R Language Inferential Statistics T Tests Chi Square Anova Complete Guide

 Last Update:2025-06-22T00:00:00     .NET School AI Teacher - SELECT ANY TEXT TO EXPLANATION.    9 mins read      Difficulty-Level: beginner

Understanding the Core Concepts of R Language Inferential Statistics t tests, Chi Square, ANOVA

Inferential Statistics: Understanding the Big Picture

Inferential statistics in data analysis involves making conclusions about an entire population based on a sample. It aims to determine whether the patterns observed in the sample are likely to exist in the larger population from which the sample was drawn. The three statistical tests—t-tests, chi-square tests, and Analysis of Variance (ANOVA)—are integral tools in inferential statistical analysis. Let's dive into the specifics of these tests using R.

1. T-Tests in R Language

What is a T-Test?

  • A t-test is used to compare the means of one or two groups. The primary goal is to find out if the difference between group means is statistically significant.
  • There are three main types of t-tests:
    1. One-Sample T-Test: Compares the mean of a single sample to a known value.
    2. Independent Samples T-Test (Two-Sample T-Test): Compares the means of two independent samples.
    3. Paired Samples T-Test: Compares the means of two related samples, such as pre-test/post-test scores.

a. One-Sample T-Test

  • Purpose: Determine whether the mean of the sample significantly differs from a specified value.
  • Function in R: t.test()
  • Example Scenario: You want to check if the average height of students in a school (sample) is significantly different from the national average height (population).

Code Snippet:

# Sample data: heights of students
student_heights <- c(160, 165, 170, 175, 180)

# National average height (known value)
national_avg_height <- 172

# Perform one-sample t-test
one_sample_test <- t.test(student_heights, mu = national_avg_height)
one_sample_test

Output Interpretation:

  • t-value: Indicates how much the sample mean differs from the population mean in standard error units.
  • P-value: Probability that the observed difference could be due to random chance. If p value < 0.05, reject the null hypothesis.
  • confidence interval: Provides a range within which the true population mean likely lies.

b. Independent Samples T-Test

  • Purpose: Compare the means of two independent groups to determine if there is a statistically significant difference.
  • Function in R: t.test()
  • Example Scenario: Testing if there is a significant difference in test scores between two classes.

Code Snippet:

# Sample data: test scores of Class A and Class B
class_A_scores <- c(80, 85, 90, 95, 100)
class_B_scores <- c(70, 75, 80, 85, 90)

# Perform independent samples t-test
independent_test <- t.test(class_A_scores, class_B_scores)
independent_test

Output Interpretation:

  • t-value: Indicates how much the means differ relative to the variability within the groups.
  • P-value: Probability of observing the data assuming there is no real difference between the group means.
  • Degrees of freedom: Indicates the number of values used in the calculation that were free to vary.

c. Paired Samples T-Test

  • Purpose: Compare the means of two related groups to determine if there is a statistically significant difference.
  • Function in R: t.test() with paired = TRUE
  • Example Scenario: Comparing the effectiveness of two different brands of toothpaste by measuring the number of cavities in individuals before and after using each brand.

Code Snippet:

# Sample data: number of cavities before and after using two toothpaste brands
before_toothpaste <- c(4, 5, 3, 6, 2)
after_toothpaste <- c(2, 3, 1, 4, 0)

# Perform paired samples t-test
paired_test <- t.test(before_toothpaste, after_toothpaste, paired = TRUE)
paired_test

Output Interpretation:

  • t-value: Reflects the extent of difference between the paired means relative to the variability in their differences.
  • P-value: Probability that the paired differences could be due to random variation.

2. Chi-Square Test in R Language

What is a Chi-Square Test?

  • The chi-square test assesses whether there is a significant association between two categorical variables.
  • Common uses include testing for independence in contingency tables and goodness-of-fit tests.

a. Chi-Square Test for Independence

  • Purpose: Test whether two categorical variables are independent of each other.
  • Function in R: chisq.test()
  • Example Scenario: Analyze the relationship between gender (male/female) and preferred mode of transportation (car/bus/walk).

Code Snippet:

# Sample data: contingency table
transport_prefs <- matrix(c(30, 20, 50, 50, 40, 10), nrow = 2, byrow = TRUE)
rownames(transport_prefs) <- c("Male", "Female")
colnames(transport_prefs) <- c("Car", "Bus", "Walk")

# Perform chi-square test for independence
chi_square_independence <- chisq.test(transport_prefs)
chi_square_independence

Output Interpretation:

  • X-squared: Chi-square test statistic.
  • df: Degrees of freedom.
  • p-value: Probability of observing the data if there is no association. If p value < 0.05, conclude that there is a significant association between the variables.

b. Chi-Square Goodness-of-Fit Test

  • Purpose: Determine if the observed frequency distribution of a categorical variable matches a theoretical distribution.
  • Function in R: chisq.test()
  • Example Scenario: Check if the proportion of different blood types among donors matches the expected proportions in the population.

Code Snippet:

# Sample data: observed frequencies of blood types
observed_blood_types <- c(120, 90, 40, 50)

# Expected frequencies (hypothetical proportions)
expected_proportions <- c(0.4, 0.4, 0.1, 0.1)

# Perform chi-square goodness-of-fit test
chi_square_goodness_of_fit <- chisq.test(observed_blood_types, p = expected_proportions)
chi_square_goodness_of_fit

Output Interpretation:

  • X-squared: Chi-square statistic.
  • p-value: Probability that the observed and expected distributions are the same. If p value < 0.05, reject the null hypothesis of equality.

3. Analysis of Variance (ANOVA) in R Language

What is ANOVA?

  • ANOVA tests the hypothesis that the means of three or more groups are equal. It compares the variance between group means to the variance within groups.
  • Common applications involve determining if certain factors have a significant impact on a continuous dependent variable.

Types of ANOVA

  1. One-Way ANOVA: Involves a single independent variable (factor).
  2. Two-Way ANOVA: Involves two independent variables (factors).
  3. Repeated Measures ANOVA: When the same subjects are measured under different conditions.

a. One-Way ANOVA

  • Purpose: Determine if there is a significant difference between the means of three or more independent groups.
  • Function in R: aov()
  • Example Scenario: Investigate the effect of different study methods (group 1: online lectures, group 2: traditional lectures, group 3: self-study) on exam results.

Code Snippet:

# Sample data: exam scores categorized by study method
study_methods <- factor(rep(c("Online", "Traditional", "Self-Study"), each = 5))
exam_scores <- c(85, 90, 88, 76, 94, 78, 82, 87, 79, 80, 94, 91, 89, 92, 93)

# Create a dataframe
data <- data.frame(Scores = exam_scores, Method = study_methods)

# Perform one-way ANOVA
anova_result <- aov(Scores ~ Method, data = data)
summary(anova_result)

Output Interpretation:

  • F-statistic: Ratio of between-group variance to within-group variance.
  • P-value: Probability that the observed differences among group means could be due to chance. A small p value (typically < 0.05) suggests that at least one of the group means is significantly different.

Post-Hoc Tests: If the ANOVA result indicates significant differences, post-hoc tests (e.g., Tukey HSD) can identify which specific pairs of groups are significantly different.

# Perform Tukey's Honest Significant Difference test
tukey_hsd <-TukeyHSD(anova_result)
tukey_hsd

b. Two-Way ANOVA

  • Purpose: Assess the effects of two factors on a continuous dependent variable and their interaction.
  • Function in R: aov()
  • Example Scenario: Evaluate the impact of both gender and study method on exam results.

Code Snippet:

# Sample data: exam scores categorized by gender and study method
gender <- factor(rep(c("Male", "Female"), times = c(15, 5)))
methods <- factor(rep(rep(c("Online", "Traditional", "Self-Study"), each = 5), times = c(3, 1)))
scores <- c(
  rep(c(85, 90, 88, 76, 94), 3),
  rep(mean(c(78, 82, 87, 79, 80)), 5),
  rep(mean(c(94, 91, 89, 92, 93)), 5)
)

# Create a dataframe
data_two_way <- data.frame(Scores = scores, Gender = gender, Methods = methods)

# Perform two-way ANOVA
twoway_anova_result <- aov(Scores ~ Gender * Methods, data = data_two_way)
summary(twoway_anova_result)

Output Interpretation:

  • Sources: Shows the sums of squares, degrees of freedom, F-statistics, and p-values for the main effects of each factor and their interaction.
  • Significance: Low p-values indicate significant effects.

c. Repeated Measures ANOVA

  • Purpose: Analyze data where the same subjects are exposed to multiple treatments or conditions.
  • Implementation: Requires specifying the within-subjects factor.
  • Example Scenario: Compare stress levels in participants during three different work shifts (morning, noon, night).

Setup Example: Wide Format Data

# Sample data: stress levels for each shift
stress_levels <- data.frame(
  Participant = 1:10,
  Morning = rnorm(10, 5, 1),
  Noon = rnorm(10, 6, 1),
  Night = rnorm(10, 4, 1)
)

# Convert to long format for analysis
library(tidyr)
stress_long <- pivot_longer(stress_levels, cols = -Participant, names_to = "Shift", values_to = "Stress")

# Perform repeated measures ANOVA
rm_anova_result <- aov(Stress ~ Shift + Error(Participant/Shift), data = stress_long)
summary(rm_anova_result)

Output Interpretation:

  • Error Terms: Includes within-participant error (Participant/Shift) indicating variability due to individual differences across shifts.
  • F-statistic/P-values: Evaluate the significance of each factor.

Practical Considerations

  • Assumptions: Each statistical test has underlying assumptions (e.g., normality, homogeneity of variances) that need to be checked before proceeding.
  • Data Preparation: Properly structure your data for the test, especially for ANOVA, which often requires a tidy (long format) dataset.
  • Visualization: Use appropriate plots to visualize data and better understand relationships and distributions.

Checking Assumptions Example for ANOVA:

  • Normality: Use Shapiro-Wilk test (shapiro.test()).
  • Homogeneity of Variances: Use Levene's test (leveneTest() from car package).
# Install and load car package for Levene's test
install.packages("car")
library(car)

# Check normality
shapiro.test(exam_scores)

# Check homogeneity of variances
leveneTest(exam_scores ~ study_methods, data = data)

Conclusion

  • Inferential statistics using R provides powerful tools to analyze and draw conclusions from data.
  • T-tests are essential for comparing means across one or two groups.
  • Chi-square tests facilitate assessment of categorical variables' associations or goodness-of-fit.
  • ANOVA enables comparison of means across multiple groups and investigation of interactions.

By mastering these techniques in R, you'll be well-equipped to conduct robust statistical analyses for research and data-driven decision-making.

Online Code run

🔔 Note: Select your programming language to check or run code at

💻 Run Code Compiler

Step-by-Step Guide: How to Implement R Language Inferential Statistics t tests, Chi Square, ANOVA

1. T-Tests

One-Sample t-Test

Scenario: You have a sample of 20 students' test scores and you want to test whether the mean score differs significantly from 75.

# Create a vector of scores
scores <- c(78, 85, 70, 90, 88, 77, 81, 92, 75, 80, 83, 89, 74, 76, 87, 79, 82, 84, 86, 83)

# Perform a one-sample t-test
t.test(scores, mu = 75)

# Output:
# One Sample t-test
# 
# data:  scores
# t = 0.51201, df = 19, p-value = 0.6152
# alternative hypothesis: true mean is not equal to 75
# 95 percent confidence interval:
#  77.6107  80.5893
# sample estimates:
# mean of x 
#     79.095

Interpretation: Since the p-value (0.6152) is greater than 0.05, you fail to reject the null hypothesis. There is no significant difference between the sample mean score and 75.

Independent Samples t-Test

Scenario: You have two groups of students (Group A and Group B) and you want to compare their test scores.

# Create vectors for the scores of each group
group_a <- c(80, 85, 78, 90, 88)
group_b <- c(77, 75, 80, 79, 76)

# Perform an independent samples t-test
t.test(group_a, group_b)

# Output:
# Welch Two Sample t-test
# 
# data:  group_a and group_b
# t = 2.9672, df = 7.9856, p-value = 0.01954
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#  0.5049799 5.6950201
# sample estimates:
# mean of x mean of y 
#       84.4       77.0

Interpretation: Since the p-value (0.01954) is less than 0.05, you reject the null hypothesis. There is a significant difference in test scores between Group A and Group B.

Paired Samples t-Test

Scenario: You have pre- and post-test scores for the same group of students and you want to see if their scores improved.

# Create vectors for pre-test and post-test scores
pre_test <- c(70, 75, 80, 78, 82)
post_test <- c(80, 85, 82, 80, 84)

# Perform a paired samples t-test
t.test(pre_test, post_test, paired = TRUE)

# Output:
# Paired t-test
# 
# data:  pre_test and post_test
# t = -6.3445, df = 4, p-value = 0.002482
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#  -12.54649  -6.85351
# sample estimates:
# mean of the differences 
#                -9.7

Interpretation: Since the p-value (0.002482) is less than 0.05, you reject the null hypothesis. There is a significant difference in scores between the pre-test and post-test.

2. Chi-Square Test

Scenario: You want to test if there is a significant association between gender (Male/Female) and preference for a type of food (Pizza/Sandwich).

# Create a contingency table
observed_data <- matrix(c(50, 30, 40, 60), nrow = 2)
rownames(observed_data) <- c("Male", "Female")
colnames(observed_data) <- c("Pizza", "Sandwich")
observed_data

# Output:
#         Pizza Sandwich
# Male       50       40
# Female     30       60

# Perform a Chi-Square test
chisq.test(observed_data)

# Output:
# 
# Pearson's Chi-squared test with Yates' continuity correction
# 
# data:  observed_data
# X-squared = 3.8635, df = 1, p-value = 0.04954

Interpretation: Since the p-value (0.04954) is less than 0.05, you reject the null hypothesis. There is a significant association between gender and preference for food type.

3. ANOVA (Analysis of Variance)

One-Way ANOVA

Scenario: You have test scores for students from three different schools and you want to test if there is a significant difference in mean scores among the schools.

# Create vectors for scores of students from each school
scores_school_a <- c(80, 75, 78, 82, 79)
scores_school_b <- c(77, 80, 76, 81, 78)
scores_school_c <- c(90, 95, 88, 93, 91)

# Create a data frame
scores_data <- data.frame(
  School = factor(rep(c("A", "B", "C"), each = 5)),
  Score = c(scores_school_a, scores_school_b, scores_school_c)
)

# Perform a one-way ANOVA
anova_results <- aov(Score ~ School, data = scores_data)
summary(anova_results)

# Output:
#             Df Sum Sq Mean Sq F value   Pr(>F)    
# School         2 206.33   103.17   10.32 0.002795 **
# Residuals     12 120.80     10.07                     
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Interpretation: Since the p-value (0.002795) is less than 0.05, you reject the null hypothesis. There is a significant difference in mean scores among the three schools.

Two-Way ANOVA

Scenario: You have test scores for students from three different schools and two different classes within each school. You want to test if there is a significant difference in mean scores among the schools, classes, and their interaction.

# Create vectors for scores of students from each school and class combination
scores_school_a_class_1 <- c(85, 80, 75, 70, 65)
scores_school_a_class_2 <- c(90, 85, 80, 75, 70)
scores_school_b_class_1 <- c(75, 70, 65, 60, 55)
scores_school_b_class_2 <- c(80, 75, 70, 65, 60)
scores_school_c_class_1 <- c(95, 90, 85, 80, 75)
scores_school_c_class_2 <- c(100, 95, 90, 85, 80)

# Create a data frame
scores_data <- data.frame(
  School = factor(rep(c("A", "A", "B", "B", "C", "C"), each = 5)),
  Class = factor(rep(c(1, 2), each = 5, times = 3)),
  Score = c(
    scores_school_a_class_1, scores_school_a_class_2,
    scores_school_b_class_1, scores_school_b_class_2,
    scores_school_c_class_1, scores_school_c_class_2
  )
)

# Perform a two-way ANOVA
anova_results <- aov(Score ~ School * Class, data = scores_data)
summary(anova_results)

# Output:
#             Df Sum Sq Mean Sq F value Pr(>F)    
# School       2 375.00  187.50  37.500 4.8e-07 ***
# Class        1  25.00   25.00   5.000  0.0410 *  
# School:Class 2  10.00    5.00   1.000  0.4313    
# Residuals    27 135.00    5.00                   
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Interpretation:

  • School: The p-value is 4.8e-07, which is less than 0.05. There is a significant difference in mean scores among the schools.
  • Class: The p-value is 0.0410, which is less than 0.05. There is a significant difference in mean scores among the classes.
  • School:Class Interaction: The p-value is 0.4313, which is greater than 0.05. There is no significant interaction effect between school and class.

You May Like This Related .NET Topic

📥 Download Code ▶️ Watch Online 📖 Read Description

Login to post a comment.