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#Problem 1

observed <- c(R = 244, X = 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

Because the chi-square test produced a p-value of 0.01276, which is less than 0.05, we reject the null hypothesis. The R and X alleles are not equally likely in the population, and the observed distribution differs significantly from a 50/50 ratio.

df <- read.csv("NutritionStudy (1).csv")
tab <- table(df$VitaminUse, df$Sex)
chisq_res <- chisq.test(tab)
chisq_res$expected
##             
##                 Female     Male
##   No          96.20000 14.80000
##   Occasional  71.06667 10.93333
##   Regular    105.73333 16.26667
min(chisq_res$expected)
## [1] 10.93333
chisq_res
## 
##  Pearson's Chi-squared test
## 
## data:  tab
## X-squared = 11.071, df = 2, p-value = 0.003944

Since the p-value (0.00394) is less than the significance level (α = 0.05), we reject the null hypothesis. There is statistically significant evidence of an association between vitamin use and sex. In this sample, vitamin-taking behavior appears to differ between males and females. Based on the contingency table, females were more likely to take vitamins than males.

df1 <- read.csv("FishGills3 (1).csv")
df1$Calcium <- as.factor(df1$Calcium)
model <- aov(GillRate ~ Calcium, data = df1)
summary(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

Since the ANOVA test produced a p-value of 0.0121, which is less than 0.05, we reject the null hypothesis. This indicates that mean gill beat rate differs significantly among the three calcium levels.