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:
H₀: p₁ = p₂ = 0.5
H₁: at least one proportion is not 0.5
H₀: The two alleles R and X are equally likely
H₁: The two alleles R and X are not equally likely
# 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?
#The p-value is 0.01276. Since the p-value is less than α = 0.05, we reject H₀Q3. Write your conclusion in plain English:
There is sufficient evidence to conclude that the R and X alleles are not equally likely in the population
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.
chisq.test(vit_table)##
## 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?The p-value is 0.003944. Since the p-value is less than α = 0.05,We reject H₀.
Q7. Write your conclusion in plain English:
There is sufficient evidence to conclude that vitamin use and sex are associated
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")
anova_result <- aov(GillRate ~ Calcium, data = fish)
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 ' ' 1State your hypotheses:
H₀: μ₁ = μ₂ = μ₃ (mean gill rate is the same across calcium levels)
H₁: at least one mean is different
H₀:The mean gill rate is the same for the low, medium,and high calcium levels.
H₁: at least one mean is different calcium level has a different mean gill rate.
# Q8. Run a one-way ANOVA testing GillRate by Calcium.
# (Hint: aov(GillRate ~ Calcium, data = fish))
fish <- read.csv("FishGills3.csv")
anova_result <- aov(GillRate ~ Calcium, data = fish)
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 ' ' 1Q9. Use summary() on the result. What is the F statistic and p-value?
F statistic = 4.648 p-value = 0.0121
Q10. At α = 0.05, do you reject H₀?
Since the p_value is 0.0121, which is less than α = 0.05, We reject H₀.
Q11. Write your conclusion in plain English:**
There is sufficient evidence to conclude that the mean gill rate is not the same across all calcium levels. At least one calcium level has a different mean gill rate.