1 - Using the t-Distribution
We know that, with a normal distribution, only 5% of values are more than 2 standard deviations away from the mean.
Calculate the probability of seeing t-distributed random variables being more than 2 in absolute value when the degrees of freedom are 3.
1 - pt(2, 3) + pt(-2, 3)
[1] 0.139326
2 - Plotting the t-distribution
Now use sapply to compute the same probability for degrees of freedom from 3 to 50.
Make a plot and notice when this probability converges to the normal distribution's 5%.
df <- 3:50
# Make a function called 'pt_func' that calculates the probability that a value is more than |2| for any degrees of freedom
pt_func <- function(n){
1 - pt(2, n) + pt(-2, n)
}
# Generate a vector 'probs' that uses the `pt_func` function to calculate the probabilities
probs <- sapply(df, pt_func)
# Plot 'df' on the x-axis and 'probs' on the y-axis
plot(df, probs)
- Load the neccessary libraries and data
library(dslabs) library(dplyr) data(heights)
- Use the sample code to generate 'x', a vector of male heights
x <- heights %>% filter(sex == "Male") %>%
.$height
- Create variables for the mean height 'mu', the sample size 'N', and the number of times the simulation should run 'B'
mu <- mean(x) N <- 15 B <- 10000
- Use the `set.seed` function to make sure your answer matches the expected result after random sampling
set.seed(1)
- Generate a logical vector 'res' that contains the results of the simulations
res <- replicate(B,{
t <- sample(x, N, replace = TRUE)
interval <- c(mean(t)-qnorm(0.975)*sd(t)/sqrt(N), mean(t)+qnorm(0.975)*sd(t)/sqrt(N))
hit <- between(mu, interval[1],interval[2])
})
3 - Sampling From the Normal Distribution
In a previous section, we repeatedly took random samples of 50 heights from a distribution of heights. We noticed that about 95% of the samples had confidence intervals spanning the true population mean.
Re-do this Monte Carlo simulation, but now instead of , use . Notice what happens to the proportion of hits.
mean(res)
[1] 0.9323
4 - Sampling from the t-Distribution
N = 15 is not that big. We know that heights are normally distributed, so the t-distribution should apply. Repeat the previous Monte Carlo simulation using the t-distribution instead of using the normal distribution to construct the confidence intervals.
What are the proportion of 95% confidence intervals that span the actual mean height now?
mu <- mean(x)
# Use the same sampling parameters as in the previous exercise.
set.seed(1)
N <- 15
B <- 10000
# Generate a logical vector 'res' that contains the results of the simulations using the t-distribution
res <- replicate(B,{
t <- sample(x, N, replace = TRUE)
interval <- c(mean(t)-qt(0.975,N-1)*sd(t)/sqrt(N), mean(t)+qt(0.975,N-1)*sd(t)/sqrt(N))
hit <- between(mu, interval[1],interval[2])
})
# Calculate the proportion of times the simulation produced values within the 95% confidence interval. Print this value to the console.
mean(res)
[1] 0.9523