#### HOMEWORK #2
#### SOFIA BELLO
#### STAT 488
### QUESTION 1
## (a) Perform the usual ANOVA F-test
## (b) Perform the permutation F-test
## (c) Perform the Kruskal-Wallis test
## (d) Discuss the results of these three tests. (i.e. What are the p-values? What is the conclusion? Are the conclusions different? How similar are the p-values? Are any assumptions violated? etc.)
group1 <- c(2.9736,0.9448,1.6394,0.0389,1.2958)
group2 <- c(0.7681,0.8027,0.2156,0.0740,1.5076)
group3 <- c(4.8249,2.2516,1.5609,2.0452,1.0959)
score <- c(group1, group2, group3)
group <- c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3)
dat1 <- data.frame(score,group)
## Part A
lmOut <- lm(score~group,data=dat1)
anovaOut<-anova(lmOut)
Fobs<-anovaOut[1,4]
pval<-anovaOut[1,5]
pval # 0.2243092
## [1] 0.2243092
Fobs # 1.627979
## [1] 1.627979
## Part B
n <- length(dat1$score)
nsim<-1000
FstatVecNull<-rep(NA,nsim)
for (i in 1:nsim)
{ print(i)
dat1Permute <- dat1
dat1Permute$score <- dat1$score[sample(1:n,n)]
lmOut <- lm(score~group,data=dat1Permute)
anovaOut <- anova(lmOut)
FstatVecNull[i]<-anovaOut[1,4] }
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pvalPerm<-sum(Fobs<FstatVecNull)/nsim
pvalPerm
## [1] 0.219
# permutation p value: 0.248
# F distribution p value: 0.2243092
## Part C
kruskal.test(list(group1,group2,group3))
##
## Kruskal-Wallis rank sum test
##
## data: list(group1, group2, group3)
## Kruskal-Wallis chi-squared = 5.78, df = 2, p-value = 0.05558
# Kruskal-Wallis chi-squared = 5.78, df = 2, p-value = 0.05558
## Part D
boxplot(score~group, data=dat1)
var(group1)
## [1] 1.150469
var(group2)
## [1] 0.3226093
var(group3)
## [1] 2.10575
# P values: Part A (ANOVA F-test) p=0.2243092, Part B (Permutation F-test) p=0.248, Part C (Kruskal Wallis) p=0.05558. P values from the ANOVA and permutation F-tests are pretty similar, the differnce in those p values is ~ 0.02. The Kruskal Wallis test, on the other hand, had a much lower p-value. Still, in all cases, we failed to reject the null hypothesis but it is worth noting that the KW p-value was only marginally greater than the α=0.05 cut off. KW is prefferred because we violated multiple assumptions including equal variances (group 1 variance=1.15, group 2 variance=0.32, group 3 variance=2.11) and normality of the data. Box plot shows heavy outliers in groups 1 and 3 and medians in groups 2 and 3.
### QUESTION 2
# Download this data set: https://bit.ly/2FKGu43
# It contains 50 observations divided equally between 5 groups. First perform the Kruskal-Wallis test and report your results. Then, if necessary, perform multiple comparisons using the Wilcoxon rank-sum test using the Bonferroni correction to control the family-wise (or experiment-wise) error rate at α=0.05. Report your results and any significant differences that you find.
dat2 <- read.csv("/Users/sofia/Desktop/SPRING 2019/488/Homework/HW2/NP_HW2.csv")
dat2$ranks<-rank(dat2$score)
g1 <- dat2[c(1:10), c(1)]
g2 <- dat2[c(11:20), c(1)]
g3 <- dat2[c(21:30), c(1)]
g4 <- dat2[c(31:40), c(1)]
g5 <- dat2[c(41:50), c(1)]
kruskal.test(list(g1,g2,g3,g4,g5))
##
## Kruskal-Wallis rank sum test
##
## data: list(g1, g2, g3, g4, g5)
## Kruskal-Wallis chi-squared = 26.033, df = 4, p-value = 3.116e-05
# Kruskal-Wallis chi-squared = 26.033, df = 4, p-value = 3.116e-05
k <- 5
alpha <- 0.05
alpha/(k*(k-1)/2)
## [1] 0.005
# α'=0.005
pairwise.wilcox.test(dat2$score,dat2$group,p.adj="bonf")
##
## Pairwise comparisons using Wilcoxon rank sum test
##
## data: dat2$score and dat2$group
##
## 1 2 3 4
## 2 1.0000 - - -
## 3 0.0021 0.0209 - -
## 4 0.0389 0.7526 0.0049 -
## 5 0.0209 0.1469 1.0000 0.0049
##
## P value adjustment method: bonferroni
# 1 2 3 4
# 2 1.0000 - - -
# 3 0.0021 0.0209 - -
# 4 0.0389 0.7526 0.0049 -
# 5 0.0209 0.1469 1.0000 0.0049
### We reject the null hypothesis and have evidence (p=0.0000311) to conclude that there are statistically significant differences in the medians between one or more groups.
## Results at Bonferroni adjusted α=0.005
## Statistically significant differences between group 1 and groups 3, 4, and 5.
## Statistically significant differences between group 4 and groups 3 and 5.
## Statistically significant differences between group 2 and 3.
pairwise.wilcox.test(dat2$score,dat2$group)
##
## Pairwise comparisons using Wilcoxon rank sum test
##
## data: dat2$score and dat2$group
##
## 1 2 3 4
## 2 1.0000 - - -
## 3 0.0021 0.0146 - -
## 4 0.0194 0.2258 0.0044 -
## 5 0.0146 0.0588 1.0000 0.0044
##
## P value adjustment method: holm
## Results at α=0.05 same as above. No values on the table happened to be lower than α=0.05 but greater than the α=0.005 cut off.
### QUESTION 3
# Perform the Kruskal-Wallis test on the data to determine if there is a difference between the median lengths for the YOY gizzard shad populations in the four Kokosing Lake sites. Make sure to stat the null and alternative hypotheses, the p-value, and your conclusion.
site1<-c(46,28,46,37,32,41,42,45,38,44)
site2<-c(42,60,32,42,45,58,27,51,42,52)
site3<-c(38,33,26,25,28,28,26,27,27,27)
site4<-c(31,30,27,29,30,25,25,24,27,30)
length <- c(site1, site2, site3, site4)
cat <- factor(rep(1:4, c(10,10,10, 10)), labels=c("Site 1", "Site 2", "Site 3", "Site 4"))
dat3 <- data.frame(length=length, category=cat)
boxplot(length~category, data=dat3)
kruskal.test(list(site1, site2, site3, site4))
##
## Kruskal-Wallis rank sum test
##
## data: list(site1, site2, site3, site4)
## Kruskal-Wallis chi-squared = 22.852, df = 3, p-value = 4.335e-05
# Kruskal-Wallis chi-squared = 22.852, df = 3, p-value = 4.335e-05
# H0: Medians are equal for all four lake sites
# HA: Medians of the four lake sites are not all equal, for 1 or more sites
# p-value = .00004335
# At a significance level of 0.05, we reject the null hypothesis and conclude that there is a statistically significant difference in the medians of one or more Kokosing Lake sites (p<.0001).