Confidence Intervals

SD07 — Sampling Distributions and Inference

Author

Samer El Kababji

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.

Code

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]
)

        z_star standard_error   margin_error          lower          upper 
      1.959964       2.688109       5.268596      60.851404      71.388596

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:

Code

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]
)

        t_star standard_error   margin_error          lower          upper 
      2.144787       2.378012       5.100328      61.019672      71.220328

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

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

Code

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]
)

     p_hat  successes   failures      lower      upper 
 0.7000000 84.0000000 36.0000000  0.6180088  0.7819912

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.

Code

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

  confidence   n margin_error
1       0.90  25    3.2897073
2       0.95  25    3.9199280
3       0.99  25    5.1516586
4       0.90 100    1.6448536
5       0.95 100    1.9599640
6       0.99 100    2.5758293
7       0.90 400    0.8224268
8       0.95 400    0.9799820
9       0.99 400    1.2879147

Repeated-interval interpretation (fictional extension)

Code

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)

[1] 0.9531

Practice questions

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

Check your answers

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