Loading libraries

library(tidyverse)

Loading data set

nutrition <- read_csv("NutritionStudy.csv")
fishgills <- read_csv("FishGills3.csv")

Problem 1

observed counts

observed <- c(244,192)

theoritical_prop <- rep(1/2,2) ##Null values

Hypothesis Test

\(H_0\):\(p_R\) = \(p_X\) = 1/2 \(H_a\): R and X alleles are not equally likely.

Checking expected values to see if we can perform chi-square test

expected_values <- theoritical_prop * sum(observed)
expected_values
## [1] 218 218

All values are more than 5 so, we are doing the chi-square test

Chi-square test

chisq.test(observed)
## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 6.2018, df = 1, p-value = 0.01276

Therefore, based on the p-value (0.01276) obtained, we reject the idea that the outcomes are equally likely and conclude that there are differences in R and X alleles.

Problem 2

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

\(H_0\) : Vitamin use is not associated with gender \(H_a\) : Vitamin use is associated with gender

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

with a p-value of 0.003944, which is less than the typical significance level of 0.05, there is sufficient evidence to reject the null hypothesis.

Therefore, we conclude that there is a significant association between vitamin use and the gender.

Problem 3

\(H_0\): All calcium level groups have the same mean gill rate

\(H_a\): At least one calcium group has a different mean gill rate

anova_result <- aov(GillRate ~ Calcium, data = fishgills)
anova_result
## Call:
##    aov(formula = GillRate ~ Calcium, data = fishgills)
## 
## Terms:
##                   Calcium Residuals
## Sum of Squares   2037.222 19064.333
## Deg. of Freedom         2        87
## 
## Residual standard error: 14.80305
## Estimated effects may be unbalanced
summary(anova_result)
##             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

The p-value is small (0.0121): indicating strong evidence against the null hypothesis. Overall, this test suggests that at least one calcium group has a different mean gill rate.