Three problems — one chi-square goodness-of-fit, one chi-square test of independence, and one one-way ANOVA. Each one follows the same pattern: state hypotheses, run the test, interpret the result.

Problem 1 — Chi-Square Goodness of Fit (ACTN3 alleles)

ACTN3 is a gene that encodes alpha-actinin-3, a protein in fast-twitch muscle fibers. The gene has two main alleles: R (functional) and X (non-functional). The R allele is linked to better performance in strength/speed/power sports; the X allele is associated with endurance.

A study of 436 people classified 244 as R and 192 as X. Does this provide evidence that the two options are NOT equally likely?

State your hypotheses:

# Q1. Run a chi-square goodness-of-fit test.
#     (Hint: observed <- c(244, 192); chisq.test(observed))
# Following the directions in the hint, I make a vector with the relevant values and input it into chisq.test.
observed <- c(244, 192)
chisq.test(observed)
## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 6.2018, df = 1, p-value = 0.01276
# Q2. What is the p-value? At α = 0.05, do you reject H₀?
# p = 0.01276, yes we reject the null hypothesis.

Q3. Write your conclusion in plain English: We have strong evidence to conclude that the R and X allele are not equally likely in the population (p=0.013). — ## Problem 2 — Chi-Square Test of Independence (Vitamin Use & Gender)

The NutritionStudy.csv dataset contains data on vitamin use (VitaminUse) and gender (Sex) for many participants. Is there a significant association between these two variables?

Download NutritionStudy.csv from the Datasets folder on Blackboard.

nutrition <- read.csv("NutritionStudy.csv")

State your hypotheses:

# Q4. Build a contingency table of VitaminUse and Sex using table().
# I use the table function to make a table and the $ operator to select the relevant columns from the df. 
nut_table <- table(nutrition$VitaminUse, nutrition$Sex)

# Q5. Run a chi-square test of independence on that table.
chisq.test(nut_table)
## 
##  Pearson's Chi-squared test
## 
## data:  nut_table
## X-squared = 11.071, df = 2, p-value = 0.003944
# Q6. What is the p-value? Do you reject H₀ at α = 0.05?
# p = 0.003944, yes, we reject the null hypothesis.

Q7. Write your conclusion in plain English: We have storng evidence to conclude that vitamin use and sex are associated (p=0.0039). —

Problem 3 — One-Way ANOVA (Fish Gills)

Researchers wanted to know how water chemistry affects fish ventilation. Fish were randomly assigned to one of three tanks with different calcium levels:

The team counted gill rates (beats per minute) for 30 fish in each tank. The data is in FishGills3.csv.

Download FishGills3.csv from the Datasets folder on Blackboard.

fish <- read.csv("FishGills3.csv")

State your hypotheses:

# Q8. Run a one-way ANOVA testing GillRate by Calcium.
#     (Hint: aov(GillRate ~ Calcium, data = fish))
# Following the instructions in the hint, I use the anova function. I put the dependent variable on the left and the group after it, seperated by ~. I set data = the relevant df. 

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
# Q9. Use summary() on the result. What is the F statistic and p-value?
# F = 4.648, p = 0.0121
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
# Q10. At α = 0.05, do you reject H₀?
# Yes, we reject the null hypothesis

Q11. Write your conclusion in plain English: We have strong evidence to conclude that the mean gill rate is not the same across all calcium levels.