Nutrition <- read.csv("/Users/danielmedlin/Downloads/NutritionStudy.csv")
Gills <- read.csv("/Users/danielmedlin/Downloads/FishGills3.csv")##Question 1 #Hypotheses
Null Hypothesis \(H_0\): The R and X alleles are equally likely.
Alternative Hypothesis \(H_1\): The R and X alleles are not equally likely.
alleles <- c(244, 192)
chisq.test( alleles, p = c(0.5, 0.5))##
## Chi-squared test for given probabilities
##
## data: alleles
## X-squared = 6.2018, df = 1, p-value = 0.01276#Conclusion Since the p-value of 0.01276 is less than the p-value of 0.05, we reject the null hypothesis since there’s not enough evidence the R and X alleles can happen equally.
##Question 2 #Hypotheses
Null Hypothesis \(H_0\): Vitamin use and Gender are not associated.
Alternative Hypothesis \(H_1\): Vitamin use and Gender are associated.
vitamin_table <- table(Nutrition$VitaminUse,
Nutrition$Sex)
vitamin_table##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13chisq.test(vitamin_table)##
## Pearson's Chi-squared test
##
## data: vitamin_table
## X-squared = 11.071, df = 2, p-value = 0.003944#Conclusion The test came back with a p-value of 0.003944, which is less than 0.05. With this, we reject the null hypothesis and conclude there is a significant association between gender and vitamin use.
##Question 3 #Hypotheses
Null \(H_0\): Mean gill rates are equal in all levels of calcium.
Alternative \(H_1\): At least on of the calcium levels have a different mean gill rate.
gill_rates <- aov(GillRate ~ Calcium, data = Gills)
summary(gill_rates)## 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#Conclusion Since the p-value of 0.0121 is less than 0.05, we reject the null hypothesis. There is enough evidence to conclude that mean gill rates do differ in different calcium levels.