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# Statistics
## Confidence intervals
### (Chapter 13)

### Christian Vedel,<br>Department of Economics<br>University of Southern Denmark

### Email: [christian-vs@sam.sdu.dk](mailto:christian-vs@sam.sdu.dk)

### Updated 2026-05-05

---
class: middle
# Today's lecture
.pull-left-wide[
**Turning point estimates into intervals that quantify uncertainty**

- **Section 1:** Confidence interval for the mean value
- **Section 2:** Confidence interval for the variance
- **Section 3:** Confidence interval for the difference in means
- **Section 4:** Confidence interval for the ratio of two variances
- **Section 5:** Determining the required sample size
]

.pull-right-narrow[
![Trees](Figures/Trees1.jpg)
]

---
class: inverse, middle, center
# Confidence interval for the mean value

---
# Why confidence intervals?

.pull-left-wide[
- Estimators such as `$\bar{X}$` are random variables — the probability that they exactly equal the population value is virtually zero
- We hope they are "close" to the true value, but how close?
]

.pull-left-wide[
- A **confidence interval** uses the sampling distribution of the estimator to construct a range of values that conveys this uncertainty:
  - a point estimate gives you one number
  - a confidence interval gives you a range and a probability that the range contains the true value
]

---
# Example: average income in a population

.pull-left-wide[
- Suppose we want to estimate the average monthly income `$\mu$` in a large population
- We cannot survey everyone — instead we draw a simple random sample of `$n = 100$` households
- We observe: `$\bar{X} = 35{,}200$` DKK and `$S = 12{,}000$` DKK
]

.pull-left-wide[
- The point estimate `$\bar{X} = 35{,}200$` DKK is our best single guess — but we know it will not equal `$\mu$` exactly
- A 95% confidence interval (using the formula we will derive) gives:
`$$\hat{I} = \left[35{,}200 - 1.96 \cdot \frac{12{,}000}{\sqrt{100}},\;\; 35{,}200 + 1.96 \cdot \frac{12{,}000}{\sqrt{100}}\right] \approx [32{,}848,\; 37{,}552] \text{ DKK}$$`
- In 95% of samples drawn this way, the resulting interval would contain the true mean income — we will now derive where this formula comes from
]

---
# Setting up the CI for `$\mu$`

.pull-left-wide[
- We want an interval `$\hat{I} = [W_l, W_u]$` centred around `$\bar{X}$`:
`$$\hat{I} = \left[\bar{X} - k,\; \bar{X} + k\right]$$`
- Because `$\bar{X}$` is random, the bounds are random — the interval will not always contain `$\mu$`
]

.pull-left-wide[
- We choose the **confidence level** `$1 - \alpha$`: the probability that the interval contains `$\mu$`

`$$1 - \alpha = P\!\left(\mu \in \hat{I}\right) = P\!\left(\bar{X} - k \leq \mu \leq \bar{X} + k\right)$$`
]

---
# Finding `$k$`: standardising `$\bar{X}$`

.pull-left-wide[
- For a simple random sample, `$\bar{X} \overset{a}{\sim} \mathcal{N}\!\left(\mu, \sigma^2/n\right)$`, so:
`$$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \overset{a}{\sim} \mathcal{N}(0, 1)$$`
]

.pull-left-wide[
- Rewriting the coverage condition in terms of `$Z$`:
`$$1 - \alpha = P\!\left(-\frac{\sqrt{n}\,k}{\sigma} \leq Z \leq \frac{\sqrt{n}\,k}{\sigma}\right) = \Phi(a) - \Phi(-a)$$`
where `$a = \sqrt{n}\,k/\sigma$`
]

.pull-left-wide[
- The **critical value** `$z_{1-\alpha/2}$` satisfies `$\Phi(z_{1-\alpha/2}) = 1 - \alpha/2$`, giving `$a = z_{1-\alpha/2}$` and:
`$$k = \frac{z_{1-\alpha/2} \cdot \sigma}{\sqrt{n}}$$`
]

---
# Illustration: standard normal and the CI for `$\mu$`

.pull-left-narrow[
- Symmetric around 0 → CI is symmetric around `$\bar{X}$`
- Equal tail areas `$\alpha/2$` on each side
- Critical values `$\pm z_{1-\alpha/2}$` are equidistant from centre
]

.pull-right-wide[

![Standard normal distribution with critical values for a 95% CI](Figures/normal-ci.png)
]

---
# CI for `$\mu$`: simple random sample

.pull-left-wide[
> **CI for `$\mu$`, `$\sigma$` known (approximate):**
`$$\hat{I} = \left[\bar{X} - z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},\;\; \bar{X} + z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\right]$$`
]

.pull-left-wide[
**Common critical values:**

| Confidence level | `$z_{1-\alpha/2}$` |
|---|---|
| 90% | 1.645 |
| 95% | 1.960 |
| 99% | 2.576 |
]

---
# CI for `$\mu$`: `$\sigma$` unknown

.pull-left-wide[
- In practice `$\sigma$` is rarely known — we replace it with the sample standard deviation `$S$`
]

.pull-left-wide[
> **CI for `$\mu$`, `$\sigma$` unknown (approximate):**
`$$\hat{I} = \left[\bar{X} - z_{1-\alpha/2} \cdot \sqrt{\frac{S^2}{n}},\;\; \bar{X} + z_{1-\alpha/2} \cdot \sqrt{\frac{S^2}{n}}\right]$$`
- Approximate because it relies on the approximate normal distribution of `$\bar{X}$`
- If we know the population distribution of `$X$`, we can do better
]

---
# Exact CI: `$X \sim \mathcal{N}(\mu, \sigma^2)$`, `$\sigma$` known

.pull-left-wide[
- If `$X$` is normally distributed and `$\sigma^2$` is known, then `$Z = \sqrt{n}(\bar{X} - \mu)/\sigma$` is **exactly** `$\mathcal{N}(0,1)$`
]

.pull-left-wide[
> **Exact CI for `$\mu$`, normal population, `$\sigma$` known:**
`$$\hat{I} = \left[\bar{X} - z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},\;\; \bar{X} + z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\right]$$`
- Same formula as before — but now exact, not approximate
]

---
# The Student's `$t$`-distribution

.pull-left[
> If `$Z \sim \mathcal{N}(0,1)$` and `$Y \sim \chi^2(k)$` independent:
`$$T = \frac{Z}{\sqrt{Y/k}} \sim t(k)$$`
- `$E(T) = 0$`, `$\quad Var(T) = \dfrac{k}{k-2}$`
- Heavier tails than normal; `$t(k) \to \mathcal{N}(0,1)$` as `$k \to \infty$`
- Used for CI when `$\sigma^2$` is unknown — which is exactly what we do next
]

.pull-right[
![Student t](Figures/t-fig.png)
]

---
# Exact CI: `$X \sim \mathcal{N}(\mu, \sigma^2)$`, `$\sigma$` unknown

.pull-left-wide[
- When `$\sigma^2$` is unknown, replacing it with `$S^2$` changes the distribution of the pivot
- It can be shown that:
`$$t = \frac{\bar{X} - \mu}{\sqrt{S^2/n}} \sim t(n-1)$$`
- The ***t*-distribution** has heavier tails than the normal; as `$n \to \infty$` it converges to `$\mathcal{N}(0,1)$`
]

.pull-left-wide[
`$$\hat{I} = \left[\bar{X} - t_{1-\alpha/2}(n-1) \cdot \sqrt{\frac{S^2}{n}},\;\; \bar{X} + t_{1-\alpha/2}(n-1) \cdot \sqrt{\frac{S^2}{n}}\right]$$`
]

---
# `$t$` vs `$z$`: when does it matter?

.pull-left-wide[
- The `$t(k)$` distribution converges to `$\mathcal{N}(0,1)$` as `$k \to \infty$` — for large samples the two are nearly identical
- **Rule of thumb:**
  - `$n \gtrsim 30$`: the normal approximation is usually acceptable in practice
  - `$n \gtrsim 100$`: safe to use `$z$` even if the population distribution is moderately skewed
]

.pull-left-wide[
- **Practical advice:** when `$\sigma^2$` is unknown, default to `$t$` — it is exact for normal populations and reduces to `$z$` for large samples
- This rule applies generally across all the CIs in this lecture
]

---
# CI for `$\mu$`: `$X \sim Bernoulli(p)$`

.pull-left-wide[
- If `$X \sim Bernoulli(p)$`, then `$E(X) = p$` and `$Var(X) = p(1-p)$`
- The sample mean `$\bar{X}$` estimates `$p$`; we estimate the variance as `$\bar{X}(1 - \bar{X})$`
- This exploits the Bernoulli structure and is more efficient than the general `$S^2$`
]

.pull-left-wide[
> **Approximate CI for `$p$`:**
`$$\hat{I} = \left[\bar{X} - z_{1-\alpha/2} \cdot \sqrt{\frac{\bar{X}(1-\bar{X})}{n}},\;\; \bar{X} + z_{1-\alpha/2} \cdot \sqrt{\frac{\bar{X}(1-\bar{X})}{n}}\right]$$`
]

---
# CI for `$\mu$`: stratified sample

.pull-left-wide[
- For a stratified sample, we use `$\bar{X}_{st} = \sum_{j=1}^m w_j \bar{X}_j$` as the estimator
- We estimate `$Var(\bar{X}_{st}) = \sum_{j=1}^m w_j^2 \sigma_j^2/n_j$` by replacing `$\sigma_j^2$` with `$S_j^2$`:
`$$\widehat{Var}(\bar{X}_{st}) = \sum_{j=1}^m w_j^2 \cdot \frac{S_j^2}{n_j}$$`
]

.pull-left-wide[
> **Approximate CI for `$\mu$`, stratified sample:**
`$$\hat{I} = \left[\bar{X}_{st} - z_{1-\alpha/2} \cdot \sqrt{\widehat{Var}(\bar{X}_{st})},\;\; \bar{X}_{st} + z_{1-\alpha/2} \cdot \sqrt{\widehat{Var}(\bar{X}_{st})}\right]$$`
]

---
# .red[Raise your hand 1: CI for the mean]

.pull-left-wide[
**Q1.** A 95% CI for `$\mu$` is `$[45, 55]$`. Which statement is correct?

- **A)** There is a 95% probability that `$\mu$` lies in `$[45, 55]$`
- **B)** 95% of future sample means will fall in `$[45, 55]$`
- **C)** If we repeated the sampling many times, 95% of the resulting intervals would contain `$\mu$`
]

.pull-left-wide[
**Q2.** When should we use the `$t$`-distribution rather than the normal for a CI for `$\mu$`?

- **A)** When the population is normal and `$\sigma^2$` is unknown — the `$t$` arises from estimating `$\sigma$` with `$S$`
- **B)** Whenever the sample size is small (`$n < 30$`)
- **C)** Whenever the population distribution is unknown
]

---
# .red[Practice 1: CI for the mean]

.pull-left-wide[
A random sample of `$n = 25$` students has mean exam score `$\bar{X} = 68$` and sample standard deviation `$S = 10$`. Assume scores are normally distributed.

1. Construct a 95% CI for the population mean `$\mu$` using the `$t$`-distribution
2. How does the CI change if `$n = 100$` instead? (Recalculate)
3. If the population standard deviation were known to be `$\sigma = 10$`, would you use `$z$` or `$t$`? Construct that CI for `$n = 25$`

*Hint:* `$t_{0.975}(24) \approx 2.064$`; `$t_{0.975}(99) \approx 1.984$`; `$z_{0.975} = 1.96$`
]

---
class: inverse, middle, center
# Confidence interval for the variance

---
# CI for the variance

.pull-left-wide[
- The sample variance `$S^2$` estimates `$\sigma^2$`; to build a CI we need its sampling distribution
- When `$X \sim \mathcal{N}(\mu, \sigma^2)$`, it can be shown that:
`$$Y = \frac{(n-1) S^2}{\sigma^2} \sim \chi^2(n-1)$$`
- The `$\chi^2$` distribution is right-skewed — so the CI will be **asymmetric** around `$S^2$`
]

.pull-left-wide[
- Trapping `$Y$` between its `$\alpha/2$` and `$1-\alpha/2$` quantiles and rearranging for `$\sigma^2$` gives:

> **CI for `$\sigma^2$`:**
`$$\hat{I} = \left[\frac{(n-1)S^2}{\chi^2_{1-\alpha/2}(n-1)},\;\; \frac{(n-1)S^2}{\chi^2_{\alpha/2}(n-1)}\right]$$`
]

---
# Illustration: `$\chi^2(n-1)$` distribution and the CI for `$\sigma^2$`

.pull-left-narrow[
- Right-skewed: longer upper tail
- `$\chi^2_{\alpha/2}$` and `$\chi^2_{1-\alpha/2}$` are **not** equidistant from the centre
- This is why the CI for `$\sigma^2$` is asymmetric — it stretches further upward than downward
]

.pull-right-wide[

![Chi-square distribution with critical values for a 95% CI](Figures/chisq-ci.png)
]

---
# .red[Raise your hand 2: CI for the variance]

.pull-left-wide[
**Q1.** Why is the CI for `$\sigma^2$` not symmetric around `$S^2$`?

- **A)** Because `$S^2$` is a biased estimator of `$\sigma^2$`
- **B)** Because the `$\chi^2$` distribution is right-skewed — the lower and upper critical values are not equidistant from the centre
- **C)** Because `$\alpha$` is split unequally between the two tails
]

.pull-left-wide[
**Q2.** The pivot `$(n-1)S^2/\sigma^2 \sim \chi^2(n-1)$` is an **exact** result. What does it require?

- **A)** That `$X$` is normally distributed in the population
- **B)** Only that `$n$` is large — the CLT makes it hold for any distribution
- **C)** That `$\sigma^2$` is known — otherwise the pivot is not well-defined
]

---
# .red[Practice 2: CI for the variance]

.pull-left-wide[
A normal population is sampled with `$n = 20$`, giving `$S^2 = 16$`.

1. Construct a 95% CI for `$\sigma^2$`
2. Is the interval symmetric around `$S^2 = 16$`? Why or why not?
3. What does it mean if `$\sigma^2 = 9$` is inside the interval?

*Hint:* `$\chi^2_{0.025}(19) \approx 8.91$`; `$\chi^2_{0.975}(19) \approx 32.85$`
]

---
class: inverse, middle, center
# Confidence interval for the difference in means

---
# Setup: two independent samples

.pull-left-wide[
- We often want to compare two groups (e.g. wage gap between men and women)
- Assume two independent simple random samples of sizes `$n_1$` and `$n_2$`
]

.pull-left-wide[
- The natural estimator of `$\mu_1 - \mu_2$` is `$\bar{X}_1 - \bar{X}_2$`:
`$$E(\bar{X}_1 - \bar{X}_2) = \mu_1 - \mu_2 \quad \text{(unbiased)}$$`
`$$Var(\bar{X}_1 - \bar{X}_2) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}$$`
]

.pull-left-wide[
- By the CLT, `$\bar{X}_1 - \bar{X}_2$` is approximately normal, so:
`$$\frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}} \overset{a}{\sim} \mathcal{N}(0,1)$$`
]

---
# CI for `$\mu_1 - \mu_2$`: `$\sigma_1^2, \sigma_2^2$` known or unknown

.pull-left-wide[
.small123[
> **Approximate CI for `$\mu_1 - \mu_2$`, `$\sigma_j^2$` known:**
`$$\hat{I} = \left[(\bar{X}_1 - \bar{X}_2) \pm z_{1-\alpha/2} \cdot \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\right]$$`
]
]

.pull-left-wide[
.small123[
> **Approximate CI for `$\mu_1 - \mu_2$`, `$\sigma_j^2$` unknown (replace with `$S_j^2$`):**
`$$\hat{I} = \left[(\bar{X}_1 - \bar{X}_2) \pm z_{1-\alpha/2} \cdot \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}\right]$$`
*or*
`$$\hat{I} = \left[(\bar{X}_1 - \bar{X}_2) \pm t_{1-\alpha/2}(n_1+n_2-2) \cdot \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}\right]$$`
]
]

---
# Pooled variance estimator

.pull-left-wide[
- If we have good reason to believe `$\sigma_1^2 = \sigma_2^2 = \sigma^2$`, we can combine information from both samples
]

.pull-left-wide[
> The **pooled variance estimator** is:
`$$S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}$$`
]

.pull-left-wide[
> **Exact CI for `$\mu_1 - \mu_2$`, equal unknown variances:**
`$$\hat{I} = \left[(\bar{X}_1 - \bar{X}_2) \pm t_{1-\alpha/2}(n_1+n_2-2) \cdot \sqrt{S_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\right]$$`
- **Use `$t$`, not `$z$`, when `$\sigma^2$` is unknown** — `$t(n_1+n_2-2)$` is exact for normal populations; `$z$` is only asymptotically valid
]

---
# .red[Raise your hand 3: CI for the difference in means]

.pull-left-wide[
**Q1.** A 95% CI for `$\mu_1 - \mu_2$` is `$[-2,\; 8]$`. What can we conclude?

- **A)** `$\mu_1 > \mu_2$` at the 95% confidence level, since the interval is mostly positive
- **B)** The data are consistent with both `$\mu_1 > \mu_2$` and `$\mu_1 \leq \mu_2$` — zero is inside the interval
- **C)** `$\mu_2 > \mu_1$` at the 95% confidence level, since the interval contains negative values
]

.pull-left-wide[
**Q2.** Why does the pooled variance estimator `$S_p^2$` give a better estimate than using `$S_1^2$` and `$S_2^2$` separately?

- **A)** Because it always produces a smaller variance estimate, narrowing the CI
- **B)** Because it avoids the need to assume normality in either group
- **C)** Because when `$\sigma_1^2 = \sigma_2^2$`, pooling combines information from both samples, reducing estimation uncertainty
]

---
# .red[Practice 3: CI for the difference in means]

.pull-left-wide[
Two independent samples of wages (€/hour):
- Group 1 (men): `$n_1 = 36$`, `$\bar{X}_1 = 28$`, `$S_1^2 = 25$`
- Group 2 (women): `$n_2 = 49$`, `$\bar{X}_2 = 25$`, `$S_2^2 = 16$`

1. Construct a 95% CI for `$\mu_1 - \mu_2$` (use `$z_{0.975} = 1.96$`, variances unknown)
2. Does the CI suggest a statistically meaningful difference between the groups?
3. Now assume equal variances. Calculate `$S_p^2$` and construct the pooled CI. Does the conclusion change?
]

---
class: inverse, middle, center
# Confidence interval for the ratio of two variances

---
# The `$F$`-distribution

.pull-left[
> If `$Y_1 \sim \chi^2(d_1)$` and `$Y_2 \sim \chi^2(d_2)$` are independent:
`$$F = \frac{Y_1/d_1}{Y_2/d_2} \sim F(d_1, d_2)$$`
- `$E(F) = \dfrac{d_2}{d_2-2}$` (for `$d_2 > 2$`)
- Right-skewed, positive values only
- We will use it for the CI for `$\sigma_2^2/\sigma_1^2$`
]

.pull-right[
![F-distribution](Figures/f-fig.png)
]

---
# When do we compare variances?

.pull-left-wide[
- Sometimes we want to know whether variability differs between two groups
- We construct a CI for the **ratio** `$\sigma_2^2 / \sigma_1^2$`
- If this ratio is 1, the variances are equal
]

.pull-left-wide[
- Assume `$X_j \sim \mathcal{N}(\mu_j, \sigma_j^2)$` in each group (`$j = 1, 2$`), independent samples
- It can be shown that:
`$$\frac{S_1^2 / \sigma_1^2}{S_2^2 / \sigma_2^2} = \frac{S_1^2 \cdot \sigma_2^2}{S_2^2 \cdot \sigma_1^2} \sim F(n_1 - 1,\; n_2 - 1)$$`
]

---
# Deriving the CI for `$\sigma_2^2 / \sigma_1^2$`

.pull-left-wide[
- Let `$F_{\alpha/2}$` and `$F_{1-\alpha/2}$` be the critical values from `$F(n_1-1, n_2-1)$`:
`$$P\!\left(F_{\alpha/2} \leq \frac{S_1^2 \cdot \sigma_2^2}{S_2^2 \cdot \sigma_1^2} \leq F_{1-\alpha/2}\right) = 1 - \alpha$$`
]

.pull-left-wide[
- Rearranging to isolate `$\sigma_2^2/\sigma_1^2$`:

> **Exact CI for `$\sigma_2^2/\sigma_1^2$`:**
`$$\hat{I} = \left[\frac{S_2^2}{S_1^2} \cdot F_{\alpha/2}(n_1-1, n_2-1),\;\; \frac{S_2^2}{S_1^2} \cdot F_{1-\alpha/2}(n_1-1, n_2-1)\right]$$`
- Also **not symmetric** around `$S_2^2/S_1^2$` because the `$F$`-distribution is skewed
]

---
# Illustration: `$F(n_1-1,\, n_2-1)$` distribution and the CI for `$\sigma_2^2/\sigma_1^2$`

.pull-left-narrow[
- Right-skewed, positive-valued
- `$F_{\alpha/2}$` and `$F_{1-\alpha/2}$` are **not** equidistant from 1
- Hence the CI for `$\sigma_2^2/\sigma_1^2$` is also asymmetric — same principle as the `$\chi^2$` case
]

.pull-right-wide[

![F-distribution with critical values for a 95% CI](Figures/f-ci.png)
]

---
# .red[Raise your hand 4: CI for the ratio of variances]

.pull-left-wide[
**Q1.** A 95% CI for `$\sigma_2^2/\sigma_1^2$` is `$[0.8,\; 2.5]$`. What can we conclude?

- **A)** `$\sigma_2 > \sigma_1$`, since most of the interval is above 1
- **B)** The data are consistent with equal variances — 1 is inside the interval — as well as with `$\sigma_2^2$` being up to 2.5 times `$\sigma_1^2$`
- **C)** `$\sigma_2^2 > \sigma_1^2$` at the 95% confidence level
]

.pull-left-wide[
**Q2.** Why does the `$F$`-distribution appear when comparing two sample variances?

- **A)** Because the ratio of two normal random variables follows an `$F$`-distribution
- **B)** Because `$F$` is always used when comparing statistics from two independent groups
- **C)** Because each `$(n_j-1)S_j^2/\sigma_j^2 \sim \chi^2(n_j-1)$`, and the ratio of two independent chi-square variables (divided by their degrees of freedom) defines an `$F$`
]

---
# .red[Practice 4: CI for the ratio of variances]

.pull-left-wide[
Two independent normal samples:
- Group 1: `$n_1 = 11$`, `$S_1^2 = 8$`
- Group 2: `$n_2 = 16$`, `$S_2^2 = 18$`

Construct a 95% CI for `$\sigma_2^2/\sigma_1^2$`.

*Hint:* `$F_{0.025}(10, 15) \approx 0.30$`; `$F_{0.975}(10, 15) \approx 3.06$`

Does the CI suggest the variances are significantly different?
]

---
class: inverse, middle, center
# Determining the required sample size

---
# Width of a CI and sample size

.pull-left-wide[
- The variance of `$\bar{X}$` is `$\sigma^2/n$` — larger `$n$` gives a narrower CI
- We can invert the CI formula to find the `$n$` needed to achieve a desired width
]

.pull-left-wide[
- The width of the CI for `$\mu$` (known `$\sigma^2$`) is `$2 z_{1-\alpha/2} \cdot \sigma/\sqrt{n}$`
- Suppose we want the half-width to be at most `$K$`:
`$$z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \leq K \quad \Longrightarrow \quad n \geq \frac{z_{1-\alpha/2}^2 \cdot \sigma^2}{K^2}$$`
]

.pull-left-wide[
> **Required sample size:**
`$$n^* = \frac{z_{1-\alpha/2}^2 \cdot \sigma^2}{K^2}$$`
- Requires knowing `$\sigma^2$` — use a pilot study if unknown
]

---
# Example: choosing `$n$` for a desired precision

.pull-left-wide[
**Setting:** we want to estimate the mean height of students with a 95% CI and half-width at most 2 cm. A previous study suggests `$\sigma \approx 10$` cm.

`$$n^* = \frac{1.96^2 \times 10^2}{2^2} = \frac{3.84 \times 100}{4} = \frac{384}{4} \approx 97$$`
]

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**What if we want twice the precision** — half-width `$\leq 1$` cm?

`$$n^* = \frac{1.96^2 \times 10^2}{1^2} \approx 385$$`

Halving the half-width requires **4× the sample size** — because width `$\propto 1/\sqrt{n}$`, so to halve it you must quadruple `$n$`
]

---
# Sample size for Bernoulli `$X$`

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- If `$X \sim Bernoulli(p)$`, then `$\sigma^2 = p(1-p)$`, which depends on the unknown `$p$`
]

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- The product `$p(1-p)$` is maximised at `$p = 0.5$`, giving `$\sigma^2 \leq 0.25$`
- Substituting this worst-case bound:
`$$n^* = \frac{z_{1-\alpha/2}^2 \cdot 0.25}{K^2}$$`
- This guarantees the desired width **regardless of the true** `$p$` — a conservative bound
]

---
# .red[Raise your hand 5: Sample size]

.pull-left-wide[
**Q1.** To halve the width of a CI for `$\mu$`, how must `$n$` change?

- **A)** Double `$n$` — width is proportional to `$1/n$`
- **B)** Quadruple `$n$` — width is proportional to `$1/\sqrt{n}$`, so halving width requires multiplying `$n$` by 4
- **C)** Triple `$n$` — halving the width requires tripling the observations
]

.pull-left-wide[
**Q2.** For `$X \sim Bernoulli(p)$`, using `$p(1-p) \leq 0.25$` in the sample size formula gives what guarantee?

- **A)** The required `$n$` is sufficient regardless of the true `$p$` — it is a conservative worst-case bound
- **B)** The CI width is exactly minimised when `$p = 0.5$`
- **C)** The formula only applies when `$p$` is close to `$0.5$`
]

---
# .red[Practice 5: Sample size]

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A researcher wants a 95% CI for the mean income in a population, with a half-width of at most €500. A pilot study suggests `$\sigma \approx 2{,}500$`.

1. Calculate the required sample size `$n^*$`
2. How does `$n^*$` change if the required half-width is tightened to €250?
3. Suppose instead `$X \sim Bernoulli(p)$` (e.g. employed/not employed) and we want half-width `$\leq 0.03$`. What is the conservative `$n^*$`?

*Hint:* `$z_{0.975} = 1.96$`
]

---
# How to (not) interpret a confidence interval

*This often causes confusion, so let's be clear about what a CI does and does not mean.*

.pull-left[
### It isn't:
- The probability that the true parameter is in the interval
  + (the parameter is fixed, not random)
- The probability that the interval contains the true parameter
  + (the interval is random, but we don't talk about "the" interval — we talk about the procedure that generates intervals, which has a certain long-run success rate)
]

.pull-right[
### It is:

- A random interval that contains the true parameter with probability `$1-\alpha$` in repeated sampling
- A range of plausible values for the parameter based on the observed data and the sampling distribution of the estimator
]

---
class: inverse, middle, center
# .orange[Advanced: Clopper-Pearson interval for a proportion]
### Not part of the curriculum — for the curious

---
# .orange[Why the standard Bernoulli CI can fail]

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- The approximate CI for `$p$` relies on the normal approximation to the binomial
- This works well when `$n$` is large and `$p$` is not too close to 0 or 1
- **But it breaks down when:**
  - `$n$` is small — the CLT has not yet kicked in
  - `$p$` is near 0 or 1 — the binomial distribution is strongly skewed
  - The resulting interval can extend **outside** `$[0,1]$`, which is logically impossible for a probability
]

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**Example:** `$n = 20$`, `$k = 2$` successes (so `$\hat{p} = 0.10$`)

`$$\hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.10 \pm 1.96\sqrt{\frac{0.10 \times 0.90}{20}} = 0.10 \pm 0.131 = [-0.031,\; 0.231]$$`

The lower bound is **negative** — the normal approximation has failed
]

---
# .orange[The Clopper-Pearson exact interval]

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- The **Clopper-Pearson interval** is constructed directly from the binomial distribution — no normal approximation
- Given `$k$` successes out of `$n$` trials, the exact `$1-\alpha$` CI is `$[p_L,\; p_U]$` where:

`$$p_L = B_{\alpha/2}(k,\; n-k+1), \qquad p_U = B_{1-\alpha/2}(k+1,\; n-k)$$`

- `$B_q(a, b)$` denotes the `$q$`-th quantile of the **Beta$(a,b)$ distribution**
- Special cases: if `$k = 0$` set `$p_L = 0$`; if `$k = n$` set `$p_U = 1$`
]

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- The interval is **conservative**: the true coverage is at least `$1-\alpha$` for every `$p$` and `$n$`
- It always stays within `$[0, 1]$` by construction
- The cost: slightly wider intervals than the approximate CI when the approximation is valid
]

---
# .orange[Example: Clopper-Pearson vs. normal approximation]

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**Setting:** `$n = 20$` households surveyed, `$k = 2$` report unemployment; `$\hat{p} = 0.10$`

|  | Lower bound | Upper bound |
|---|---|---|
| Normal approximation | `$-0.031$` | `$0.231$` |
| Clopper-Pearson (exact) | `$0.012$` | `$0.317$` |

*Note:* `$p_L = B_{0.025}(2,\,19) \approx 0.012$`; `$p_U = B_{0.975}(3,\,18) \approx 0.317$`
]

.pull-left-wide[
- The exact interval is fully inside `$[0,1]$` and provides a meaningful lower bound
- The approximation fails badly here: `$n$` is small and `$p$` is near 0
- In practice: use the normal approximation when `$n\hat{p} \geq 5$` and `$n(1-\hat{p}) \geq 5$`; use Clopper-Pearson otherwise
]

---
# Key takeaways

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**CI for `$\mu$`:** use `$z$` when `$\sigma$` known; use `$t(n-1)$` when `$\sigma$` unknown and `$X$` normal — `$t$` is exact, `$z$` is only asymptotic

**CI for `$\sigma^2$`:** `$\chi^2(n-1)$` pivot — asymmetric interval; exact only when `$X$` is normal
]

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**CI for `$\mu_1 - \mu_2$`:** approximate with `$z$` and separate `$S_j^2$`; exact with `$t(n_1+n_2-2)$` and pooled `$S_p^2$` when `$\sigma_1^2 = \sigma_2^2$` and `$X$` normal

**CI for `$\sigma_2^2/\sigma_1^2$`:** `$F(n_1-1, n_2-1)$` pivot — also asymmetric; exact only when `$X$` is normal
]

.pull-left-wide[
**Sample size:** `$n^* = z_{1-\alpha/2}^2 \sigma^2 / K^2$`; width `$\propto 1/\sqrt{n}$` — halving the width requires 4× the observations; for Bernoulli use `$\sigma^2 = 0.25$` as a conservative bound

**Rule of thumb:** `$n \gtrsim 30$` for `$z$` to be adequate; `$n \gtrsim 100$` if the population is skewed
]

---
# Before next time
.pull-left[
- Read the assigned reading
- Next time: Hypothesis testing `$\rightarrow$` Chapter 14
]

.pull-right[
![Trees](Figures/Trees1.jpg)
]