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))
chisq.test(observed <- c(244, 192))
## 
##  Chi-squared test for given probabilities
## 
## data:  observed <- c(244, 192)
## 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-value = 0.01276 and reject it


**Q3. 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 observed distribution significantly differs from a 50/50 split.
---

## 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.


``` r
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,so yes reject it is under 0.05


**Q7. Write your conclusion in plain English:**
There is sufficient evidence of an association between VitaminUse and Sex. This suggests that vitamin use patterns differ between males and females.
---

## 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:

- Low (0.71 mg/L)
- Medium (5.24 mg/L)
- High (18.24 mg/L)

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.


``` r
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_anova <- aov(GillRate ~ Calcium, data = fish)

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

Since 0.0121 < 0.05, Reject H₀

Q11. Write your conclusion in plain English: There is sufficient evidence that mean gill rate differs among at least one of the calcium levels. This suggests that calcium concentration has a statistically signifcant effect on fish gill rate