Chi-Square Test of Independence

This analysis is for research scenario 2. It tests to determine whether there is an association between uniform design preference and parent type.

Hypotheses

Null Hypothesis (H₀): There is no association between parent type and uniform design preference.
Alternative Hypothesis (H₁): There is an association between parent type and uniform design preference.

Results Paragraph

A Chi-Square Test of Independence was conducted to examine the association between parent type (mother, father) and uniform design preference (Design A, Design B) among 100 participants. The results indicated that there was no statistically significant association between parent type and uniform design preference, χ²(1, N = 100) = 3.25, p = .072. Although the result was not statistically significant, the effect size was small (Cramér’s V = 0.18).

R Code

```r # Load required packages library(readxl) library(lsr)

Import the dataset

RQ2 <- read_excel(“C:/Users/dadal/Downloads/RQ2.xlsx”)

Create contingency table

contingencytable <- table(RQ2\(Parent, RQ2\)Preferred_Design)

Run Chi-Square Test of Independence

chisq_indep <- chisq.test(contingencytable) print(chisq_indep)

## Pearson’s Chi-squared test with Yates’ continuity correction ## ## data: contingencytable ## X-squared = 3.2452, df = 1, p-value = 0.07163

Calculate Cramér’s V

cramers_v <- cramersV(contingencytable) cat(“Cramer’s V (effect size):”, round(cramers_v, 3), “”)