Homework #3
Ross Enlow
9/24/2019
Problem 2
a) It is the standard error of the bootstrap distribution that will be approximately the same as the standard error of the original sample, not the standard deviation.
b) The bootstrap distribution IS created WITH replacement from the original sample.
c) False, it is best to use samples larger than the size of the original sample.
d) False, the bootstrap distributioni s created by resampling with replacement from the SAMPLE population.
Problem 3
a) 57 + 61 = 118, 118/2 = 59. 42 + 62 + 41 + 28 = 173, 173/4 = 43.25. 59 - 43.25 = 15.75.
b)
SRS<-c(57, 61, 42, 62, 41, 28)
sample(SRS, 6, replace = FALSE)## [1] 61 41 28 62 42 57Treatment group: 41, 61. Mean = 51.5
Control group: 42, 62, 41, 28. Mean = 43.25
Difference in means = 8.25
c)
SRS<-c(57, 61, 42, 62, 41, 28)
replicate(20, (sample(SRS, 6, replace = FALSE)))## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
## [1,] 28 62 57 57 61 61 57 28 28 41 57 61 28
## [2,] 41 42 62 28 42 41 41 61 62 62 61 57 61
## [3,] 42 41 61 61 62 42 62 42 42 57 28 28 42
## [4,] 62 57 42 42 41 57 42 62 41 42 41 62 62
## [5,] 57 28 28 41 57 28 28 41 57 61 42 42 57
## [6,] 61 61 41 62 28 62 61 57 61 28 62 41 41
## [,14] [,15] [,16] [,17] [,18] [,19] [,20]
## [1,] 42 61 28 57 62 61 41
## [2,] 28 62 57 62 61 42 62
## [3,] 61 41 61 41 42 57 42
## [4,] 57 42 41 61 57 28 61
## [5,] 41 28 62 42 41 41 28
## [6,] 62 57 42 28 28 62 57hist(c(49-48.25, 51.5-47, 35-55.25, 44.5-50.5, 51.5-47, 35-55.25, 42.5-51.5, 49-48.25, 35-55.25, 49-48.25, 59-43.25, 52-46.75, 45-50.25, 59-43.25, 34.5-55.5, 49.5-47.75, 51-47.25, 51.5-47, 52-46.75, 35-55.25))
d)
x<-c(49-48.25, 51.5-47, 35-55.25, 44.5-50.5, 51.5-47, 35-55.25, 42.5-51.5, 49-48.25, 35-55.25, 49-48.25, 59-43.25, 52-46.75, 45-50.25, 59-43.25, 34.5-55.5, 49.5-47.75, 51-47.25, 51.5-47, 52-46.75, 35-55.25)
x[x>=15.75]## [1] 15.75 15.7510% of scores were greater than or equal to 15.75.
e)
x<-c(49-48.25, 51.5-47, 35-55.25, 44.5-50.5, 51.5-47, 35-55.25, 42.5-51.5, 49-48.25, 35-55.25, 49-48.25, 59-43.25, 52-46.75, 45-50.25, 59-43.25, 34.5-55.5, 49.5-47.75, 51-47.25, 51.5-47, 52-46.75, 35-55.25)
x[x>=15.75]## [1] 15.75 15.752/15## [1] 0.1333333The exact p-vaue is 0.13. Thi is 0.03 off of my estimate.
Problem 4
Part 1
a)
library(readr)
calls80 <- read_csv("calls80.csv")## Parsed with column specification:
## cols(
## length = col_double()
## )calls<-(calls80$length)
hist(calls)
#### The distribution is skewed to the right.
b)
bootstrapCalls<-function(data, nsim){
n<- length(data)
bootCI<-c()
for (i in 1:nsim){
bootSamp=sample(1:n, n, replace=TRUE)
thisXbar<-mean(data[bootSamp])
bootCI=c(bootCI, thisXbar)
}
return(bootCI)
}
hist(bootstrapCalls(calls,1000))
c)
qqnorm(bootstrapCalls(calls,1000))
#### After assessing the qqPlot it appears that the tails containe values greater than would be expected in a normal distribution.
Part 2
d)
callsSRS=c(104, 102, 35, 211, 56, 325, 67, 9, 179, 59)
bootstrapCalls<-function(data, nsim){
n<- length(data)
bootCI<-c()
for (i in 1:nsim){
bootSamp=sample(1:n, n, replace=TRUE)
thisXbar<-mean(data[bootSamp])
bootCI=c(bootCI, thisXbar)
}
return(bootCI)
}
hist(bootstrapCalls(callsSRS,1000))
qqnorm(bootstrapCalls(callsSRS,1000))
#### This distribution appears to be farther away from normal when assessing the qqPlot.
e)
sd(bootstrapCalls(calls,10000))## [1] 37.615sd(bootstrapCalls(callsSRS,10000))## [1] 29.16442I’m confused, the se for SRS sample is smaller, not larger. We would expect that the standard error of the smaller SRS would be larger because of the much smaller sample size = 10, as opposed to the sample of 80. However, R seems/a simulation error seems to disagree. Please let me know if you’re able to detect an error in the code! Thank you.
Problem 5
library(readr)
nspines <- read_csv("/Volumes/raenlow/MATH239/nspines.csv")## Parsed with column specification:
## cols(
## ns = col_character(),
## dbh = col_double()
## )a)
Npine= nspines[1:30, ]
Spine =nspines[31:60, ]
#names(nspines)
library(tidyverse)
ggplot(nspines, aes(dbh))+
geom_histogram(bins=7)+
facet_wrap(~ns)
ggplot(nspines, aes(y=dbh, x=ns, fill=ns))+
geom_boxplot()
#### Neither distribution appears to be approximately normal and the sample sizes of n=30 are barely of sufficient size if going by the rule of n=30. From these observations it may not be reasonable to use standard t procedures.
b)
meanspine=mean(Spine$dbh)
meannpine=mean(Npine$dbh)
meannpine-meanspine## [1] -10.83333c)
bootStrapCI2<-function(data1, data2, nsim){
n1<-length(data1)
n2<-length(data2)
bootCI2<-c()
for(i in 1:nsim){
bootSamp1<-sample(1:n1, n1, replace=TRUE)
bootSamp2<-sample(1:n2, n2, replace=TRUE)
thisXbar<-mean(data1[bootSamp1])-mean(data2[bootSamp2])
bootCI2<-c(bootCI2, thisXbar)
}
return(bootCI2)
}
pinebootCI=bootStrapCI2(Npine$dbh, Spine$dbh, 1000)
hist(pinebootCI)
d)
Quantile Method
quantile(pinebootCI, c(0.025, 0.975))## 2.5% 97.5%
## -18.57533 -2.77775Hybrid Method
bootSE=sd(pinebootCI)
(mean(Npine$dbh)-mean(Spine$dbh))+c(-1,1)*qt(0.975, df = 58)*bootSE## [1] -18.749693 -2.916974e)
mean(pinebootCI)-(meannpine-meanspine)## [1] -0.1196967The bias of 0.09 seems to be relatively small and the distribution does appear approxmiately normal. The conditions for the hybrid method appear to be met and thus I do believe this interal is reliable.
f)
t.test(Npine$dbh, Npine$dbh)##
## Welch Two Sample t-test
##
## data: Npine$dbh and Npine$dbh
## t = 0, df = 58, p-value = 1
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -9.044798 9.044798
## sample estimates:
## mean of x mean of y
## 23.7 23.7