Confidence Intervals

General Form

$\text{statistic} \pm \text{critical value} \times \text{standard error}$ $\text{statistic} \pm \text{margin of error}$

Confidence Interval for a Proportion ($\hat{p}$)

$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Conditions:

1. Random sample.
2. $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$ (Large Counts).
3. Independence (10% condition: $n \leq 0.1N$).

Confidence Interval for a Mean ($\bar{x}$)

$\bar{x} \pm t^* \frac{s}{\sqrt{n}}$ with $df = n - 1$.

Conditions:

1. Random sample.
2. Normality: population normal, or $n \geq 30$ (CLT), or no strong skew/outliers for small $n$.
3. Independence (10% condition).

Interpreting Confidence Level

"We are 95% confident that the true [parameter] lies between [lower, upper]."

The confidence level means: if we repeated this procedure many times, about 95% of the intervals would contain the true parameter.

Margin of Error

$ME = z^* \cdot SE \quad \text{or} \quad ME = t^* \cdot SE$

To reduce ME: increase $n$ or decrease confidence level.

Sample Size Determination (Proportion)

$n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p})$

Use $\hat{p} = 0.5$ for the most conservative estimate.

Common Critical Values ($z^*$)

Problem: In a random sample of 200 voters, 120 support a candidate. Construct a 95% CI for the true proportion.

Solution:

$\hat{p} = 120/200 = 0.6$. Check conditions: random ✓, $200(0.6) = 120 \geq 10$ and $200(0.4) = 80 \geq 10$ ✓.

$SE = \sqrt{\frac{0.6(0.4)}{200}} = \sqrt{0.0012} = 0.0346$

$0.6 \pm 1.96(0.0346) = 0.6 \pm 0.068$

95% CI: $(0.532, 0.668)$.

We are 95% confident the true proportion of voters supporting the candidate is between 53.2% and 66.8%.

• CI = statistic ± margin of error.
• Always check conditions: random, normal/large counts, independence.
• Larger samples and lower confidence levels give narrower intervals.
• Interpret: "We are C% confident the true parameter is in [interval]."