Quiz 2

Week 8.2: Confidence Intervals

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# Week 8.2: Confidence Intervals > **Prerequisites:** Sampling Distributions ([Week 6.2: Sampling Distributions](/notes/01-foundation-bsma1004-stats-2-week06-19-sampling-distributions)), Properties ([Week 8.1: Properties of Estimators](/notes/01-foundation-bsma1004-stats-2-week08-22-properties-estimators)) > **Cross...

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 θ^\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 μ\mu lies in (Xˉ−1.96⋅SE,Xˉ+1.96⋅SE)(\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 θ\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)

If X1,…,Xn∼N(μ,σ2)X_1,\dots,X_n \sim \mathcal{N}(\mu, \sigma^2) or nn large (CLT):
Xˉ−μσ/n∼N(0,1).\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1).
A 100(1−α)%100(1-\alpha)\% CI:
Xˉ±zα/2⋅σn,\bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},
where zα/2z_{\alpha/2} is the N(0,1)\mathcal{N}(0,1) quantile with P(Z>zα/2)=α/2\mathbb{P}(Z > z_{\alpha/2}) = \alpha/2. Common zz values:
  • 90% CI: z0.05=1.645z_{0.05} = 1.645
  • 95% CI: z0.025=1.96z_{0.025} = 1.96
  • 99% CI: z0.005=2.576z_{0.005} = 2.576

3. CI for Mean (Unknown Variance)

When σ\sigma is unknown, estimate with SS and use tt-distribution:
Xˉ−μS/n∼tn−1.\frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{n-1}.
A 100(1−α)%100(1-\alpha)\% CI:
Xˉ±tα/2,n−1⋅Sn.\bar{X} \pm t_{\alpha/2, n-1} \cdot \frac{S}{\sqrt{n}}.
Note: As n→∞n \to \infty, tα/2,n−1→zα/2t_{\alpha/2, n-1} \to z_{\alpha/2}, so the tt and zz intervals coincide for large nn.

4. CI for Variance (Normal Data)

Using (n−1)S2/σ2∼χn−12(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}: A 100(1−α)%100(1-\alpha)\% CI for σ2\sigma^2:
((n−1)S2χα/2,n−12,(n−1)S2χ1−α/2,n−12).\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 Xi∼Bernoulli(p)X_i \sim \text{Bernoulli}(p) and large nn:
p^±zα/2⋅p^(1−p^)n.\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.
Conditions: np^≥10n\hat{p} \geq 10 and n(1−p^)≥10n(1-\hat{p}) \geq 10.

6. Summary Table

ParameterCIDistribution
μ\mu (known σ\sigma )Xˉ±zα/2⋅σn\bar{X} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}N(0,1)\mathcal{N}(0,1)
μ\mu (unknown σ\sigma )Xˉ±tα/2,n−1⋅Sn\bar{X} \pm t_{\alpha/2, n-1} \cdot \frac{S}{\sqrt{n}}tn−1t_{n-1}
σ2\sigma^2((n−1)S2χα/22,(n−1)S2χ1−α/22)\left(\frac{(n-1)S^2}{\chi^2_{\alpha/2}}, \frac{(n-1)S^2}{\chi^2_{1-\alpha/2}}\right)χn−12\chi^2_{n-1}
ppp^±zα/2p^(1−p^)n\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}N(0,1)\mathcal{N}(0,1) approx

7. Practice Questions

Q1 (Easy)

n=100n=100, xˉ=50\bar{x}=50, σ=10\sigma=10. Find 95% CI for μ\mu.
Full Solution
SE=10/100=1\text{SE} = 10/\sqrt{100} = 1. z0.025=1.96z_{0.025} = 1.96.
95% CI: 50±1.96⋅1=(48.04,51.96)50 \pm 1.96 \cdot 1 = (48.04, 51.96).

Q2 (Medium)

n=25n=25, xˉ=100\bar{x}=100, s=15s=15. Find 95% CI for μ\mu.
Full Solution
SE=15/25=3\text{SE} = 15/\sqrt{25} = 3. t0.025,24=2.064t_{0.025,24} = 2.064.
95% CI: 100±2.064⋅3=(93.81,106.19)100 \pm 2.064 \cdot 3 = (93.81, 106.19).

Q3 (Hard)

From n=20n=20 normal samples, s2=16s^2=16. Find a 90% CI for σ2\sigma^2.
Full Solution
χ0.05,192=30.144\chi^2_{0.05,19} = 30.144, χ0.95,192=10.117\chi^2_{0.95,19} = 10.117.
90% CI: (19⋅1630.144,19⋅1610.117)=(10.08,30.05)\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
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