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)
gof <- chisq.test(observed)
gof##
## 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?
gof$p.value## [1] 0.01276179# p-value ≈ 0.013 < 0.05, so we REJECT H0.Q3. Write your conclusion in plain English:
The chi-square statistic is about 6.20 with a p-value of about 0.013, which is below 0.05, so we reject the null hypothesis. There is significant evidence that the R and X alleles are not equally likely — in this sample the R allele (244) occurs more often than the X allele (192).
The NutritionStudy.csv dataset contains data on vitamin
use (VitaminUse) and gender (Sex) for many
participants. Is there a significant association between these two
variables?
Download NutritionStudy.csv from the Datasets folder on
Blackboard.
nutrition <- read.csv("NutritionStudy.csv")State your hypotheses:
# Q4. Build a contingency table of VitaminUse and Sex using table().
vit_table <- table(nutrition$VitaminUse, nutrition$Sex)
vit_table##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13# Q5. Run a chi-square test of independence on that table.
ind_test <- chisq.test(vit_table)
ind_test##
## Pearson's Chi-squared test
##
## data: vit_table
## X-squared = 11.071, df = 2, p-value = 0.003944# Q6. What is the p-value? Do you reject H₀ at α = 0.05?
ind_test$p.value## [1] 0.003944277# p-value ≈ 0.004 < 0.05, so we REJECT H0.Q7. Write your conclusion in plain English:
The chi-square test of independence gives a statistic of about 11.07 (df = 2) with a p-value of about 0.004, which is well below 0.05, so we reject the null hypothesis. There is a statistically significant association between vitamin use and gender — the pattern of vitamin use differs between males and females in this dataset.
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# F statistic ≈ 4.65, p-value ≈ 0.012.
# Q10. At α = 0.05, do you reject H₀?
# p-value ≈ 0.012 < 0.05, so we REJECT H0.Q11. Write your conclusion in plain English:
The ANOVA gives an F statistic of about 4.65 with a p-value of about 0.012, which is below 0.05, so we reject the null hypothesis. There is significant evidence that the mean gill (ventilation) rate is not the same across the three calcium levels — at least one calcium level produces a different average gill rate, so water calcium concentration does appear to affect fish ventilation.