Problem 1: Chi-Square Goodness-of-Fit Test (ACTN3 Gene)

Research Question

Does the sample provide evidence that the ACTN3 alleles R and X are not equally likely?

Hypotheses

  • Null hypothesis (H₀): The R and X alleles are equally likely (50% each).
  • Alternative hypothesis (H₁): The R and X alleles are not equally likely.

Observed Data

  • R allele: 244
  • X allele: 192
  • Total: 436

Expected Counts

If equally likely: - R = 218 - X = 218

Chi-Square Test

observed <- c(244, 192)
expected <- c(218, 218)

chisq.test(x = 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 p-value ≈ 0.0128

###Conclusion Since the p-value is less than 0.05, we reject the null hypothesis. There is statistically significant evidence that the ACTN3 alleles R and X are not equally distributed in this sample.

##Problem 2: Chi-Square Test of Association (Vitamin Use and Gender) Research Question Is there an association between Vitamin Use and Gender?

###Hypotheses Null hypothesis (H₀): Vitamin use and gender are independent.

Alternative hypothesis (H₁): Vitamin use and gender are associated.

###Data Preparation

nutrition <- read.csv("NutritionStudy.csv")

table_vitamin_gender <- table(nutrition$VitaminUse, nutrition$Sex)
table_vitamin_gender
##             
##              Female Male
##   No             87   24
##   Occasional     77    5
##   Regular       109   13
#Chi-Square Test

chisq.test(table_vitamin_gender)
## 
##  Pearson's Chi-squared test
## 
## data:  table_vitamin_gender
## X-squared = 11.071, df = 2, p-value = 0.003944

P-value p-value ≈ 0.003

###Conclusion Because the p-value is less than 0.05, we reject the null hypothesis. There is a significant association between gender and vitamin use, indicating that males and females differ in vitamin consumption behavior.

##Problem 3: One-Way ANOVA (Gill Rate and Calcium Level) Research Question Does the mean gill beat rate differ based on calcium level in water?

###Hypotheses Null hypothesis (H₀): Mean gill rate is the same for all calcium levels.

###Alternative hypothesis (H₁): At least one calcium level has a different mean gill rate.

###Data Preparation

# Read data
fish <- read.csv("FishGills3.csv")
str(fish)
## 'data.frame':    90 obs. of  2 variables:
##  $ Calcium : chr  "Low" "Low" "Low" "Low" ...
##  $ GillRate: int  55 63 78 85 65 98 68 84 44 87 ...
# Convert Calcium to factor
fish$Calcium <- as.factor(fish$Calcium)

# One-Way ANOVA
anova_model <- aov(GillRate ~ Calcium, data = fish)
summary(anova_model)
##             Df Sum Sq Mean Sq F value Pr(>F)  
## Calcium      2   2037  1018.6   4.648 0.0121 *
## Residuals   87  19064   219.1                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

P-value p-value ≈ 0.012

###Conclusion Since the p-value is less than 0.05, we reject the null hypothesis. There is statistically significant evidence that the mean gill beat rate differs depending on the calcium level in the water.