---
title: "Sample Means and the Central Limit Theorem"
subtitle: "SD07 — Sampling Distributions and Inference"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
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        href: SD07-01-sample-means-and-central-limit-theorem.Rmd?download=1
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, message = FALSE, warning = TRUE)
```

**Course:** Statistics and Probability for Data Science (DS14270).  
**Source deck:** Samer El Kababji, *SD07 — Sampling Distributions and Inference*.

This lesson distinguishes observations from sample means, derives the sample
mean's center and spread, and uses the central limit theorem (CLT).

## The sample mean is a random variable

For an i.i.d. sample $X_1,\ldots,X_n$ from a population with mean $\mu$ and
variance $\sigma^2$,

$$
\overline X=\frac1n\sum_{i=1}^nX_i.
$$

Different random samples produce different values of $\overline X$. Its
distribution across all possible samples is the **sampling distribution**.

```{r one-weibull-sample}
set.seed(14270)
sample_size <- 10
one_sample <- rweibull(sample_size, shape = 2, scale = 5)

c(sample_mean = mean(one_sample), sample_sd = sd(one_sample))
```

The deck introduces a Weibull population with shape 2 and scale 5. Its
population moments are

$$
\mu=5\Gamma(1+1/2)\approx4.4311,
$$

$$
\sigma^2=5^2\left[\Gamma(2)-\Gamma(1.5)^2\right]\approx5.365.
$$

```{r weibull-population-moments}
shape <- 2
scale <- 5
mu <- scale * gamma(1 + 1 / shape)
variance <- scale^2 * (
  gamma(1 + 2 / shape) - gamma(1 + 1 / shape)^2
)

c(mean = mu, variance = variance, sd = sqrt(variance))
```

## Mean, variance, and standard error

Linearity of expectation gives

$$
E(\overline X)
=\frac1n\sum_{i=1}^nE(X_i)=\mu.
$$

Independence makes the variances add:

$$
\operatorname{Var}(\overline X)
=\frac1{n^2}\sum_{i=1}^n\operatorname{Var}(X_i)
=\frac{\sigma^2}{n}.
$$

The standard deviation of the sampling distribution is the **standard
error**:

$$
\operatorname{SE}(\overline X)=\frac{\sigma}{\sqrt n}.
$$

```{r standard-error-by-sample-size}
n <- c(1, 5, 10, 30, 100)
data.frame(
  sample_size = n,
  expected_sample_mean = mu,
  standard_error = sqrt(variance) / sqrt(n)
)
```

## Exact result for a normal population

If every $X_i\sim N(\mu,\sigma^2)$ independently, then for every sample size

$$
\overline X\sim N\left(\mu,\frac{\sigma^2}{n}\right).
$$

```{r exact-normal-sample-mean}
mu_normal <- 20
sigma_normal <- 6
n <- 9

c(
  sample_mean_mean = mu_normal,
  sample_mean_variance = sigma_normal^2 / n,
  sample_mean_sd = sigma_normal / sqrt(n)
)
```

## Central limit theorem

For i.i.d. observations with finite mean and variance, as $n$ increases,

$$
\frac{\overline X-\mu}{\sigma/\sqrt n}
\overset{d}{\longrightarrow}N(0,1).
$$

Equivalently, for sufficiently large $n$,

$$
\overline X\ \dot\sim\
N\left(\mu,\frac{\sigma^2}{n}\right),
$$

even when the population is not normal. The deck uses $n\geq30$ as a common
rule of thumb, not a guarantee for every population.

```{r clt-weibull-simulation, fig.width=8, fig.height=4}
set.seed(14270)
repetitions <- 20000
sample_sizes <- c(2, 30)

old_par <- par(mfrow = c(1, 2))
for (n in sample_sizes) {
  sample_means <- replicate(
    repetitions,
    mean(rweibull(n, shape = 2, scale = 5))
  )
  hist(
    sample_means, probability = TRUE, breaks = 40,
    col = "grey80", border = "white",
    main = paste("Weibull means, n =", n), xlab = "Sample mean"
  )
  curve(
    dnorm(x, mean = mu, sd = sqrt(variance / n)),
    add = TRUE, col = "steelblue", lwd = 2
  )
}
par(old_par)
```

## Deck example: impurity in chemical batches

The population has mean 4.0 g and standard deviation 1.5 g. For $n=50$,

$$
\operatorname{SE}(\overline X)=\frac{1.5}{\sqrt{50}}\approx0.2121.
$$

The CLT gives

$$
P(3.5\leq\overline X\leq3.8)
\approx
\Phi\left(\frac{3.8-4}{1.5/\sqrt{50}}\right)
-\Phi\left(\frac{3.5-4}{1.5/\sqrt{50}}\right).
$$

```{r impurity-clt-example}
mu <- 4
sigma <- 1.5
n <- 50
se <- sigma / sqrt(n)
z_lower <- (3.5 - mu) / se
z_upper <- (3.8 - mu) / se

c(
  standard_error = se,
  z_lower = z_lower,
  z_upper = z_upper,
  probability = pnorm(z_upper) - pnorm(z_lower),
  rounded_table_probability = pnorm(-0.94) - pnorm(-2.36)
)
```

R's unrounded calculation is about $0.1649$; the deck's rounded z-table
calculation is $0.1645$.

## Practice questions

1. A population has $\mu=80$, $\sigma=12$. For samples of size 36, find
   $E(\overline X)$ and $\operatorname{SE}(\overline X)$.
2. Approximate $P(78\leq\overline X\leq83)$.
3. What happens to the standard error if $n$ is multiplied by 4?
4. When is the sampling distribution of the mean exactly normal rather than
   approximately normal?

::: {.callout-note collapse="true"}
### Check your answers

1. The mean is 80 and the standard error is $12/\sqrt{36}=2$.
2. `pnorm(83, 80, 2) - pnorm(78, 80, 2)` gives approximately $0.7745$.
3. It is divided by $\sqrt4=2$.
4. When the population itself is normal, assuming an independent sample.
:::