# INDEPENDENT T-TEST & MANN-WHITNEY U TEST
# HYPOTHESIS TESTED:
#To determine if Medication A is more effective at reducing headaches than the current medication on the market (Medication B).
# H0:There is no difference in the mean number of headache days between participants taking Medication A and those taking Medication B
# H1: Medication A is more effective than Medication B, meaning the mean number of headache days for the Medication A group is less than that for the Medication B group
options(repos = c(CRAN = "https://cloud.r-project.org"))
install.packages("readxl")
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# LOAD THE PACKAGE
library(readxl)
A6R1 <- read_excel("C:/Users/chris/Downloads/A6R1.xlsx")
# DESCRIPTIVE STATISTICS
install.packages("dplyr")
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# LOAD THE PACKAGE
library(dplyr)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
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## filter, lag
## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, union
A6R1 %>%
group_by(Medication) %>%
summarise(
Mean = mean(HeadacheDays, na.rm = TRUE),
Median = median(HeadacheDays, na.rm = TRUE),
SD = sd(HeadacheDays, na.rm = TRUE),
N = n()
)
## # A tibble: 2 × 5
## Medication Mean Median SD N
## <chr> <dbl> <dbl> <dbl> <int>
## 1 A 8.1 8 2.81 50
## 2 B 12.6 12.5 3.59 50
# HISTOGRAMS
# CREATE THE HISTOGRAMS
# Replace "dataset" with your dataset name (without .xlsx)
# Replace "score" with your dependent variable R code name (example: USD)
# Replace "group" with your independent variable R code name (example: Country)
# Replace "Group1" with the R code name for your first group (example: USA)
# Replace "Group2" with the R code name for your second group (example: India)
hist(A6R1$HeadacheDays[A6R1$Medication == "A"],
main = "Histogram of Group 1 Scores",
xlab = "Value",
ylab = "Frequency",
col = "lightblue",
border = "black",
breaks = 20)
hist(A6R1$HeadacheDays[A6R1$Medication == "B"],
main = "Histogram of Group 2 Scores",
xlab = "Value",
ylab = "Frequency",
col = "lightgreen",
border = "black",
breaks = 20)
# 1)The SKEWNESS of the VARIABLE 1 histogram.
# The histogram of VARIABLE 1 (Group 1 Scores) appears fairly symmetrical with a single peak.
# Therefore, it is likely to have skewness close to zero and looks symmetrical.
# 2 The KURTOSIS of the VARIABLE 1 histogram.
# The histogram has a moderate peak and tails, suggesting a mesokurtic distribution.
# It resembles a proper bell curve.
# 3 The SKEWNESS of the VARIABLE 2 histogram.
# The histogram of VARIABLE 2 (Group 2 Scores) shows multiple peaks and a slight right tail.
# This suggests it is slightly positively skewed.
# 4 The KURTOSIS of the VARIABLE 2 histogram.
# The histogram has sharper peaks and heavier tails, indicating a leptokurtic distribution.
# It looks too tall compared to a normal bell curve.
# SHAPIRO-WILK TEST
# Purpose: Check the normality for each group's score statistically.
# The Shapiro-Wilk Test is a test that checks skewness and kurtosis at the same time.
# The test is checking "Is this variable the SAME as normal data (null hypothesis) or DIFFERENT from normal data (alternate hypothesis)?"
# For this test, if p is GREATER than .05 (p > .05), the data is NORMAL.
# If p is LESS than .05 (p < .05), the data is NOT normal.
# CONDUCT THE SHAPIRO-WILK TEST
# Replace "dataset" with your dataset name (without .xlsx)
# Replace "score" with your dependent variable R code name (example: USD)
# Replace "group" with your independent variable R code name (example: Country)
# Replace "Group1" with the R code name for your first group (example: USA)
# Replace "Group2" with the R code name for your second group (example: India)
shapiro.test(A6R1$HeadacheDays[A6R1$Medication == "A"])
##
## Shapiro-Wilk normality test
##
## data: A6R1$HeadacheDays[A6R1$Medication == "A"]
## W = 0.97852, p-value = 0.4913
shapiro.test(A6R1$HeadacheDays[A6R1$Medication == "B"])
##
## Shapiro-Wilk normality test
##
## data: A6R1$HeadacheDays[A6R1$Medication == "B"]
## W = 0.98758, p-value = 0.8741
# The data normally distributed for Variable 1
# The data normally distributed for Variable 2
#box-plot test below.
# BOXPLOT
# INSTALL REQUIRED PACKAGE
install.packages("ggplot2")
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install.packages("ggpubr")
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# LOAD THE PACKAGE
# Always reload the package you want to use.
library(ggplot2)
library(ggpubr)
# CREATE THE BOXPLOT
# Replace "dataset" with your dataset name (without .xlsx)
# Replace "score" with your dependent variable R code name (example: USD)
# Replace "group" with your independent variable R code name (example: Country)
ggboxplot(A6R1, x = "Medication", y = "HeadacheDays",
color = "Medication",
palette = "jco",
add = "jitter")
# Q1) Were there any dots outside of the boxplots? These dots represent participants with extreme scores.
# Yes, both boxplots (for Medication A and B) show several dots outside the boxes, indicating the presence of outliers.
# Q2) If there are outliers, in your opinion are the scores of those dots changing the mean so much that the mean no longer accurately represents the average score?
# Although there are outliers in both groups, they are relatively few compared to the total number of data points.
# Therefore, the mean still appears to be a reasonably accurate representation of the average score.
# If there were no extreme outliers, this means the data is NORMAL. Continue to the Independent t-test.
# INDEPENDENT T-TEST
#============================================================================
# PURPOSE: Test if there was a difference between the means of the two groups.
# Replace "dataset" with your dataset name (without .xlsx)
# Replace "score" with your dependent variable R code name (example: USD)
# Replace "group" with your independent variable R code name (example: Country)
t.test(HeadacheDays ~ Medication, data = A6R1, var.equal = TRUE)
##
## Two Sample t-test
##
## data: HeadacheDays by Medication
## t = -6.9862, df = 98, p-value = 3.431e-10
## alternative hypothesis: true difference in means between group A and group B is not equal to 0
## 95 percent confidence interval:
## -5.778247 -3.221753
## sample estimates:
## mean in group A mean in group B
## 8.1 12.6
# DETERMINE STATISTICAL SIGNIFICANCE
# Results were statistically significant (p < .05), continue to effect size section below.
# EFFECT-SIZE
install.packages("effectsize")
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library(effectsize)
# COHEN’S D
cohen_d_result <- cohens_d(HeadacheDays ~ Medication, data = A6R1, pooled_sd = TRUE)
print(cohen_d_result)
## Cohen's d | 95% CI
## --------------------------
## -1.40 | [-1.83, -0.96]
##
## - Estimated using pooled SD.
# Q1) What is the size of the effect?
# very large =-1.40 indicates the difference between the group A&B
# Q2) Group B had the higher average score?
# WRITTEN REPORT FOR INDEPENDENT T-TEST
# OUTPUT
# 1. The name of the inferential test used (Independent t-test)
# 2. The names of the IV- Medication and DV-HeadacheDays
# 3. The sample size for each group =50
# 4. The inferential test results were statistically significant (p < .05)
# 5. mean 8.1 and SD 2.81 for A
# mean 12.6 and SD 3.59 for B
# 7. Degrees of freedom =98
# 8. t-value = -6.9862
# 9. EXACT p-value to three decimals= p < .001
# 10. Effect size (Cohen’s d) =-1.40
# 2) REPORT
# REPORT
# An Independent t-test was conducted to compare
# the number of headache days between participants who took Medication A (n = 50) and those who took Medication B (n = 50).
# Participants taking Medication A reported significantly fewer headache days (M = 8.10, SD = 2.81) than
# those taking Medication B (M = 12.60, SD = 3.59), t(98) = -6.99, p < .001.
# The effect size was very large (d = -1.40), indicating a substantial difference in headache reduction between the two medications.
# Overall, Medication A was more effective at reducing headache days than Medication B..