A6Q3

Open the Installed Packages

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)

Import and Name the Dataset

A6Q3 <- read_excel("C:/Users/krish/Downloads/A6Q3-2.xlsx")

Calculate the Descriptive Statistics

A6Q3 %>%
  group_by(Exercise) %>%
  summarise(
    Mean = mean(Weight, na.rm = TRUE),
    Median = median(Weight, na.rm = TRUE),
    SD = sd(Weight, na.rm = TRUE),
    N = n()
  )

## # A tibble: 2 × 5
##   Exercise  Mean Median    SD     N
##   <chr>    <dbl>  <dbl> <dbl> <int>
## 1 cardio    74.7   73.3  7.57    25
## 2 nocardio  70.8   69.5  7.35    25

Create Histograms (Normality Check #1)

hist(A6Q3$Weight[A6Q3$Exercise == "cardio"],
     main = "Histogram of Cardio",
     xlab = "Value",
     ylab = "Frequency",
     col = "lightyellow",
     border = "black",
     breaks = 10)

hist(A6Q3$Weight[A6Q3$Exercise == "nocardio"],
     main = "Histogram of Nocardio",
     xlab = "Value",
     ylab = "Frequency",
     col = "brown",
     border = "black",
     breaks = 10)

Interpret the Histograms

Group 1: cardio

The first variable looks normally distributed.

The data is symmetrical.

The data has a proper bell curve.

Group 2: nocardio

The second variable looks normally distributed.

The data is symmetrical.

The data has a proper bell curve.

Create Boxplots for Outliers (Normality Check #2)

ggboxplot(A6Q3, x = "Exercise", y = "Weight",
          color = "Exercise",
          palette = "jco",
          add = "jitter")

Interpret the Boxplots

Boxplot 1: Nocardio

There are no dots outside the boxplot.

Boxplot 2: Cardio

There are no dots outside the boxplot.

Shapiro-Wilk Tests (Normality Check #3)

shapiro.test(A6Q3$Weight[A6Q3$Exercise == "cardio"])

## 
##  Shapiro-Wilk normality test
## 
## data:  A6Q3$Weight[A6Q3$Exercise == "cardio"]
## W = 0.96745, p-value = 0.5812

shapiro.test(A6Q3$Weight[A6Q3$Exercise == "nocardio"])

## 
##  Shapiro-Wilk normality test
## 
## data:  A6Q3$Weight[A6Q3$Exercise == "nocardio"]
## W = 0.97686, p-value = 0.8166

Interpret the Shapiro-Wilk Test Group 1: Exercise

The first group is normally distributed, p >.05.

Group 2: Weight

The second group is normally distributed, p >.05.

Conduct the Independent T-Test

t.test(Weight ~ Exercise, data = A6Q3, var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  Weight by Exercise
## t = 1.8552, df = 48, p-value = 0.06971
## alternative hypothesis: true difference in means between group cardio and group nocardio is not equal to 0
## 95 percent confidence interval:
##  -0.3280454  8.1605622
## sample estimates:
##   mean in group cardio mean in group nocardio 
##               74.73336               70.81710

Report the Independent T-Test

An Independent T-Test was conducted to determine if there was a difference in Weight between cardio and nocardio.

Cardio scores (M = 74.7, SD = 7.57) were significantly different from nocardio scores (M = 70.8, SD = 7.35), t(48) = 1.85, p >0.05.