Confidence Intervals

Overview

A confidence interval provides a range of plausible values for a population parameter, along with a level of confidence that the interval contains the true parameter.

Interpretation

A 95% confidence interval means: If we repeated sampling many times, about 95% of the intervals would contain the true parameter.

Important: It does NOT mean there's a 95% probability the parameter is in this specific interval.

General Formula

$$\text{Point Estimate} \pm (\text{Critical Value}) \times (\text{Standard Error})$$

Or:

$$CI = \hat{\theta} \pm z_{\alpha/2} \times SE(\hat{\theta})$$

Confidence Interval for Mean

When $\sigma$ is Known

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

When $\sigma$ is Unknown (Use t)

$$\bar{x} \pm t_{\alpha/2, df} \times \frac{s}{\sqrt{n}}$$

Where $df = n - 1$

Common Critical Values

Confidence Level	$z_{\alpha/2}$	Two-tailed $\alpha$
90%	1.645	0.10
95%	1.96	0.05
99%	2.576	0.01

Confidence Interval for Proportion

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

Condition: $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$

Margin of Error

The margin of error (ME) is half the interval width:

$$ME = z_{\alpha/2} \times SE$$

The CI can be written as:

$$CI = (\text{Point Estimate} - ME, \text{Point Estimate} + ME)$$

Factors Affecting CI Width

Factor	Effect on Width
Larger $n$	Narrower
Smaller $\sigma$	Narrower
Higher confidence	Wider
Lower confidence	Narrower

Examples

Example 1: CI for Mean ($\sigma$ known)

$n = 100$, $\bar{x} = 75$, $\sigma = 12$, 95% confidence

$$ME = 1.96 \times \frac{12}{\sqrt{100}} = 1.96 \times 1.2 = 2.35$$ $$\text{95\% CI: } 75 \pm 2.35 = (72.65, 77.35)$$

Example 2: CI for Mean ($\sigma$ unknown)

$n = 25$, $\bar{x} = 85$, $s = 10$, 95% confidence

$$df = 24, \quad t_{0.025, 24} = 2.064$$ $$ME = 2.064 \times \frac{10}{\sqrt{25}} = 2.064 \times 2 = 4.13$$ $$\text{95\% CI: } 85 \pm 4.13 = (80.87, 89.13)$$

Example 3: CI for Proportion

$n = 400$ voters, 220 support candidate ($\hat{p} = 0.55$), 95% confidence

$$SE = \sqrt{\frac{(0.55)(0.45)}{400}} = \sqrt{0.000619} = 0.0249$$ $$ME = 1.96 \times 0.0249 = 0.049$$ $$\text{95\% CI: } 0.55 \pm 0.049 = (0.501, 0.599)$$

We're 95% confident 50.1% to 59.9% support the candidate.

Example 4: Required Sample Size

Find $n$ for $ME = 2$ when estimating mean, $\sigma = 15$, 95% confidence:

$$ME = z \times \frac{\sigma}{\sqrt{n}}$$ $$2 = 1.96 \times \frac{15}{\sqrt{n}}$$ $$\sqrt{n} = \frac{1.96 \times 15}{2} = 14.7$$ $$n = 216.1 \Rightarrow n = 217$$

Confidence vs Precision Trade-off

$$\text{Higher confidence} \to \text{Wider interval} \to \text{Less precise}$$ $$\text{Lower confidence} \to \text{Narrower interval} \to \text{More precise}$$

To be more confident AND more precise → increase $n$