Confidence Intervals Summary

Confidence Intervals

Definition and Interval Estimate

• Interval estimation involves creating an interval to estimate population parameters with a measure of uncertainty.
• A random sample $X$ is taken to estimate specific values for parameters of interest like $\mu$ (mean) and $\sigma^2$ (variance).

Confidence Interval (CI) Definition

• CI is defined by a lower bound $L = L(X<em>1, \dots, X</em>n)$ and an upper bound $U = U(X<em>1, \dots, X</em>n)$.
• $P(L \le \theta \le U) = 1 - \alpha$, where $\theta$ is the parameter of interest, and $1 - \alpha$ is the desired level of confidence (e.g., 99% when $\alpha = 1\%$.

Parameters for CI Construction

• CIs can be built for:
  • $\mu$ (mean)
  • $\sigma^2$ (variance)
  • $\mu<em>X - \mu</em>Y$ (difference of means)
  • $\frac{\sigma^2<em>X}{\sigma^2</em>Y}$ (ratio of variances)
  • $p$ (proportion)

Normal Distribution Reminder

• For normally distributed random variables $X_i \sim N(\mu, \sigma^2)$, where $i = 1, 2, \dots, n$:
  • Sample mean: $\overline{X} = \frac{1}{n} \sum<em>{i=1}^{n} X</em>i$
  • Sample variance: $S^2 = \frac{1}{n-1} \sum<em>{i=1}^{n} (X</em>i - \overline{X})^2$
  • $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$
  • $\frac{\overline{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$

CI for $\mu$ from Normal Distribution ($\sigma^2$ known)

• CI for $\mu$ is calculated as: $\overline{X} \pm z_{1-\alpha/2} \frac{\sigma}{\sqrt{n}}$
• For a 95% CI: $\overline{X} \pm 1.96 \frac{\sigma}{\sqrt{n}}$

Confidence Interval Comments

• $L$ and $U$ are random variables.
• The width of the CI is $U - L$. For example, $U = \overline{X} + 1.96 \frac{\sigma}{\sqrt{n}}$, $L = \overline{X} - 1.96 \frac{\sigma}{\sqrt{n}}$
• Higher confidence implies a wider interval.

Confidence Interval Interpretation

• Before sampling: A 95% CI contains $\mu$ with 95% probability.
• After sampling: There is 95% confidence that the calculated interval contains $\mu$.

Sample Size Calculation

• For normally distributed data with unknown $\mu$ but known $\sigma^2$, the required sample size $n$ for a given confidence level $(1 - \alpha)$ and desired width $\omega_0$ is:
  • $n = \frac{4 z^2<em>{1-\alpha/2} \sigma^2}{\omega^2</em>0}$