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:
The proportion of the R allele and the X allele are both equal to 0.5
# 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?Q3. Write your conclusion in plain English: p-value = 0.01276
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")
dim(nutrition)## [1] 315 17head(nutrition)## ID Age Smoke Quetelet Vitamin Calories Fat Fiber Alcohol Cholesterol
## 1 1 64 No 21.4838 1 1298.8 57.0 6.3 0.0 170.3
## 2 2 76 No 23.8763 1 1032.5 50.1 15.8 0.0 75.8
## 3 3 38 No 20.0108 2 2372.3 83.6 19.1 14.1 257.9
## 4 4 40 No 25.1406 3 2449.5 97.5 26.5 0.5 332.6
## 5 5 72 No 20.9850 1 1952.1 82.6 16.2 0.0 170.8
## 6 6 40 No 27.5214 3 1366.9 56.0 9.6 1.3 154.6
## BetaDiet RetinolDiet BetaPlasma RetinolPlasma Sex VitaminUse PriorSmoke
## 1 1945 890 200 915 Female Regular 2
## 2 2653 451 124 727 Female Regular 1
## 3 6321 660 328 721 Female Occasional 2
## 4 1061 864 153 615 Female No 2
## 5 2863 1209 92 799 Female Regular 1
## 6 1729 1439 148 654 Female No 2State your hypotheses:
# Q4. Build a contingency table of VitaminUse and Sex using table().
table(nutrition$Vitamin, nutrition$Sex) ##
## Female Male
## 1 109 13
## 2 77 5
## 3 87 24# Q5. Run a chi-square test of independence on that table.
chisq.test(nutrition$Vitamin, nutrition$Sex)##
## Pearson's Chi-squared test
##
## data: nutrition$Vitamin and nutrition$Sex
## X-squared = 11.071, df = 2, p-value = 0.003944# Q6. What is the p-value? Do you reject H₀ at α = 0.05?Q7. Write your conclusion in plain English: The p-value is p-value = 0.003944; therefore, we would not accept the Null Hypothesis —
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))
aov(GillRate ~ Calcium, data = fish)## Call:
## aov(formula = GillRate ~ Calcium, data = fish)
##
## Terms:
## Calcium Residuals
## Sum of Squares 2037.222 19064.333
## Deg. of Freedom 2 87
##
## Residual standard error: 14.80305
## Estimated effects may be unbalanced# Q9. Use summary() on the result. What is the F statistic and p-value?
summary(aov(GillRate ~ Calcium, data = fish))## 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?Q11. Write your conclusion in plain English: No we will not reject the Null Hypothesis