Hypothesis Testing Using R

Hypothesis testing is a fundamental statistical technique used to make inferences about population parameters based on sample data. In this tutorial, we will explore how to perform hypothesis testing in R using common statistical tests, such as the t-test, chi-square test, and ANOVA.

Understanding Hypothesis Testing

Hypothesis testing involves the following steps:

1. Formulate Hypotheses:

Null Hypothesis (H₀): The default assumption (e.g., no effect or no difference).
Alternative Hypothesis (H₁): The claim we want to test (e.g., there is an effect or difference).

2. Choose a Significance Level (α): Commonly set to 0.05 (5%).

3. Calculate the Test Statistic: A value computed from the sample data.

4. Determine the p-value: The probability of observing the test statistic under the null hypothesis.

Make a Decision:

If p-value < α, reject the null hypothesis.
If p-value ≥ α, fail to reject the null hypothesis.

Load Required Libraries

For this tutorial, we will use base R functions, so no additional libraries are required. However, we will use the ggplot2 package for visualization:

Install and load ggplot2 (for visualization)

install.packages(“ggplot2”)

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 4.4.2

One-Sample t-Test

A one-sample t-test is used to compare the mean of a sample to a known value.

Example:

Suppose we have a sample of 30 students’ test scores, and we want to test if the mean score is significantly different from 70.

Sample data

set.seed(123)
scores <- rnorm(30, mean = 75, sd = 10)

One-sample t-test

t_test_result <- t.test(scores, mu = 70)
t_test_result

## 
##  One Sample t-test
## 
## data:  scores
## t = 2.5286, df = 29, p-value = 0.01715
## alternative hypothesis: true mean is not equal to 70
## 95 percent confidence interval:
##  70.86573 78.19219
## sample estimates:
## mean of x 
##  74.52896

Interpretation:

Null Hypothesis (H₀): The mean score is 70.
Alternative Hypothesis (H₁): The mean score is not 70.
If the p-value < 0.05, we reject H₀ and conclude that the mean score is significantly different from 70.

Since the p-value (0.01715) is less than the significance level (α = 0.05), we reject the null hypothesis. This means there is sufficient evidence to conclude that the sample mean is significantly different from 70.

Additionally, the 95% confidence interval (70.86573 to 78.19219) does not include the hypothesized mean (70), further supporting the conclusion that the true population mean is not 70.

Two-Sample t-Test

A two-sample t-test is used to compare the means of two independent groups.

Example:

Suppose we have test scores for two groups of students (Group A and Group B) and want to test if their means are significantly different.

Sample data

set.seed(123)
group_a <- rnorm(30, mean = 75, sd = 10)
group_b <- rnorm(30, mean = 80, sd = 10)

Two-sample t-test

t_test_result <- t.test(group_a, group_b)
t_test_result

## 
##  Welch Two Sample t-test
## 
## data:  group_a and group_b
## t = -3.0841, df = 56.559, p-value = 0.003156
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -11.965426  -2.543416
## sample estimates:
## mean of x mean of y 
##  74.52896  81.78338

Interpretation:

Null Hypothesis (H₀): The means of Group A and Group B are equal.
Alternative Hypothesis (H₁): The means of Group A and Group B are not equal.
If the p-value < 0.05, we reject H₀ and conclude that the means are significantly different.

Since the p-value (0.003156) is less than the significance level (α = 0.05), we reject the null hypothesis. This means there is sufficient evidence to conclude that the means of the two groups are significantly different.

Additionally, the 95% confidence interval (-11.965426 to -2.543416) does not include zero, further supporting the conclusion that the true difference in means is not zero.

Paired t-Test

A paired t-test is used to compare the means of two related groups (e.g., before and after measurements).

Example:

Suppose we have blood pressure measurements for 20 patients before and after a treatment.

Sample data

set.seed(123)
before <- rnorm(20, mean = 120, sd = 10)
after <- rnorm(20, mean = 115, sd = 10)

Paired t-test

t_test_result <- t.test(before, after, paired = TRUE)
t_test_result

## 
##  Paired t-test
## 
## data:  before and after
## t = 2.3206, df = 19, p-value = 0.03159
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##   0.679524 13.178095
## sample estimates:
## mean difference 
##         6.92881

Interpretation:

Null Hypothesis (H₀): The mean difference between before and after measurements is zero.
Alternative Hypothesis (H₁): The mean difference is not zero.
If the p-value < 0.05, we reject H₀ and conclude that the treatment had a significant effect.

Since the p-value (0.03159) is less than the significance level (α = 0.05), we reject the null hypothesis. This means there is sufficient evidence to conclude that the mean difference between the paired observations is significantly different from zero.

Additionally, the 95% confidence interval (0.679524 to 13.178095) does not include zero, further supporting the conclusion that the true mean difference is not zero.

Chi-Square Test

A chi-square test is used to test the association between categorical variables.

Example:

Suppose we have a contingency table showing the distribution of gender (Male, Female) and preference for a product (Like, Dislike).

Contingency table

data <- matrix(c(30, 10, 20, 40), nrow = 2, byrow = TRUE)
rownames(data) <- c("Male", "Female")
colnames(data) <- c("Like", "Dislike")

Chi-square test

chi_test_result <- chisq.test(data)
chi_test_result

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  data
## X-squared = 15.042, df = 1, p-value = 0.0001052

Interpretation:

Null Hypothesis (H₀): There is no association between gender and product preference.
Alternative Hypothesis (H₁): There is an association between gender and product preference.
If the p-value < 0.05, we reject H₀ and conclude that there is a significant association.

Since the p-value (0.0001052) is less than the significance level (α = 0.05), we reject the null hypothesis. This means there is sufficient evidence to conclude that there is a significant association between the two categorical variables.

ANOVA (Analysis of Variance)

ANOVA is used to compare the means of three or more groups.

Example:

Suppose we have test scores for three groups of students (Group A, Group B, Group C) and want to test if their means are significantly different.

Sample data

set.seed(123)
group_a <- rnorm(30, mean = 75, sd = 10)
group_b <- rnorm(30, mean = 80, sd = 10)
group_c <- rnorm(30, mean = 85, sd = 10)

Combine data into a data frame

data <- data.frame(
    score = c(group_a, group_b, group_c),
    group = factor(rep(c("A", "B", "C"), each = 30))
)

ANOVA

anova_result <- aov(score ~ group, data = data)
summary(anova_result)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## group        2   1794   897.1   11.14 4.94e-05 ***
## Residuals   87   7008    80.5                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Interpretation:

Null Hypothesis (H₀): The means of all groups are equal.
Alternative Hypothesis (H₁): At least one group mean is different.
If the p-value < 0.05, we reject H₀ and conclude that there are significant differences between the groups.

Since the p-value (0.0000494) is less than the significance level (α = 0.05), we reject the null hypothesis. This means there is sufficient evidence to conclude that at least one group mean is significantly different from the others.

Post-Hoc Analysis (Tukey’s HSD)

If ANOVA results are significant, perform post-hoc analysis to identify which groups differ.

Tukey’s HSD test

tukey_result <- TukeyHSD(anova_result)
tukey_result

##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = score ~ group, data = data)
## 
## $group
##          diff       lwr       upr     p adj
## B-A  7.254421  1.728917 12.779925 0.0066511
## C-A 10.715241  5.189738 16.240745 0.0000384
## C-B  3.460821 -2.064683  8.986324 0.2990254

Interpretation:

The output shows pairwise comparisons between groups and their p-values.
Identify pairs with p-values < 0.05 as significantly different.

Significant Differences:

B vs. A: There is a statistically significant difference between group B and group A (p < 0.05).
C vs. A: There is a highly statistically significant difference between group C and group A (p < 0.001).

No Significant Difference:

C vs. B: There is no statistically significant difference between group C and group B (p > 0.05).

Visualization

Visualize the results using boxplots or bar plots.

Boxplot for ANOVA:

ggplot(data, aes(x = group, y = score, fill = group)) +
    geom_boxplot() +
    theme_minimal() +
    labs(title = "Test Scores by Group", x = "Group", y = "Score")

Summary

In this tutorial, we covered:

One-sample t-test: Compare a sample mean to a known value.

Two-sample t-test: Compare means of two independent groups.

Paired t-test: Compare means of two related groups.

Chi-square test: Test association between categorical variables.

ANOVA: Compare means of three or more groups.

Post-hoc analysis: Identify specific group differences after ANOVA.

HYPOTHESIS TESTING

Oumo David

2025-02-26

Hypothesis Testing Using R

Understanding Hypothesis Testing

1. Formulate Hypotheses:

2. Choose a Significance Level (α): Commonly set to 0.05 (5%).

3. Calculate the Test Statistic: A value computed from the sample data.

4. Determine the p-value: The probability of observing the test statistic under the null hypothesis.

Make a Decision:

Load Required Libraries

One-Sample t-Test

Interpretation:

Two-Sample t-Test

Interpretation:

Paired t-Test

Interpretation:

Chi-Square Test

Interpretation:

ANOVA (Analysis of Variance)

ANOVA

Interpretation:

Post-Hoc Analysis (Tukey’s HSD)

Tukey’s HSD test

Interpretation:

Visualization

Summary