The purpose of this exercise is to get you more familiar with the Central Limit Theorem (CLT), and how it works with real values. Recall that the CLT states:

Let $X_1, X_2, \dots, X_n$ be random variables from some distribution with mean $\mu$ and variance $\sigma^2$. Then $\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i$ will be approximately normally distributed and $\frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \sim N(0, 1)$. Let’s take a closer took at what that means for a sampling distribution.

Start by executing the following code cell to generate $5000$ samples from a Weibull distribution, save them as data, and plot a histogram of the values. For the remainder of this notebook, this will be our underlying population distribution.

set.seed(12345)
# Generate random values from a Weibull distribution
data = rweibull(5000, 1.5, scale=1)
# Plot a histogram of the values
breaks = seq(0, 4, 0.25)
hist(data, breaks=breaks)

Does this distribution look particularly normal? No, it has a long right tail and seems to have a hard cutoff at $x=0$. Let’s see how this impacts the CLT.

But first, what is the population mean? Use the mean() function to find out.

mean(data)

0.902832458756059

In reality, we won’t know the underlying population’s distribution. The best we can do is pull a sample from the population and test that. However, that leads to some questions like:

  • Is the sample mean a good estimate for the population mean?
  • How many samples do I need for my sample mean to be a good estimate of the population mean?

Those are some of the questions we’re going to answer. Finish the function below to return the mean of $n$ sample points from the data. That is, call sample(data, n) to generate $n$ samples from the population, then call mean() on those samples.

sample.mean <- function(data, n){
    # Finish this function
    # Sample n datapoints from the population
    # Return the mean of those n samples.
    return(mean(sample(data,n)))
}

Using the function you just defined, generate the mean of a sample with a sample size $n=5$. Is it a good approximation of the population mean?

sample.mean(data,5)

0.832583089764267

Notice that the sample mean $\bar{X}$ is a random variable, and will produce a different value each time we call sample.mean(). That means $\bar{X}$ has some distribution, which we can observe by generating many independent samples and plotting a histogram of the values.

To do this, we can use the replicate() function, which allows us to call a function multiple times and create a vector containing all the results. Try replicate(3, sample.mean(data, 5)) to create a vector of 3 sample means.

replicate(3, sample.mean(data, 5))

<ol class=list-inline><li>0.54969696130119</li><li>0.892258557283775</li><li>1.02802820781037</li></ol>

Generate 3 different vectors of sample means of sizes 10, 100 and 1000 respectively. For now, keep the sample size as $n=5$ for all the vectors. Plot your results with histograms. Note that it may be helpful to have a fixed x-axis, so you can see how the histograms compare on the same scale. To do this, include xlim=c(0,2) as a parameter in your call to hist(). What do you notice?

# Generate 10 sample means with n=5 and plot the results
hist(replicate(10, sample.mean(data, 5)),xlim=c(0,2))


# Generate 100 sample means with n=5 and plot the results
hist(replicate(100, sample.mean(data, 5)),xlim=c(0,2))


# Generate 1000 sample means with n=5 and plot the results
hist(replicate(1000, sample.mean(data, 5)),xlim=c(0,2))

Now change the sample size and see how that affects the results. Generate 3 vectors of 100 samples means, coming from samples of sample sizes 5, 50, and 500. Plot your results. Again, it may be helpful to have a set x-limit on your histograms. What do you notice?

# Generate 100 sample means with n=5 and plot the results
hist(replicate(100, sample.mean(data, 5)),xlim=c(0,2))


# Generate 100 sample means with n=50 and plot the results
# Generate 100 sample means with n=5 and plot the results
hist(replicate(100, sample.mean(data, 50)),xlim=c(0,2))


# Generate 100 sample means with n=500 and plot the results
# Generate 100 sample means with n=5 and plot the results
hist(replicate(100, sample.mean(data, 500)),xlim=c(0,2))

The important things to notice are:

  • The distribution of $\bar{X}$ is approximately normally distributed, even though our underlying population is not normal.
  • The mean of the sample distribution is about $0.9$, which is approximately the same as the mean for the population distribution.
  • The variance of the sample distribution decreases as the sample size increases.

These ideas are the Central Limit Theorem!