In your markdown answer the following problems. Include the following: Your hypotheses. P-value Conclusion
ACTN3 is a gene that encodes alpha-actinin-3, a protein in fast-twitch muscle fibers, important for activities like sprinting and weightlifting. The gene has two main alleles: R (functional) and X (non-functional). The R allele is linked to better performance in strength, speed, and power sports, while the X allele is associated with endurance due to a greater reliance on slow-twitch fibers. However, athletic performance is influenced by various factors, including training, environment, and other genes, making the ACTN3 genotype just one contributing factor.
A study examines the ACTN3 genetic alleles R and X, also associated with fast-twitch muscles. Of the 436 people in this sample, 244 were classified as R, and 192 were classified as X. Does the sample provide evidence that the two options are not equally likely? Conduct the test using a chi-square goodness-of-fit test
\(H_o\): \(p_1\) = \(p_2\) = 1/2, where \(p_1\) is the proportion of those classified with R and \(p_2\) is the proportion of those classified with X.
\(H_a\): At least one \(p_i\) \(\neq\) 1/3
observed <- c(244,192)
# Null values
theoritical_prop <- rep(1/2, 2) # repeat 1/2, 3 timesexpected_values <- theoritical_prop*sum(observed)
expected_values## [1] 218 218Both are greater than 5. We can conduct the Chi-Square test.
chisq.test(observed)##
## Chi-squared test for given probabilities
##
## data: observed
## X-squared = 6.2018, df = 1, p-value = 0.01276Our p-value is 0.01276, which is less than 0.05. Reject the Null, there is evidence showing at least one proportion is not equal to 1/2. This means that the ACTN3 genetic alleles R and X are not equally likely to happen.
Who Is More Likely to Take Vitamins: Males or Females? The dataset NutritionStudy contains, among other things, information about vitamin use and the gender of the participants. Is there a significant association between these two variables? Use the variables VitaminUse and Gender to conduct a chi-square analysis and give the results. (Test for Association)
Load the Dataset
setwd("C:/Users/sarah/Downloads")
nutrition <- read.csv("NutritionStudy.csv")\(H_o\): There is no significant association between the information about vitamin use and the gender.
\(H_a\): There is a significant association between the information about vitamin use and the gender.
observed_dataset <- table(nutrition$VitaminUse, nutrition$Sex)
observed_dataset##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13chisq.test(nutrition$VitaminUse, nutrition$Sex)$expected## nutrition$Sex
## nutrition$VitaminUse Female Male
## No 96.20000 14.80000
## Occasional 71.06667 10.93333
## Regular 105.73333 16.26667All counts are above 5. We can perform a Chi-Square Test.
chisq.test(observed_dataset)##
## Pearson's Chi-squared test
##
## data: observed_dataset
## X-squared = 11.071, df = 2, p-value = 0.003944With a p-value of 0.003944, which is less than 0.05, we reject the null. There is a significant association between the information about vitamin use and the gender.
Most fish use gills for respiration in water, and researchers can observe how fast a fish’s gill cover beats to study ventilation, much like we might observe a person’s breathing rate. Professor Brad Baldwin is interested in how water chemistry might affect gill beat rates. In one experiment, he randomly assigned fish to tanks with different calcium levels. One tank was low in calcium (0.71 mg/L), the second tank had a medium amount (5.24 mg/L), and the third tank had water with a high calcium level (18.24 mg/L). His research team counted gill rates (beats per minute) for samples of 30 fish in each tank. The results are stored in FishGills3. Perform ANOVA test to see if the mean gill rate differs depending on the calcium level of the water.
Load the Dataset
setwd("C:/Users/sarah/Downloads")
fishGills <- read.csv("FishGills3.csv")\(H_o\): \(μ_l\) = \(μ_m\) = \(μ_h\), where \(μ_l\) is low calcium levels, \(μ_m\) is medium calcium levels, and \(μ_h\) is high calcium levels.
\(H_a\): Atleast one \(μ_i\) is different.
anova_test <- aov(GillRate ~ Calcium, data = fishGills) # Numeric ~ Categorical
summary(anova_test)## 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 ' ' 1With a p-value of 0.0121, which is less than 0.05, we reject the null. There is sufficient evidence to condclude that at least one of the mean calcium levels is significantly different among the different calcium levels.