DATA_101_CHI_ANOVA
Prasanna Sadashivaiah
Problem 1
# Observed counts
observed <- c(244, 192)
# Chi-Square goodness-of-fit test
chisq.test(observed, p= c(0.5,0.5))##
## Chi-squared test for given probabilities
##
## data: observed
## X-squared = 6.2018, df = 1, p-value = 0.01276# P-Value
chisq.test(observed, p = c(0.5, 0.5))$p.value## [1] 0.01276179Conclusion: Assuming a standard significance level of Alpha = 0.05, we reject the null hypothesis because the p-value (0.013) is less than 0.05, P=0.013 < 0.05. Therefore, we reject the null hypothesis. There is sufficient statistical evidence to conclude that the functional R allele and non-functional X allele are not equally likely to occur within this population.The chi-square goodness-of-fit test yields a test statistic of X^2 ~ 6.20 and a p-value of 0.013, which indicates that the R and X alleles are not distributed with equal probabilities in the studied population.
Problem 2
NutritionStudy <- read.csv("NutritionStudy.csv")
# Create a contingency table
vit_gender <- table(NutritionStudy$VitaminUse, NutritionStudy$Sex)
# view the table
vit_gender##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13# Chi-square test of association
chi_test <- chisq.test(vit_gender)
# Display results
chi_test##
## Pearson's Chi-squared test
##
## data: vit_gender
## X-squared = 11.071, df = 2, p-value = 0.003944# Extract p-value
chi_test$p.value## [1] 0.003944277# Display expected counts
chi_test$expected##
## Female Male
## No 96.20000 14.80000
## Occasional 71.06667 10.93333
## Regular 105.73333 16.26667# show percentages within each gender
round(prop.table(vit_gender, margin =2) * 100, 2)##
## Female Male
## No 31.87 57.14
## Occasional 28.21 11.90
## Regular 39.93 30.95# Test statistics
chi_test$parameter## df
## 2# Degree of freedom
chi_test$parameter## df
## 2#p-value
chi_test$p.value## [1] 0.003944277Conclusion: Since the p-value (0.0039) is much lower than the standard significance level of alpha = 0.05, we reject the null hypothesis. There is strong empirical evidence to conclude that a statistically significant association exists between gender and vitamin use in this population, with females demonstrating a higher likelihood of taking vitamins overall.
Problem 3
#Read the data
fish <- read.csv ("FishGills3.csv")
# view the first few rows
head(fish)## Calcium GillRate
## 1 Low 55
## 2 Low 63
## 3 Low 78
## 4 Low 85
## 5 Low 65
## 6 Low 98# Convert Calcium to a factor
fish$Calcium <- as.factor(fish$Calcium)
# Summary statistics by group
aggregate(GillRate ~ Calcium, data = fish, FUN = function(x) c(mean = mean(x), sd = sd(x), n = length(x)))## Calcium GillRate.mean GillRate.sd GillRate.n
## 1 High 58.16667 13.77675 30.00000
## 2 Low 68.50000 16.23481 30.00000
## 3 Medium 58.66667 14.28366 30.00000# One-way ANOVA
gill_aov <- aov(GillRate ~ Calcium, data = fish)
# Check Assumption
# 1. Normality of residuals
shapiro.test(residuals(gill_aov))##
## Shapiro-Wilk normality test
##
## data: residuals(gill_aov)
## W = 0.96502, p-value = 0.01593# Q-Q Plot
qqnorm(residuals(gill_aov))
qqline(residuals(gill_aov))
# Homegeneity of vaiences
bartlett.test(GillRate ~ Calcium, data = fish)##
## Bartlett test of homogeneity of variances
##
## data: GillRate by Calcium
## Bartlett's K-squared = 0.87077, df = 2, p-value = 0.647TukeyHSD(gill_aov)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = GillRate ~ Calcium, data = fish)
##
## $Calcium
## diff lwr upr p adj
## Low-High 10.333333 1.219540 19.4471264 0.0222533
## Medium-High 0.500000 -8.613793 9.6137931 0.9906108
## Medium-Low -9.833333 -18.947126 -0.7195402 0.0313247# Boxplot for visualization
boxplot(GillRate ~ Calcium,data = fish,
col = c("lightblue", "lightgreen", "lightpink"),
main = "Gill Rate by Calcium Level", xlab = "Calcium Level",
ylab = "Gill Rate (beats/min)")
#Conclusion: Reject Ho. There is significant evidence that the mean fish gill rate varies with calcium level in the water.