A6R3

Hypothesis:

NULL HYPOTHESIS (H0): There is no difference in communication abilities between the PreTraining and PostTraining.

ALTERNATE HYPOTHESIS (H1): There is a difference in communication abilities between the PreTraining and PostTraining.

Load the Package

library(readxl)

Import the Excel File

A6R3 <- read_excel("C:\\Users\\Abhigna Thirakanam\\Downloads\\A6R3.xlsx")

Calculating the different scores

Before <- A6R3$PreTraining
After <- A6R3$PostTraining
Differences <- After - Before

Histogram

hist(Differences,
     main = "Histogram of Difference Scores",
     xlab = "Value",
     ylab = "Frequency",
     col = "blue",
     border = "black",
     breaks = 20)

Questions

1: Is the histograms symmetrical, positively skewed, or negatively skewed?

ANSWER: Approximately Symmetrical

2: Did the histogram look too flat, too tall, or did it have a proper bell curve?

ANSWER: Proper Bell Curve

Shapiro Wilk Test

shapiro.test(Differences)

## 
##  Shapiro-Wilk normality test
## 
## data:  Differences
## W = 0.98773, p-value = 0.21

Box Plot

boxplot(Differences,
        main = "Distribution of Score Differences (After - Before)",
        ylab = "Difference in Scores",
        col = "blue",
        border = "darkblue")

Questions

1: How many dots are in your boxplot?

ANSWER: B - One or two dots

2: Where are the dots in your boxplot?

ANSWER: C - Far from the whiskers

3: Based on the dots and there location, is the data normal?

ANSWER: Data is normal or abnormal

Descriptive Statistics for PreTraining

mean(Before, na.rm = TRUE)

## [1] 59.73333

median(Before, na.rm = TRUE)

## [1] 60

sd(Before, na.rm = TRUE)

## [1] 7.966091

length(Before)

## [1] 150

Descriptive Statistics for PostTraining

mean(After, na.rm = TRUE)

## [1] 69.24

median(After, na.rm = TRUE)

## [1] 69.5

sd(After, na.rm = TRUE)

## [1] 9.481653

length(After)

## [1] 150

Dependent T-Test

t.test(Before, After, paired = TRUE)

## 
##  Paired t-test
## 
## data:  Before and After
## t = -23.285, df = 149, p-value < 2.2e-16
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -10.313424  -8.699909
## sample estimates:
## mean difference 
##       -9.506667

Load the Package

library(effectsize)

Calculating Cohen’s D

cohens_d(Before, After, paired = TRUE)

## For paired samples, 'repeated_measures_d()' provides more options.

## Cohen's d |         95% CI
## --------------------------
## -1.90     | [-2.17, -1.63]

Questions

1: What is the size of the effect?

ANSWER: A Cohen’s D of 1.90 indicates the diffrence between the group averages is very large

2: Which group had the higher average score?

ANSWER: The PostTraining group have the higher average score.

Research Report on Results: A dependent t-test was conducted to compare employees communication skill scores PreTraining and PostTraining among 150 employees. Results showed that PostTraining communication scores (M = 69.24, SD = 9.48) were significantly higher than PreTraining scores (M = 59.73, SD = 7.97), t(149) = –23.29, p < .001.The effect size was Cohen’s d = 1.90, indicating a very large effect. These results suggest that the communication training program led to a substantial improvement in employees communication abilities.