Research Scenario 6.1: Tutoring and Exam Scores

library(readxl)
library(ggpubr)
## Loading required package: ggplot2
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(effectsize)
library(effsize)
Dataset6.1 <- read_excel("/Users/asfia/Desktop/Dataset6.1.xlsx")

STEP 4: CALCULATE DESCRIPTIVE STATISTICS

Dataset6.1 %>%
  group_by(Group) %>%
  summarise(
    Mean = mean(Exam_Score),
    Median = median(Exam_Score),
    SD = sd(Exam_Score),
    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

STEP 5 & 6: NORMALITY VISUALIZATIONS

hist(Dataset6.1$Exam_Score[Dataset6.1$Group == "Tutoring"],
     main = "Histogram of Exam Scores - Tutoring Group",
     xlab = "Exam Score",
     ylab = "Frequency",
     col = "lightblue",
     border = "blue",
     breaks = 10)

#### Skewness: symmetrical , Kurtosis: Proper Bell Curve

hist(Dataset6.1$Exam_Score[Dataset6.1$Group == "No Tutoring"],
     main = "Histogram of Exam Scores - No Tutoring Group",
     xlab = "Exam Score",
     ylab = "Frequency",
     col = "lightgreen",
     border = "darkgreen",
     breaks = 10)

#### Skewness: symmetrical , Kurtosis: Proper Bell Curve

Based on Reports we use the independent T-test

Method 2: Boxplots

ggboxplot(Dataset6.1, x = "Group", y = "Exam_Score",
          color = "Group",
          palette = c("blue", "green"),
          add = "jitter",
          title = "Exam Scores by Group",
          xlab = "Group",
          ylab = "Exam Score")

#### Based on the reports and the box plot we use the independent T-test”

STEP 7: SHAPIRO-WILK TEST

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

As both the tests look normal so we use the independent T-test

STEP 8: CONDUCT 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

As the value of p < 0.05 (p = 0.000233), this means the results were SIGNIFICANT.

STEP 9: CALCULATE THE EFFECT SIZE

cohens_d_result <- cohens_d(Exam_Score ~ Group, data = Dataset6.1, pooled_sd = TRUE)
print(cohens_d_result)
## Cohen's d |         95% CI
## --------------------------
## -0.86     | [-1.32, -0.40]
## 
## - Estimated using pooled SD.

As the size of the effect is (-0.86), this means the effect is ‘Large’.

#STEP 10: FINAL RESULTS REPORT

No-Tutoring Group (M = 71.94, SD = 7.67) was significantly different from Tutoring Group (M = 78.36, SD = 7.1) in exam scores, t(78) = -3.85, p = .0002. The effect size was large (Cohen’s d = -0.86).