---
title: "Confidence Intervals"
subtitle: "SD07 — Sampling Distributions and Inference"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
      - text: "Download R Markdown"
        icon: file-earmark-arrow-down
        href: SD07-03-confidence-intervals.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*.

A confidence interval combines a point estimate with a margin of error. This
lesson implements the deck's z, t, and proportion intervals in base R.

## Confidence is a property of the method

A two-sided interval has the form

$$
\text{estimate}\ \pm\ \text{critical value}\times\text{standard error}.
$$

A 95% confidence procedure captures the fixed population parameter in 95% of
intervals over repeated samples under its assumptions. After one interval is
computed, it either contains the parameter or it does not.

## Mean with known population standard deviation

When $\sigma$ is known, a $(1-\alpha)$ confidence interval for $\mu$ is

$$
\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt n},
$$

where $P(Z\leq z_{\alpha/2})=1-\alpha/2$.

The deck gives $\bar x=66.12$, $\sigma=10.411$, $n=15$, and 95% confidence.

```{r known-sigma-z-interval}
x_bar <- 66.12
sigma <- 10.411
n <- 15
confidence <- 0.95
alpha <- 1 - confidence

z_star <- qnorm(1 - alpha / 2)
standard_error <- sigma / sqrt(n)
margin_error <- z_star * standard_error
z_interval <- x_bar + c(-1, 1) * margin_error

c(
  z_star = z_star,
  standard_error = standard_error,
  margin_error = margin_error,
  lower = z_interval[1],
  upper = z_interval[2]
)
```

R gives about $(60.85,71.39)$, matching the deck after rounding.

## Mean with unknown population standard deviation

When $\sigma$ is unknown and replaced by the sample standard deviation $s$,
use the t distribution with $n-1$ degrees of freedom:

$$
\bar x\pm t_{\alpha/2,n-1}\frac{s}{\sqrt n}.
$$

For $\bar x=66.12$, $s=9.21$, $n=15$, and 95% confidence:

```{r unknown-sigma-t-interval}
x_bar <- 66.12
s <- 9.21
n <- 15
alpha <- 0.05

t_star <- qt(1 - alpha / 2, df = n - 1)
standard_error <- s / sqrt(n)
margin_error <- t_star * standard_error
t_interval <- x_bar + c(-1, 1) * margin_error

c(
  t_star = t_star,
  standard_error = standard_error,
  margin_error = margin_error,
  lower = t_interval[1],
  upper = t_interval[2]
)
```

The result is about $(61.02,71.22)$. The t critical value is larger than 1.96
because estimating $\sigma$ introduces uncertainty.

## Confidence interval for a population proportion

For $x$ successes in $n$ independent observations,

$$
\widehat p=\frac xn.
$$

The deck's large-sample z interval is

$$
\widehat p\pm z_{\alpha/2}
\sqrt{\frac{\widehat p(1-\widehat p)}n}.
$$

The normal approximation condition is

$$
n\widehat p\geq10
\quad\text{and}\quad
n(1-\widehat p)\geq10.
$$

```{r proportion-z-interval}
successes <- 84
n <- 120
p_hat <- successes / n
z_star <- qnorm(0.975)
standard_error <- sqrt(p_hat * (1 - p_hat) / n)
margin_error <- z_star * standard_error
proportion_interval <- p_hat + c(-1, 1) * margin_error

c(
  p_hat = p_hat,
  successes = n * p_hat,
  failures = n * (1 - p_hat),
  lower = proportion_interval[1],
  upper = proportion_interval[2]
)
```

This interval is sometimes called the Wald interval. Near 0 or 1, especially
with small samples, alternatives such as the Wilson interval perform better.

## What changes interval width?

For a fixed confidence level, the margin of error shrinks like $1/\sqrt n$.
For fixed $n$, demanding more confidence increases the critical value and
widens the interval.

```{r interval-width-comparison}
confidence_levels <- c(0.90, 0.95, 0.99)
sample_sizes <- c(25, 100, 400)

comparison <- expand.grid(
  confidence = confidence_levels,
  n = sample_sizes
)
comparison$margin_error <- with(
  comparison,
  qnorm(1 - (1 - confidence) / 2) * 10 / sqrt(n)
)
comparison
```

## Repeated-interval interpretation (fictional extension)

```{r confidence-coverage-simulation}
set.seed(14270)
mu <- 50
sigma <- 10
n <- 25
repetitions <- 10000

sample_means <- replicate(
  repetitions,
  mean(rnorm(n, mean = mu, sd = sigma))
)
margin_error <- qnorm(0.975) * sigma / sqrt(n)
covered <- (sample_means - margin_error <= mu) &
  (mu <= sample_means + margin_error)

mean(covered)
```

## Practice questions

1. Find a 95% z interval when $\bar x=40$, $\sigma=6$, and $n=36$.
2. Find a 95% t interval when $\bar x=40$, $s=6$, and $n=36$.
3. In a sample of 200 people, 130 respond yes. Find the 95% proportion
   interval and check the approximation condition.
4. What happens to a z-interval's margin of error if $n$ is quadrupled?

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

1. $40\pm1.96(1)$, or approximately $(38.04,41.96)$.
2. With 35 degrees of freedom, `qt(0.975, 35)` is about 2.030, giving
   approximately $(37.97,42.03)$.
3. $\widehat p=0.65$ and the interval is approximately $(0.584,0.716)$.
   There are 130 successes and 70 failures, so both counts exceed 10.
4. It is halved because $\sqrt{4n}=2\sqrt n$.
:::