative = "greater")

    One Sample t-test

data:  sample_data
t = 5.3201, df = 29, p-value = 5.21e-06
alternative hypothesis: true mean is greater than 40
95 percent confidence interval:
 46.48564      Inf
sample estimates:
mean of x 
 49.52896

Hypothesis:
\(H_0: \mu >= 58\)
\(H_1: \mu < 58\)

t.test(sample_data, mu = 58, alternative = "less")

    One Sample t-test

data:  sample_data
t = -4.7295, df = 29, p-value = 2.689e-05
alternative hypothesis: true mean is less than 58
95 percent confidence interval:
     -Inf 52.57228
sample estimates:
mean of x 
 49.52896 
ggplot(data.frame(Value = sample_data), aes(x = Value)) +
  geom_histogram(aes(y = after_stat(density)), bins = 15, fill = "blue", alpha = 0.5) +
  geom_density(color = "red", linewidth = 1) +
  labs(title = "Sample Data Distribution", x = "Value", y = "Density") +
  theme_minimal()

Two-samples t-test

Use built-in dataset: mtcars (comparing mpg for automatic vs manual cars):

data(mtcars)

Split into two groups based on transmission type”:

auto_mpg <- mtcars$mpg[mtcars$am == 0]    # Automatic
manual_mpg <- mtcars$mpg[mtcars$am == 1]  # Manual

\(H_0\): Automatic cars and manual cars have equal average mpg.
\(H_1\): Automatic cars and manual cars have unequal average mpg.

mean(auto_mpg)
[1] 17.14737
mean(manual_mpg)
[1] 24.39231
var(auto_mpg)
[1] 14.6993
var(manual_mpg)
[1] 38.02577

Normality test for both groups:

shapiro.test(auto_mpg)

    Shapiro-Wilk normality test

data:  auto_mpg
W = 0.97677, p-value = 0.8987
shapiro.test(manual_mpg)

    Shapiro-Wilk normality test

data:  manual_mpg
W = 0.9458, p-value = 0.5363

Check variance homogeneity (Levene’s test):

leveneTest(mpg ~ factor(am), data = mtcars, center = "mean")
Levene's Test for Homogeneity of Variance (center = "mean")
      Df F value  Pr(>F)  
group  1   5.921 0.02113 *
      30                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Perform two-sample t-test:

t.test(auto_mpg, manual_mpg, var.equal = FALSE) 

    Welch Two Sample t-test

data:  auto_mpg and manual_mpg
t = -3.7671, df = 18.332, p-value = 0.001374
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -11.280194  -3.209684
sample estimates:
mean of x mean of y 
 17.14737  24.39231 
# Visualize the data
ggplot(mtcars, aes(x = factor(am), y = mpg, fill = factor(am))) +
  geom_boxplot(alpha = 0.6) +
  geom_jitter(width = 0.2, alpha = 0.7) +
  labs(title = "MPG Comparison: Automatic vs Manual", 
       x = "Transmission (0 = Auto, 1 = Manual)", 
       y = "Miles Per Gallon") +
  scale_fill_manual(values = c("blue", "red"), 
                    labels = c("Automatic", "Manual")) +
  theme_minimal()

Paired samples t-test

Generate 30 observations:

set.seed(123)
before <- round(rnorm(30, mean = 300, sd = 10), 0)
summary(before)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  280.0   293.2   299.5   299.6   305.0   318.0 
# after a course
after <- before + round(rnorm(30, mean = 5, sd = 5), 0) # Simulating a increase
summary(after)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  283.0   298.2   305.0   305.5   312.2   325.0 
df <- data.frame(
  ID = 1:30,
  Before = before,
  After = after
)
mean(after) - mean(before)
[1] 5.933333

\(H_0:\) Before and after the course the true GRE average score of the students stays the same.
\(H_1:\) Before and after the course the true GRE average score of the students does not remain the same.

Perform Paired t-test

t.test(x = after, y = before, paired = TRUE, alternative = "two.sided")

    Paired t-test

data:  after and before
t = 7.7811, df = 29, p-value = 1.398e-08
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 4.373779 7.492888
sample estimates:
mean difference 
       5.933333 
# Visualize the differences
df <- data.frame(
  ID = 1:30,
  Before = before, 
  After = after
  )
df_long <- melt(df, id.vars = "ID")
ggplot(df_long, aes(x = variable, y = value, group = ID)) +
  geom_point(aes(color = variable), size = 3) +
  geom_line() +
  labs(title = "Paired Samples (Before vs. After)", 
       y = "Values", x = "Condition") +
  theme_minimal()

ggpaired(df_long, 
         x = "variable", 
         y = "value",
         color = "variable", 
         line.color = "gray", 
         line.size = 0.4,
         palette = "jco") +
  stat_compare_means(paired = TRUE, method = "t.test")