researchq1

Step 1: Load Packages

library(readxl)
library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

library(ggpubr)

## Loading required package: ggplot2

library(effectsize)
library(effsize)

Peer Tutoring vs No Tutoring

Import dataset

Dataset6_1 <- read_excel("/Users/karim/Desktop/Dataset6.1.xlsx")

Descriptive Statistics

 Dataset6_1 %>%
  group_by(Group) %>%
  summarise(
    Mean = mean(Exam_Score, na.rm = TRUE),
    Median = median(Exam_Score, na.rm = TRUE),
    SD = sd(Exam_Score, na.rm = TRUE),
    N = n()
  )

## # A tibble: 2 × 5
##   Group        Mean Median    SD     N
##   <chr>       <dbl>  <dbl> <dbl> <int>
## 1 No Tutoring  71.9   71.5  7.68    40
## 2 Tutoring     78.4   78.7  7.18    40

5–7: Normality Checks

Histogram for Tutoring Group

hist(Dataset6_1$Exam_Score[Dataset6_1$Group == "Tutoring"],
     main = "Histogram of Tutoring Scores",
     xlab = "Exam_Score",
     ylab = "Frequency",
     col = "lightblue",
     border = "black",
     breaks = 10)

Histogram for No Tutoring Group

hist(Dataset6_1$Exam_Score[Dataset6_1$Group == "No Tutoring"],
     main = "Histogram of No Tutoring Scores",
     xlab = "Exam_Score",
     ylab = "Frequency",
     col = "lightgreen",
     border = "black",
     breaks = 10)

The histograms appear approximately symmetrical and bell-shaped.

Boxplot

 ggboxplot(Dataset6_1, x = "Group", y = "Exam_Score",
          color = "Group",
          palette = "jco",
          add = "jitter")

There were no extreme outliers.

Shapiro-Wilk

 shapiro.test(Dataset6_1$Exam_Score[Dataset6_1$Group == "Tutoring"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Dataset6_1$Exam_Score[Dataset6_1$Group == "Tutoring"]
## W = 0.98859, p-value = 0.953

shapiro.test(Dataset6_1$Exam_Score[Dataset6_1$Group == "No Tutoring"])

## 
##  Shapiro-Wilk normality test
## 
## data:  Dataset6_1$Exam_Score[Dataset6_1$Group == "No Tutoring"]
## W = 0.98791, p-value = 0.9398

both p > .05 → use Independent t-test

Inferential Test Independent t-test

t.test(Exam_Score ~ Group, data = Dataset6_1, var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  Exam_Score by Group
## t = -3.8593, df = 78, p-value = 0.000233
## alternative hypothesis: true difference in means between group No Tutoring and group Tutoring is not equal to 0
## 95 percent confidence interval:
##  -9.724543 -3.105845
## sample estimates:
## mean in group No Tutoring    mean in group Tutoring 
##                  71.94627                  78.36147

Students who participated in tutoring (M = 78.36, SD = 7.18) scored significantly higher than students who did not participate (M = 71.95, SD = 7.68), t(78) = 3.86, p < .001. The effect size was large (Cohen’s d = 0.86).

Interpretation: p < .05 = significant difference between groups

For t-test

 cohens_d(Exam_Score ~ Group, data = Dataset6_1)

## Cohen's d |         95% CI
## --------------------------
## -0.86     | [-1.32, -0.40]
## 
## - Estimated using pooled SD.

The effect size was large (Cohen’s d = 0.86).

Reporting results

There was a significant difference in exam scores between students who participated in peer tutoring (M = 78.36, SD = 7.18) and students who did not participate (M = 71.95, SD = 7.68), t(78) = 3.86, p < .001. The effect size was large (Cohen’s d = 0.86).