Problem 1: ACTN3 Gene Alleles

Hypotheses

Let \(p_R\) = proportion of R allele in the population

Let \(p_X\) = proportion of X allele in the population

\[H_0: p_R = p_X\] \[H_a: p_R \neq p_X\]

observed <- c(244, 192)
names(observed) <- c("R", "X")

expected <- c(218, 218)

chisq_test1 <- chisq.test(observed, p = c(0.5, 0.5))

print(chisq_test1)
## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 6.2018, df = 1, p-value = 0.01276

Conclusion

Since the p-value (0.0128) is less than 0.05, we reject the null hypothesis.

There is sufficient evidence to conclude that the two ACTN3 alleles (R and X) are not equally likely in the population.

Problem 2: Vitamin Use and Gender

Hypotheses

Let \(p_{male}\) = proportion of males who use vitamins regularly

Let \(p_{female}\) = proportion of females who use vitamins regularly

\[H_0: p_{male} = p_{female}\] \[H_a: p_{male} \neq p_{female}\]

Analysis

nutrition <- read_csv("NutritionStudy.csv")
## Rows: 315 Columns: 17
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr  (3): Smoke, Sex, VitaminUse
## dbl (14): ID, Age, Quetelet, Vitamin, Calories, Fat, Fiber, Alcohol, Cholest...
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
contingency_table <- table(nutrition$Sex, nutrition$VitaminUse)
print(contingency_table)
##         
##           No Occasional Regular
##   Female  87         77     109
##   Male    24          5      13
chisq_test2 <- chisq.test(contingency_table)

# Display results
print(chisq_test2)
## 
##  Pearson's Chi-squared test
## 
## data:  contingency_table
## X-squared = 11.071, df = 2, p-value = 0.003944

Conclusion

Since the p-value (0.0039) is less than 0.05, we reject the null hypothesis.

There is sufficient evidence to conclude that there is a significant association between vitamin use and gender. Males and females differ in their vitamin use patterns.

Problem 3: Fish Gill Rates and Calcium Levels

Hypotheses

Let \(\mu_{low}\) = mean gill rate for low calcium level

Let \(\mu_{medium}\) = mean gill rate for medium calcium level

Let \(\mu_{high}\) = mean gill rate for high calcium level

\[H_0: \mu_{low} = \mu_{medium} = \mu_{high}\] \[H_a: \text{At least one mean differs from the others}\]

Analysis

fish <- read_csv("FishGills3.csv")
## Rows: 90 Columns: 2
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (1): Calcium
## dbl (1): GillRate
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
anova_model <- aov(GillRate ~ Calcium, data = fish)
anova_result <- summary(anova_model)
print(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

Conclusion

Since the p-value (0.0121) is less than 0.05, we reject the null hypothesis.

There is sufficient evidence to conclude that the mean gill rate differs significantly depending on the calcium level of the water. At least one calcium level produces a different mean gill rate than the others.