nutrition <- read.csv("NutritionStudy.csv")
fish <- read.csv("FishGills3.csv")
\(H_0\):\(p_1\) = \(p_2\) = 1/2
\(H_a\): at least one \(p_i\) \(\neq\) 1/2
# Observed Counts
observed <- c(244, 192)
# Null values
theoritical_prop <- rep(1/2, 2)
# Make sure Chi-square test can be performed
expected_values <- theoritical_prop*sum(observed)
expected_values
## [1] 218 218
chisq.test(observed)
##
## Chi-squared test for given probabilities
##
## data: observed
## X-squared = 6.2018, df = 1, p-value = 0.01276
P Value: 0.01276
Conclusion: There is statistically significant evidence to reject the null hypothesis that the alleles are equally likely.
\(H_0\) : Vitamin use is not
associated with gender
\(H_a\) :
Vitamin use is associated with gender
observed_dataset<- table(nutrition$VitaminUse, nutrition$Sex)
observed_dataset
##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13
chisq.test(observed_dataset)
##
## Pearson's Chi-squared test
##
## data: observed_dataset
## X-squared = 11.071, df = 2, p-value = 0.003944
P value: 0.003944
Conclusion: There is statistically significant evidence to reject the null hypothesis that vitamin use is not associated with gender.
\(H_0\): \(\mu_l\) = \(\mu_m\) = \(\mu_h\)
\(H_a\): not all \(\mu_i\) are equal
anova_result <- aov(GillRate ~ Calcium, data = fish)
anova_result
## Call:
## aov(formula = GillRate ~ Calcium, data = fish)
##
## 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
P-value: 0.0121
Conclusion:
There is statistically sufficient evidence to reject the null hypothesis
that the mean gill rates are equal, showing that there are significant
differences among different calcium levels.