Question

R problem 1: The reason that the t distribution is important is that the sampling distribution of the standardized sample mean is different depending on whether we use the true population standard deviation or one estimated from sample data. This problem addresses this issue. 1. Generate 10,000 samples of size n- 4 from a normal distribution with mean 100 and standard deviation σ = 12, Find the 10,000 sample means and find the 10,000 sample standard deviations. What are the mean and the standard deviation of the distribution of random sample means? What are the mean and the standard deviation of the distribution of random sample standard deviations? 2. Graph the sampling distribution of the sample means and of the sample standard deviations. 3. For each sample mean, calculate the z-score using the exact standard error ơ/vn. So, What are the mean and standard deviation of these z-scores? 4. For each sample mean, calculate the t-statistic using the estimated standard error s/Vn. So, x-μ What are the mean and standard deviation of the distribution of t-statistics? 5. What fraction of the sample means are within 1.96 exact standard errors of μ? This will be the number of z-scores between1.96 and 1.96. 6, what fraction of the sample means are within 1.96 estimated standard errors of μ? This will be the fraction of the t-statistics between -1.96 and 1.96. How does this compare to the answer for z-scores? 7. Would a confidence interval made by tl.96s/vn have about a 95% chance of containing ? If not, how confident should we expect to be? 8. Repeat the simulation when n-25. What fraction of the t-statistics are between -1.96 and 1.96 now?

Answer 1

1) The R code to generate 10000 samples of size $n=4$ from normal distribution with and calculating the mean and sd of sample means, mean and sd of sample sd is given belo.

μ = 100, σ 12

set.seed(1031)
n <- 4
mu <- 100
sigma <- 12
N <- 10000
X_mean <- array(dim = c(N))
X_sd <- array(dim = c(N))

for ( i in 1:N)
{
X <- rnorm(n, mean = mu, sd = sigma)
X_mean[i] <- mean(X)
X_sd[i] <- sd(X)
}

mean(X_mean)
sd(X_mean)
mean(X_sd)
sd(X_sd)

The output are:

mean of sample means = 99.86272 close to population mean $\mu =100$
sd of sample means = 5.9676 close to sd sample mean $\sigma/\sqrt{n} =6$
mean of sample sd = 11.02363
sd of sample sd = 4.633651

2) The graph of sampling distribution of sample mean is given below:

hist(X_mean, xlab = "Sample mean", ylim=c(0,0.065),border = "blue",main = "Empirical & Theoretical distribution of sample mean", prob= T)
curve(dnorm(x,mean = mu , sd = sigma/sqrt(n)), col = "darkblue", lwd = 2, add = TRUE)

The graph of sampling distribution of sample standard deviations is given below:

hist(X_sd, xlab = "Sample standard deviation",border = "blue",main = "Empirical distribution of sample standard deviation", prob= FALSE)

3) The Z- scores of each sample is calculated below.

set.seed(1031)
n <- 4
mu <- 100
sigma <- 12
N <- 10000
X_mean <- array(dim = c(N))
X_sd <- array(dim = c(N))

for ( i in 1:N)
{
X <- rnorm(n, mean = mu, sd = sigma)
X_mean[i] <- mean(X)
X_sd[i] <- sd(X)
}
Z <- (X_mean-mu)/sigma*sqrt(n)
mean(Z)
sd(Z)

The mean of Z-scores is -0.02288082 close 0

The sd of Z-scores is 0.9946 close 1

I have answered 5 parts.

We are required to solve only one question. Please post the remaining questions as another post. We do not get any additional amount for solving more.