1. Consider the population \(\{3, 6, 7, 9, 11, 14\}\). For samples of size 3 without replacement, find (and plot) the sampling distribution of the minimum. What is the mean of the sampling distribution? The statistic is an estimate of some parameter—what is the value of the parameter?
# Your code here
X ~ Unif(3, 14)
X ~ Unif(3, 14)
pop <- c(3, 6, 7, 9, 11, 14)
my.min <- numeric(1000)
for(i in  1:1000){
  y <- runif(3, 3, 14)
  my.min[i] <- min(y)
}
hist(my.min)

Answer: The mean of the sampling distribution is 3.0022201. The value of the parameter is 3.

  1. Let \(X_1, X_2, \ldots, X_n\) be a random sample from some distribution and suppose \(Y = T(X_1, X_2, \ldots, X_n)\) is a statistic. Suppose the sampling distribution of \(Y\) has pdf \(f(y) = (3/8)y^2\) for \(0 \leq y \leq 2.\) Find \(P(0 \leq Y \leq 1/5)\).
# Your code here
f <- function(x){3 * x ^ 2/ 8}
value <- integrate(f, 0, 1/5)$value

\(P(0 \leq Y \leq 1/5) =\) 0.001

  1. A friend claim that she has drawn a random sample of size 30, from the exponential distribution with \(\lambda = 1/10\). The mean of her sample is 12.

    1. What is the expected value of a sample mean?
    2. Run a simulation by drawing 10,000 random samples, each of size 30, from \(\text{Exp}(\lambda = 1/10)\) and then compute the mean. What proportion of the sample means are as large as or larger than 12?
    3. Is a mean of 12 unusual for a sample of size 30 from \(\text{Exp}(\lambda = 1/10)\)?
library(tidyverse)
set.seed(123)
# Your code here
meanA <- mean(rexp(30, 1/10))


size <- 10000
x <- numeric(size)
for(i in 1:size){
  x[i] <- mean(rexp(30, 1/10))
}
meanB <- mean(x >= 12)
  1. The expected value of the sample mean is 9.1123961
  2. The proportion of the sample means that are as large or larger than 12 is 0.1342
  3. Because the proportion is greater than 0.05, the mean of 12 is not unusual

  1. Let \(X_1, X_2, \ldots , X_{10} \overset{i.i.d}\sim N(20, 8)\) and \(Y_1, Y_2, \ldots, Y_{15} \overset{i.i.d}\sim N(16, 7)\). Let \(W = \bar{X} + \bar{Y}.\)

    1. Give the exact sampling distribution of \(W\)
    2. Simulate the sampling distribution in R and plot your results. Check that the simulated mean and standard error are close to the theoretical mean and standard error.
    3. Use your simulation to find \(P(W < 40).\) Calculate an exact answer and compare.
size <- 1000
xDis <- numeric(size)
yDis <- numeric(size)
for(i in 1:size){
  xDis[i] <- mean(rnorm(10, 20, 8))
  yDis[i] <- mean(rnorm(15, 16, 7))
}
wDis <- xDis + yDis

  1. The exact sampling distribution of W is \(\bar{X} + \bar{Y}\) ~N(36, 15)

  2. Mean is 35.9047789; Standard deviation is: 10.0005578

hist(wDis)

C) The probability that W is less than 40 is: 0.9017837

  1. Let \(X_1, X_2, \ldots , X_{9} \overset{i.i.d}\sim N(7, 3)\) and \(Y_1, Y_2, \ldots, Y_{12} \overset{i.i.d}\sim N(10, 5)\). Let \(W = \bar{X} - \bar{Y}.\)

    1. GIve the exact sampling distribution of \(W\).
    2. Simulate the sampling distribution of \(W\) in R and plot your results using ggplot2. Check that the simulated mean and the standard error are close to the theoretical mean and the standard error.
    3. Use your simulation to find \(P(W < -1.5)\). Calculate an exact answer and compare.
  1. (-3,-2)
# part b.
library(ggplot2)
size <- 1000
xDis <- numeric(1000)
yDis <- numeric(1000)
for(i in 1:size){
  xDis[i] <- mean(rnorm(9, 7, 3))
  yDis[i] <- mean(rnorm(12, 10, 5))
}
wDis <- xDis - yDis
# part c.
# simulated answer
simMean <- mean(wDis < -1.5)

# Exact answer
exMean <- pnorm(-1.5, mean(wDis), sd(wDis))

B)Mean = -2.9673439 Standard Deviation = 2.8387381

  1. Let \(X_1, X_2, \ldots , X_n\) be a random sample from \(N(0, 1)\). Let \(W = X_1^2 + X_2^2 + \cdots + X_N^2.\) Describe the sampling distribution of \(W\) by running a simulation, using \(n = 2.\) What is the mean and variance of the sampling distribution of \(W\)? Repeat using \(n = 4, n = 5.\) What observations or conjectures do you have for general \(n\)?
# Your code here
size <- 1000
w2 <- numeric(size)
w4 <- numeric(size)
w5 <- numeric(size)
for(i in 1:size){
  w2[i] <- sum(rnorm(2) ^ 2)
  w4[i] <- sum(rnorm(4) ^ 2)
  w5[i] <- sum(rnorm(5) ^ 2)
}

W2: The mean is 2.1408701. The variance is 4.8589464. W4: The mean is 4.0637784. The variance is 9.2395798. W5: The mean is 4.9899208. The variance is 9.8786346.

The mean is typically \(n\), the variance is typically \(2n\)

  1. Let \(X\) be a uniform random variable on the interval \([40, 60]\) and \(Y\) a uniform random variable on \([45, 80].\) Assume that \(X\) and \(Y\) are independent.

    1. Compute the expected value and variance of \(X + Y.\)
    2. Simulate the sampling distribution of \(X + Y.\) Desribe the graph of the distribution of \(X + Y\). Compute the mean and variance of the sampling distribution and compare this to the theoretical mean and variance.
    3. Suppose the time (in minutes) Jack takes to complete his statistics homework is \(\text{Unif}[40, 60]\) and the time JIll takes is \(\text{Unif}[40, 60].\) Assume they work independently. One day they announce that their total time to finish an assignment was less than 90 min. How likely is this?
# Your code here
xEX <- (40 + 60) / 2
xVX <- ((60 - 40) ^2) / 2
yEX <- (45 + 80) / 2
yVX <- ((80 - 45) ^ 2) / 2
EX <- xEX + yEX
VX <- xVX + yEX

size <- 1000
X <- runif(size, 40, 60)
Y <- runif(size, 45, 80)
W <- X + Y
wEX <- mean(W)
wVX <- var(W)
  1. EX is 112.5 and VX is 262.5

B)EX is 112.2695294, the VX is 140.3661209 \(P(X + Y < 90) =?\)

# part c
# P(W < 90)
mean <- mean(W < 90)

\(P(X + Y < 90) =0.026\) C) Probability is less than 0.05, it is unlikely