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Week 8.2: Confidence Intervals

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Week 8.2: Confidence Intervals

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Now · 1. Intuition: From Point to Interval

Week 8.2: Confidence Intervals

Prerequisites: Sampling Distributions (Week 6.2: Sampling Distributions), Properties (Week 8.1: Properties of Estimators) Cross-links: BSMA3012 (Linear Stat Models) — regression intervals Core question: How do we express uncertainty in our estimate with an interval?

1. Intuition: From Point to Interval

A point estimate

\hat{\theta}

alone tells us nothing about precision. A confidence interval (CI) gives a range of plausible values and quantifies uncertainty:

"We are 95% confident that $\mu$ lies in $(\bar{X} - 1.96 \cdot \text{SE}, \bar{X} + 1.96 \cdot \text{SE})$ ." Interpretation: If we repeated the sampling many times and computed a 95% CI each time, 95% of those intervals would contain the true parameter $\theta$ . Crucial: The parameter is fixed (not random). The interval is random. "95% confidence" refers to the procedure, not the specific interval.

2. CI for Mean (Known Variance)

X_1,\dots,X_n \sim \mathcal{N}(\mu, \sigma^2)

n

large (CLT):

\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1).

100(1-\alpha)\%

CI:

\bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},

where

z_{\alpha/2}

is the

\mathcal{N}(0,1)

quantile with

\mathbb{P}(Z > z_{\alpha/2}) = \alpha/2

. Common $z$ values:

90% CI: $z_{0.05} = 1.645$
95% CI: $z_{0.025} = 1.96$
99% CI: $z_{0.005} = 2.576$

3. CI for Mean (Unknown Variance)

When

\sigma

is unknown, estimate with

S

and use

t

-distribution:

\frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{n-1}.

100(1-\alpha)\%

CI:

\bar{X} \pm t_{\alpha/2, n-1} \cdot \frac{S}{\sqrt{n}}.

Note: As $n \to \infty$ , $t_{\alpha/2, n-1} \to z_{\alpha/2}$ , so the $t$ and $z$ intervals coincide for large $n$ .

4. CI for Variance (Normal Data)

Using

(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}

: A

100(1-\alpha)\%

CI for

\sigma^2

\left( \frac{(n-1)S^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)S^2}{\chi^2_{1-\alpha/2, n-1}} \right).

5. CI for Proportion (Large Sample)

For

X_i \sim \text{Bernoulli}(p)

and large

n

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

Conditions:

n\hat{p} \geq 10

and

n(1-\hat{p}) \geq 10

6. Summary Table

Parameter	CI	Distribution
$\mu$ (known $\sigma$ )	$\bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$	$\mathcal{N}(0,1)$
$\mu$ (unknown $\sigma$ )	$\bar{X} \pm t_{\alpha/2, n-1} \cdot \frac{S}{\sqrt{n}}$	$t_{n-1}$
$\sigma^2$	$\left(\frac{(n-1)S^2}{\chi^2_{\alpha/2}}, \frac{(n-1)S^2}{\chi^2_{1-\alpha/2}}\right)$	$\chi^2_{n-1}$
$p$	$\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$	$\mathcal{N}(0,1)$ approx

7. Practice Questions

Q1 (Easy)

n=100

\bar{x}=50

\sigma=10

. Find 95% CI for

\mu

Full Solution

$\text{SE} = 10/\sqrt{100} = 1$ . $z_{0.025} = 1.96$ .

95% CI: $50 \pm 1.96 \cdot 1 = (48.04, 51.96)$ .

Q2 (Medium)

n=25

\bar{x}=100

s=15

. Find 95% CI for

\mu

Full Solution

$\text{SE} = 15/\sqrt{25} = 3$ . $t_{0.025,24} = 2.064$ .

95% CI: $100 \pm 2.064 \cdot 3 = (93.81, 106.19)$ .

Q3 (Hard)

From

n=20

normal samples,

s^2=16

. Find a 90% CI for

\sigma^2

Full Solution

$\chi^2_{0.05,19} = 30.144$ , $\chi^2_{0.95,19} = 10.117$ .

90% CI: $\left(\frac{19 \cdot 16}{30.144}, \frac{19 \cdot 16}{10.117}\right) = (10.08, 30.05)$ .

Next topic: Week 9: Bayesian Estimation — Bayesian Estimation. Join Discord PreviousWeek 8.1: Properties of EstimatorsNextBayesian Estimation

Week 8.2: Confidence Intervals

Week 8.2: Confidence Intervals

1. Intuition: From Point to Interval

2. CI for Mean (Known Variance)

3. CI for Mean (Unknown Variance)

4. CI for Variance (Normal Data)

5. CI for Proportion (Large Sample)

6. Summary Table

7. Practice Questions

Q1 (Easy)

Q2 (Medium)

Q3 (Hard)

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