Please note: All the Answers have been rounded to four decimal points

Question 1

Question 1.1. Approximate the bootstrap standard error and bias of the population coefficient of variance

#Reading in the dataset for question 1 and defining some basic values
datset1 <- c(10.1,14.8,5.0,10.2,12.4,12.2,2.0,11.5,17.8,4.0)
n <- length(datset1)
B <- 1000
R <- 100

#Doing the bootstrap
stats <- rep(0,B)
for(i in 1:B){
        boot <- sample(datset1, n, replace = TRUE)
        stats[i] <- sd(boot)/abs(mean(boot))
}
std <- sd(stats)
bias <- mean(stats) - sd(datset1)/abs(mean(datset1))
Answer: The approximation for the bootstrap standard error is 0.1223 and for the bias is -0.0242

Question 1.2 Approximating the bootstrap standard error of the estimated bootstrap standard error of the estimated coefficient of variance

stats <- rep(0,B)
for(i in 1:B) {
        boot <- sample(datset1, n, replace = TRUE)
        stats2 <- rep(0,R)
        for(j in 1:R) {
                boot2 <- sample(boot, n, replace = TRUE)
                stats2[j] <- sd(boot2)/abs(mean(boot2))
        }
        stats[i] <- sd(stats2)
}
std2 <- sd(stats)
Answer: The approximate value for the bootstrap standard error of the estimated bootstrap standard error of the coefficient of variance is 0.0275

Question 1.3. Approximating the minimum of the median of the estimated coefficient of variance.

stats <- rep(0,B)
for(i in 1:B) {
        boot <- sample(datset1, n, replace = TRUE)
        stats2 <- rep(0,R)
        for(j in 1:R) {
                boot2 <- sample(boot, n, replace = TRUE)
                stats2[j] <- sd(boot2)/abs(mean(boot2))
        }
        stats[i] <- median(stats2)
}
ans <- min(stats)
Answer: The required estimate is 0.0816

Question 2

Question 2.1. Constructing a 90% confidence interval for the population variance using three different methods

a) Using the basic percentile method
#Setting the Type 1 error level
alpha <- 0.1

#Defining expressions for "r" and "s" to use in determining interval quantiles
r <- floor((B+1) * alpha/2)
s <- floor((B+1) * (1-(alpha/2)))

#Determining the confidence intervals
stats <- rep(0,B)
for(i in 1:B) {
        boot <- sample(datset1, n, replace = TRUE)
        stats[i] <- var(boot)
}
stats_sorted <- sort(stats)
ci_basic <- c(stats_sorted[r], stats_sorted[s])
Answer: The 90% confidence interval using the basic percentile method would be 9.7427, 35.9093
b) Using the bias-corrected (of the basic) percentile method
#Doing the bootstrap
stats <- rep(0,B)
for(i in 1:B) {
        boot <- sample(datset1, n, replace = TRUE)
        stats[i] <- var(boot)
}

#Determining the estimated proportion and "z0"
g <- mean(stats <= var(datset1))
z0 <- qnorm(g)

#Defining the interval quantiles "r" and "s"
r <- floor((B+1) * pnorm(2*z0 - qnorm(1-alpha/2)))
s <- floor((B+1) * pnorm(2*z0 + qnorm(1-alpha/2)))

stats.sorted <- sort(stats)
ci_basic_BC <- c(stats.sorted[r],stats.sorted[s])
Answer: The 90% confidence interval using the B-C method would be 14.3672, 43.5093
c) Using the bootstrap-t method
#Redifining "r" and "s" values
r <- floor((B+1) * (1-(alpha/2)))
s <- floor((B+1) * alpha/2)

#Doing the double boostrap
Tstats <- rep(0,B)
for(i in 1:B) {
        boot <- sample(datset1, n, replace = TRUE)
        stats2 <- rep(0,R)
        for(j in 1:R) {
                boot2 <- sample(boot, n, replace = TRUE)
                stats2[j] <- var(boot2)
        }
        se_boot <- sd(stats2)
        Tstats[i] <- (var(boot) - var(datset1))/se_boot
}
Tstats_sorted <- sort(Tstats)
ci_t <- c(var(datset1) - Tstats_sorted[r]* sd(datset1), var(datset1) - Tstats_sorted[s]*sd(datset1))
Answer: The 90% confidence interval using the bootstrap-t method would be 18.6482, 46.2356

Question 2.2: Constructing a 95% basic percentile confidence interval for the population third quartile using the parametric bootstrap.

# Redefining "r", "s", alpha and an expression for the third quartile of a general n = 7 dataset
r <- floor((B+1) * (alpha/2))
s <- floor((B+1) * (1-(alpha/2)))
alpha <- 0.05
q3 <- floor(n*0.75)
a) Assuming a normal distribution
stats <- rep(0,B)
for(i in 1:B) {
        boot <- rnorm(n = n, mean = mean(datset1), sd = sd(datset1))
        boot_sorted <- sort(boot)
        stats[i] <- boot_sorted[q3]
}
stats_sorted <- sort(stats)
ci_norm <- c(stats_sorted[r], stats_sorted[s])
Answer: The 95% confidence interval of the population third quartile, assuming normal distribution is 8.9315, 15.0377
b) Assuming a uniform distribution
stats <- rep(0,B)
for(i in 1:B) {
        boot <- runif(n = n, min = min(datset1), max = max(datset1))
        boot_sorted <- sort(boot)
        stats[i] <- boot_sorted[q3]
}
stats_sorted <- sort(stats)
ci_unif <- c(stats_sorted[r], stats_sorted[s])
Answer: The 95% confidence interval of the population third quartile, assuming uniform distribution is 8.3053, 15.2923

Question 3

#Reading in the data
x1 <- c(2.6,3.4,3.6,3.2,3.5,2.9,3.3)
x2 <- c(5,4,10,5,6,7,8)
y <- c(3300, 3600, 4000, 3500, 3900, 3600, 3800)

Question 3.1. Using the non-parametric bootstrap, testing hypothesis of beta2 = 30 at 10% significance

#Fitting a model and defining its summary
fit <- lm(y ~ x1 + x2)
sumfit <- summary(fit)

#Defining the vector of sample residuals and centering them
res <- fit$residuals
res0 <- res - mean(res)

#Defining the given null hypothesis value of beta2
beta2_null <- 30

#Defining the observed test statistic
t_stat <- (fit$coefficients[3] - beta2_null)/sumfit$coefficients[3,2]

#Defining additional parameters
n <- length(x1)
B <- 1000

#Doing the bootstrap
stats <- rep(0,B)
for(i in 1:B) {
        boot_res0 <- sample(res0, n, replace = TRUE)
        vstars <- fit$coefficients[1] + fit$coefficients[2]*x1 + beta2_null*x2 + boot_res0
        boot_fit <- lm(vstars ~ x1 + x2)
        sumbootfit <- summary(boot_fit)
        stats[i] <- (boot_fit$coefficients[3] - beta2_null)/sumbootfit$coefficients[3,2] 
}
stats_sorted <- sort(stats)

#Defining alpha, the critical value quantile "s", and the critical value
alpha <- 0.1
s <- floor(B * (1-alpha))
c_value <- stats_sorted[s]

#Considering if the observed t-statistic is larger than the critical value:
if(t_stat >= c_value) {result <- "reject the null hypothesis"} else {result <- "fail to reject the null hypothesis"}

#And calculating the bootstrap p-value
p_val <- mean(stats >= t_stat)

if(p_val <= alpha) {p_result <- "reject the null hypothesis"} else {p_result <- "failure to reject the null hypothesis"}
Answer: given the bootstrap estimate of the critical value 1.597 and the observed test statistic of 1.4264, the result of the hypothesis test at a 10% significance level is to fail to reject the null hypothesis. And, given a p-value of 0.119, which when compared with an alpha of 0.1 confirms the outcome of a failure to reject the null hypothesis.

Question 3.2. Determining a 90% confidence interval for beta1, using the hybrid method and assuming heteroscedasticity.

#Given the heteroskedasticity of the data, apply a Wild bootstrap to the centered residuals
stats <- rep(0,B)
for(i in 1:B) {
        e_boot <- rep(0,n)
        for(j in 1:n) {
                k <- sample(c(1,-1), 1)
                e_boot[j] <- res0[j] * k
        }
        ystars <- fit$coefficients[1] + fit$coefficients[2]*x1 + fit$coefficients[3]*x2 + e_boot
        boot_fit <- lm(ystars ~ x1 + x2)
        stats[i] <- boot_fit$coefficients[2]
}

#Applying order statistics, setting the value of alpha and defining "r" and "s" in context of hybrid method.
stats_sorted <- sort(stats)
alpha <- 0.1
r <- floor(B * (1-(alpha/2)))
s <- floor(B * (alpha/2))

#Calculating the required confidence interval using hybrid method
confint <- c(2 * fit$coefficients[2] - stats_sorted[r], 2 * fit$coefficients[2] - stats_sorted[s])
Answer: The 90% confidence interval for beta1 assuming heteroscedasticity and using the hybrid method is: 372.327, 568.9212

Question 4

Testing at a 5% significance level whether the population mean value of houses with four bedrooms is more than the population mean value of houses with three bedrooms.
#Reading in the data
datset3 <- read.table("data.set3.txt", header = TRUE)

#Subsetting to three and four bedrooms only:
set.3bed <- datset3[datset3$bed == 3,1]
set.4bed <- datset3[datset3$bed == 4,1]

#Setting some basic values:
n <- length(set.3bed)
m <- length(set.4bed)
alpha <- 0.05
B <- 1000
s <- floor(B * (1-alpha))

#Transforming the data to ensure the sets mimic the null hypotheses of equal means.
v3 <- set.3bed - mean(set.3bed)
v4 <- set.4bed - mean(set.4bed)

#Defining the observed t-value
t_obs <- (mean(set.4bed) - mean(set.3bed))/sqrt(var(set.4bed)/m + var(set.3bed)/n)

#Doing the bootstrap
t_stats <- rep(0,B)
for(i in 1:B) {
        boot_v3 <- sample(v3, n, replace = TRUE)
        boot_v4 <- sample(v4, m, replace = TRUE)
        t_stats[i] <- (mean(boot_v4) - mean(boot_v3))/sqrt(var(boot_v4)/m + var(boot_v3)/n)
}

#Doing the order statistics and identifying the estimate of the critical value
t_stats_sorted <- sort(t_stats)
c_val <- t_stats_sorted[s]

#Comparing the observed vs critical values
if(t_obs >= c_val) {result <- "reject the null hypothesis"} else {result <- "fail to reject the null hypothesis"}

#Considering the p-value
p_value <- mean(t_stats >= t_obs)
if(p_value <= alpha) {p_result <- "rejection of the null hypothesis"} else {p_result <- "failure to reject the null hypothesis"}
Answer: Given a critical value of 1.4845 and an observed test statistic of 2.4273 we need to reject the null hypothesis. And, given a p-value of 0.004 against an alpha of 0.05, the rejection of the null hypothesis can be confirmed.

END