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.
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))
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?
# X-squared = 6.2018, df = 1, p-value = 0.01276
Q3. Write your conclusion in plain English: There is enough evidence to conclude that the R and X alleles are not equally common in this sample. The observed proportions differ significantly from a 50/50 distribution. —
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().
table_vit <- table(nutrition$VitaminUse, nutrition$Sex)
table_vit
##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13
# Q5. Run a chi-square test of independence on that table.
chisq.test(table_vit)
##
## Pearson's Chi-squared test
##
## data: table_vit
## 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-value < 0.05 → Reject H₀. |> p-value = 0.003944 we can reject the null
#p-value ≥ 0.05 → Fail to reject H₀.
Q7. Write your conclusion in plain English: There is sufficient evidence to conclude that vitamin use is associated with gender. —
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))
fish_aov <- aov(GillRate ~ Calcium, data = fish)
# Q9. Use summary() on the result. What is the F statistic and p-value?
summary(fish_aov)
## 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?
# p-value < 0.05, reject H₀.
Q11. Write your conclusion in plain English: There is sufficient evidence to conclude that calcium level affects the mean gill rate of the fish.